Deck 13: Simple Linear Regression

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Question
TABLE 13-10
The management of a chain electronic store would like to develop a model for predicting the weekly sales (in thousand of dollars) for individual stores based on the number of customers who made purchases. A random sample of 12 stores yields the following results:
 Customers  Sales (Thousands of Dollars) 90711.2092611.057138.217419.217809.4289810.085106.735297.024606.128729.526507.536037.25\begin{array} { | l | l | } \hline \text { Customers } & \text { Sales (Thousands of Dollars) } \\\hline 907 & 11.20 \\\hline 926 & 11.05 \\\hline 713 & 8.21 \\\hline 741 & 9.21 \\\hline 780 & 9.42 \\\hline 898 & 10.08 \\\hline 510 & 6.73 \\\hline 529 & 7.02 \\\hline 460 & 6.12 \\\hline 872 & 9.52 \\\hline 650 & 7.53 \\\hline 603 & 7.25 \\\hline\end{array}

-Referring to Table 13-10, which is the correct null hypothesis for testing whether the number of customers who make purchase affects weekly sales?

A) H0 : ?0 = 0
B) H0 : µ = 0
C) H0 : ?1 = 0
D) H0 : ?= 0
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Question
TABLE 13-7
An investment specialist claims that if one holds a portfolio that moves in the opposite direction to the market index like the S&P 500, then it is possible to reduce the variability of the portfolio's return. In other words, one can create a portfolio with positive returns but less exposure to risk. A sample of 26 years of S&P 500 index and a portfolio consisting of stocks of private prisons, which are believed to be negatively related to the S&P 500 index, is collected. A regression analysis was performed by regressing the returns of the prison stocks portfolio (Y) on the returns of S&P 500 index (X) to prove that the prison stocks portfolio is negatively related to the S&P 500 index at a 5% level of significance. The results are given in the following EXCEL output.
 Coefficients  Standard Error T Stat  p-value  Intercept 4.8660042580.3574360913.613634418.7932E13 S& P 0.5025135060.0715971527.01862425294942E07\begin{array} { l c c c c } \hline & \text { Coefficients } & \text { Standard Error } &T \text { Stat } & \text { p-value } \\\hline \text { Intercept } & 4.866004258 & 0.35743609 & 13.61363441 & 8.7932 \mathrm { E } - 13 \\\text { S\& P } & - 0.502513506 & 0.071597152 & - 7.01862425 & 294942 \mathrm { E } - 07 \\\hline\end{array}

-Referring to Table 13-7, to test whether the prison stocks portfolio is negatively related to the S&P 500 index, the measured value of the test statistic is

A) 0.357.
B) 0.072.
C) -0.503.
D) -7.019.
Question
TABLE 13-2
A candy bar manufacturer is interested in trying to estimate how sales are influenced by the price of their product. To do this, the company randomly chooses 6 small cities and offers the candy bar at different prices. Using candy bar sales as the dependent variable, the company will conduct a simple linear regression on the data below:
 City  Price ($) Sales  River Falls 1.30100 Hudson 1.6090 Ellsworth 1.8090 Prescott 2.0040 Rock Elm 2.4038 Stillwater 2.9032\begin{array} { l l c } \hline \text { City } & \text { Price } ( \$ ) & \text { Sales } \\\hline \text { River Falls } & 1.30 & 100 \\\text { Hudson } & 1.60 & 90 \\\text { Ellsworth } & 1.80 & 90 \\\text { Prescott } & 2.00 & 40 \\\text { Rock Elm } & 2.40 & 38 \\\text { Stillwater } & 2.90 & 32 \\\hline\end{array}

-Referring to Table 13-2, what is ?(XXˉ)2?(X -\bar X )^2 for these data?

A) 2.54
B) 0
C) 1.66
D) 25.66
Question
TABLE 13-2
A candy bar manufacturer is interested in trying to estimate how sales are influenced by the price of their product. To do this, the company randomly chooses 6 small cities and offers the candy bar at different prices. Using candy bar sales as the dependent variable, the company will conduct a simple linear regression on the data below:
 City  Price ($) Sales  River Falls 1.30100 Hudson 1.6090 Ellsworth 1.8090 Prescott 2.0040 Rock Elm 2.4038 Stillwater 2.9032\begin{array} { l l c } \hline \text { City } & \text { Price } ( \$ ) & \text { Sales } \\\hline \text { River Falls } & 1.30 & 100 \\\text { Hudson } & 1.60 & 90 \\\text { Ellsworth } & 1.80 & 90 \\\text { Prescott } & 2.00 & 40 \\\text { Rock Elm } & 2.40 & 38 \\\text { Stillwater } & 2.90 & 32 \\\hline\end{array}

-Referring to Table 13-2, what is the coefficient of correlation for these data?

A) - 0.7839
B) 0.8854
C) 0.7839
D) - 0.8854
Question
TABLE 13-12
The manager of the purchasing department of a large banking organization would like to develop a model to predict the amount of time (measured in hours) it takes to process invoices. Data are collected from a sample of 30 days, and the number of invoices processed and completion time in hours is recorded. Below is the regression output:
 Regression Statistics  Multiple R 0.9947 R Square 0.8924 Adjusted R Square 0.8886 Standard Error 0.3342 Observations 30\begin{array}{lc}\text { Regression Statistics } \\\hline \text { Multiple R } & 0.9947 \\\text { R Square } & 0.8924 \\\text { Adjusted R Square } & 0.8886 \\\text { Standard Error } & 0.3342 \\\text { Observations } & 30\end{array}


 ANOVA \text { ANOVA }
df SS  MS F Significance F Regression 125.943825.9438232.22004.3946E15 Residual 283.12820.1117 Total 2929.072\begin{array}{lccccc} & d f & \text { SS } & \text { MS } & F & \text { Significance } F \\\text { Regression } & 1 & 25.9438 & 25.9438 & 232.2200 & 4.3946 \mathrm{E}-15 \\\text { Residual } & 28 & 3.1282 & 0.1117 & & \\\text { Total } & 29 & 29.072 & & \\\hline\end{array}


 Coeffcients  StandardError t Stat p-value  Lower 95%  Upper 95%  Invoices 0.40240.12363.25590.00300.14920.6555 Processed 0.01260.000815.23884.3946E150.01090.0143\begin{array}{lrrrrrr}\hline & \text { Coeffcients } & \text { StandardError } & t \text { Stat } & p \text {-value } & \text { Lower 95\% } & \text { Upper 95\% } \\ \hline \text { Invoices } & 0.4024 & 0.1236 & 3.2559 & 0.0030 & 0.1492 & 0.6555 \\\text { Processed } & 0.0126 & 0.0008 & 15.2388 & 4.3946 \mathrm{E}-15 & 0.0109 & 0.0143 \\\hline\end{array}
 <strong>TABLE 13-12 The manager of the purchasing department of a large banking organization would like to develop a model to predict the amount of time (measured in hours) it takes to process invoices. Data are collected from a sample of 30 days, and the number of invoices processed and completion time in hours is recorded. Below is the regression output:  \begin{array}{lc} \text { Regression Statistics } \\ \hline \text { Multiple R } & 0.9947 \\ \text { R Square } & 0.8924 \\ \text { Adjusted R Square } & 0.8886 \\ \text { Standard Error } & 0.3342 \\ \text { Observations } & 30 \end{array}     \text { ANOVA }   \begin{array}{lccccc} & d f & \text { SS } & \text { MS } & F & \text { Significance } F \\ \text { Regression } & 1 & 25.9438 & 25.9438 & 232.2200 & 4.3946 \mathrm{E}-15 \\ \text { Residual } & 28 & 3.1282 & 0.1117 & & \\ \text { Total } & 29 & 29.072 & & \\ \hline \end{array}     \begin{array}{lrrrrrr} \hline & \text { Coeffcients } & \text { StandardError } & t \text { Stat } & p \text {-value } & \text { Lower 95\% } & \text { Upper 95\% } \\ \hline \text { Invoices } & 0.4024 & 0.1236 & 3.2559 & 0.0030 & 0.1492 & 0.6555 \\ \text { Processed } & 0.0126 & 0.0008 & 15.2388 & 4.3946 \mathrm{E}-15 & 0.0109 & 0.0143 \\ \hline \end{array}       -Referring to Table 13-12, the degrees of freedom for the F test on whether the number of invoices processed affects the amount of time are</strong> A) 1, 29. B) 28, 1. C) 29, 1. D) 1, 28. <div style=padding-top: 35px>   <strong>TABLE 13-12 The manager of the purchasing department of a large banking organization would like to develop a model to predict the amount of time (measured in hours) it takes to process invoices. Data are collected from a sample of 30 days, and the number of invoices processed and completion time in hours is recorded. Below is the regression output:  \begin{array}{lc} \text { Regression Statistics } \\ \hline \text { Multiple R } & 0.9947 \\ \text { R Square } & 0.8924 \\ \text { Adjusted R Square } & 0.8886 \\ \text { Standard Error } & 0.3342 \\ \text { Observations } & 30 \end{array}     \text { ANOVA }   \begin{array}{lccccc} & d f & \text { SS } & \text { MS } & F & \text { Significance } F \\ \text { Regression } & 1 & 25.9438 & 25.9438 & 232.2200 & 4.3946 \mathrm{E}-15 \\ \text { Residual } & 28 & 3.1282 & 0.1117 & & \\ \text { Total } & 29 & 29.072 & & \\ \hline \end{array}     \begin{array}{lrrrrrr} \hline & \text { Coeffcients } & \text { StandardError } & t \text { Stat } & p \text {-value } & \text { Lower 95\% } & \text { Upper 95\% } \\ \hline \text { Invoices } & 0.4024 & 0.1236 & 3.2559 & 0.0030 & 0.1492 & 0.6555 \\ \text { Processed } & 0.0126 & 0.0008 & 15.2388 & 4.3946 \mathrm{E}-15 & 0.0109 & 0.0143 \\ \hline \end{array}       -Referring to Table 13-12, the degrees of freedom for the F test on whether the number of invoices processed affects the amount of time are</strong> A) 1, 29. B) 28, 1. C) 29, 1. D) 1, 28. <div style=padding-top: 35px>

-Referring to Table 13-12, the degrees of freedom for the F test on whether the number of invoices processed affects the amount of time are

A) 1, 29.
B) 28, 1.
C) 29, 1.
D) 1, 28.
Question
TABLE 13-2
A candy bar manufacturer is interested in trying to estimate how sales are influenced by the price of their product. To do this, the company randomly chooses 6 small cities and offers the candy bar at different prices. Using candy bar sales as the dependent variable, the company will conduct a simple linear regression on the data below:
 City  Price ($) Sales  River Falls 1.30100 Hudson 1.6090 Ellsworth 1.8090 Prescott 2.0040 Rock Elm 2.4038 Stillwater 2.9032\begin{array} { l l c } \hline \text { City } & \text { Price } ( \$ ) & \text { Sales } \\\hline \text { River Falls } & 1.30 & 100 \\\text { Hudson } & 1.60 & 90 \\\text { Ellsworth } & 1.80 & 90 \\\text { Prescott } & 2.00 & 40 \\\text { Rock Elm } & 2.40 & 38 \\\text { Stillwater } & 2.90 & 32 \\\hline\end{array}

-The sample correlation coefficient between X and Y is 0.375. It has been found out that the p- value is 0.744 when testing H0 : ? = 0 against the one- sided alternative H0 : ? < 0. To test H0 : ? = 0 against the two- sided alternative H0 : ?? 0 at a significance level of 0.2, the p- value is

A) (1 - 0.744)(2).
B) (0.744)(2).
C) 0.744/2.
D) 1 - 0.744.
Question
TABLE 13-9
It is believed that, the average numbers of hours spent studying per day (HOURS) during undergraduate education should have a positive linear relationship with the starting salary (SALARY, measured in thousands of dollars per month) after graduation. Given below is the Excel output from regressing starting salary on number of hours spent studying per day for a sample of 51 students. NOTE: Some of the numbers in the output are purposely erased.
 Regression Statistics  Multiple R 0.8857 R Square 0.7845 Adjusted R Square 0.7801 Standard Error 1.3704 Observations 51\begin{array}{l}\text { Regression Statistics }\\\begin{array} { l c } \hline \text { Multiple R } & 0.8857 \\\text { R Square } & 0.7845 \\\text { Adjusted R Square } & 0.7801 \\\text { Standard Error } & 1.3704 \\\text { Observations } & 51 \\\hline\end{array}\end{array}
 ANOVA \text { ANOVA }
 df  SS  MS  Significance F  Regression 1335.0472335.0473178.3859 Residual 1.8782 Total 50427.0798\begin{array}{llcc}\hline&\text { df }&\text { SS } &\text { MS }&\text {F }&\text { Significance F } \\\text { Regression } & 1 & 335.0472 & 335.0473 & 178.3859 \\\text { Residual } & & & 1.8782 & \\\text { Total } & 50 & 427.0798 & &\end{array}  <strong>TABLE 13-9 It is believed that, the average numbers of hours spent studying per day (HOURS) during undergraduate education should have a positive linear relationship with the starting salary (SALARY, measured in thousands of dollars per month) after graduation. Given below is the Excel output from regressing starting salary on number of hours spent studying per day for a sample of 51 students. NOTE: Some of the numbers in the output are purposely erased.  \begin{array}{l} \text { Regression Statistics }\\ \begin{array} { l c } \hline \text { Multiple R } & 0.8857 \\ \text { R Square } & 0.7845 \\ \text { Adjusted R Square } & 0.7801 \\ \text { Standard Error } & 1.3704 \\ \text { Observations } & 51 \\ \hline \end{array} \end{array}   \text { ANOVA }   \begin{array}{llcc} \hline&\text { df }&\text { SS } &\text { MS }&\text {F }&\text { Significance F } \\ \text { Regression } & 1 & 335.0472 & 335.0473 & 178.3859 \\ \text { Residual } & & & 1.8782 & \\ \text { Total } & 50 & 427.0798 & & \end{array}     -Referring to Table 13-9, the 90% confidence interval for the average change in SALARY (in thousands of dollars) as a result of spending an extra hour per day studying is</strong> A) wider than [-2.70159, -1.08654]. B) wider than [0.8321927, 1.12697]. C) narrower than [0.8321927, 1.12697]. D) narrower than [-2.70159, -1.08654]. <div style=padding-top: 35px>

-Referring to Table 13-9, the 90% confidence interval for the average change in SALARY (in thousands of dollars) as a result of spending an extra hour per day studying is

A) wider than [-2.70159, -1.08654].
B) wider than [0.8321927, 1.12697].
C) narrower than [0.8321927, 1.12697].
D) narrower than [-2.70159, -1.08654].
Question
Which of the following assumptions concerning the probability distribution of the random error term is stated incorrectly?

A) The variance of the distribution increases as X increases.
B) The distribution is normal.
C) The errors are independent.
D) The mean of the distribution is 0.
Question
TABLE 13-9
It is believed that, the average numbers of hours spent studying per day (HOURS) during undergraduate education should have a positive linear relationship with the starting salary (SALARY, measured in thousands of dollars per month) after graduation. Given below is the Excel output from regressing starting salary on number of hours spent studying per day for a sample of 51 students. NOTE: Some of the numbers in the output are purposely erased.
 Regression Statistics  Multiple R 0.8857 R Square 0.7845 Adjusted R Square 0.7801 Standard Error 1.3704 Observations 51\begin{array}{l}\text { Regression Statistics }\\\begin{array} { l c } \hline \text { Multiple R } & 0.8857 \\\text { R Square } & 0.7845 \\\text { Adjusted R Square } & 0.7801 \\\text { Standard Error } & 1.3704 \\\text { Observations } & 51 \\\hline\end{array}\end{array}
 ANOVA \text { ANOVA }
 df  SS  MS  Significance F  Regression 1335.0472335.0473178.3859 Residual 1.8782 Total 50427.0798\begin{array}{llcc}\hline&\text { df }&\text { SS } &\text { MS }&\text {F }&\text { Significance F } \\\text { Regression } & 1 & 335.0472 & 335.0473 & 178.3859 \\\text { Residual } & & & 1.8782 & \\\text { Total } & 50 & 427.0798 & &\end{array}

-Referring to Table 13-9, the value of the measured t-test statistic to test whether average SALARY depends linearly on HOURS is  Coefficients  Standaad Error t Stat p-value  Lower 95% Upper 95%  Intercept 1.89400.40184.71342.051E052.70151.0865 Hours 0.97950.073313.35615.944E180.83211.1269\begin{array} { l r c c c r r } & \text { Coefficients } & \text { Standaad Error } &t \text { Stat } & p \text {-value } & \text { Lower 95\%} & \text { Upper 95\% } \\\hline \text { Intercept } & - 1.8940 & 0.4018 & - 4.7134 & 2.051 \mathrm { E } - 05 & - 2.7015 & - 1.0865 \\\text { Hours } & 0.9795 & 0.0733 & 13.3561 & 5.944 \mathrm { E } - 18 & 0.8321 & 1.1269 \\\hline\end{array}

A) 13.3561.
B) 0.9795.
C) -4.7134.
D) -1.8940.
Question
TABLE 13-2
A candy bar manufacturer is interested in trying to estimate how sales are influenced by the price of their product. To do this, the company randomly chooses 6 small cities and offers the candy bar at different prices. Using candy bar sales as the dependent variable, the company will conduct a simple linear regression on the data below:
 City  Price ($) Sales  River Falls 1.30100 Hudson 1.6090 Ellsworth 1.8090 Prescott 2.0040 Rock Elm 2.4038 Stillwater 2.9032\begin{array} { l l c } \hline \text { City } & \text { Price } ( \$ ) & \text { Sales } \\\hline \text { River Falls } & 1.30 & 100 \\\text { Hudson } & 1.60 & 90 \\\text { Ellsworth } & 1.80 & 90 \\\text { Prescott } & 2.00 & 40 \\\text { Rock Elm } & 2.40 & 38 \\\text { Stillwater } & 2.90 & 32 \\\hline\end{array}

-Referring to Table 13-2, what is the percentage of the total variation in candy bar sales explained by the regression model?

A) 78.39%
B) 100%
C) 48.19%
D) 88.54%
Question
TABLE 13-9
It is believed that, the average numbers of hours spent studying per day (HOURS) during undergraduate education should have a positive linear relationship with the starting salary (SALARY, measured in thousands of dollars per month) after graduation. Given below is the Excel output from regressing starting salary on number of hours spent studying per day for a sample of 51 students. NOTE: Some of the numbers in the output are purposely erased.
 Regression Statistics  Multiple R 0.8857 R Square 0.7845 Adjusted R Square 0.7801 Standard Error 1.3704 Observations 51\begin{array}{lc}\text { Regression Statistics } & \\\hline \text { Multiple R } & 0.8857 \\\text { R Square } & 0.7845 \\\text { Adjusted R Square } & 0.7801 \\\text { Standard Error } & 1.3704 \\\text { Observations } & 51\end{array}
 ANOVA \text { ANOVA }
 df  SS  MS  Significance F  Regression 1335.0472335.0473178.3859 Residual 1.8782 Total 50427.0798\begin{array}{llcc}\hline&\text { df }&\text { SS } &\text { MS }&\text {F }&\text { Significance F } \\\text { Regression } & 1 & 335.0472 & 335.0473 & 178.3859 \\\text { Residual } & & & 1.8782 & \\\text { Total } & 50 & 427.0798 & &\end{array}

-Referring to Table 13-9, the degrees of freedom for the F test on whether HOURS affects SALARY are  Coefficients  Standaad Error t Stat p-value  Lower 95%  Upper 95%  Intercept 1.89400.40184.71342.051E052.70151.0865 Hours 0.97950.073313.35615.944E180.83211.1269\begin{array} { l r c c c r r } & \text { Coefficients } & \text { Standaad Error } &t \text { Stat } & p \text {-value } & \text { Lower 95\% } & \text { Upper 95\% } \\\hline \text { Intercept } & - 1.8940 & 0.4018 & - 4.7134 & 2.051 \mathrm { E } - 05 & - 2.7015 & - 1.0865 \\\text { Hours } & 0.9795 & 0.0733 & 13.3561 & 5.944 \mathrm { E } - 18 & 0.8321 & 1.1269 \\\hline\end{array}

A) 50, 1.
B) 1, 50.
C) 49, 1.
D) 1, 49.
Question
If you wanted to find out if alcohol consumption (measured in fluid oz.) and grade point average on a 4-point scale are linearly related, you would perform a

A) a t test for a correlation coefficient.
B) ?2 test for independence.
C) ?2 test for the difference in two proportions.
D) a Z test for the difference in two proportions.
Question
Testing for the existence of correlation is equivalent to

A) the confidence interval estimate for predicting Y.
B) testing for the existence of the slope (?1).
C) testing for the existence of the Y-intercept (?2).
D) none of the above
Question
TABLE 13-9
It is believed that, the average numbers of hours spent studying per day (HOURS) during undergraduate education should have a positive linear relationship with the starting salary (SALARY, measured in thousands of dollars per month) after graduation. Given below is the Excel output from regressing starting salary on number of hours spent studying per day for a sample of 51 students. NOTE: Some of the numbers in the output are purposely erased.
 Regression Statistics  Multiple R 0.8857 R Square 0.7845 Adjusted R Square 0.7801 Standard Error 1.3704 Observations 51\begin{array}{lc}\text { Regression Statistics } & \\\hline \text { Multiple R } & 0.8857 \\\text { R Square } & 0.7845 \\\text { Adjusted R Square } & 0.7801 \\\text { Standard Error } & 1.3704 \\\text { Observations } & 51\end{array}
 ANOVA \text { ANOVA }
 df  SS  MS  Significance F  Regression 1335.0472335.0473178.3859 Residual 1.8782 Total 50427.0798\begin{array}{llcc}\hline&\text { df }&\text { SS } &\text { MS }&\text {F }&\text { Significance F } \\\text { Regression } & 1 & 335.0472 & 335.0473 & 178.3859 \\\text { Residual } & & & 1.8782 & \\\text { Total } & 50 & 427.0798 & &\end{array}  <strong>TABLE 13-9 It is believed that, the average numbers of hours spent studying per day (HOURS) during undergraduate education should have a positive linear relationship with the starting salary (SALARY, measured in thousands of dollars per month) after graduation. Given below is the Excel output from regressing starting salary on number of hours spent studying per day for a sample of 51 students. NOTE: Some of the numbers in the output are purposely erased.  \begin{array}{lc} \text { Regression Statistics } & \\ \hline \text { Multiple R } & 0.8857 \\ \text { R Square } & 0.7845 \\ \text { Adjusted R Square } & 0.7801 \\ \text { Standard Error } & 1.3704 \\ \text { Observations } & 51 \end{array}   \text { ANOVA }   \begin{array}{llcc} \hline&\text { df }&\text { SS } &\text { MS }&\text {F }&\text { Significance F } \\ \text { Regression } & 1 & 335.0472 & 335.0473 & 178.3859 \\ \text { Residual } & & & 1.8782 & \\ \text { Total } & 50 & 427.0798 & & \end{array}     -Referring to Table 13-9, the estimated average change in salary (in thousands of dollars) as a result of spending an extra hour per day studying is</strong> A) 0.9795. B) -1.8940. C) 0.7845. D) 335.0473. <div style=padding-top: 35px>

-Referring to Table 13-9, the estimated average change in salary (in thousands of dollars) as a result of spending an extra hour per day studying is

A) 0.9795.
B) -1.8940.
C) 0.7845.
D) 335.0473.
Question
TABLE 13-12
The manager of the purchasing department of a large banking organization would like to develop a model to predict the amount of time (measured in hours) it takes to process invoices. Data are collected from a sample of 30 days, and the number of invoices processed and completion time in hours is recorded. Below is the regression output:
 Regression  Statistics  Multiple R 0.9947 R Square 0.8924 Adjusted R Square 0.8886 Standard Error 0.3342 Observations 30\begin{array} { l c } { \text { Regression } \text { Statistics } } \\\hline \text { Multiple R } & 0.9947 \\\text { R Square } & 0.8924 \\\text { Adjusted R Square } & 0.8886 \\\text { Standard Error } & 0.3342 \\\text { Observations } & 30 \\\hline\end{array}  ANOVA df SS  MS F Significance F Regression 125.943825.9438232.22004.3946E15 Residual 283.12820.1117 Total 2929.072\begin{array}{l}\text { ANOVA }\\\begin{array} { l c c c c c } & d f & \text { SS } & \text { MS } & F & \text { Significance } F \\\text { Regression } & 1 & 25.9438&25 .9438 & 232.2200&4 .3946 \mathrm { E } - 15 \\\text { Residual } & 28 & 3.12820 .1117 & \\\text { Total } & 29 & 29.072 \\\hline\end{array}\end{array}  Coefficients  Standard Enror t Stat p-value  Lower 95%  Upper 95%  Invoices 0.40240.12363.25590.00300.14920.6555 Processed 0.01260.000815.23884.3946E150.01090.0143\begin{array} { l r r r r r r } \hline & \text { Coefficients } & \text { Standard Enror } & t \text { Stat } & p \text {-value } & \text { Lower 95\% } & \text { Upper 95\% } \\\hline \text { Invoices } & 0.4024 & 0.1236 & 3.2559 & 0.0030 & 0.1492 & 0.6555 \\\text { Processed } & 0.0126 & 0.0008 & 15.2388 & 4.3946 \mathrm { E } 15 & 0.0109 & 0.0143\end{array}  Coefficients  Standard Enor  t Stat p-value  Lower 95%  Upper 95%  Invoices 0.40240.12363.25590.00300.14920.6555\begin{array}{rrrrrrr} & \text { Coefficients } & \text { Standard Enor } & \text { t Stat } & p \text {-value } & \text { Lower 95\% } & \text { Upper 95\% } \\\hline \text { Invoices } & 0.4024 & 0.1236 & 3.2559 & 0.0030 & 0.1492 & 0.6555\end{array}
 <strong>TABLE 13-12 The manager of the purchasing department of a large banking organization would like to develop a model to predict the amount of time (measured in hours) it takes to process invoices. Data are collected from a sample of 30 days, and the number of invoices processed and completion time in hours is recorded. Below is the regression output:  \begin{array} { l c } { \text { Regression } \text { Statistics } } \\ \hline \text { Multiple R } & 0.9947 \\ \text { R Square } & 0.8924 \\ \text { Adjusted R Square } & 0.8886 \\ \text { Standard Error } & 0.3342 \\ \text { Observations } & 30 \\ \hline \end{array}   \begin{array}{l} \text { ANOVA }\\ \begin{array} { l c c c c c } & d f & \text { SS } & \text { MS } & F & \text { Significance } F \\ \text { Regression } & 1 & 25.9438&25 .9438 & 232.2200&4 .3946 \mathrm { E } - 15 \\ \text { Residual } & 28 & 3.12820 .1117 & \\ \text { Total } & 29 & 29.072 \\ \hline \end{array} \end{array}   \begin{array} { l r r r r r r } \hline & \text { Coefficients } & \text { Standard Enror } & t \text { Stat } & p \text {-value } & \text { Lower 95\% } & \text { Upper 95\% } \\ \hline \text { Invoices } & 0.4024 & 0.1236 & 3.2559 & 0.0030 & 0.1492 & 0.6555 \\ \text { Processed } & 0.0126 & 0.0008 & 15.2388 & 4.3946 \mathrm { E } 15 & 0.0109 & 0.0143 \end{array}   \begin{array}{rrrrrrr} & \text { Coefficients } & \text { Standard Enor } & \text { t Stat } & p \text {-value } & \text { Lower 95\% } & \text { Upper 95\% } \\ \hline \text { Invoices } & 0.4024 & 0.1236 & 3.2559 & 0.0030 & 0.1492 & 0.6555 \end{array}       -Referring to Table 13-12, the 90% confidence interval for the average change in the amount of time needed as a result of processing one additional invoice is</strong> A) narrower than [0.0109, 0.0143]. B) wider than [0.0109, 0.0143]. C) narrower than [0.1492, 0.6555]. D) wider than [0.1492, 0.6555]. <div style=padding-top: 35px>   <strong>TABLE 13-12 The manager of the purchasing department of a large banking organization would like to develop a model to predict the amount of time (measured in hours) it takes to process invoices. Data are collected from a sample of 30 days, and the number of invoices processed and completion time in hours is recorded. Below is the regression output:  \begin{array} { l c } { \text { Regression } \text { Statistics } } \\ \hline \text { Multiple R } & 0.9947 \\ \text { R Square } & 0.8924 \\ \text { Adjusted R Square } & 0.8886 \\ \text { Standard Error } & 0.3342 \\ \text { Observations } & 30 \\ \hline \end{array}   \begin{array}{l} \text { ANOVA }\\ \begin{array} { l c c c c c } & d f & \text { SS } & \text { MS } & F & \text { Significance } F \\ \text { Regression } & 1 & 25.9438&25 .9438 & 232.2200&4 .3946 \mathrm { E } - 15 \\ \text { Residual } & 28 & 3.12820 .1117 & \\ \text { Total } & 29 & 29.072 \\ \hline \end{array} \end{array}   \begin{array} { l r r r r r r } \hline & \text { Coefficients } & \text { Standard Enror } & t \text { Stat } & p \text {-value } & \text { Lower 95\% } & \text { Upper 95\% } \\ \hline \text { Invoices } & 0.4024 & 0.1236 & 3.2559 & 0.0030 & 0.1492 & 0.6555 \\ \text { Processed } & 0.0126 & 0.0008 & 15.2388 & 4.3946 \mathrm { E } 15 & 0.0109 & 0.0143 \end{array}   \begin{array}{rrrrrrr} & \text { Coefficients } & \text { Standard Enor } & \text { t Stat } & p \text {-value } & \text { Lower 95\% } & \text { Upper 95\% } \\ \hline \text { Invoices } & 0.4024 & 0.1236 & 3.2559 & 0.0030 & 0.1492 & 0.6555 \end{array}       -Referring to Table 13-12, the 90% confidence interval for the average change in the amount of time needed as a result of processing one additional invoice is</strong> A) narrower than [0.0109, 0.0143]. B) wider than [0.0109, 0.0143]. C) narrower than [0.1492, 0.6555]. D) wider than [0.1492, 0.6555]. <div style=padding-top: 35px>

-Referring to Table 13-12, the 90% confidence interval for the average change in the amount of time needed as a result of processing one additional invoice is

A) narrower than [0.0109, 0.0143].
B) wider than [0.0109, 0.0143].
C) narrower than [0.1492, 0.6555].
D) wider than [0.1492, 0.6555].
Question
TABLE 13-2
A candy bar manufacturer is interested in trying to estimate how sales are influenced by the price of their product. To do this, the company randomly chooses 6 small cities and offers the candy bar at different prices. Using candy bar sales as the dependent variable, the company will conduct a simple linear regression on the data below:
 City  Price ($) Sales  River Falls 1.30100 Hudson 1.6090 Ellsworth 1.8090 Prescott 2.0040 Rock Elm 2.4038 Stillwater 2.9032\begin{array} { l l c } \hline \text { City } & \text { Price } ( \$ ) & \text { Sales } \\\hline \text { River Falls } & 1.30 & 100 \\\text { Hudson } & 1.60 & 90 \\\text { Ellsworth } & 1.80 & 90 \\\text { Prescott } & 2.00 & 40 \\\text { Rock Elm } & 2.40 & 38 \\\text { Stillwater } & 2.90 & 32 \\\hline\end{array}

-Referring to Table 13-2, if the price of the candy bar is set at $2, the estimated average sales will be

A) 30.
B) 90.
C) 100.
D) 65.
Question
What do we mean when we say that a simple linear regression model is "statistically" useful?

A) The model is an excellent predictor of Y.
B) The model is a better predictor of Y than the sample mean, Y.
C) The model is "practically" useful for predicting Y.
D) All the statistics computed from the sample make sense.
Question
The sample correlation coefficient between X and Y is 0.375. It has been found out that the p- value is 0.256 when testing H0 : q = 0 against the one- sided alternative H0 : q > 0. To test H0 : q = 0 against the two- sided alternative H0 : q × 0 at a significance level of 0.2, the p- value is

A) (0.256)(2).
B) 1 - 0.256/2.
C) 0.256/2.
D) 1 - 0.256.
Question
TABLE 13-12
The manager of the purchasing department of a large banking organization would like to develop a model to predict the amount of time (measured in hours) it takes to process invoices. Data are collected from a sample of 30 days, and the number of invoices processed and completion time in hours is recorded. Below is the regression output:
 Regression  Statistics  Multiple R 0.9947 R Square 0.8924 Adjusted R Square 0.8886 Standard Error 0.3342 Observations 30\begin{array} { l c } { \text { Regression } \text { Statistics } } \\\hline \text { Multiple R } & 0.9947 \\\text { R Square } & 0.8924 \\\text { Adjusted R Square } & 0.8886 \\\text { Standard Error } & 0.3342 \\\text { Observations } & 30 \\\hline\end{array} ANOVA\text{ANOVA}

 dfSS M S F Significance F Regression 125.943825.9438232.22004.3946E15Residual 283.12820.1117 Total 2929.072\begin{array}{llcc}&\text { df} &\text {SS } &\text {M S}&\text { F}&\text { Significance F}\\\text { Regression } &1&25.9438&25 .9438&232 .2200&4 .3946 \mathrm{E}-15\\\text {Residual }&28 &3.12820 .1117 \\\text { Total } & 29 & 29.072\end{array}
 <strong>TABLE 13-12 The manager of the purchasing department of a large banking organization would like to develop a model to predict the amount of time (measured in hours) it takes to process invoices. Data are collected from a sample of 30 days, and the number of invoices processed and completion time in hours is recorded. Below is the regression output:  \begin{array} { l c } { \text { Regression } \text { Statistics } } \\ \hline \text { Multiple R } & 0.9947 \\ \text { R Square } & 0.8924 \\ \text { Adjusted R Square } & 0.8886 \\ \text { Standard Error } & 0.3342 \\ \text { Observations } & 30 \\ \hline \end{array}   \text{ANOVA}    \begin{array}{llcc} &\text { df} &\text {SS } &\text {M S}&\text { F}&\text { Significance F}\\ \text { Regression } &1&25.9438&25 .9438&232 .2200&4 .3946 \mathrm{E}-15\\ \text {Residual }&28 &3.12820 .1117 \\ \text { Total } & 29 & 29.072 \end{array}         -Referring to Table 13-12, the degrees of freedom for the t test on whether the number of invoices processed affects the amount of time are</strong> A) 1. B) 28. C) 30. D) 29. <div style=padding-top: 35px>
 <strong>TABLE 13-12 The manager of the purchasing department of a large banking organization would like to develop a model to predict the amount of time (measured in hours) it takes to process invoices. Data are collected from a sample of 30 days, and the number of invoices processed and completion time in hours is recorded. Below is the regression output:  \begin{array} { l c } { \text { Regression } \text { Statistics } } \\ \hline \text { Multiple R } & 0.9947 \\ \text { R Square } & 0.8924 \\ \text { Adjusted R Square } & 0.8886 \\ \text { Standard Error } & 0.3342 \\ \text { Observations } & 30 \\ \hline \end{array}   \text{ANOVA}    \begin{array}{llcc} &\text { df} &\text {SS } &\text {M S}&\text { F}&\text { Significance F}\\ \text { Regression } &1&25.9438&25 .9438&232 .2200&4 .3946 \mathrm{E}-15\\ \text {Residual }&28 &3.12820 .1117 \\ \text { Total } & 29 & 29.072 \end{array}         -Referring to Table 13-12, the degrees of freedom for the t test on whether the number of invoices processed affects the amount of time are</strong> A) 1. B) 28. C) 30. D) 29. <div style=padding-top: 35px>   <strong>TABLE 13-12 The manager of the purchasing department of a large banking organization would like to develop a model to predict the amount of time (measured in hours) it takes to process invoices. Data are collected from a sample of 30 days, and the number of invoices processed and completion time in hours is recorded. Below is the regression output:  \begin{array} { l c } { \text { Regression } \text { Statistics } } \\ \hline \text { Multiple R } & 0.9947 \\ \text { R Square } & 0.8924 \\ \text { Adjusted R Square } & 0.8886 \\ \text { Standard Error } & 0.3342 \\ \text { Observations } & 30 \\ \hline \end{array}   \text{ANOVA}    \begin{array}{llcc} &\text { df} &\text {SS } &\text {M S}&\text { F}&\text { Significance F}\\ \text { Regression } &1&25.9438&25 .9438&232 .2200&4 .3946 \mathrm{E}-15\\ \text {Residual }&28 &3.12820 .1117 \\ \text { Total } & 29 & 29.072 \end{array}         -Referring to Table 13-12, the degrees of freedom for the t test on whether the number of invoices processed affects the amount of time are</strong> A) 1. B) 28. C) 30. D) 29. <div style=padding-top: 35px>

-Referring to Table 13-12, the degrees of freedom for the t test on whether the number of invoices processed affects the amount of time are

A) 1.
B) 28.
C) 30.
D) 29.
Question
The least squares method minimizes which of the following?

A) SSR
B) SST
C) SSE
D) all of the above
Question
TABLE 13-5
The managing partner of an advertising agency believes that his company's sales are related to the industry sales. He uses Microsoft Excel's Data Analysis tool to analyze the last 4 years of quarterly data with the following results:
 Regression Statistics  Multiple R 0.802 R Square 0.643 Adjusted R Square 0.618 Standard Error SYX0.9224 Observations 16\begin{array}{l}\text { Regression Statistics }\\\begin{array} { l c } \hline \text { Multiple R } & 0.802 \\\text { R Square } & 0.643 \\\text { Adjusted R Square } & 0.618 \\\text { Standard Error } S Y X & 0.9224 \\\text { Observations } & 16 \\\hline\end{array}\end{array}  ANOVA df SS  MS F Sig. F Regression 121.49721.49725.270.000 Error 1411.9120.851 Total 1533.409\begin{array}{l}\text { ANOVA }\\\begin{array} { l r l r l r } \hline & d f & \text { SS } & \text { MS } & F & \text { Sig. } F \\\hline \text { Regression } & 1 & 21.497 & 21.497 & 25.27 & 0.000 \\\text { Error } & 14 & 11.912 & 0.851 & & \\\text { Total } & 15 & 33.409 & & &\end{array}\end{array}  Predictor  Coefficients Standard Error t Stat p-value  Intercept 3.9621.4402.750.016 Industry 0.0404510.0080485.030.000\begin{array}{l}\text { Predictor }&\text { Coefficients}&\text { Standard Error }& t \text { Stat } &p \text {-value }\\\text { Intercept } & 3.962 & 1.440 & 2.75 & 0.016\\\text { Industry } & 0.040451 & 0.008048 & 5.03 & 0.000\end{array} Durbin- Watson Statistic 1.59

-Referring to Table 13-5, the partner wants to test for autocorrelation using the Durbin-Watson statistic. Using a level of significance of 0.05, the decision he should make is

A) there is no evidence of autocorrelation.
B) there is not enough information to perform the test.
C) there is evidence of autocorrelation.
D) the test is unable to make a definite conclusion.
Question
TABLE 13-2
A candy bar manufacturer is interested in trying to estimate how sales are influenced by the price of their product. To do this, the company randomly chooses 6 small cities and offers the candy bar at different prices. Using candy bar sales as the dependent variable, the company will conduct a simple linear regression on the data below:
 City  Price ($) Sales  River Falls 1.30100 Hudson 1.6090 Ellsworth 1.8090 Prescott 2.0040 Rock Elm 2.4038 Stillwater 2.9032\begin{array} { l l c } \hline \text { City } & \text { Price } ( \$ ) & \text { Sales } \\\hline \text { River Falls } & 1.30 & 100 \\\text { Hudson } & 1.60 & 90 \\\text { Ellsworth } & 1.80 & 90 \\\text { Prescott } & 2.00 & 40 \\\text { Rock Elm } & 2.40 & 38 \\\text { Stillwater } & 2.90 & 32 \\\hline\end{array}

-Referring to Table 13-2, to test whether a change in price will have any impact on average sales, what would be the critical values? Use ? = 0.05.

A) ±2.7765
B) ±2.5706
C) ±3.1634
D) ±3.4954
Question
If the correlation coefficient (r) = 1.00, then

A) there is no explained variation.
B) there is no unexplained variation.
C) the Y-intercept (b0) must equal 0.
D) the explained variation equals the unexplained variation.
Question
If the plot of the residuals is fan shaped, which assumption is violated?

A) independence of errors
B) normality
C) homoscedasticity
D) No assumptions are violated; the graph should resemble a fan.
Question
TABLE 13-12
The manager of the purchasing department of a large banking organization would like to develop a model to predict the amount of time (measured in hours) it takes to process invoices. Data are collected from a sample of 30 days, and the number of invoices processed and completion time in hours is recorded. Below is the regression output:
 Regression Statistics  Multiple R 0.9947 R Square 0.8924 Adjusted R Square 0.8886 Standard Error 0.3342 ations 30\begin{array}{l}\text { Regression Statistics }\\\begin{array} { l c } \hline \text { Multiple R } & 0.9947 \\\text { R Square } & 0.8924 \\\text { Adjusted R Square } & 0.8886 \\\text { Standard Error } & 0.3342 \\\text { ations } & 30 \\\hline\end{array}\end{array}

 d f  SS MS F  Significance F Regression125.943825.9438232.22004.3946E15Residual 283.12820.1117Total 2929.072\begin{array}{lrrccc}\hline & \text { d f } & \text { SS } & \text {MS} & \text { F } & \text { Significance F } \\\hline \text {Regression} & 1 & 25.9438 & 25.9438 & 232.2200 & 4.3946 \mathrm{E}-15 \\ \text {Residual }& 28 & 3.1282 & 0.1117 & & \\ \text {Total }& 29 & 29.072 & & & \\\hline\end{array}

 Coefficients  Standard Error  t Stat  p -valueLower 95%Upper 95% Invoices 0.40240.12363.25590.00300.14920.6555Processed 0.01260.000815.23884.3946E150.01090.0143\begin{array}{lrrrrrr}\hline & \text { Coefficients }& \text { Standard Error }& \text { t Stat }& \text { p -value}& \text {Lower 95\%} & \text {Upper 95\%} \\\hline \text { Invoices }& 0.4024 & 0.1236 & 3.2559 & 0.0030 & 0.1492 & 0.6555 \\ \text {Processed }& 0.0126 & 0.0008 & 15.2388 & 4.3946 \mathrm{E}-15 & 0.0109 & 0.0143 \\\hline\end{array}

 <strong>TABLE 13-12 The manager of the purchasing department of a large banking organization would like to develop a model to predict the amount of time (measured in hours) it takes to process invoices. Data are collected from a sample of 30 days, and the number of invoices processed and completion time in hours is recorded. Below is the regression output:  \begin{array}{l} \text { Regression Statistics }\\ \begin{array} { l c } \hline \text { Multiple R } & 0.9947 \\ \text { R Square } & 0.8924 \\ \text { Adjusted R Square } & 0.8886 \\ \text { Standard Error } & 0.3342 \\ \text { ations } & 30 \\ \hline \end{array} \end{array}    \begin{array}{lrrccc} \hline & \text { d f } &  \text { SS } &  \text {MS} &  \text { F } & \text { Significance  F } \\ \hline  \text {Regression} & 1 & 25.9438 & 25.9438 & 232.2200 & 4.3946 \mathrm{E}-15  \\ \text {Residual }& 28 & 3.1282 & 0.1117 & & \\ \text {Total }& 29 & 29.072 & & & \\ \hline \end{array}    \begin{array}{lrrrrrr} \hline & \text { Coefficients }& \text { Standard Error }& \text { t Stat }&  \text { p -value}& \text {Lower 95\%} &  \text {Upper 95\%} \\ \hline \text { Invoices }& 0.4024 & 0.1236 & 3.2559 & 0.0030 & 0.1492 & 0.6555 \\ \text {Processed }& 0.0126 & 0.0008 & 15.2388 &  4.3946 \mathrm{E}-15  & 0.0109 & 0.0143 \\ \hline \end{array}          -Referring to Table 13-12, the p-value of the measured t-test statistic to test whether the number of invoices processed affects the amount of time is</strong> A) (0.0030)/2. B) 0.0030. C) 4.3946E-15. D) (4.3946E-15)/2. <div style=padding-top: 35px>

 <strong>TABLE 13-12 The manager of the purchasing department of a large banking organization would like to develop a model to predict the amount of time (measured in hours) it takes to process invoices. Data are collected from a sample of 30 days, and the number of invoices processed and completion time in hours is recorded. Below is the regression output:  \begin{array}{l} \text { Regression Statistics }\\ \begin{array} { l c } \hline \text { Multiple R } & 0.9947 \\ \text { R Square } & 0.8924 \\ \text { Adjusted R Square } & 0.8886 \\ \text { Standard Error } & 0.3342 \\ \text { ations } & 30 \\ \hline \end{array} \end{array}    \begin{array}{lrrccc} \hline & \text { d f } &  \text { SS } &  \text {MS} &  \text { F } & \text { Significance  F } \\ \hline  \text {Regression} & 1 & 25.9438 & 25.9438 & 232.2200 & 4.3946 \mathrm{E}-15  \\ \text {Residual }& 28 & 3.1282 & 0.1117 & & \\ \text {Total }& 29 & 29.072 & & & \\ \hline \end{array}    \begin{array}{lrrrrrr} \hline & \text { Coefficients }& \text { Standard Error }& \text { t Stat }&  \text { p -value}& \text {Lower 95\%} &  \text {Upper 95\%} \\ \hline \text { Invoices }& 0.4024 & 0.1236 & 3.2559 & 0.0030 & 0.1492 & 0.6555 \\ \text {Processed }& 0.0126 & 0.0008 & 15.2388 &  4.3946 \mathrm{E}-15  & 0.0109 & 0.0143 \\ \hline \end{array}          -Referring to Table 13-12, the p-value of the measured t-test statistic to test whether the number of invoices processed affects the amount of time is</strong> A) (0.0030)/2. B) 0.0030. C) 4.3946E-15. D) (4.3946E-15)/2. <div style=padding-top: 35px>


-Referring to Table 13-12, the p-value of the measured t-test statistic to test whether the number of invoices processed affects the amount of time is

A) (0.0030)/2.
B) 0.0030.
C) 4.3946E-15.
D) (4.3946E-15)/2.
Question
Assuming a linear relationship between X and Y, if the coefficient of correlation (r) equals - 0.30,

A) the slope (b1) is negative.
B) there is no correlation.
C) variable X is larger than variable Y.
D) the variance of X is negative.
Question
TABLE 13-01
A large national bank charges local companies for using their services. A bank official reported the results of a regression analysis designed to predict the bank's charges (Y) -- measured in dollars per month -- for services rendered to local companies. One independent variable used to predict service charge to a company is the company's sales revenue (X) -- measured in millions of dollars. Data for 21 companies who use the bank's services were used to fit the model:
E(Y)=β0+β1XE(Y)=\beta_{0}+\beta_{1} X
The results of the simple linear regression are provided below.
Y^=2,700+20X, SYX=65, two-tailed p value =0.034( for testing β1)\hat{Y}=-2,700+20 X, \mathrm{~S}_{Y X}=65, \text { two-tailed } p \text { value }=0.034\left(\text { for testing } \beta_{1}\right)


-Referring to Table 13-1, interpret the p-value for testing whether þ1 exceeds 0.

A) There is sufficient evidence (at the ? = 0.05) to conclude that sales revenue (X) is a useful linear predictor of service charge (Y).
B) For every $1 million increase in sales revenue, we expect a service charge to increase $0.034.
C) There is insufficient evidence (at the ? = 0.10) to conclude that sales revenue (X) is a useful linear predictor of service charge (Y).
D) Sales revenue (X) is a poor predictor of service charge (Y).
Question
TABLE 13-2
A candy bar manufacturer is interested in trying to estimate how sales are influenced by the price of their product. To do this, the company randomly chooses 6 small cities and offers the candy bar at different prices. Using candy bar sales as the dependent variable, the company will conduct a simple linear regression on the data below:
 City  Price ($) Sales  River Falls 1.30100 Hudson 1.6090 Ellsworth 1.8090 Prescott 2.0040 Rock Elm 2.4038 Stillwater 2.9032\begin{array} { l l c } \hline \text { City } & \text { Price } ( \$ ) & \text { Sales } \\\hline \text { River Falls } & 1.30 & 100 \\\text { Hudson } & 1.60 & 90 \\\text { Ellsworth } & 1.80 & 90 \\\text { Prescott } & 2.00 & 40 \\\text { Rock Elm } & 2.40 & 38 \\\text { Stillwater } & 2.90 & 32 \\\hline\end{array}

-Referring to Table 13-2, what percentage of the total variation in candy bar sales is explained by prices?

A) 100%
B) 78.39%
C) 88.54%
D) 48.19%
Question
The Y-intercept (b0) represents the

A) change in estimated average Y per unit change in X.
B) variation around the sample regression line.
C) predicted value of Y.
D) estimated average Y when X = 0.
Question
TABLE 13- 11
A company that has the distribution rights to home video sales of previously released movies would like to use the box office gross (in millions of dollars) to estimate the number of units (in thousands of units) that it can expect to sell. Following is the output from a simple linear regression along with the residual plot and normal probability plot obtained from a data set of 30 different movie titles:
 Regression Statistics  Multiple R 0.8531 RSquare 0.7278 Adjusted R Square 0.7180 Standard Error 47.8668 Observations 30\begin{array}{l}\text { Regression Statistics }\\\begin{array} { l c } \hline \text { Multiple R } & 0.8531 \\\text { RSquare } & 0.7278 \\\text { Adjusted R Square } & 0.7180 \\\text { Standard Error } & 47.8668 \\\text { Observations } & 30\end{array}\end{array}
ANOVA
 d f SS  MS Significance FRegression 1171499.78171499.7874.85052.1259E09Residual2864154.422291.23Total29235654.20\begin{array}{lrrrrr}\hline &\text { d f}& \text { SS } & \text { MS } & \text {F }& \text {Significance F} \\\hline \text {Regression }& 1 & 171499.78 & 171499.78 & 74.8505 & 2.1259E-09 \\\text {Residual} & 28 & 64154.42 & 2291.23 & & \\\text {Total} & 29 & 235654.20 & & & \\\hline\end{array}

Coefficients  Standard Error t Stat  p -value Lower 95% Upper 95%  Intercept 76.535111.83186.46865.24E0752.2987100.7716Gross4.33310.50088.65162.13E093.30725.3590\begin{array}{lrrrrrr}\hline & \text {Coefficients }& \text { Standard Error}& \text { t Stat }& \text { p -value }& \text {Lower 95\% }& \text {Upper 95\% }\\\hline \text { Intercept }& 76.5351 & 11.8318 & 6.4686 & 5.24 \mathrm{E}-07& 52.2987 & 100.7716 \\ \text {Gross} & 4.3331 & 0.5008 & 8.6516 & 2.13 \mathrm{E}-09 & 3.3072 & 5.3590 \\\hline\end{array}

 <strong>TABLE 13- 11 A company that has the distribution rights to home video sales of previously released movies would like to use the box office gross (in millions of dollars) to estimate the number of units (in thousands of units) that it can expect to sell. Following is the output from a simple linear regression along with the residual plot and normal probability plot obtained from a data set of 30 different movie titles:  \begin{array}{l} \text { Regression Statistics }\\ \begin{array} { l c } \hline \text { Multiple R } & 0.8531 \\ \text { RSquare } & 0.7278 \\ \text { Adjusted R Square } & 0.7180 \\ \text { Standard Error } & 47.8668 \\ \text { Observations } & 30 \end{array} \end{array}  ANOVA  \begin{array}{lrrrrr} \hline &\text { d f}& \text { SS } & \text { MS } & \text {F }& \text {Significance F}  \\ \hline \text {Regression }& 1 & 171499.78 & 171499.78 & 74.8505 & 2.1259E-09 \\ \text {Residual} & 28 & 64154.42 & 2291.23 & & \\ \text {Total} & 29 & 235654.20 & & & \\ \hline\end{array}    \begin{array}{lrrrrrr} \hline &  \text {Coefficients }& \text { Standard Error}& \text { t  Stat }&  \text { p -value }&  \text {Lower 95\% }& \text {Upper 95\% }\\ \hline \text { Intercept }& 76.5351 & 11.8318 & 6.4686 & 5.24 \mathrm{E}-07& 52.2987 & 100.7716 \\ \text {Gross} & 4.3331 & 0.5008 & 8.6516 & 2.13 \mathrm{E}-09 & 3.3072 & 5.3590 \\ \hline \end{array}        -Referring to Table 13-11, which of the following is the correct interpretation for the slope coefficient?</strong> A) For each increase of 1 dollar in box office gross, expected home video units sold are estimated to increase by 4.3331 thousand units. B) For each increase of 1 million dollars in box office gross, expected home video units sold are estimated to increase by 4.3331 units. C) For each increase of 1 dollar in box office gross, expected home video units sold are estimated to increase by 4.3331 units. D) For each increase of 1 million dollars in box office gross, expected home video units sold are estimated to increase by 4.3331 thousand units. <div style=padding-top: 35px>   <strong>TABLE 13- 11 A company that has the distribution rights to home video sales of previously released movies would like to use the box office gross (in millions of dollars) to estimate the number of units (in thousands of units) that it can expect to sell. Following is the output from a simple linear regression along with the residual plot and normal probability plot obtained from a data set of 30 different movie titles:  \begin{array}{l} \text { Regression Statistics }\\ \begin{array} { l c } \hline \text { Multiple R } & 0.8531 \\ \text { RSquare } & 0.7278 \\ \text { Adjusted R Square } & 0.7180 \\ \text { Standard Error } & 47.8668 \\ \text { Observations } & 30 \end{array} \end{array}  ANOVA  \begin{array}{lrrrrr} \hline &\text { d f}& \text { SS } & \text { MS } & \text {F }& \text {Significance F}  \\ \hline \text {Regression }& 1 & 171499.78 & 171499.78 & 74.8505 & 2.1259E-09 \\ \text {Residual} & 28 & 64154.42 & 2291.23 & & \\ \text {Total} & 29 & 235654.20 & & & \\ \hline\end{array}    \begin{array}{lrrrrrr} \hline &  \text {Coefficients }& \text { Standard Error}& \text { t  Stat }&  \text { p -value }&  \text {Lower 95\% }& \text {Upper 95\% }\\ \hline \text { Intercept }& 76.5351 & 11.8318 & 6.4686 & 5.24 \mathrm{E}-07& 52.2987 & 100.7716 \\ \text {Gross} & 4.3331 & 0.5008 & 8.6516 & 2.13 \mathrm{E}-09 & 3.3072 & 5.3590 \\ \hline \end{array}        -Referring to Table 13-11, which of the following is the correct interpretation for the slope coefficient?</strong> A) For each increase of 1 dollar in box office gross, expected home video units sold are estimated to increase by 4.3331 thousand units. B) For each increase of 1 million dollars in box office gross, expected home video units sold are estimated to increase by 4.3331 units. C) For each increase of 1 dollar in box office gross, expected home video units sold are estimated to increase by 4.3331 units. D) For each increase of 1 million dollars in box office gross, expected home video units sold are estimated to increase by 4.3331 thousand units. <div style=padding-top: 35px>

-Referring to Table 13-11, which of the following is the correct interpretation for the slope coefficient?

A) For each increase of 1 dollar in box office gross, expected home video units sold are estimated to increase by 4.3331 thousand units.
B) For each increase of 1 million dollars in box office gross, expected home video units sold are estimated to increase by 4.3331 units.
C) For each increase of 1 dollar in box office gross, expected home video units sold are estimated to increase by 4.3331 units.
D) For each increase of 1 million dollars in box office gross, expected home video units sold are estimated to increase by 4.3331 thousand units.
Question
TABLE 13-01
A large national bank charges local companies for using their services. A bank official reported the results of a regression analysis designed to predict the bank's charges
E(Y)=β0+β1XE(Y)=\beta_{0}+\beta_{1} X
The results of the simple linear regression are provided below.
Y^=2,700+20X, SYX=65, two-tailed p value =0.034( for testing β1)\hat{Y}=-2,700+20 X, \mathrm{~S}_{Y X}=65, \text { two-tailed } p \text { value }=0.034\left(\text { for testing } \beta_{1}\right)

(Y) -- measured in dollars per month -- for services rendered to local companies. One independent variable used to predict service charge to a company is the company's sales revenue (X) -- measured in millions of dollars. Data for 21 companies who use the bank's services were used to fit the model:



-Referring to Table 13-1, interpret the estimate of a, the standard deviation of the random error term (standard error of the estimate) in the model.

A) About 95% of the observed service charges fall within $130 of the least squares line.
B) For every $1 million increase in sales revenue, we expect a service charge to increase $65.
C) About 95% of the observed service charges fall within $65 of the least squares line.
D) About 95% of the observed service charges equal their corresponding predicted values.
Question
TABLE 13-01
A large national bank charges local companies for using their services. A bank official reported the results of a regression analysis designed to predict the bank's charges (Y) -- measured in dollars per month -- for services rendered to local companies. One independent variable used to predict service charge to a company is the company's sales revenue (X) -- measured in millions of dollars. Data for 21 companies who use the bank's services were used to fit the model:
E(Y) =?0 + ?1X
The results of the simple linear regression are provided below.

Y^=2,700+20X,SYX=65 \hat{Y}=-2,700+20 X, S_{Y X}=65 , two-tailed p p value =0.034( =0.034\left(\right. for testing β1) \left.\beta_{1}\right)


-Referring to Table 13-1, a 95% confidence interval for ?1 is (15, 30). Interpret the interval.

A) At the ? = 0.05 level, there is no evidence of a linear relationship between service charge (Y) and sales revenue (X).
B) We are 95% confident that the mean service charge will fall between $15 and $30 per month.
C) We are 95% confident that average service charge (Y) will increase between $15 and $30 for every $1 million increase in sales revenue (X).
D) We are 95% confident that the sales revenue (X) will increase between $15 and $30 million for every $1 increase in service charge (Y).
Question
The slope (b1) represents

A) the estimated average change in Y per unit change in X.
B) predicted value of Y when X = 0.
C) the predicted value of Y.
D) variation around the line of regression.
Question
TABLE 13-7
An investment specialist claims that if one holds a portfolio that moves in the opposite direction to the market index like the S&P 500, then it is possible to reduce the variability of the portfolio's return. In other words, one can create a portfolio with positive returns but less exposure to risk. A sample of 26 years of S&P 500 index and a portfolio consisting of stocks of private prisons, which are believed to be negatively related to the S&P 500 index, is collected. A regression analysis was performed by regressing the returns of the prison stocks portfolio (Y) on the returns of S&P 500 index (X) to prove that the prison stocks portfolio is negatively related to the S&P 500 index at a 5% level of significance. The results are given in the following EXCEL output.
 Coefficients  Standard Enor  T Stat p-value  Intercept 4.8660042580.3574360913.613634418.7932E13 S&P 0.5025135060.0715971527.018624252.94942E07\begin{array} { l c c c c } \hline & \text { Coefficients } & \text { Standard Enor } & \text { T Stat } & p \text {-value } \\\hline \text { Intercept } & 4.866004258 & 0.35743609 & 13.61363441 & 8.7932 \mathrm { E } - 13 \\\text { S\&P } & - 0.502513506 & 0.071597152 -& 7.01862425 & 2.94942 \mathrm { E } - 07 \\\hline\end{array}

-Referring to Table 13-7, which of the following will be a correct conclusion?

A) We can reject the null hypothesis and, therefore, conclude that there is sufficient evidence to show that the prisons stock portfolio and S&P 500 index are negatively related.
B) We can reject the null hypothesis and conclude that there is not sufficient evidence to show that the prisons stock portfolio and S&P 500 index are negatively related.
C) We cannot reject the null hypothesis and, therefore, conclude that there is not sufficient evidence to show that the prisons stock portfolio and S&P 500 index are negatively related.
D) We cannot reject the null hypothesis and, therefore, conclude that there is sufficient evidence to show that the prisons stock portfolio and S&P 500 index are negatively related.
Question
 TABLE 13-10 \text { TABLE 13-10 } The management of a chain electronic store would like to develop a model for predicting the weekly sales (in thousand of dollars) for individual stores based on the number of customers who made purchases. A random sample of 12 stores yields the following results:
 Cugtomera  5aleg (Thougardaof Dollara) 90711.2092611.057138.217419.217809.4289810.085106.735297.024606.128729.526507.536037.25\begin{array} { | c | c | } \hline \text { Cugtomera } & \text { 5aleg (Thougardaof Dollara) } \\\hline \mathbf { 9 0 7 } & 11.20 \\\hline \mathbf { 9 2 6 } & 11.05 \\\hline 713 & \mathbf { 8 . 2 1 } \\\hline 741 & \mathbf { 9 . 2 1 } \\\hline 780 & \mathbf { 9 . 4 2 } \\\hline \mathbf { 8 9 8 } & \mathbf { 1 0 . 0 8 } \\\hline 510 & \mathbf { 6 . 7 3 } \\\hline 529 & 7.02 \\\hline 460 & \mathbf { 6 . 1 2 } \\\hline \mathbf { 8 7 2 } & \mathbf { 9 . 5 2 } \\\hline \mathbf { 6 5 0 } & 7.53 \\\hline \mathbf { 6 0 3 } & 7.25 \\\hline\end{array}

-Referring to Table 13-10, the residual plot indicates possible violation of which assumptions?

A) normality
B) autocorrelation
C) linearity of the relationship
D) homoscedasticity
Question
In a simple linear regression problem, r and b1

A) must have the same sign.
B) may have opposite signs.
C) must have opposite signs.
D) are equal.
Question
The standard error of the estimate is a measure of

A) the variation of the X variable.
B) total variation of the Y variable.
C) the variation around the sample regression line.
D) explained variation.
Question
TABLE 13-9
It is believed that, the average numbers of hours spent studying per day (HOURS) during undergraduate education should have a positive linear relationship with the starting salary (SALARY, measured in thousands of dollars per month) after graduation. Given below is the Excel output from regressing starting salary on number of hours spent studying per day for a sample of 51 students. NOTE: Some of the numbers in the output are purposely erased.
 Regression Stedistics  Multiple R 0.8857 RSquare 0.7845 Adjusted R Square 0.7801 Standard Error 1.3704\begin{array}{ll}\text { Regression Stedistics } \\\hline \text { Multiple R } & 0.8857 \\\text { RSquare } & 0.7845 \\\text { Adjusted R Square } & 0.7801 \\\text { Standard Error } & 1.3704\end{array}

 Observations 51\text { Observations } \quad 51

ANOVAANOVA
dfSSMSF Significance F Regression 1335.0472335.0473178.3859 Residual 1.8782 Total 50427.0798\begin{array}{lccrcc}\hline & d f & S S &{M S} & F & \text { Significance } F \\\hline \text { Regression } & 1 & 335.0472 & 335.0473 & 178.3859 \\\text { Residual } & & 1.8782 & \\\text { Total } & 50 & 427.0798 & & \\\hline\end{array}


 Coeffcients  Standard Error t Stat p-value  Lower 95% Upper 95%  Intercept 1.89400.40184.71342.051E052.70151.0865 Hours 0.97950.073313.35615.944E180.83211.1269\begin{array}{lccccrr}\hline & \text { Coeffcients } & \text { Standard Error }& t \text { Stat } & p \text {-value } & \text { Lower 95\%} & \text { Upper 95\% } \\\hline \text { Intercept } & -1.8940 & 0.4018 & -4.7134 & 2.051 \mathrm{E}-05 & -2.7015 & -1.0865 \\\text { Hours } & 0.9795 & 0.0733 & 13.3561 & 5.944 \mathrm{E}-18 & 0.8321 & 1.1269 \\\hline\end{array}



-Referring to Table 13-9, to test the claim that average SALARY depends positively on HOURS against the null hypothesis that average SALARY does not depend linearly on HOURS, the p-value of the test statistic is

A) (2.051E-05)/2.
B) 5.944E-18.
C) (5.944E-18)/2.
D) 2.051E-05.
Question
The strength of the linear relationship between two numerical variables may be measured by the

A) slope.
B) Y-intercept.
C) coefficient of correlation.
D) scatter diagram.
Question
TABLE 13-8
It is believed that GPA (grade point average, based on a four point scale) should have a positive linear relationship with ACT scores. Given below is the Excel output from regressing GPA on ACT scores using a data set of 8 randomly chosen students from a Big-Ten university.
 Regressing GPA on ACT Regression Statistics  Multiple R 0.7598 R Square 0.5774 Adjusted R Square 0.5069 Standard E rror 0.2691 Observations 8\begin{array}{l}\text { Regressing GPA on } \mathrm { ACT }\\\begin{array} { l c } \hline & \text { Regression Statistics } \\\hline \text { Multiple R } & 0.7598 \\\text { R Square } & 0.5774 \\\text { Adjusted R Square } & 0.5069 \\\text { Standard E rror } & 0.2691 \\\text { Observations } & 8\end{array}\end{array}

ANOVA
d fSS MS  F  Significance FRegression 10.59400.59408.19860.0286Residual 60.43470.0724Total 71.0287\begin{array}{lcllcc}\hline & \text {d f} & \text {SS} & \text { MS }& \text { F } & \text { Significance F} \\\hline \text {Regression }&1 & 0.5940 & 0.5940 & 8.1986 & 0.0286 \\ \text {Residual }& 6 & 0.4347 & 0.0724 & & \\ \text {Total }& 7 & 1.0287 & & & \\\hline\end{array}

Coefficients Standard Error T Stat  p -value Lower 95% Upper 95%  Intercept0.56810.92840.61190.56301.70362.8398ACT 0.10210.03562.86330.02860.01480.1895\begin{array}{lrrrrrr}\hline & \text {Coefficients }& \text {Standard Error }& \text {T Stat }& \text { p -value }& \text {Lower 95\%}& \text { Upper 95\% } \\\hline \text { Intercept} & 0.5681 & 0.9284 & 0.6119 & 0.5630 & -1.7036 & 2.8398 \\ \text {ACT }& 0.1021 & 0.0356 & 2.8633 & 0.0286 & 0.0148 & 0.1895 \\\hline\end{array}


-Referring to Table 13-8, the value of the measured (observed) test statistic of the F-test for H0 : ?1 = 0 versus H1 : ?1 ? 0

A) is always positive.
B) has the same sign as the corresponding t test statistic.
C) may be negative.
D) is always negative.
Question
The coefficient of determination (r2) tells us

A) the proportion of total variation that is explained.
B) that we should not partition the total variation.
C) whether r has any significance.
D) that the coefficient of correlation (r) is larger than 1.
Question
TABLE 13-9
It is believed that, the average numbers of hours spent studying per day (HOURS) during undergraduate education should have a positive linear relationship with the starting salary (SALARY, measured in thousands of dollars per month) after graduation. Given below is the Excel output from regressing starting salary on number of hours spent studying per day for a sample of 51 students. NOTE: Some of the numbers in the output are purposely erased.
 Regression Stedistics  Multiple R 0.8857 R Square 0.7845 Adjusted R Square 0.7801 Standard Error 1.3704 Observations 51\begin{array}{l}\text { Regression Stedistics }\\\begin{array} { l c } \hline \text { Multiple R } & 0.8857 \\\text { R Square } & 0.7845 \\ \text { Adjusted R Square } & 0.7801 \\\text { Standard Error } & 1.3704 \\\text { Observations } & 51\end{array}\end{array}  ANOVA \text { ANOVA }
dfSSMSF Significance F Regression 1335.0472335.0473178.3859 Residual 1.8782 Total 50427.0798\begin{array}{lccrcc}\hline & d f & S S &{M S} & F & \text { Significance } F \\\hline \text { Regression } & 1 & 335.0472 & 335.0473 & 178.3859 \\\text { Residual } & & & 1.8782 & \\\text { Total } & 50 & 427.0798 & &\end{array}  Coefficients  Standard Enror t Stat p-value  Lower 95%  Upper 95%  Intercept 1.89400.40184.71342.051E052.70151.0865 Hours 0.97950.073313.35615.944E180.83211.1269\begin{array} { l r c c c r r } & \text { Coefficients } & \text { Standard Enror } & t \text { Stat } & { p \text {-value } } & \text { Lower 95\% } & \text { Upper 95\% } \\\hline \text { Intercept } & - 1.8940 & 0.4018 & - 4.7134 & 2.051 \mathrm { E } - 05 & - 2.7015 & - 1.0865 \\\text { Hours } & 0.9795 & 0.0733 & 13.3561 & 5.944 \mathrm { E } - 18 & 0.8321 & 1.1269 \\\hline\end{array} \begin{array} { l r r r r r r } \end{array}

-Referring to Table 13-9, the error sum of squares (SSE) of the above regression is

A) 1.878215.
B) 335.047257.
C) 92.0325465.
D) 427.079804.
Question
TABLE 13-7
An investment specialist claims that if one holds a portfolio that moves in the opposite direction to the market index like the S&P 500, then it is possible to reduce the variability of the portfolio's return. In other words, one can create a portfolio with positive returns but less exposure to risk. A sample of 26 years of S&P 500 index and a portfolio consisting of stocks of private prisons, which are believed to be negatively related to the S&P 500 index, is collected. A regression analysis was performed by regressing the returns of the prison stocks portfolio (Y) on the returns of S&P 500 index (X) to prove that the prison stocks portfolio is negatively related to the S&P 500 index at a 5% level of significance. The results are given in the following EXCEL output.
 Coefficients  Standard Error  T Stat p-value  Intercept 4.8660042580.3574360913.613634418.7932E13 S& P 0.5025135060.0715971527.01862425294942E07\begin{array} { l c c c c } \hline & \text { Coefficients } & \text { Standard Error } & \text { T Stat } & { p \text {-value } } \\\hline \text { Intercept } & 4.866004258 & 0.35743609 & 13.61363441 & 8.7932 \mathrm { E } - 13 \\\text { S\& P } & - 0.502513506 & 0.071597152 &- 7.01862425 & 294942 \mathrm { E } - 07 \\\hline\end{array}

-Referring to Table 13-7, to test whether the prison stocks portfolio is negatively related to the S&P 500 index, the p-value of the associated test statistic is

A) 8.7932E- 13.
B) 2.94942E- 07.
C) (2.94942E- 07)2.
D) 2.94942E- 07/2.
Question
TABLE 13-2
A candy bar manufacturer is interested in trying to estimate how sales are influenced by the price of their product. To do this, the company randomly chooses 6 small cities and offers the candy bar at different prices. Using candy bar sales as the dependent variable, the company will conduct a simple linear regression on the data below:
 City  Price ($) Sales  River Falls 1.30100 Hudson 1.6090 Ellsworth 1.8090 Prescott 2.0040 Rock Elm 2.4038 Stillwater 2.9032\begin{array} { l l c } \hline \text { City } & \text { Price } ( \$ ) & \text { Sales } \\\hline \text { River Falls } & 1.30 & 100 \\\text { Hudson } & 1.60 & 90 \\\text { Ellsworth } & 1.80 & 90 \\\text { Prescott } & 2.00 & 40 \\\text { Rock Elm } & 2.40 & 38 \\\text { Stillwater } & 2.90 & 32 \\\hline\end{array}

-Referring to Table 13-2, what is the standard error of the regression slope estimate, Sb1?

A) 12.650
B) 0.784
C) 0.885
D) 16.299
Question
TABLE 13-2
A candy bar manufacturer is interested in trying to estimate how sales are influenced by the price of their product. To do this, the company randomly chooses 6 small cities and offers the candy bar at different prices. Using candy bar sales as the dependent variable, the company will conduct a simple linear regression on the data below:
 City  Price ($) Sales  River Falls 1.30100 Hudson 1.6090 Ellsworth 1.8090 Prescott 2.0040 Rock Elm 2.4038 Stillwater 2.9032\begin{array} { l l c } \hline \text { City } & \text { Price } ( \$ ) & \text { Sales } \\\hline \text { River Falls } & 1.30 & 100 \\\text { Hudson } & 1.60 & 90 \\\text { Ellsworth } & 1.80 & 90 \\\text { Prescott } & 2.00 & 40 \\\text { Rock Elm } & 2.40 & 38 \\\text { Stillwater } & 2.90 & 32 \\\hline\end{array}

-Referring to Table 13-2, what is the standard error of the estimate, SYX, for the data?

A) 0.885
B) 16.299
C) 0.784
D) 12.650
Question
The sample correlation coefficient between X and Y is 0.375. It has been found out that the p-value is 0.256 when testing H0 : ? = 0 against the two-sided alternative H0 :? ? 0. To test H0 : ? = 0 against the one-sided alternative H0 : ? < 0 at a significance level of 0.2, the p-value is

A) 1 - 0.256/2.
B) (0.256)2.
C) 1 - 0.256.
D) 0.256/2.
Question
The width of the prediction interval for the predicted value of Y is dependent on

A) the standard error of the estimate.
B) the sample size.
C) the value of X for which the prediction is being made.
D) all of the above
Question
The sample correlation coefficient between X and Y is 0.375. It has been found out that the p- value is 0.256 when testing H0 : ? = 0 against the two- sided alternative H0: ?? 0. To test H0 :? = 0 against the one- sided alternative H0 : ? > 0 at a significance level of 0.2, the p- value is

A) (0.256)2.
B) 1 - 0.256/2.
C) 0.256/2.
D) 1 - 0.256.
Question
Based on the residual plot below, you will conclude that there might be a violation of which of the following assumptions?
<strong>Based on the residual plot below, you will conclude that there might be a violation of which of the following assumptions?  </strong> A) Homoscedasticity B) Independence of errors C) Normality of errors D) Linearity of the relationship <div style=padding-top: 35px>

A) Homoscedasticity
B) Independence of errors
C) Normality of errors
D) Linearity of the relationship
Question
TABLE 13-12
The manager of the purchasing department of a large banking organization would like to develop a model to predict the amount of time (measured in hours) it takes to process invoices. Data are collected from a sample of 30 days, and the number of invoices processed and completion time in hours is recorded. Below is the regression output:
 Regression Statistics  Multiple R 0.9947 R Square 0.8924 Adjusted R Square 0.8886 Standard Error 0.3342 ations 30\begin{array}{l}\text { Regression Statistics }\\\begin{array} { l c } \hline \text { Multiple R } & 0.9947 \\\text { R Square } & 0.8924 \\\text { Adjusted R Square } & 0.8886 \\\text { Standard Error } & 0.3342 \\\text { ations } & 30 \\\hline\end{array}\end{array}

 d f  SS MS F  Significance F Regression125.943825.9438232.22004.3946E15Residual 283.12820.1117Total 2929.072\begin{array}{lrrccc}\hline & \text { d f } & \text { SS } & \text {MS} & \text { F } & \text { Significance F } \\\hline \text {Regression} & 1 & 25.9438 & 25.9438 & 232.2200 & 4.3946 \mathrm{E}-15 \\ \text {Residual }& 28 & 3.1282 & 0.1117 & & \\ \text {Total }& 29 & 29.072 & & & \\\hline\end{array}

 Coefficients  Standard Error  t Stat  p -valueLower 95%Upper 95% Invoices 0.40240.12363.25590.00300.14920.6555Processed 0.01260.000815.23884.3946E150.01090.0143\begin{array}{lrrrrrr}\hline & \text { Coefficients }& \text { Standard Error }& \text { t Stat }& \text { p -value}& \text {Lower 95\%} & \text {Upper 95\%} \\\hline \text { Invoices }& 0.4024 & 0.1236 & 3.2559 & 0.0030 & 0.1492 & 0.6555 \\ \text {Processed }& 0.0126 & 0.0008 & 15.2388 & 4.3946 \mathrm{E}-15 & 0.0109 & 0.0143 \\\hline\end{array}

 <strong>TABLE 13-12 The manager of the purchasing department of a large banking organization would like to develop a model to predict the amount of time (measured in hours) it takes to process invoices. Data are collected from a sample of 30 days, and the number of invoices processed and completion time in hours is recorded. Below is the regression output:  \begin{array}{l} \text { Regression Statistics }\\ \begin{array} { l c } \hline \text { Multiple R } & 0.9947 \\ \text { R Square } & 0.8924 \\ \text { Adjusted R Square } & 0.8886 \\ \text { Standard Error } & 0.3342 \\ \text { ations } & 30 \\ \hline \end{array} \end{array}    \begin{array}{lrrccc} \hline & \text { d f } &  \text { SS } &  \text {MS} &  \text { F } & \text { Significance  F } \\ \hline  \text {Regression} & 1 & 25.9438 & 25.9438 & 232.2200 & 4.3946 \mathrm{E}-15  \\ \text {Residual }& 28 & 3.1282 & 0.1117 & & \\ \text {Total }& 29 & 29.072 & & & \\ \hline \end{array}    \begin{array}{lrrrrrr} \hline & \text { Coefficients }& \text { Standard Error }& \text { t Stat }&  \text { p -value}& \text {Lower 95\%} &  \text {Upper 95\%} \\ \hline \text { Invoices }& 0.4024 & 0.1236 & 3.2559 & 0.0030 & 0.1492 & 0.6555 \\ \text {Processed }& 0.0126 & 0.0008 & 15.2388 &  4.3946 \mathrm{E}-15  & 0.0109 & 0.0143 \\ \hline \end{array}          -Referring to Table 13-12, the p-value of the measured F-test statistic to test whether the number of invoices processed affects the amount of time is</strong> A) (4.3946E-15)/2. B) (0.0030)/2. C) 0.0030. D) 4.3946E-15. <div style=padding-top: 35px>

 <strong>TABLE 13-12 The manager of the purchasing department of a large banking organization would like to develop a model to predict the amount of time (measured in hours) it takes to process invoices. Data are collected from a sample of 30 days, and the number of invoices processed and completion time in hours is recorded. Below is the regression output:  \begin{array}{l} \text { Regression Statistics }\\ \begin{array} { l c } \hline \text { Multiple R } & 0.9947 \\ \text { R Square } & 0.8924 \\ \text { Adjusted R Square } & 0.8886 \\ \text { Standard Error } & 0.3342 \\ \text { ations } & 30 \\ \hline \end{array} \end{array}    \begin{array}{lrrccc} \hline & \text { d f } &  \text { SS } &  \text {MS} &  \text { F } & \text { Significance  F } \\ \hline  \text {Regression} & 1 & 25.9438 & 25.9438 & 232.2200 & 4.3946 \mathrm{E}-15  \\ \text {Residual }& 28 & 3.1282 & 0.1117 & & \\ \text {Total }& 29 & 29.072 & & & \\ \hline \end{array}    \begin{array}{lrrrrrr} \hline & \text { Coefficients }& \text { Standard Error }& \text { t Stat }&  \text { p -value}& \text {Lower 95\%} &  \text {Upper 95\%} \\ \hline \text { Invoices }& 0.4024 & 0.1236 & 3.2559 & 0.0030 & 0.1492 & 0.6555 \\ \text {Processed }& 0.0126 & 0.0008 & 15.2388 &  4.3946 \mathrm{E}-15  & 0.0109 & 0.0143 \\ \hline \end{array}          -Referring to Table 13-12, the p-value of the measured F-test statistic to test whether the number of invoices processed affects the amount of time is</strong> A) (4.3946E-15)/2. B) (0.0030)/2. C) 0.0030. D) 4.3946E-15. <div style=padding-top: 35px>


-Referring to Table 13-12, the p-value of the measured F-test statistic to test whether the number of invoices processed affects the amount of time is

A) (4.3946E-15)/2.
B) (0.0030)/2.
C) 0.0030.
D) 4.3946E-15.
Question
If the correlation coefficient (r) = 1.00, then

A) all the data points must fall exactly on a straight line with a slope that equals 1.00.
B) all the data points must fall exactly on a straight line with a negative slope.
C) all the data points must fall exactly on a horizontal straight line with a zero slope.
D) all the data points must fall exactly on a straight line with a positive slope.
Question
TABLE 13- 11
A company that has the distribution rights to home video sales of previously released movies would like to use the box office gross (in millions of dollars) to estimate the number of units (in thousands of units) that it can expect to sell. Following is the output from a simple linear regression along with the residual plot and normal probability plot obtained from a data set of 30 different movie titles:
 Regression Statistics  Multiple R 0.8531 RSquare 0.7278 Adjusted R Square 0.7180 Standard Error 47.8668 Observations 30\begin{array}{l}\text { Regression Statistics }\\\begin{array} { l c } \hline \text { Multiple R } & 0.8531 \\\text { RSquare } & 0.7278 \\\text { Adjusted R Square } & 0.7180 \\\text { Standard Error } & 47.8668 \\\text { Observations } & 30\end{array}\end{array}
ANOVA
 d f SS  MS Significance FRegression 1171499.78171499.7874.85052.1259E09Residual2864154.422291.23Total29235654.20\begin{array}{lrrrrr}\hline &\text { d f}& \text { SS } & \text { MS } & \text {F }& \text {Significance F} \\\hline \text {Regression }& 1 & 171499.78 & 171499.78 & 74.8505 & 2.1259E-09 \\\text {Residual} & 28 & 64154.42 & 2291.23 & & \\\text {Total} & 29 & 235654.20 & & & \\\hline\end{array}

Coefficients  Standard Error t Stat  p -value Lower 95% Upper 95%  Intercept 76.535111.83186.46865.24E0752.2987100.7716Gross4.33310.50088.65162.13E093.30725.3590\begin{array}{lrrrrrr}\hline & \text {Coefficients }& \text { Standard Error}& \text { t Stat }& \text { p -value }& \text {Lower 95\% }& \text {Upper 95\% }\\\hline \text { Intercept }& 76.5351 & 11.8318 & 6.4686 & 5.24 \mathrm{E}-07& 52.2987 & 100.7716 \\ \text {Gross} & 4.3331 & 0.5008 & 8.6516 & 2.13 \mathrm{E}-09 & 3.3072 & 5.3590 \\\hline\end{array}

 <strong>TABLE 13- 11 A company that has the distribution rights to home video sales of previously released movies would like to use the box office gross (in millions of dollars) to estimate the number of units (in thousands of units) that it can expect to sell. Following is the output from a simple linear regression along with the residual plot and normal probability plot obtained from a data set of 30 different movie titles:  \begin{array}{l} \text { Regression Statistics }\\ \begin{array} { l c } \hline \text { Multiple R } & 0.8531 \\ \text { RSquare } & 0.7278 \\ \text { Adjusted R Square } & 0.7180 \\ \text { Standard Error } & 47.8668 \\ \text { Observations } & 30 \end{array} \end{array}  ANOVA  \begin{array}{lrrrrr} \hline &\text { d f}& \text { SS } & \text { MS } & \text {F }& \text {Significance F}  \\ \hline \text {Regression }& 1 & 171499.78 & 171499.78 & 74.8505 & 2.1259E-09 \\ \text {Residual} & 28 & 64154.42 & 2291.23 & & \\ \text {Total} & 29 & 235654.20 & & & \\ \hline\end{array}    \begin{array}{lrrrrrr} \hline &  \text {Coefficients }& \text { Standard Error}& \text { t  Stat }&  \text { p -value }&  \text {Lower 95\% }& \text {Upper 95\% }\\ \hline \text { Intercept }& 76.5351 & 11.8318 & 6.4686 & 5.24 \mathrm{E}-07& 52.2987 & 100.7716 \\ \text {Gross} & 4.3331 & 0.5008 & 8.6516 & 2.13 \mathrm{E}-09 & 3.3072 & 5.3590 \\ \hline \end{array}        -Referring to Table 13-11, which of the following assumptions appears to have been violated?</strong> A) homoscedasticity B) independence of errors C) normality of error D) none of the above <div style=padding-top: 35px>   <strong>TABLE 13- 11 A company that has the distribution rights to home video sales of previously released movies would like to use the box office gross (in millions of dollars) to estimate the number of units (in thousands of units) that it can expect to sell. Following is the output from a simple linear regression along with the residual plot and normal probability plot obtained from a data set of 30 different movie titles:  \begin{array}{l} \text { Regression Statistics }\\ \begin{array} { l c } \hline \text { Multiple R } & 0.8531 \\ \text { RSquare } & 0.7278 \\ \text { Adjusted R Square } & 0.7180 \\ \text { Standard Error } & 47.8668 \\ \text { Observations } & 30 \end{array} \end{array}  ANOVA  \begin{array}{lrrrrr} \hline &\text { d f}& \text { SS } & \text { MS } & \text {F }& \text {Significance F}  \\ \hline \text {Regression }& 1 & 171499.78 & 171499.78 & 74.8505 & 2.1259E-09 \\ \text {Residual} & 28 & 64154.42 & 2291.23 & & \\ \text {Total} & 29 & 235654.20 & & & \\ \hline\end{array}    \begin{array}{lrrrrrr} \hline &  \text {Coefficients }& \text { Standard Error}& \text { t  Stat }&  \text { p -value }&  \text {Lower 95\% }& \text {Upper 95\% }\\ \hline \text { Intercept }& 76.5351 & 11.8318 & 6.4686 & 5.24 \mathrm{E}-07& 52.2987 & 100.7716 \\ \text {Gross} & 4.3331 & 0.5008 & 8.6516 & 2.13 \mathrm{E}-09 & 3.3072 & 5.3590 \\ \hline \end{array}        -Referring to Table 13-11, which of the following assumptions appears to have been violated?</strong> A) homoscedasticity B) independence of errors C) normality of error D) none of the above <div style=padding-top: 35px>

-Referring to Table 13-11, which of the following assumptions appears to have been violated?

A) homoscedasticity
B) independence of errors
C) normality of error
D) none of the above
Question
TABLE 13-8
It is believed that GPA (grade point average, based on a four point scale) should have a positive linear relationship with ACT scores. Given below is the Excel output from regressing GPA on ACT scores using a data set of 8 randomly chosen students from a Big-Ten university.
Regressing GPA on ACT\text {Regressing GPA on \(\mathrm { ACT }\)}
 Regressing GPA on ACT Regression Statistics  Multiple R 0.7598 R Square 0.5774 Adjusted R Square 0.5069 Standard E rror 0.2691 Observations 8\begin{array}{l}\text { Regressing GPA on } \mathrm { ACT }\\\begin{array} { l c } \hline & \text { Regression Statistics } \\\hline \text { Multiple R } & 0.7598 \\\text { R Square } & 0.5774 \\\text { Adjusted R Square } & 0.5069 \\\text { Standard E rror } & 0.2691 \\\text { Observations } & 8\end{array}\end{array}

ANOVA
d fSS MS  F  Significance FRegression 10.59400.59408.19860.0286Residual 60.43470.0724Total 71.0287\begin{array}{lcllcc}\hline & \text {d f} & \text {SS} & \text { MS }& \text { F } & \text { Significance F} \\\hline \text {Regression }&1 & 0.5940 & 0.5940 & 8.1986 & 0.0286 \\ \text {Residual }& 6 & 0.4347 & 0.0724 & & \\ \text {Total }& 7 & 1.0287 & & & \\\hline\end{array}

Coefficients Standard Error T Stat  p -value Lower 95% Upper 95%  Intercept0.56810.92840.61190.56301.70362.8398ACT 0.10210.03562.86330.02860.01480.1895\begin{array}{lrrrrrr}\hline & \text {Coefficients }& \text {Standard Error }& \text {T Stat }& \text { p -value }& \text {Lower 95\%}& \text { Upper 95\% } \\\hline \text { Intercept} & 0.5681 & 0.9284 & 0.6119 & 0.5630 & -1.7036 & 2.8398 \\ \text {ACT }& 0.1021 & 0.0356 & 2.8633 & 0.0286 & 0.0148 & 0.1895 \\\hline\end{array}



-Referring to Table 13-8, the interpretation of the coefficient of determination in this regression is

A) ACT scores account for 57.74% of the total fluctuation in GPA.
B) 57.74% of the total variation of ACT scores can be explained by GPA.
C) GPA accounts for 57.74% of the variability of ACT scores.
D) none of the above
Question
TABLE 13-6
The following EXCEL tables are obtained when "Score received on an exam (measured in percentage points)" (Y) is regressed on "percentage attendance" (X) for 22 students in a Statistics for Business and Economics course.
 Regression Statistics  Multiple R 0.142620229 R Square 0.02034053 Adjusted R Square 0.028642444 Standard Error20.25979924 Observations 22\begin{array}{lc}\text { Regression Statistics } \\\text { Multiple R } & 0.142620229 \\\text { R Square } & 0.02034053 \\\text { Adjusted R Square } & -0.028642444 \\\text { Standard Error} &20.25979924 \\\text { Observations } & 22 \\\hline\end{array}

 Coefficients  Standard Enor  T Stat p-value  Intercept 39.3902730937.243476591.0576422160.302826622 Attendance 0.3405835730.528524520.6444044890.526635689\begin{array} { l c r c c } \hline & \text { Coefficients } & \text { Standard Enor } & \text { T Stat } & p \text {-value } \\\hline \text { Intercept } & 39.39027309 & 37.24347659 & 1.057642216 & 0.302826622 \\\text { Attendance } & 0.340583573 & 0.52852452 & 0.644404489 & 0.526635689 \\\hline\end{array}

-Referring to Table 13-6, which of the following statements is true?

A) If attendance increases by 1%, the estimated average score received will increase by 39.39 percentage points.
B) If the score received increases by 39.39%, the estimated average attendance will go up by 1%.
C) If attendance increases by 1%, the estimated average score received will increase by 0.341 percentage points.
D) If attendance increases by 0.341%, the estimated average score received will increase by 1 percentage point.
Question
TABLE 13-12
The manager of the purchasing department of a large banking organization would like to develop a model to predict the amount of time (measured in hours) it takes to process invoices. Data are collected from a sample of 30 days, and the number of invoices processed and completion time in hours is recorded. Below is the regression output:
 Regression Statistics  Multiple R 0.9947 R Square 0.8924 Adjusted R Square 0.8886 Standard Error 0.3342 ations 30\begin{array}{l}\text { Regression Statistics }\\\begin{array} { l c } \hline \text { Multiple R } & 0.9947 \\\text { R Square } & 0.8924 \\\text { Adjusted R Square } & 0.8886 \\\text { Standard Error } & 0.3342 \\\text { ations } & 30 \\\hline\end{array}\end{array}

 d f  SS MS F  Significance F Regression125.943825.9438232.22004.3946E15Residual 283.12820.1117Total 2929.072\begin{array}{lrrccc}\hline & \text { d f } & \text { SS } & \text {MS} & \text { F } & \text { Significance F } \\\hline \text {Regression} & 1 & 25.9438 & 25.9438 & 232.2200 & 4.3946 \mathrm{E}-15 \\ \text {Residual }& 28 & 3.1282 & 0.1117 & & \\ \text {Total }& 29 & 29.072 & & & \\\hline\end{array}

 Coefficients  Standard Error  t Stat  p -valueLower 95%Upper 95% Invoices 0.40240.12363.25590.00300.14920.6555Processed 0.01260.000815.23884.3946E150.01090.0143\begin{array}{lrrrrrr}\hline & \text { Coefficients }& \text { Standard Error }& \text { t Stat }& \text { p -value}& \text {Lower 95\%} & \text {Upper 95\%} \\\hline \text { Invoices }& 0.4024 & 0.1236 & 3.2559 & 0.0030 & 0.1492 & 0.6555 \\ \text {Processed }& 0.0126 & 0.0008 & 15.2388 & 4.3946 \mathrm{E}-15 & 0.0109 & 0.0143 \\\hline\end{array}

 <strong>TABLE 13-12 The manager of the purchasing department of a large banking organization would like to develop a model to predict the amount of time (measured in hours) it takes to process invoices. Data are collected from a sample of 30 days, and the number of invoices processed and completion time in hours is recorded. Below is the regression output:  \begin{array}{l} \text { Regression Statistics }\\ \begin{array} { l c } \hline \text { Multiple R } & 0.9947 \\ \text { R Square } & 0.8924 \\ \text { Adjusted R Square } & 0.8886 \\ \text { Standard Error } & 0.3342 \\ \text { ations } & 30 \\ \hline \end{array} \end{array}    \begin{array}{lrrccc} \hline & \text { d f } &  \text { SS } &  \text {MS} &  \text { F } & \text { Significance  F } \\ \hline  \text {Regression} & 1 & 25.9438 & 25.9438 & 232.2200 & 4.3946 \mathrm{E}-15  \\ \text {Residual }& 28 & 3.1282 & 0.1117 & & \\ \text {Total }& 29 & 29.072 & & & \\ \hline \end{array}    \begin{array}{lrrrrrr} \hline & \text { Coefficients }& \text { Standard Error }& \text { t Stat }&  \text { p -value}& \text {Lower 95\%} &  \text {Upper 95\%} \\ \hline \text { Invoices }& 0.4024 & 0.1236 & 3.2559 & 0.0030 & 0.1492 & 0.6555 \\ \text {Processed }& 0.0126 & 0.0008 & 15.2388 &  4.3946 \mathrm{E}-15  & 0.0109 & 0.0143 \\ \hline \end{array}          -Referring to Table 13-12, the error sum of squares (SSE) of the above regression is</strong> A) 0.1117. B) 29.0720. C) 25.9438. D) 3.1282. <div style=padding-top: 35px>

 <strong>TABLE 13-12 The manager of the purchasing department of a large banking organization would like to develop a model to predict the amount of time (measured in hours) it takes to process invoices. Data are collected from a sample of 30 days, and the number of invoices processed and completion time in hours is recorded. Below is the regression output:  \begin{array}{l} \text { Regression Statistics }\\ \begin{array} { l c } \hline \text { Multiple R } & 0.9947 \\ \text { R Square } & 0.8924 \\ \text { Adjusted R Square } & 0.8886 \\ \text { Standard Error } & 0.3342 \\ \text { ations } & 30 \\ \hline \end{array} \end{array}    \begin{array}{lrrccc} \hline & \text { d f } &  \text { SS } &  \text {MS} &  \text { F } & \text { Significance  F } \\ \hline  \text {Regression} & 1 & 25.9438 & 25.9438 & 232.2200 & 4.3946 \mathrm{E}-15  \\ \text {Residual }& 28 & 3.1282 & 0.1117 & & \\ \text {Total }& 29 & 29.072 & & & \\ \hline \end{array}    \begin{array}{lrrrrrr} \hline & \text { Coefficients }& \text { Standard Error }& \text { t Stat }&  \text { p -value}& \text {Lower 95\%} &  \text {Upper 95\%} \\ \hline \text { Invoices }& 0.4024 & 0.1236 & 3.2559 & 0.0030 & 0.1492 & 0.6555 \\ \text {Processed }& 0.0126 & 0.0008 & 15.2388 &  4.3946 \mathrm{E}-15  & 0.0109 & 0.0143 \\ \hline \end{array}          -Referring to Table 13-12, the error sum of squares (SSE) of the above regression is</strong> A) 0.1117. B) 29.0720. C) 25.9438. D) 3.1282. <div style=padding-top: 35px>


-Referring to Table 13-12, the error sum of squares (SSE) of the above regression is

A) 0.1117.
B) 29.0720.
C) 25.9438.
D) 3.1282.
Question
In performing a regression analysis involving two numerical variables, we are assuming

A) the variation around the line of regression is the same for each X value.
B) that X and Y are independent.
C) the variances of X and Y are equal.
D) all of the above
Question
TABLE 13- 11
A company that has the distribution rights to home video sales of previously released movies would like to use the box office gross (in millions of dollars) to estimate the number of units (in thousands of units) that it can expect to sell. Following is the output from a simple linear regression along with the residual plot and normal probability plot obtained from a data set of 30 different movie titles:
 Regression Statistics  Multiple R 0.8531 R Square 0.7278 Adjusted R Square 0.7180 Standard Error 47.8668 Observations 30\begin{array} { l c } & \text { Regression Statistics } \\\hline \text { Multiple R } & 0.8531 \\\text { R Square } & 0.7278 \\\text { Adjusted R Square } & 0.7180 \\\text { Standard Error } & 47.8668 \\\text { Observations } & 30\end{array} ANOVA
 d f SS  MS Significance FRegression 1171499.78171499.7874.85052.1259E09Residual2864154.422291.23Total29235654.20\begin{array}{lrrrrr}\hline &\text { d f}& \text { SS } & \text { MS } & \text {F }& \text {Significance F} \\\hline \text {Regression }& 1 & 171499.78 & 171499.78 & 74.8505 & 2.1259E-09 \\\text {Residual} & 28 & 64154.42 & 2291.23 & & \\\text {Total} & 29 & 235654.20 & & & \\\hline\end{array}

Coefficients  Standard Error t Stat  p -value Lower 95% Upper 95%  Intercept 76.535111.83186.46865.24E0752.2987100.7716Gross4.33310.50088.65162.13E093.30725.3590\begin{array}{lrrrrrr}\hline & \text {Coefficients }& \text { Standard Error}& \text { t Stat }& \text { p -value }& \text {Lower 95\% }& \text {Upper 95\% }\\\hline \text { Intercept }& 76.5351 & 11.8318 & 6.4686 & 5.24 \mathrm{E}-07& 52.2987 & 100.7716 \\ \text {Gross} & 4.3331 & 0.5008 & 8.6516 & 2.13 \mathrm{E}-09 & 3.3072 & 5.3590 \\\hline\end{array}
 <strong>TABLE 13- 11 A company that has the distribution rights to home video sales of previously released movies would like to use the box office gross (in millions of dollars) to estimate the number of units (in thousands of units) that it can expect to sell. Following is the output from a simple linear regression along with the residual plot and normal probability plot obtained from a data set of 30 different movie titles:  \begin{array} { l c } & \text { Regression Statistics } \\ \hline \text { Multiple R } & 0.8531 \\ \text { R Square } & 0.7278 \\ \text { Adjusted R Square } & 0.7180 \\ \text { Standard Error } & 47.8668 \\ \text { Observations } & 30 \end{array}  ANOVA  \begin{array}{lrrrrr} \hline &\text { d f}& \text { SS } & \text { MS } & \text {F }& \text {Significance F}  \\ \hline \text {Regression }& 1 & 171499.78 & 171499.78 & 74.8505 & 2.1259E-09 \\ \text {Residual} & 28 & 64154.42 & 2291.23 & & \\ \text {Total} & 29 & 235654.20 & & & \\ \hline\end{array}    \begin{array}{lrrrrrr} \hline &  \text {Coefficients }& \text { Standard Error}& \text { t  Stat }&  \text { p -value }&  \text {Lower 95\% }& \text {Upper 95\% }\\ \hline \text { Intercept }& 76.5351 & 11.8318 & 6.4686 & 5.24 \mathrm{E}-07& 52.2987 & 100.7716 \\ \text {Gross} & 4.3331 & 0.5008 & 8.6516 & 2.13 \mathrm{E}-09 & 3.3072 & 5.3590 \\ \hline \end{array}        -Referring to Table 13-11, which of the following is the correct alternative hypothesis for testing whether there is a linear relationship between box office gross and home video unit sales?</strong> A)   H_{1}: \beta_{1}=0   B)   H_{1}: b_{1}=0   C)   H_{1}: \beta_{1} \neq 0   D)   H_{1}: b_{1} \neq 0   <div style=padding-top: 35px>
 <strong>TABLE 13- 11 A company that has the distribution rights to home video sales of previously released movies would like to use the box office gross (in millions of dollars) to estimate the number of units (in thousands of units) that it can expect to sell. Following is the output from a simple linear regression along with the residual plot and normal probability plot obtained from a data set of 30 different movie titles:  \begin{array} { l c } & \text { Regression Statistics } \\ \hline \text { Multiple R } & 0.8531 \\ \text { R Square } & 0.7278 \\ \text { Adjusted R Square } & 0.7180 \\ \text { Standard Error } & 47.8668 \\ \text { Observations } & 30 \end{array}  ANOVA  \begin{array}{lrrrrr} \hline &\text { d f}& \text { SS } & \text { MS } & \text {F }& \text {Significance F}  \\ \hline \text {Regression }& 1 & 171499.78 & 171499.78 & 74.8505 & 2.1259E-09 \\ \text {Residual} & 28 & 64154.42 & 2291.23 & & \\ \text {Total} & 29 & 235654.20 & & & \\ \hline\end{array}    \begin{array}{lrrrrrr} \hline &  \text {Coefficients }& \text { Standard Error}& \text { t  Stat }&  \text { p -value }&  \text {Lower 95\% }& \text {Upper 95\% }\\ \hline \text { Intercept }& 76.5351 & 11.8318 & 6.4686 & 5.24 \mathrm{E}-07& 52.2987 & 100.7716 \\ \text {Gross} & 4.3331 & 0.5008 & 8.6516 & 2.13 \mathrm{E}-09 & 3.3072 & 5.3590 \\ \hline \end{array}        -Referring to Table 13-11, which of the following is the correct alternative hypothesis for testing whether there is a linear relationship between box office gross and home video unit sales?</strong> A)   H_{1}: \beta_{1}=0   B)   H_{1}: b_{1}=0   C)   H_{1}: \beta_{1} \neq 0   D)   H_{1}: b_{1} \neq 0   <div style=padding-top: 35px>


-Referring to Table 13-11, which of the following is the correct alternative hypothesis for testing whether there is a linear relationship between box office gross and home video unit sales?

A) H1:β1=0 H_{1}: \beta_{1}=0
B) H1:b1=0 H_{1}: b_{1}=0
C) H1:β10 H_{1}: \beta_{1} \neq 0
D) H1:b10 H_{1}: b_{1} \neq 0
Question
If the Durbin-Watson statistic has a value close to 4, which assumption is violated?

A) normality of the errors
B) homoscedasticity
C) independence of errors
D) none of the above
Question
TABLE 13-2
A candy bar manufacturer is interested in trying to estimate how sales are influenced by the price of their product. To do this, the company randomly chooses 6 small cities and offers the candy bar at different prices. Using candy bar sales as the dependent variable, the company will conduct a simple linear regression on the data below:
 City  Price ($) Sales  River Falls 1.30100 Hudson 1.6090 Ellsworth 1.8090 Prescott 2.0040 Rock Elm 2.4038 Stillwater 2.9032\begin{array} { l l c } \hline \text { City } & \text { Price } ( \$ ) & \text { Sales } \\\hline \text { River Falls } & 1.30 & 100 \\\text { Hudson } & 1.60 & 90 \\\text { Ellsworth } & 1.80 & 90 \\\text { Prescott } & 2.00 & 40 \\\text { Rock Elm } & 2.40 & 38 \\\text { Stillwater } & 2.90 & 32 \\\hline\end{array}

-Referring to Table 13-2, what is the estimated slope parameter for the candy bar price and sales data?

A) 161.386
B) - 48.193
C) 0.784
D) - 3.810
Question
TABLE 13-8
It is believed that GPA (grade point average, based on a four point scale) should have a positive linear relationship with ACT scores. Given below is the Excel output from regressing GPA on ACT scores using a data set of 8 randomly chosen students from a Big-Ten university.
Regressing GPA on ACT\text {Regressing GPA on \(\mathrm { ACT }\)}
 Regressing GPA on ACT Regression Statistics  Multiple R 0.7598 R Square 0.5774 Adjusted R Square 0.5069 Standard E rror 0.2691 Observations 8\begin{array}{l}\text { Regressing GPA on } \mathrm { ACT }\\\begin{array} { l c } \hline & \text { Regression Statistics } \\\hline \text { Multiple R } & 0.7598 \\\text { R Square } & 0.5774 \\\text { Adjusted R Square } & 0.5069 \\\text { Standard E rror } & 0.2691 \\\text { Observations } & 8\end{array}\end{array}

ANOVA
d fSS MS  F  Significance FRegression 10.59400.59408.19860.0286Residual 60.43470.0724Total 71.0287\begin{array}{lcllcc}\hline & \text {d f} & \text {SS} & \text { MS }& \text { F } & \text { Significance F} \\\hline \text {Regression }&1 & 0.5940 & 0.5940 & 8.1986 & 0.0286 \\ \text {Residual }& 6 & 0.4347 & 0.0724 & & \\ \text {Total }& 7 & 1.0287 & & & \\\hline\end{array}

Coefficients Standard Error T Stat  p -value Lower 95% Upper 95%  Intercept0.56810.92840.61190.56301.70362.8398ACT 0.10210.03562.86330.02860.01480.1895\begin{array}{lrrrrrr}\hline & \text {Coefficients }& \text {Standard Error }& \text {T Stat }& \text { p -value }& \text {Lower 95\%}& \text { Upper 95\% } \\\hline \text { Intercept} & 0.5681 & 0.9284 & 0.6119 & 0.5630 & -1.7036 & 2.8398 \\ \text {ACT }& 0.1021 & 0.0356 & 2.8633 & 0.0286 & 0.0148 & 0.1895 \\\hline\end{array}



-Referring to Table 13-8, what is the predicted average value of GPA when ACT = 20?

A) 2.66
B) 2.61
C) 3.12
D) 2.80
Question
TABLE 13-8
It is believed that GPA (grade point average, based on a four point scale) should have a positive linear relationship with ACT scores. Given below is the Excel output from regressing GPA on ACT scores using a data set of 8 randomly chosen students from a Big-Ten university.
 Regressing GPA on ACT Regression Statistics  Multiple R 0.7598 R Square 0.5774 Adjusted R Square 0.5069 Standard E rror 0.2691 Observations 8\begin{array}{l}\text { Regressing GPA on } \mathrm { ACT }\\\begin{array} { l c } \hline & \text { Regression Statistics } \\\hline \text { Multiple R } & 0.7598 \\\text { R Square } & 0.5774 \\\text { Adjusted R Square } & 0.5069 \\\text { Standard E rror } & 0.2691 \\\text { Observations } & 8\end{array}\end{array}

ANOVA
d fSS MS  F  Significance FRegression 10.59400.59408.19860.0286Residual 60.43470.0724Total 71.0287\begin{array}{lcllcc}\hline & \text {d f} & \text {SS} & \text { MS }& \text { F } & \text { Significance F} \\\hline \text {Regression }&1 & 0.5940 & 0.5940 & 8.1986 & 0.0286 \\ \text {Residual }& 6 & 0.4347 & 0.0724 & & \\ \text {Total }& 7 & 1.0287 & & & \\\hline\end{array}

Coefficients Standard Error T Stat  p -value Lower 95% Upper 95%  Intercept0.56810.92840.61190.56301.70362.8398ACT 0.10210.03562.86330.02860.01480.1895\begin{array}{lrrrrrr}\hline & \text {Coefficients }& \text {Standard Error }& \text {T Stat }& \text { p -value }& \text {Lower 95\%}& \text { Upper 95\% } \\\hline \text { Intercept} & 0.5681 & 0.9284 & 0.6119 & 0.5630 & -1.7036 & 2.8398 \\ \text {ACT }& 0.1021 & 0.0356 & 2.8633 & 0.0286 & 0.0148 & 0.1895 \\\hline\end{array}


-Referring to Table 13-8, what are the decision and conclusion on testing whether there is any linear relationship at 1% level of significance between GPA and ACT scores?

A) Do not reject the null hypothesis; hence there is sufficient evidence to show that ACT scores and GPA are linearly related.
B) Reject the null hypothesis; hence there is not sufficient evidence to show that ACT scores and GPA are linearly related.
C) Reject the null hypothesis; hence there is sufficient evidence to show that ACT scores and GPA are linearly related.
D) Do not reject the null hypothesis; hence there is not sufficient evidence to show that ACT scores and GPA are linearly related.
Question
TABLE 13-2
A candy bar manufacturer is interested in trying to estimate how sales are influenced by the price of their product. To do this, the company randomly chooses 6 small cities and offers the candy bar at different prices. Using candy bar sales as the dependent variable, the company will conduct a simple linear regression on the data below:
 City  Price ($) Sales  River Falls 1.30100 Hudson 1.6090 Ellsworth 1.8090 Prescott 2.0040 Rock Elm 2.4038 Stillwater 2.9032\begin{array} { l l c } \hline \text { City } & \text { Price } ( \$ ) & \text { Sales } \\\hline \text { River Falls } & 1.30 & 100 \\\text { Hudson } & 1.60 & 90 \\\text { Ellsworth } & 1.80 & 90 \\\text { Prescott } & 2.00 & 40 \\\text { Rock Elm } & 2.40 & 38 \\\text { Stillwater } & 2.90 & 32\end{array}

-Referring to Table 13-2, to test that the regression coefficient, þ1, is not equal to 0, what would be the critical values? Use ? = 0.05.

A) ±3.1634
B) ±2.5706
C) ±2.7764
D) ±3.4954
Question
TABLE 13-2
A candy bar manufacturer is interested in trying to estimate how sales are influenced by the price of their product. To do this, the company randomly chooses 6 small cities and offers the candy bar at different prices. Using candy bar sales as the dependent variable, the company will conduct a simple linear regression on the data below:
 City  Price ($) Sales  River Falls 1.30100 Hudson 1.6090 Ellsworth 1.8090 Prescott 2.0040 Rock Elm 2.4038 Stillwater 2.9032\begin{array} { l l c } \hline \text { City } & \text { Price } ( \$ ) & \text { Sales } \\\hline \text { River Falls } & 1.30 & 100 \\\text { Hudson } & 1.60 & 90 \\\text { Ellsworth } & 1.80 & 90 \\\text { Prescott } & 2.00 & 40 \\\text { Rock Elm } & 2.40 & 38 \\\text { Stillwater } & 2.90 & 32\end{array}

-Referring to Table 13-2, if the price of the candy bar is set at $2, the predicted sales will be

A) 90.
B) 65.
C) 100.
D) 30.
Question
The Y-intercept (b0) represents the

A) change in estimated average Y per unit change in X.
B) predicted value of Y.
C) predicted value of Y when X = 0.
D) variation around the sample regression line.
Question
The residuals represent

A) the difference between the actual Y values and the predicted Y values.
B) the square root of the slope.
C) the predicted value of Y for the average X value.
D) the difference between the actual Y values and the mean of Y.
Question
TABLE 13-2
A candy bar manufacturer is interested in trying to estimate how sales are influenced by the price of their product. To do this, the company randomly chooses 6 small cities and offers the candy bar at different prices. Using candy bar sales as the dependent variable, the company will conduct a simple linear regression on the data below:
 City  Price ($) Sales  River Falls 1.30100 Hudson 1.6090 Ellsworth 1.8090 Prescott 2.0040 Rock Elm 2.4038 Stillwater 2.9032\begin{array} { l l c } \hline \text { City } & \text { Price } ( \$ ) & \text { Sales } \\\hline \text { River Falls } & 1.30 & 100 \\\text { Hudson } & 1.60 & 90 \\\text { Ellsworth } & 1.80 & 90 \\\text { Prescott } & 2.00 & 40 \\\text { Rock Elm } & 2.40 & 38 \\\text { Stillwater } & 2.90 & 32 \\\hline\end{array}

-Referring to Table 13-2, what is the estimated average change in the sales of the candy bar if price goes up by $1.00?

A) - 3.810
B) - 48.193
C) 161.386
D) 0.784
Question
TABLE 13-12
The manager of the purchasing department of a large banking organization would like to develop a model to predict the amount of time (measured in hours) it takes to process invoices. Data are collected from a sample of 30 days, and the number of invoices processed and completion time in hours is recorded. Below is the regression output:
 Regression Statistics  Multiple R 0.9947 R Square 0.8924 Adjusted R Square 0.8886 Standard Error 0.3342 ations 30\begin{array}{l}\text { Regression Statistics }\\\begin{array} { l c } \hline \text { Multiple R } & 0.9947 \\\text { R Square } & 0.8924 \\\text { Adjusted R Square } & 0.8886 \\\text { Standard Error } & 0.3342 \\\text { ations } & 30 \\\hline\end{array}\end{array}

 d f  SS MS F  Significance F Regression125.943825.9438232.22004.3946E15Residual 283.12820.1117Total 2929.072\begin{array}{lrrccc}\hline & \text { d f } & \text { SS } & \text {MS} & \text { F } & \text { Significance F } \\\hline \text {Regression} & 1 & 25.9438 & 25.9438 & 232.2200 & 4.3946 \mathrm{E}-15 \\ \text {Residual }& 28 & 3.1282 & 0.1117 & & \\ \text {Total }& 29 & 29.072 & & & \\\hline\end{array}

 Coefficients  Standard Error  t Stat  p -valueLower 95%Upper 95% Invoices 0.40240.12363.25590.00300.14920.6555Processed 0.01260.000815.23884.3946E150.01090.0143\begin{array}{lrrrrrr}\hline & \text { Coefficients }& \text { Standard Error }& \text { t Stat }& \text { p -value}& \text {Lower 95\%} & \text {Upper 95\%} \\\hline \text { Invoices }& 0.4024 & 0.1236 & 3.2559 & 0.0030 & 0.1492 & 0.6555 \\ \text {Processed }& 0.0126 & 0.0008 & 15.2388 & 4.3946 \mathrm{E}-15 & 0.0109 & 0.0143 \\\hline\end{array}

 <strong>TABLE 13-12 The manager of the purchasing department of a large banking organization would like to develop a model to predict the amount of time (measured in hours) it takes to process invoices. Data are collected from a sample of 30 days, and the number of invoices processed and completion time in hours is recorded. Below is the regression output:  \begin{array}{l} \text { Regression Statistics }\\ \begin{array} { l c } \hline \text { Multiple R } & 0.9947 \\ \text { R Square } & 0.8924 \\ \text { Adjusted R Square } & 0.8886 \\ \text { Standard Error } & 0.3342 \\ \text { ations } & 30 \\ \hline \end{array} \end{array}    \begin{array}{lrrccc} \hline & \text { d f } &  \text { SS } &  \text {MS} &  \text { F } & \text { Significance  F } \\ \hline  \text {Regression} & 1 & 25.9438 & 25.9438 & 232.2200 & 4.3946 \mathrm{E}-15  \\ \text {Residual }& 28 & 3.1282 & 0.1117 & & \\ \text {Total }& 29 & 29.072 & & & \\ \hline \end{array}    \begin{array}{lrrrrrr} \hline & \text { Coefficients }& \text { Standard Error }& \text { t Stat }&  \text { p -value}& \text {Lower 95\%} &  \text {Upper 95\%} \\ \hline \text { Invoices }& 0.4024 & 0.1236 & 3.2559 & 0.0030 & 0.1492 & 0.6555 \\ \text {Processed }& 0.0126 & 0.0008 & 15.2388 &  4.3946 \mathrm{E}-15  & 0.0109 & 0.0143 \\ \hline \end{array}          -Referring to Table 13-12, the estimated average amount of time it takes to process one additional invoice is</strong> A) 0.0126 more hours. B) 0.0126 fewer hours. C) 0.4024 more hours. D) 0.4024 fewer hours. <div style=padding-top: 35px>

 <strong>TABLE 13-12 The manager of the purchasing department of a large banking organization would like to develop a model to predict the amount of time (measured in hours) it takes to process invoices. Data are collected from a sample of 30 days, and the number of invoices processed and completion time in hours is recorded. Below is the regression output:  \begin{array}{l} \text { Regression Statistics }\\ \begin{array} { l c } \hline \text { Multiple R } & 0.9947 \\ \text { R Square } & 0.8924 \\ \text { Adjusted R Square } & 0.8886 \\ \text { Standard Error } & 0.3342 \\ \text { ations } & 30 \\ \hline \end{array} \end{array}    \begin{array}{lrrccc} \hline & \text { d f } &  \text { SS } &  \text {MS} &  \text { F } & \text { Significance  F } \\ \hline  \text {Regression} & 1 & 25.9438 & 25.9438 & 232.2200 & 4.3946 \mathrm{E}-15  \\ \text {Residual }& 28 & 3.1282 & 0.1117 & & \\ \text {Total }& 29 & 29.072 & & & \\ \hline \end{array}    \begin{array}{lrrrrrr} \hline & \text { Coefficients }& \text { Standard Error }& \text { t Stat }&  \text { p -value}& \text {Lower 95\%} &  \text {Upper 95\%} \\ \hline \text { Invoices }& 0.4024 & 0.1236 & 3.2559 & 0.0030 & 0.1492 & 0.6555 \\ \text {Processed }& 0.0126 & 0.0008 & 15.2388 &  4.3946 \mathrm{E}-15  & 0.0109 & 0.0143 \\ \hline \end{array}          -Referring to Table 13-12, the estimated average amount of time it takes to process one additional invoice is</strong> A) 0.0126 more hours. B) 0.0126 fewer hours. C) 0.4024 more hours. D) 0.4024 fewer hours. <div style=padding-top: 35px>


-Referring to Table 13-12, the estimated average amount of time it takes to process one additional invoice is

A) 0.0126 more hours.
B) 0.0126 fewer hours.
C) 0.4024 more hours.
D) 0.4024 fewer hours.
Question
TABLE 13- 11
A company that has the distribution rights to home video sales of previously released movies would like to use the box office gross (in millions of dollars) to estimate the number of units (in thousands of units) that it can expect to sell. Following is the output from a simple linear regression along with the residual plot and normal probability plot obtained from a data set of 30 different movie titles:
 Regression Stadistics  Multiple R 0.8531 R Square 0.7278 Adjusted R Square 0.7180 Standard Error 47.8668 Observations 30\begin{array} { l c } \hline & \text { Regression Stadistics } \\\hline \text { Multiple R } & 0.8531 \\\text { R Square } & 0.7278 \\\text { Adjusted R Square } & 0.7180 \\\text { Standard Error } & 47.8668 \\\text { Observations } & 30\end{array} ANOVA
 d f SS  MS Significance FRegression 1171499.78171499.7874.85052.1259E09Residual2864154.422291.23Total29235654.20\begin{array}{lrrrrr}\hline &\text { d f}& \text { SS } & \text { MS } & \text {F }& \text {Significance F} \\\hline \text {Regression }& 1 & 171499.78 & 171499.78 & 74.8505 & 2.1259E-09 \\\text {Residual} & 28 & 64154.42 & 2291.23 & & \\\text {Total} & 29 & 235654.20 & & & \\\hline\end{array}

Coefficients  Standard Error t Stat  p -value Lower 95% Upper 95%  Intercept 76.535111.83186.46865.24E0752.2987100.7716Gross4.33310.50088.65162.13E093.30725.3590\begin{array}{lrrrrrr}\hline & \text {Coefficients }& \text { Standard Error}& \text { t Stat }& \text { p -value }& \text {Lower 95\% }& \text {Upper 95\% }\\\hline \text { Intercept }& 76.5351 & 11.8318 & 6.4686 & 5.24 \mathrm{E}-07& 52.2987 & 100.7716 \\ \text {Gross} & 4.3331 & 0.5008 & 8.6516 & 2.13 \mathrm{E}-09 & 3.3072 & 5.3590 \\\hline\end{array}

 TABLE 13- 11 A company that has the distribution rights to home video sales of previously released movies would like to use the box office gross (in millions of dollars) to estimate the number of units (in thousands of units) that it can expect to sell. Following is the output from a simple linear regression along with the residual plot and normal probability plot obtained from a data set of 30 different movie titles:  \begin{array} { l c } \hline & \text { Regression Stadistics } \\ \hline \text { Multiple R } & 0.8531 \\ \text { R Square } & 0.7278 \\ \text { Adjusted R Square } & 0.7180 \\ \text { Standard Error } & 47.8668 \\ \text { Observations } & 30 \end{array}  ANOVA  \begin{array}{lrrrrr} \hline &\text { d f}& \text { SS } & \text { MS } & \text {F }& \text {Significance F}  \\ \hline \text {Regression }& 1 & 171499.78 & 171499.78 & 74.8505 & 2.1259E-09 \\ \text {Residual} & 28 & 64154.42 & 2291.23 & & \\ \text {Total} & 29 & 235654.20 & & & \\ \hline\end{array}    \begin{array}{lrrrrrr} \hline &  \text {Coefficients }& \text { Standard Error}& \text { t  Stat }&  \text { p -value }&  \text {Lower 95\% }& \text {Upper 95\% }\\ \hline \text { Intercept }& 76.5351 & 11.8318 & 6.4686 & 5.24 \mathrm{E}-07& 52.2987 & 100.7716 \\ \text {Gross} & 4.3331 & 0.5008 & 8.6516 & 2.13 \mathrm{E}-09 & 3.3072 & 5.3590 \\ \hline \end{array}        -Referring to Table 13-11, what is the standard deviation around the regression line?<div style=padding-top: 35px>   TABLE 13- 11 A company that has the distribution rights to home video sales of previously released movies would like to use the box office gross (in millions of dollars) to estimate the number of units (in thousands of units) that it can expect to sell. Following is the output from a simple linear regression along with the residual plot and normal probability plot obtained from a data set of 30 different movie titles:  \begin{array} { l c } \hline & \text { Regression Stadistics } \\ \hline \text { Multiple R } & 0.8531 \\ \text { R Square } & 0.7278 \\ \text { Adjusted R Square } & 0.7180 \\ \text { Standard Error } & 47.8668 \\ \text { Observations } & 30 \end{array}  ANOVA  \begin{array}{lrrrrr} \hline &\text { d f}& \text { SS } & \text { MS } & \text {F }& \text {Significance F}  \\ \hline \text {Regression }& 1 & 171499.78 & 171499.78 & 74.8505 & 2.1259E-09 \\ \text {Residual} & 28 & 64154.42 & 2291.23 & & \\ \text {Total} & 29 & 235654.20 & & & \\ \hline\end{array}    \begin{array}{lrrrrrr} \hline &  \text {Coefficients }& \text { Standard Error}& \text { t  Stat }&  \text { p -value }&  \text {Lower 95\% }& \text {Upper 95\% }\\ \hline \text { Intercept }& 76.5351 & 11.8318 & 6.4686 & 5.24 \mathrm{E}-07& 52.2987 & 100.7716 \\ \text {Gross} & 4.3331 & 0.5008 & 8.6516 & 2.13 \mathrm{E}-09 & 3.3072 & 5.3590 \\ \hline \end{array}        -Referring to Table 13-11, what is the standard deviation around the regression line?<div style=padding-top: 35px>

-Referring to Table 13-11, what is the standard deviation around the regression line?
Question
TABLE 13-6
The following EXCEL tables are obtained when "Score received on an exam (measured in percentage points)" (Y) is regressed on "percentage attendance" (X) for 22 students in a Statistics for Business and Economics course.
Regression Statistics  Multiple R0.142620229R Square0.02034053Adjusted R Square0.028642444Standard Error20.25979924Observations22\begin{array}{ll}\hline \text {Regression Statistics } \\\hline \text { Multiple R} & 0.142620229 \\ \text {R Square} & 0.02034053 \\ \text {Adjusted R Square} & -0.028642444 \\ \text {Standard Error} & 20.25979924 \\ \text {Observations} & 22 \\\hline\end{array}

 Coefficients  Standard Enor  T Stat p-value  Intercept 39.3902730937.243476591.0576422160.302826622 Attendance 0.3405835730.528524520.6444044890.526635689\begin{array}{lcrrc} & \text { Coefficients } & \text { Standard Enor } & {\text { T Stat }} & p \text {-value } \\\hline \text { Intercept } & 39.39027309 & 37.24347659 & 1.057642216 & 0.302826622 \\\text { Attendance } & 0.340583573 & 0.52852452 & 0.644404489 & 0.526635689 \\\hline\end{array}


-Referring to Table 13-6, which of the following statements is true?

A) 2% of the total variability in percentage attendance can be explained by score received.
B) -2.86% of the total variability in percentage attendance can be explained by score received.
C) -2.86% of the total variability in score received can be explained by percentage attendance.
D) 2% of the total variability in score received can be explained by percentage attendance.
Question
TABLE 13-9
It is believed that, the average numbers of hours spent studying per day (HOURS) during undergraduate education should have a positive linear relationship with the starting salary (SALARY, measured in thousands of dollars per month) after graduation. Given below is the Excel output from regressing starting salary on number of hours spent studying per day for a sample of 51 students. NOTE: Some of the numbers in the output are purposely erased.
Regression Stedistics
 Multiple R 0.8857 RSquare 0.7845 Adjusted R Square 0.7801 Standard Error 1.3704\begin{array} { l l } \text { Multiple R } & 0.8857 \\ \text { RSquare } & 0.7845 \\ \text { Adjusted R Square } & 0.7801 \\ \text { Standard Error } & 1.3704 \end{array}
Observations 51

ANOVA

dfSSMSF Significance F Regression 1335.0472335.0473178.3859 Residual 1.8782 Total 50427.0798\begin{array}{lccrcc}\hline & d f & S S &{M S} & F & \text { Significance } F \\\hline \text { Regression } & 1 & 335.0472 & 335.0473 & 178.3859 \\\text { Residual } & & & 1.8782 & \\\text { Total } & 50 & 427.0798 & &\end{array}

 Coeffcients  Standard Error t Stat p-value  Lower 95%  Upper 95%  Intercept 1.89400.40184.71342.051E052.70151.0865 Hours 0.97950.073313.35615.944E180.83211.1269\begin{array}{lrccccr} & \text { Coeffcients } & \text { Standard Error } &t \text { Stat } & p-{\text {-value }} & \text { Lower 95\% } & \text { Upper 95\% } \\\hline \text { Intercept } & -1.8940 & 0.4018 & -4.7134 & 2.051 \mathrm{E}-05 & -2.7015 & -1.0865 \\\text { Hours } & 0.9795 & 0.0733 & 13.3561 & 5.944 \mathrm{E}-18 & 0.8321 & 1.1269 \\\hline\end{array}


-Referring to Table 13-9, the p-value of the measured F-test statistic to test whether HOURS affects SALARY is

A) 2.051E-05.
B) (2.051E-05)/2.
C) 5.944E-18.
D) (5.944E-18)/2.
Question
TABLE 13-3
The director of cooperative education at a state college wants to examine the effect of cooperative education job experience on marketability in the work place. She takes a random sample of 4 students. For these 4, she finds out how many times each had a cooperative education job and how many job offers they received upon graduation. These data are presented in the table below.
 Student  CoopJobs  JobOffer 114226313401\begin{array} { l l l } \hline \text { Student } & \text { CoopJobs } & \text { JobOffer } \\\hline 1 & 1 & 4 \\2 & 2 & 6 \\3 & 1 & 3 \\4 & 0 & 1\end{array}

-Referring to Table 13-3, the total sum of squares (SST) is _____.
Question
TABLE 13-01
A large national bank charges local companies for using their services. A bank official reported the results of a regression analysis designed to predict the bank's charges (Y) -- measured in dollars per month -- for services rendered to local companies. One independent variable used to predict service charge to a company is the company's sales revenue (X) -- measured in millions of dollars. Data for 21 companies who use the bank's services were used to fit the model:
E(Y)=β0+β1XE(Y)=\beta_{0}+\beta_{1} X
The results of the simple linear regression are provided below.
Y^=2,700+20X, SYX=65, two-tailed p value =0.034( for testing β1)\hat{Y}=-2,700+20 X, \mathrm{~S}_{Y X}=65, \text { two-tailed } p \text { value }=0.034\left(\text { for testing } \beta_{1}\right)


-Referring to Table 13-1, interpret the estimate of þ0, the Y-intercept of the line.

A) About 95% of the observed service charges fall within $2,700 of the least squares line.
B) For every $1 million increase in sales revenue, we expect a service charge to decrease $2,700.
C) All companies will be charged at least $2,700 by the bank.
D) There is no practical interpretation since a sales revenue of $0 is a nonsensical value.
Question
TABLE 13- 11
A company that has the distribution rights to home video sales of previously released movies would like to use the box office gross (in millions of dollars) to estimate the number of units (in thousands of units) that it can expect to sell. Following is the output from a simple linear regression along with the residual plot and normal probability plot obtained from a data set of 30 different movie titles:
 Regression Statistics  Multiple R 0.8531 RSquare 0.7278 Adjusted R Square 0.7180 Standard Error 47.8668 Observations 30\begin{array}{l}\text { Regression Statistics }\\\begin{array} { l c } \hline \text { Multiple R } & 0.8531 \\\text { RSquare } & 0.7278 \\\text { Adjusted R Square } & 0.7180 \\\text { Standard Error } & 47.8668 \\\text { Observations } & 30\end{array}\end{array}
ANOVA
 d f SS  MS Significance FRegression 1171499.78171499.7874.85052.1259E09Residual2864154.422291.23Total29235654.20\begin{array}{lrrrrr}\hline &\text { d f}& \text { SS } & \text { MS } & \text {F }& \text {Significance F} \\\hline \text {Regression }& 1 & 171499.78 & 171499.78 & 74.8505 & 2.1259E-09 \\\text {Residual} & 28 & 64154.42 & 2291.23 & & \\\text {Total} & 29 & 235654.20 & & & \\\hline\end{array}

Coefficients  Standard Error t Stat  p -value Lower 95% Upper 95%  Intercept 76.535111.83186.46865.24E0752.2987100.7716Gross4.33310.50088.65162.13E093.30725.3590\begin{array}{lrrrrrr}\hline & \text {Coefficients }& \text { Standard Error}& \text { t Stat }& \text { p -value }& \text {Lower 95\% }& \text {Upper 95\% }\\\hline \text { Intercept }& 76.5351 & 11.8318 & 6.4686 & 5.24 \mathrm{E}-07& 52.2987 & 100.7716 \\ \text {Gross} & 4.3331 & 0.5008 & 8.6516 & 2.13 \mathrm{E}-09 & 3.3072 & 5.3590 \\\hline\end{array}

 <strong>TABLE 13- 11 A company that has the distribution rights to home video sales of previously released movies would like to use the box office gross (in millions of dollars) to estimate the number of units (in thousands of units) that it can expect to sell. Following is the output from a simple linear regression along with the residual plot and normal probability plot obtained from a data set of 30 different movie titles:  \begin{array}{l} \text { Regression Statistics }\\ \begin{array} { l c } \hline \text { Multiple R } & 0.8531 \\ \text { RSquare } & 0.7278 \\ \text { Adjusted R Square } & 0.7180 \\ \text { Standard Error } & 47.8668 \\ \text { Observations } & 30 \end{array} \end{array}  ANOVA  \begin{array}{lrrrrr} \hline &\text { d f}& \text { SS } & \text { MS } & \text {F }& \text {Significance F}  \\ \hline \text {Regression }& 1 & 171499.78 & 171499.78 & 74.8505 & 2.1259E-09 \\ \text {Residual} & 28 & 64154.42 & 2291.23 & & \\ \text {Total} & 29 & 235654.20 & & & \\ \hline\end{array}    \begin{array}{lrrrrrr} \hline &  \text {Coefficients }& \text { Standard Error}& \text { t  Stat }&  \text { p -value }&  \text {Lower 95\% }& \text {Upper 95\% }\\ \hline \text { Intercept }& 76.5351 & 11.8318 & 6.4686 & 5.24 \mathrm{E}-07& 52.2987 & 100.7716 \\ \text {Gross} & 4.3331 & 0.5008 & 8.6516 & 2.13 \mathrm{E}-09 & 3.3072 & 5.3590 \\ \hline \end{array}        -Referring to Table 13-11, which of the following is the correct interpretation for the coefficient of determination?</strong> A) 72.8% of the variation in the box office gross can be explained by the variation in the video unit sales. B) 71.8% of the variation in the video unit sales can be explained by the variation in the box office gross. C) 71.8% of the variation in the box office gross can be explained by the variation in the video unit sales. D) 72.8% of the variation in the video unit sales can be explained by the variation in the box office gross. <div style=padding-top: 35px>   <strong>TABLE 13- 11 A company that has the distribution rights to home video sales of previously released movies would like to use the box office gross (in millions of dollars) to estimate the number of units (in thousands of units) that it can expect to sell. Following is the output from a simple linear regression along with the residual plot and normal probability plot obtained from a data set of 30 different movie titles:  \begin{array}{l} \text { Regression Statistics }\\ \begin{array} { l c } \hline \text { Multiple R } & 0.8531 \\ \text { RSquare } & 0.7278 \\ \text { Adjusted R Square } & 0.7180 \\ \text { Standard Error } & 47.8668 \\ \text { Observations } & 30 \end{array} \end{array}  ANOVA  \begin{array}{lrrrrr} \hline &\text { d f}& \text { SS } & \text { MS } & \text {F }& \text {Significance F}  \\ \hline \text {Regression }& 1 & 171499.78 & 171499.78 & 74.8505 & 2.1259E-09 \\ \text {Residual} & 28 & 64154.42 & 2291.23 & & \\ \text {Total} & 29 & 235654.20 & & & \\ \hline\end{array}    \begin{array}{lrrrrrr} \hline &  \text {Coefficients }& \text { Standard Error}& \text { t  Stat }&  \text { p -value }&  \text {Lower 95\% }& \text {Upper 95\% }\\ \hline \text { Intercept }& 76.5351 & 11.8318 & 6.4686 & 5.24 \mathrm{E}-07& 52.2987 & 100.7716 \\ \text {Gross} & 4.3331 & 0.5008 & 8.6516 & 2.13 \mathrm{E}-09 & 3.3072 & 5.3590 \\ \hline \end{array}        -Referring to Table 13-11, which of the following is the correct interpretation for the coefficient of determination?</strong> A) 72.8% of the variation in the box office gross can be explained by the variation in the video unit sales. B) 71.8% of the variation in the video unit sales can be explained by the variation in the box office gross. C) 71.8% of the variation in the box office gross can be explained by the variation in the video unit sales. D) 72.8% of the variation in the video unit sales can be explained by the variation in the box office gross. <div style=padding-top: 35px>

-Referring to Table 13-11, which of the following is the correct interpretation for the coefficient of determination?

A) 72.8% of the variation in the box office gross can be explained by the variation in the video unit sales.
B) 71.8% of the variation in the video unit sales can be explained by the variation in the box office gross.
C) 71.8% of the variation in the box office gross can be explained by the variation in the video unit sales.
D) 72.8% of the variation in the video unit sales can be explained by the variation in the box office gross.
Question
TABLE 13-8
It is believed that GPA (grade point average, based on a four point scale) should have a positive linear relationship with ACT scores. Given below is the Excel output from regressing GPA on ACT scores using a data set of 8 randomly chosen students from a Big-Ten university.
Regressing GPA on ACTA C T
 Regression Statistics  Multiple R 0.7598 R Square 0.5774 Adjusted R Square 0.5069 Standard Error 0.2691 Observations 8\begin{array}{lc}\text { Regression Statistics } \\\hline \text { Multiple R } & 0.7598 \\\text { R Square } & 0.5774 \\\text { Adjusted R Square } & 0.5069 \\\text { Standard Error } & 0.2691 \\\text { Observations } & 8\end{array}
ANOVA
dfSSMSF Significance F Regression 1 0.59400.59408.19860.0286 Residual 60.43470.0724 Total 71.0287\begin{array}{lccccc} & d f & S S & M S & F & \text { Significance } F \\\hline \text { Regression 1 } & 0.5940 & 0.5940 & 8.1986 & 0.0286 \\\text { Residual } & 6 & 0.4347 & 0.0724 & & \\\text { Total } & 7 & 1.0287 & & & \\\hline\end{array}

Coefficients Standard Error T Stat  p -value Lower 95% Upper 95%  Intercept0.56810.92840.61190.56301.70362.8398ACT 0.10210.03562.86330.02860.01480.1895\begin{array}{lrrrrrr}\hline & \text {Coefficients }& \text {Standard Error }& \text {T Stat }& \text { p -value }& \text {Lower 95\%}& \text { Upper 95\% } \\\hline \text { Intercept} & 0.5681 & 0.9284 & 0.6119 & 0.5630 & -1.7036 & 2.8398 \\ \text {ACT }& 0.1021 & 0.0356 & 2.8633 & 0.0286 & 0.0148 & 0.1895 \\\hline\end{array}



-Referring to Table 13-8, the value of the measured test statistic to test whether there is any linear relationship between GPA and ACT is

A) 0.7598.
B) 0.0356.
C) 0.1021.
D) 2.8633.
Question
TABLE 13-10
The management of a chain electronic store would like to develop a model for predicting the weekly sales (in thousand of dollars) for individual stores based on the number of customers who made purchases. A random sample of 12 stores yields the following results:
 Customers  Sales (Thousands of Dollars) 90711.2092611.057138.217419.217809.4289810.085106.735297.024606.128729.526507.536037.25\begin{array} { | l | l | } \hline \text { Customers } & \text { Sales (Thousands of Dollars) } \\\hline 907 & 11.20 \\\hline 926 & 11.05 \\\hline 713 & 8.21 \\\hline 741 & 9.21 \\\hline 780 & 9.42 \\\hline 898 & 10.08 \\\hline 510 & 6.73 \\\hline 529 & 7.02 \\\hline 460 & 6.12 \\\hline 872 & 9.52 \\\hline 650 & 7.53 \\\hline 603 & 7.25 \\\hline\end{array}

-Referring to Table 13-10, what is the value of the coefficient of correlation?
Question
TABLE 13-12
The manager of the purchasing department of a large banking organization would like to develop a model to predict the amount of time (measured in hours) it takes to process invoices. Data are collected from a sample of 30 days, and the number of invoices processed and completion time in hours is recorded. Below is the regression output:
 Regression Statistics  Multiple R 0.9947 R Square 0.8924 Adjusted R Square 0.8886 Standard Error 0.3342 ations 30\begin{array}{l}\text { Regression Statistics }\\\begin{array} { l c } \hline \text { Multiple R } & 0.9947 \\\text { R Square } & 0.8924 \\\text { Adjusted R Square } & 0.8886 \\\text { Standard Error } & 0.3342 \\\text { ations } & 30 \\\hline\end{array}\end{array}

 d f  SS MS F  Significance F Regression125.943825.9438232.22004.3946E15Residual 283.12820.1117Total 2929.072\begin{array}{lrrccc}\hline & \text { d f } & \text { SS } & \text {MS} & \text { F } & \text { Significance F } \\\hline \text {Regression} & 1 & 25.9438 & 25.9438 & 232.2200 & 4.3946 \mathrm{E}-15 \\ \text {Residual }& 28 & 3.1282 & 0.1117 & & \\ \text {Total }& 29 & 29.072 & & & \\\hline\end{array}

 Coefficients  Standard Error  t Stat  p -valueLower 95%Upper 95% Invoices 0.40240.12363.25590.00300.14920.6555Processed 0.01260.000815.23884.3946E150.01090.0143\begin{array}{lrrrrrr}\hline & \text { Coefficients }& \text { Standard Error }& \text { t Stat }& \text { p -value}& \text {Lower 95\%} & \text {Upper 95\%} \\\hline \text { Invoices }& 0.4024 & 0.1236 & 3.2559 & 0.0030 & 0.1492 & 0.6555 \\ \text {Processed }& 0.0126 & 0.0008 & 15.2388 & 4.3946 \mathrm{E}-15 & 0.0109 & 0.0143 \\\hline\end{array}

 <strong>TABLE 13-12 The manager of the purchasing department of a large banking organization would like to develop a model to predict the amount of time (measured in hours) it takes to process invoices. Data are collected from a sample of 30 days, and the number of invoices processed and completion time in hours is recorded. Below is the regression output:  \begin{array}{l} \text { Regression Statistics }\\ \begin{array} { l c } \hline \text { Multiple R } & 0.9947 \\ \text { R Square } & 0.8924 \\ \text { Adjusted R Square } & 0.8886 \\ \text { Standard Error } & 0.3342 \\ \text { ations } & 30 \\ \hline \end{array} \end{array}    \begin{array}{lrrccc} \hline & \text { d f } &  \text { SS } &  \text {MS} &  \text { F } & \text { Significance  F } \\ \hline  \text {Regression} & 1 & 25.9438 & 25.9438 & 232.2200 & 4.3946 \mathrm{E}-15  \\ \text {Residual }& 28 & 3.1282 & 0.1117 & & \\ \text {Total }& 29 & 29.072 & & & \\ \hline \end{array}    \begin{array}{lrrrrrr} \hline & \text { Coefficients }& \text { Standard Error }& \text { t Stat }&  \text { p -value}& \text {Lower 95\%} &  \text {Upper 95\%} \\ \hline \text { Invoices }& 0.4024 & 0.1236 & 3.2559 & 0.0030 & 0.1492 & 0.6555 \\ \text {Processed }& 0.0126 & 0.0008 & 15.2388 &  4.3946 \mathrm{E}-15  & 0.0109 & 0.0143 \\ \hline \end{array}          -Referring to Table 13-12, the value of the measured t-test statistic to test whether the amount of time depends linearly on the number of invoices processed is</strong> A) 0.8924. B) 15.2388. C) 232.2200. D) 3.2559. <div style=padding-top: 35px>

 <strong>TABLE 13-12 The manager of the purchasing department of a large banking organization would like to develop a model to predict the amount of time (measured in hours) it takes to process invoices. Data are collected from a sample of 30 days, and the number of invoices processed and completion time in hours is recorded. Below is the regression output:  \begin{array}{l} \text { Regression Statistics }\\ \begin{array} { l c } \hline \text { Multiple R } & 0.9947 \\ \text { R Square } & 0.8924 \\ \text { Adjusted R Square } & 0.8886 \\ \text { Standard Error } & 0.3342 \\ \text { ations } & 30 \\ \hline \end{array} \end{array}    \begin{array}{lrrccc} \hline & \text { d f } &  \text { SS } &  \text {MS} &  \text { F } & \text { Significance  F } \\ \hline  \text {Regression} & 1 & 25.9438 & 25.9438 & 232.2200 & 4.3946 \mathrm{E}-15  \\ \text {Residual }& 28 & 3.1282 & 0.1117 & & \\ \text {Total }& 29 & 29.072 & & & \\ \hline \end{array}    \begin{array}{lrrrrrr} \hline & \text { Coefficients }& \text { Standard Error }& \text { t Stat }&  \text { p -value}& \text {Lower 95\%} &  \text {Upper 95\%} \\ \hline \text { Invoices }& 0.4024 & 0.1236 & 3.2559 & 0.0030 & 0.1492 & 0.6555 \\ \text {Processed }& 0.0126 & 0.0008 & 15.2388 &  4.3946 \mathrm{E}-15  & 0.0109 & 0.0143 \\ \hline \end{array}          -Referring to Table 13-12, the value of the measured t-test statistic to test whether the amount of time depends linearly on the number of invoices processed is</strong> A) 0.8924. B) 15.2388. C) 232.2200. D) 3.2559. <div style=padding-top: 35px>


-Referring to Table 13-12, the value of the measured t-test statistic to test whether the amount of time depends linearly on the number of invoices processed is

A) 0.8924.
B) 15.2388.
C) 232.2200.
D) 3.2559.
Question
TABLE 13-3
The director of cooperative education at a state college wants to examine the effect of cooperative education job experience on marketability in the work place. She takes a random sample of 4 students. For these 4, she finds out how many times each had a cooperative education job and how many job offers they received upon graduation. These data are presented in the table below.
 Student  CoopJobs  JobOffer 114226313401\begin{array} { l l l } \hline \text { Student } & \text { CoopJobs } & \text { JobOffer } \\\hline 1 & 1 & 4 \\2 & 2 & 6 \\3 & 1 & 3 \\4 & 0 & 1\end{array}

-Referring to Table 13-3, suppose the director of cooperative education wants to obtain a 95% confidence-interval estimate for the mean number of job offers received by people who have had exactly one cooperative education job. The t critical value she would use is_______
Question
If the Durbin-Watson statistic has a value close to 0, which assumption is violated?

A) homoscedasticity
B) normality of the errors
C) independence of errors
D) none of the above
Question
TABLE 13-7
An investment specialist claims that if one holds a portfolio that moves in the opposite direction to the market index like the S&P 500, then it is possible to reduce the variability of the portfolio's return. In other words, one can create a portfolio with positive returns but less exposure to risk. A sample of 26 years of S&P 500 index and a portfolio consisting of stocks of private prisons, which are believed to be negatively related to the S&P 500 index, is collected. A regression analysis was performed by regressing the returns of the prison stocks portfolio (Y) on the returns of S&P 500 index (X) to prove that the prison stocks portfolio is negatively related to the S&P 500 index at a 5% level of significance. The results are given in the following EXCEL output.
 Coefficients  Standard Error  T Stat p-value  Intercept 4.8660042580.3574360913.613634418.7932E13 S& P 0.5025135060.0715971527.01862425294942E07\begin{array} { l c c c c } \hline & \text { Coefficients } & \text { Standard Error } & \text { T Stat } & { p \text {-value } } \\\hline \text { Intercept } & 4.866004258 & 0.35743609 & 13.61363441 & 8.7932 \mathrm { E } - 13 \\\text { S\& P } & - 0.502513506 & 0.071597152 &- 7.01862425 & 294942 \mathrm { E } - 07 \\\hline\end{array}

-Referring to Table 13-7, to test whether the prison stocks portfolio is negatively related to the S&P 500 index, the appropriate null and alternative hypotheses are, respectively,

A) H0:r0 H_{0}: r \leq 0 versus H1:r>0 H_{1}: r>0 .
B) H0:r0 H_{0}: r \geq 0 versus H1:r<0 H_{1}: r<0 .
C) H0:ρ0 H_{0}: \rho \geq 0 versus H1:ρ<0 H_{1}: \rho<0 .
D) H0:ρ0 H_{0}: \rho \leq 0 versus H1:ρ>0 H_{1}: \rho>0 .
Question
TABLE 13-5
The managing partner of an advertising agency believes that his company's sales are related to the industry sales. He uses Microsoft Excel's Data Analysis tool to analyze the last 4 years of quarterly data with the following results:
 Regression Sttuistics  Multiple R 0.802 R Square 0.643 Adjusted RSquare 0.618 Standard Error SYX 0.9224\begin{array}{ll}\text { Regression Sttuistics } \\\hline \text { Multiple R } & 0.802 \\\text { R Square } & 0.643 \\\text { Adjusted RSquare } & 0.618 \\\text { Standard Error SYX } & 0.9224\end{array}

Observations 16
ANOVA

df SS  MS F Sig.F  Regression 121.49721.49725.270.000 Error 1411.9120.851 Total 1533.409\begin{array}{lrlrlr} & d f & \text { SS } & \text { MS } & F & \text { Sig.F } \\\hline \text { Regression } & 1 & 21.497 & 21.497 & 25.27 & 0.000 \\\text { Error } & 14 & 11.912 & 0.851 & & \\\text { Total } & 15 & 33.409 & & &\end{array}

 Predictor Coeffcients  Standard Error  tStat  p-value  Intercept 3.9621.4402.750.016 Industry 0.0404510.0080485.030.000 Durbin- Watson Statistic 1.59\begin{array}{l}\begin{array}{lllrl}\hline \text { Predictor}& \text { Coeffcients } & \text { Standard Error } & \text { tStat } & \text { p-value } \\\text { Intercept } & 3.962 & 1.440 & 2.75 & 0.016 \\\text { Industry } & 0.040451 & 0.008048 & 5.03 & 0.000 \\\hline\end{array}\\\text { Durbin- Watson Statistic } 1.59\end{array}


-Referring to Table 13-5, the standard error of the estimated slope coefficient is________ .
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Deck 13: Simple Linear Regression
1
TABLE 13-10
The management of a chain electronic store would like to develop a model for predicting the weekly sales (in thousand of dollars) for individual stores based on the number of customers who made purchases. A random sample of 12 stores yields the following results:
 Customers  Sales (Thousands of Dollars) 90711.2092611.057138.217419.217809.4289810.085106.735297.024606.128729.526507.536037.25\begin{array} { | l | l | } \hline \text { Customers } & \text { Sales (Thousands of Dollars) } \\\hline 907 & 11.20 \\\hline 926 & 11.05 \\\hline 713 & 8.21 \\\hline 741 & 9.21 \\\hline 780 & 9.42 \\\hline 898 & 10.08 \\\hline 510 & 6.73 \\\hline 529 & 7.02 \\\hline 460 & 6.12 \\\hline 872 & 9.52 \\\hline 650 & 7.53 \\\hline 603 & 7.25 \\\hline\end{array}

-Referring to Table 13-10, which is the correct null hypothesis for testing whether the number of customers who make purchase affects weekly sales?

A) H0 : ?0 = 0
B) H0 : µ = 0
C) H0 : ?1 = 0
D) H0 : ?= 0
H0 : ?1 = 0
2
TABLE 13-7
An investment specialist claims that if one holds a portfolio that moves in the opposite direction to the market index like the S&P 500, then it is possible to reduce the variability of the portfolio's return. In other words, one can create a portfolio with positive returns but less exposure to risk. A sample of 26 years of S&P 500 index and a portfolio consisting of stocks of private prisons, which are believed to be negatively related to the S&P 500 index, is collected. A regression analysis was performed by regressing the returns of the prison stocks portfolio (Y) on the returns of S&P 500 index (X) to prove that the prison stocks portfolio is negatively related to the S&P 500 index at a 5% level of significance. The results are given in the following EXCEL output.
 Coefficients  Standard Error T Stat  p-value  Intercept 4.8660042580.3574360913.613634418.7932E13 S& P 0.5025135060.0715971527.01862425294942E07\begin{array} { l c c c c } \hline & \text { Coefficients } & \text { Standard Error } &T \text { Stat } & \text { p-value } \\\hline \text { Intercept } & 4.866004258 & 0.35743609 & 13.61363441 & 8.7932 \mathrm { E } - 13 \\\text { S\& P } & - 0.502513506 & 0.071597152 & - 7.01862425 & 294942 \mathrm { E } - 07 \\\hline\end{array}

-Referring to Table 13-7, to test whether the prison stocks portfolio is negatively related to the S&P 500 index, the measured value of the test statistic is

A) 0.357.
B) 0.072.
C) -0.503.
D) -7.019.
-7.019.
3
TABLE 13-2
A candy bar manufacturer is interested in trying to estimate how sales are influenced by the price of their product. To do this, the company randomly chooses 6 small cities and offers the candy bar at different prices. Using candy bar sales as the dependent variable, the company will conduct a simple linear regression on the data below:
 City  Price ($) Sales  River Falls 1.30100 Hudson 1.6090 Ellsworth 1.8090 Prescott 2.0040 Rock Elm 2.4038 Stillwater 2.9032\begin{array} { l l c } \hline \text { City } & \text { Price } ( \$ ) & \text { Sales } \\\hline \text { River Falls } & 1.30 & 100 \\\text { Hudson } & 1.60 & 90 \\\text { Ellsworth } & 1.80 & 90 \\\text { Prescott } & 2.00 & 40 \\\text { Rock Elm } & 2.40 & 38 \\\text { Stillwater } & 2.90 & 32 \\\hline\end{array}

-Referring to Table 13-2, what is ?(XXˉ)2?(X -\bar X )^2 for these data?

A) 2.54
B) 0
C) 1.66
D) 25.66
1.66
4
TABLE 13-2
A candy bar manufacturer is interested in trying to estimate how sales are influenced by the price of their product. To do this, the company randomly chooses 6 small cities and offers the candy bar at different prices. Using candy bar sales as the dependent variable, the company will conduct a simple linear regression on the data below:
 City  Price ($) Sales  River Falls 1.30100 Hudson 1.6090 Ellsworth 1.8090 Prescott 2.0040 Rock Elm 2.4038 Stillwater 2.9032\begin{array} { l l c } \hline \text { City } & \text { Price } ( \$ ) & \text { Sales } \\\hline \text { River Falls } & 1.30 & 100 \\\text { Hudson } & 1.60 & 90 \\\text { Ellsworth } & 1.80 & 90 \\\text { Prescott } & 2.00 & 40 \\\text { Rock Elm } & 2.40 & 38 \\\text { Stillwater } & 2.90 & 32 \\\hline\end{array}

-Referring to Table 13-2, what is the coefficient of correlation for these data?

A) - 0.7839
B) 0.8854
C) 0.7839
D) - 0.8854
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5
TABLE 13-12
The manager of the purchasing department of a large banking organization would like to develop a model to predict the amount of time (measured in hours) it takes to process invoices. Data are collected from a sample of 30 days, and the number of invoices processed and completion time in hours is recorded. Below is the regression output:
 Regression Statistics  Multiple R 0.9947 R Square 0.8924 Adjusted R Square 0.8886 Standard Error 0.3342 Observations 30\begin{array}{lc}\text { Regression Statistics } \\\hline \text { Multiple R } & 0.9947 \\\text { R Square } & 0.8924 \\\text { Adjusted R Square } & 0.8886 \\\text { Standard Error } & 0.3342 \\\text { Observations } & 30\end{array}


 ANOVA \text { ANOVA }
df SS  MS F Significance F Regression 125.943825.9438232.22004.3946E15 Residual 283.12820.1117 Total 2929.072\begin{array}{lccccc} & d f & \text { SS } & \text { MS } & F & \text { Significance } F \\\text { Regression } & 1 & 25.9438 & 25.9438 & 232.2200 & 4.3946 \mathrm{E}-15 \\\text { Residual } & 28 & 3.1282 & 0.1117 & & \\\text { Total } & 29 & 29.072 & & \\\hline\end{array}


 Coeffcients  StandardError t Stat p-value  Lower 95%  Upper 95%  Invoices 0.40240.12363.25590.00300.14920.6555 Processed 0.01260.000815.23884.3946E150.01090.0143\begin{array}{lrrrrrr}\hline & \text { Coeffcients } & \text { StandardError } & t \text { Stat } & p \text {-value } & \text { Lower 95\% } & \text { Upper 95\% } \\ \hline \text { Invoices } & 0.4024 & 0.1236 & 3.2559 & 0.0030 & 0.1492 & 0.6555 \\\text { Processed } & 0.0126 & 0.0008 & 15.2388 & 4.3946 \mathrm{E}-15 & 0.0109 & 0.0143 \\\hline\end{array}
 <strong>TABLE 13-12 The manager of the purchasing department of a large banking organization would like to develop a model to predict the amount of time (measured in hours) it takes to process invoices. Data are collected from a sample of 30 days, and the number of invoices processed and completion time in hours is recorded. Below is the regression output:  \begin{array}{lc} \text { Regression Statistics } \\ \hline \text { Multiple R } & 0.9947 \\ \text { R Square } & 0.8924 \\ \text { Adjusted R Square } & 0.8886 \\ \text { Standard Error } & 0.3342 \\ \text { Observations } & 30 \end{array}     \text { ANOVA }   \begin{array}{lccccc} & d f & \text { SS } & \text { MS } & F & \text { Significance } F \\ \text { Regression } & 1 & 25.9438 & 25.9438 & 232.2200 & 4.3946 \mathrm{E}-15 \\ \text { Residual } & 28 & 3.1282 & 0.1117 & & \\ \text { Total } & 29 & 29.072 & & \\ \hline \end{array}     \begin{array}{lrrrrrr} \hline & \text { Coeffcients } & \text { StandardError } & t \text { Stat } & p \text {-value } & \text { Lower 95\% } & \text { Upper 95\% } \\ \hline \text { Invoices } & 0.4024 & 0.1236 & 3.2559 & 0.0030 & 0.1492 & 0.6555 \\ \text { Processed } & 0.0126 & 0.0008 & 15.2388 & 4.3946 \mathrm{E}-15 & 0.0109 & 0.0143 \\ \hline \end{array}       -Referring to Table 13-12, the degrees of freedom for the F test on whether the number of invoices processed affects the amount of time are</strong> A) 1, 29. B) 28, 1. C) 29, 1. D) 1, 28.   <strong>TABLE 13-12 The manager of the purchasing department of a large banking organization would like to develop a model to predict the amount of time (measured in hours) it takes to process invoices. Data are collected from a sample of 30 days, and the number of invoices processed and completion time in hours is recorded. Below is the regression output:  \begin{array}{lc} \text { Regression Statistics } \\ \hline \text { Multiple R } & 0.9947 \\ \text { R Square } & 0.8924 \\ \text { Adjusted R Square } & 0.8886 \\ \text { Standard Error } & 0.3342 \\ \text { Observations } & 30 \end{array}     \text { ANOVA }   \begin{array}{lccccc} & d f & \text { SS } & \text { MS } & F & \text { Significance } F \\ \text { Regression } & 1 & 25.9438 & 25.9438 & 232.2200 & 4.3946 \mathrm{E}-15 \\ \text { Residual } & 28 & 3.1282 & 0.1117 & & \\ \text { Total } & 29 & 29.072 & & \\ \hline \end{array}     \begin{array}{lrrrrrr} \hline & \text { Coeffcients } & \text { StandardError } & t \text { Stat } & p \text {-value } & \text { Lower 95\% } & \text { Upper 95\% } \\ \hline \text { Invoices } & 0.4024 & 0.1236 & 3.2559 & 0.0030 & 0.1492 & 0.6555 \\ \text { Processed } & 0.0126 & 0.0008 & 15.2388 & 4.3946 \mathrm{E}-15 & 0.0109 & 0.0143 \\ \hline \end{array}       -Referring to Table 13-12, the degrees of freedom for the F test on whether the number of invoices processed affects the amount of time are</strong> A) 1, 29. B) 28, 1. C) 29, 1. D) 1, 28.

-Referring to Table 13-12, the degrees of freedom for the F test on whether the number of invoices processed affects the amount of time are

A) 1, 29.
B) 28, 1.
C) 29, 1.
D) 1, 28.
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6
TABLE 13-2
A candy bar manufacturer is interested in trying to estimate how sales are influenced by the price of their product. To do this, the company randomly chooses 6 small cities and offers the candy bar at different prices. Using candy bar sales as the dependent variable, the company will conduct a simple linear regression on the data below:
 City  Price ($) Sales  River Falls 1.30100 Hudson 1.6090 Ellsworth 1.8090 Prescott 2.0040 Rock Elm 2.4038 Stillwater 2.9032\begin{array} { l l c } \hline \text { City } & \text { Price } ( \$ ) & \text { Sales } \\\hline \text { River Falls } & 1.30 & 100 \\\text { Hudson } & 1.60 & 90 \\\text { Ellsworth } & 1.80 & 90 \\\text { Prescott } & 2.00 & 40 \\\text { Rock Elm } & 2.40 & 38 \\\text { Stillwater } & 2.90 & 32 \\\hline\end{array}

-The sample correlation coefficient between X and Y is 0.375. It has been found out that the p- value is 0.744 when testing H0 : ? = 0 against the one- sided alternative H0 : ? < 0. To test H0 : ? = 0 against the two- sided alternative H0 : ?? 0 at a significance level of 0.2, the p- value is

A) (1 - 0.744)(2).
B) (0.744)(2).
C) 0.744/2.
D) 1 - 0.744.
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7
TABLE 13-9
It is believed that, the average numbers of hours spent studying per day (HOURS) during undergraduate education should have a positive linear relationship with the starting salary (SALARY, measured in thousands of dollars per month) after graduation. Given below is the Excel output from regressing starting salary on number of hours spent studying per day for a sample of 51 students. NOTE: Some of the numbers in the output are purposely erased.
 Regression Statistics  Multiple R 0.8857 R Square 0.7845 Adjusted R Square 0.7801 Standard Error 1.3704 Observations 51\begin{array}{l}\text { Regression Statistics }\\\begin{array} { l c } \hline \text { Multiple R } & 0.8857 \\\text { R Square } & 0.7845 \\\text { Adjusted R Square } & 0.7801 \\\text { Standard Error } & 1.3704 \\\text { Observations } & 51 \\\hline\end{array}\end{array}
 ANOVA \text { ANOVA }
 df  SS  MS  Significance F  Regression 1335.0472335.0473178.3859 Residual 1.8782 Total 50427.0798\begin{array}{llcc}\hline&\text { df }&\text { SS } &\text { MS }&\text {F }&\text { Significance F } \\\text { Regression } & 1 & 335.0472 & 335.0473 & 178.3859 \\\text { Residual } & & & 1.8782 & \\\text { Total } & 50 & 427.0798 & &\end{array}  <strong>TABLE 13-9 It is believed that, the average numbers of hours spent studying per day (HOURS) during undergraduate education should have a positive linear relationship with the starting salary (SALARY, measured in thousands of dollars per month) after graduation. Given below is the Excel output from regressing starting salary on number of hours spent studying per day for a sample of 51 students. NOTE: Some of the numbers in the output are purposely erased.  \begin{array}{l} \text { Regression Statistics }\\ \begin{array} { l c } \hline \text { Multiple R } & 0.8857 \\ \text { R Square } & 0.7845 \\ \text { Adjusted R Square } & 0.7801 \\ \text { Standard Error } & 1.3704 \\ \text { Observations } & 51 \\ \hline \end{array} \end{array}   \text { ANOVA }   \begin{array}{llcc} \hline&\text { df }&\text { SS } &\text { MS }&\text {F }&\text { Significance F } \\ \text { Regression } & 1 & 335.0472 & 335.0473 & 178.3859 \\ \text { Residual } & & & 1.8782 & \\ \text { Total } & 50 & 427.0798 & & \end{array}     -Referring to Table 13-9, the 90% confidence interval for the average change in SALARY (in thousands of dollars) as a result of spending an extra hour per day studying is</strong> A) wider than [-2.70159, -1.08654]. B) wider than [0.8321927, 1.12697]. C) narrower than [0.8321927, 1.12697]. D) narrower than [-2.70159, -1.08654].

-Referring to Table 13-9, the 90% confidence interval for the average change in SALARY (in thousands of dollars) as a result of spending an extra hour per day studying is

A) wider than [-2.70159, -1.08654].
B) wider than [0.8321927, 1.12697].
C) narrower than [0.8321927, 1.12697].
D) narrower than [-2.70159, -1.08654].
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8
Which of the following assumptions concerning the probability distribution of the random error term is stated incorrectly?

A) The variance of the distribution increases as X increases.
B) The distribution is normal.
C) The errors are independent.
D) The mean of the distribution is 0.
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9
TABLE 13-9
It is believed that, the average numbers of hours spent studying per day (HOURS) during undergraduate education should have a positive linear relationship with the starting salary (SALARY, measured in thousands of dollars per month) after graduation. Given below is the Excel output from regressing starting salary on number of hours spent studying per day for a sample of 51 students. NOTE: Some of the numbers in the output are purposely erased.
 Regression Statistics  Multiple R 0.8857 R Square 0.7845 Adjusted R Square 0.7801 Standard Error 1.3704 Observations 51\begin{array}{l}\text { Regression Statistics }\\\begin{array} { l c } \hline \text { Multiple R } & 0.8857 \\\text { R Square } & 0.7845 \\\text { Adjusted R Square } & 0.7801 \\\text { Standard Error } & 1.3704 \\\text { Observations } & 51 \\\hline\end{array}\end{array}
 ANOVA \text { ANOVA }
 df  SS  MS  Significance F  Regression 1335.0472335.0473178.3859 Residual 1.8782 Total 50427.0798\begin{array}{llcc}\hline&\text { df }&\text { SS } &\text { MS }&\text {F }&\text { Significance F } \\\text { Regression } & 1 & 335.0472 & 335.0473 & 178.3859 \\\text { Residual } & & & 1.8782 & \\\text { Total } & 50 & 427.0798 & &\end{array}

-Referring to Table 13-9, the value of the measured t-test statistic to test whether average SALARY depends linearly on HOURS is  Coefficients  Standaad Error t Stat p-value  Lower 95% Upper 95%  Intercept 1.89400.40184.71342.051E052.70151.0865 Hours 0.97950.073313.35615.944E180.83211.1269\begin{array} { l r c c c r r } & \text { Coefficients } & \text { Standaad Error } &t \text { Stat } & p \text {-value } & \text { Lower 95\%} & \text { Upper 95\% } \\\hline \text { Intercept } & - 1.8940 & 0.4018 & - 4.7134 & 2.051 \mathrm { E } - 05 & - 2.7015 & - 1.0865 \\\text { Hours } & 0.9795 & 0.0733 & 13.3561 & 5.944 \mathrm { E } - 18 & 0.8321 & 1.1269 \\\hline\end{array}

A) 13.3561.
B) 0.9795.
C) -4.7134.
D) -1.8940.
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10
TABLE 13-2
A candy bar manufacturer is interested in trying to estimate how sales are influenced by the price of their product. To do this, the company randomly chooses 6 small cities and offers the candy bar at different prices. Using candy bar sales as the dependent variable, the company will conduct a simple linear regression on the data below:
 City  Price ($) Sales  River Falls 1.30100 Hudson 1.6090 Ellsworth 1.8090 Prescott 2.0040 Rock Elm 2.4038 Stillwater 2.9032\begin{array} { l l c } \hline \text { City } & \text { Price } ( \$ ) & \text { Sales } \\\hline \text { River Falls } & 1.30 & 100 \\\text { Hudson } & 1.60 & 90 \\\text { Ellsworth } & 1.80 & 90 \\\text { Prescott } & 2.00 & 40 \\\text { Rock Elm } & 2.40 & 38 \\\text { Stillwater } & 2.90 & 32 \\\hline\end{array}

-Referring to Table 13-2, what is the percentage of the total variation in candy bar sales explained by the regression model?

A) 78.39%
B) 100%
C) 48.19%
D) 88.54%
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11
TABLE 13-9
It is believed that, the average numbers of hours spent studying per day (HOURS) during undergraduate education should have a positive linear relationship with the starting salary (SALARY, measured in thousands of dollars per month) after graduation. Given below is the Excel output from regressing starting salary on number of hours spent studying per day for a sample of 51 students. NOTE: Some of the numbers in the output are purposely erased.
 Regression Statistics  Multiple R 0.8857 R Square 0.7845 Adjusted R Square 0.7801 Standard Error 1.3704 Observations 51\begin{array}{lc}\text { Regression Statistics } & \\\hline \text { Multiple R } & 0.8857 \\\text { R Square } & 0.7845 \\\text { Adjusted R Square } & 0.7801 \\\text { Standard Error } & 1.3704 \\\text { Observations } & 51\end{array}
 ANOVA \text { ANOVA }
 df  SS  MS  Significance F  Regression 1335.0472335.0473178.3859 Residual 1.8782 Total 50427.0798\begin{array}{llcc}\hline&\text { df }&\text { SS } &\text { MS }&\text {F }&\text { Significance F } \\\text { Regression } & 1 & 335.0472 & 335.0473 & 178.3859 \\\text { Residual } & & & 1.8782 & \\\text { Total } & 50 & 427.0798 & &\end{array}

-Referring to Table 13-9, the degrees of freedom for the F test on whether HOURS affects SALARY are  Coefficients  Standaad Error t Stat p-value  Lower 95%  Upper 95%  Intercept 1.89400.40184.71342.051E052.70151.0865 Hours 0.97950.073313.35615.944E180.83211.1269\begin{array} { l r c c c r r } & \text { Coefficients } & \text { Standaad Error } &t \text { Stat } & p \text {-value } & \text { Lower 95\% } & \text { Upper 95\% } \\\hline \text { Intercept } & - 1.8940 & 0.4018 & - 4.7134 & 2.051 \mathrm { E } - 05 & - 2.7015 & - 1.0865 \\\text { Hours } & 0.9795 & 0.0733 & 13.3561 & 5.944 \mathrm { E } - 18 & 0.8321 & 1.1269 \\\hline\end{array}

A) 50, 1.
B) 1, 50.
C) 49, 1.
D) 1, 49.
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12
If you wanted to find out if alcohol consumption (measured in fluid oz.) and grade point average on a 4-point scale are linearly related, you would perform a

A) a t test for a correlation coefficient.
B) ?2 test for independence.
C) ?2 test for the difference in two proportions.
D) a Z test for the difference in two proportions.
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13
Testing for the existence of correlation is equivalent to

A) the confidence interval estimate for predicting Y.
B) testing for the existence of the slope (?1).
C) testing for the existence of the Y-intercept (?2).
D) none of the above
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14
TABLE 13-9
It is believed that, the average numbers of hours spent studying per day (HOURS) during undergraduate education should have a positive linear relationship with the starting salary (SALARY, measured in thousands of dollars per month) after graduation. Given below is the Excel output from regressing starting salary on number of hours spent studying per day for a sample of 51 students. NOTE: Some of the numbers in the output are purposely erased.
 Regression Statistics  Multiple R 0.8857 R Square 0.7845 Adjusted R Square 0.7801 Standard Error 1.3704 Observations 51\begin{array}{lc}\text { Regression Statistics } & \\\hline \text { Multiple R } & 0.8857 \\\text { R Square } & 0.7845 \\\text { Adjusted R Square } & 0.7801 \\\text { Standard Error } & 1.3704 \\\text { Observations } & 51\end{array}
 ANOVA \text { ANOVA }
 df  SS  MS  Significance F  Regression 1335.0472335.0473178.3859 Residual 1.8782 Total 50427.0798\begin{array}{llcc}\hline&\text { df }&\text { SS } &\text { MS }&\text {F }&\text { Significance F } \\\text { Regression } & 1 & 335.0472 & 335.0473 & 178.3859 \\\text { Residual } & & & 1.8782 & \\\text { Total } & 50 & 427.0798 & &\end{array}  <strong>TABLE 13-9 It is believed that, the average numbers of hours spent studying per day (HOURS) during undergraduate education should have a positive linear relationship with the starting salary (SALARY, measured in thousands of dollars per month) after graduation. Given below is the Excel output from regressing starting salary on number of hours spent studying per day for a sample of 51 students. NOTE: Some of the numbers in the output are purposely erased.  \begin{array}{lc} \text { Regression Statistics } & \\ \hline \text { Multiple R } & 0.8857 \\ \text { R Square } & 0.7845 \\ \text { Adjusted R Square } & 0.7801 \\ \text { Standard Error } & 1.3704 \\ \text { Observations } & 51 \end{array}   \text { ANOVA }   \begin{array}{llcc} \hline&\text { df }&\text { SS } &\text { MS }&\text {F }&\text { Significance F } \\ \text { Regression } & 1 & 335.0472 & 335.0473 & 178.3859 \\ \text { Residual } & & & 1.8782 & \\ \text { Total } & 50 & 427.0798 & & \end{array}     -Referring to Table 13-9, the estimated average change in salary (in thousands of dollars) as a result of spending an extra hour per day studying is</strong> A) 0.9795. B) -1.8940. C) 0.7845. D) 335.0473.

-Referring to Table 13-9, the estimated average change in salary (in thousands of dollars) as a result of spending an extra hour per day studying is

A) 0.9795.
B) -1.8940.
C) 0.7845.
D) 335.0473.
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15
TABLE 13-12
The manager of the purchasing department of a large banking organization would like to develop a model to predict the amount of time (measured in hours) it takes to process invoices. Data are collected from a sample of 30 days, and the number of invoices processed and completion time in hours is recorded. Below is the regression output:
 Regression  Statistics  Multiple R 0.9947 R Square 0.8924 Adjusted R Square 0.8886 Standard Error 0.3342 Observations 30\begin{array} { l c } { \text { Regression } \text { Statistics } } \\\hline \text { Multiple R } & 0.9947 \\\text { R Square } & 0.8924 \\\text { Adjusted R Square } & 0.8886 \\\text { Standard Error } & 0.3342 \\\text { Observations } & 30 \\\hline\end{array}  ANOVA df SS  MS F Significance F Regression 125.943825.9438232.22004.3946E15 Residual 283.12820.1117 Total 2929.072\begin{array}{l}\text { ANOVA }\\\begin{array} { l c c c c c } & d f & \text { SS } & \text { MS } & F & \text { Significance } F \\\text { Regression } & 1 & 25.9438&25 .9438 & 232.2200&4 .3946 \mathrm { E } - 15 \\\text { Residual } & 28 & 3.12820 .1117 & \\\text { Total } & 29 & 29.072 \\\hline\end{array}\end{array}  Coefficients  Standard Enror t Stat p-value  Lower 95%  Upper 95%  Invoices 0.40240.12363.25590.00300.14920.6555 Processed 0.01260.000815.23884.3946E150.01090.0143\begin{array} { l r r r r r r } \hline & \text { Coefficients } & \text { Standard Enror } & t \text { Stat } & p \text {-value } & \text { Lower 95\% } & \text { Upper 95\% } \\\hline \text { Invoices } & 0.4024 & 0.1236 & 3.2559 & 0.0030 & 0.1492 & 0.6555 \\\text { Processed } & 0.0126 & 0.0008 & 15.2388 & 4.3946 \mathrm { E } 15 & 0.0109 & 0.0143\end{array}  Coefficients  Standard Enor  t Stat p-value  Lower 95%  Upper 95%  Invoices 0.40240.12363.25590.00300.14920.6555\begin{array}{rrrrrrr} & \text { Coefficients } & \text { Standard Enor } & \text { t Stat } & p \text {-value } & \text { Lower 95\% } & \text { Upper 95\% } \\\hline \text { Invoices } & 0.4024 & 0.1236 & 3.2559 & 0.0030 & 0.1492 & 0.6555\end{array}
 <strong>TABLE 13-12 The manager of the purchasing department of a large banking organization would like to develop a model to predict the amount of time (measured in hours) it takes to process invoices. Data are collected from a sample of 30 days, and the number of invoices processed and completion time in hours is recorded. Below is the regression output:  \begin{array} { l c } { \text { Regression } \text { Statistics } } \\ \hline \text { Multiple R } & 0.9947 \\ \text { R Square } & 0.8924 \\ \text { Adjusted R Square } & 0.8886 \\ \text { Standard Error } & 0.3342 \\ \text { Observations } & 30 \\ \hline \end{array}   \begin{array}{l} \text { ANOVA }\\ \begin{array} { l c c c c c } & d f & \text { SS } & \text { MS } & F & \text { Significance } F \\ \text { Regression } & 1 & 25.9438&25 .9438 & 232.2200&4 .3946 \mathrm { E } - 15 \\ \text { Residual } & 28 & 3.12820 .1117 & \\ \text { Total } & 29 & 29.072 \\ \hline \end{array} \end{array}   \begin{array} { l r r r r r r } \hline & \text { Coefficients } & \text { Standard Enror } & t \text { Stat } & p \text {-value } & \text { Lower 95\% } & \text { Upper 95\% } \\ \hline \text { Invoices } & 0.4024 & 0.1236 & 3.2559 & 0.0030 & 0.1492 & 0.6555 \\ \text { Processed } & 0.0126 & 0.0008 & 15.2388 & 4.3946 \mathrm { E } 15 & 0.0109 & 0.0143 \end{array}   \begin{array}{rrrrrrr} & \text { Coefficients } & \text { Standard Enor } & \text { t Stat } & p \text {-value } & \text { Lower 95\% } & \text { Upper 95\% } \\ \hline \text { Invoices } & 0.4024 & 0.1236 & 3.2559 & 0.0030 & 0.1492 & 0.6555 \end{array}       -Referring to Table 13-12, the 90% confidence interval for the average change in the amount of time needed as a result of processing one additional invoice is</strong> A) narrower than [0.0109, 0.0143]. B) wider than [0.0109, 0.0143]. C) narrower than [0.1492, 0.6555]. D) wider than [0.1492, 0.6555].   <strong>TABLE 13-12 The manager of the purchasing department of a large banking organization would like to develop a model to predict the amount of time (measured in hours) it takes to process invoices. Data are collected from a sample of 30 days, and the number of invoices processed and completion time in hours is recorded. Below is the regression output:  \begin{array} { l c } { \text { Regression } \text { Statistics } } \\ \hline \text { Multiple R } & 0.9947 \\ \text { R Square } & 0.8924 \\ \text { Adjusted R Square } & 0.8886 \\ \text { Standard Error } & 0.3342 \\ \text { Observations } & 30 \\ \hline \end{array}   \begin{array}{l} \text { ANOVA }\\ \begin{array} { l c c c c c } & d f & \text { SS } & \text { MS } & F & \text { Significance } F \\ \text { Regression } & 1 & 25.9438&25 .9438 & 232.2200&4 .3946 \mathrm { E } - 15 \\ \text { Residual } & 28 & 3.12820 .1117 & \\ \text { Total } & 29 & 29.072 \\ \hline \end{array} \end{array}   \begin{array} { l r r r r r r } \hline & \text { Coefficients } & \text { Standard Enror } & t \text { Stat } & p \text {-value } & \text { Lower 95\% } & \text { Upper 95\% } \\ \hline \text { Invoices } & 0.4024 & 0.1236 & 3.2559 & 0.0030 & 0.1492 & 0.6555 \\ \text { Processed } & 0.0126 & 0.0008 & 15.2388 & 4.3946 \mathrm { E } 15 & 0.0109 & 0.0143 \end{array}   \begin{array}{rrrrrrr} & \text { Coefficients } & \text { Standard Enor } & \text { t Stat } & p \text {-value } & \text { Lower 95\% } & \text { Upper 95\% } \\ \hline \text { Invoices } & 0.4024 & 0.1236 & 3.2559 & 0.0030 & 0.1492 & 0.6555 \end{array}       -Referring to Table 13-12, the 90% confidence interval for the average change in the amount of time needed as a result of processing one additional invoice is</strong> A) narrower than [0.0109, 0.0143]. B) wider than [0.0109, 0.0143]. C) narrower than [0.1492, 0.6555]. D) wider than [0.1492, 0.6555].

-Referring to Table 13-12, the 90% confidence interval for the average change in the amount of time needed as a result of processing one additional invoice is

A) narrower than [0.0109, 0.0143].
B) wider than [0.0109, 0.0143].
C) narrower than [0.1492, 0.6555].
D) wider than [0.1492, 0.6555].
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16
TABLE 13-2
A candy bar manufacturer is interested in trying to estimate how sales are influenced by the price of their product. To do this, the company randomly chooses 6 small cities and offers the candy bar at different prices. Using candy bar sales as the dependent variable, the company will conduct a simple linear regression on the data below:
 City  Price ($) Sales  River Falls 1.30100 Hudson 1.6090 Ellsworth 1.8090 Prescott 2.0040 Rock Elm 2.4038 Stillwater 2.9032\begin{array} { l l c } \hline \text { City } & \text { Price } ( \$ ) & \text { Sales } \\\hline \text { River Falls } & 1.30 & 100 \\\text { Hudson } & 1.60 & 90 \\\text { Ellsworth } & 1.80 & 90 \\\text { Prescott } & 2.00 & 40 \\\text { Rock Elm } & 2.40 & 38 \\\text { Stillwater } & 2.90 & 32 \\\hline\end{array}

-Referring to Table 13-2, if the price of the candy bar is set at $2, the estimated average sales will be

A) 30.
B) 90.
C) 100.
D) 65.
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17
What do we mean when we say that a simple linear regression model is "statistically" useful?

A) The model is an excellent predictor of Y.
B) The model is a better predictor of Y than the sample mean, Y.
C) The model is "practically" useful for predicting Y.
D) All the statistics computed from the sample make sense.
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18
The sample correlation coefficient between X and Y is 0.375. It has been found out that the p- value is 0.256 when testing H0 : q = 0 against the one- sided alternative H0 : q > 0. To test H0 : q = 0 against the two- sided alternative H0 : q × 0 at a significance level of 0.2, the p- value is

A) (0.256)(2).
B) 1 - 0.256/2.
C) 0.256/2.
D) 1 - 0.256.
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19
TABLE 13-12
The manager of the purchasing department of a large banking organization would like to develop a model to predict the amount of time (measured in hours) it takes to process invoices. Data are collected from a sample of 30 days, and the number of invoices processed and completion time in hours is recorded. Below is the regression output:
 Regression  Statistics  Multiple R 0.9947 R Square 0.8924 Adjusted R Square 0.8886 Standard Error 0.3342 Observations 30\begin{array} { l c } { \text { Regression } \text { Statistics } } \\\hline \text { Multiple R } & 0.9947 \\\text { R Square } & 0.8924 \\\text { Adjusted R Square } & 0.8886 \\\text { Standard Error } & 0.3342 \\\text { Observations } & 30 \\\hline\end{array} ANOVA\text{ANOVA}

 dfSS M S F Significance F Regression 125.943825.9438232.22004.3946E15Residual 283.12820.1117 Total 2929.072\begin{array}{llcc}&\text { df} &\text {SS } &\text {M S}&\text { F}&\text { Significance F}\\\text { Regression } &1&25.9438&25 .9438&232 .2200&4 .3946 \mathrm{E}-15\\\text {Residual }&28 &3.12820 .1117 \\\text { Total } & 29 & 29.072\end{array}
 <strong>TABLE 13-12 The manager of the purchasing department of a large banking organization would like to develop a model to predict the amount of time (measured in hours) it takes to process invoices. Data are collected from a sample of 30 days, and the number of invoices processed and completion time in hours is recorded. Below is the regression output:  \begin{array} { l c } { \text { Regression } \text { Statistics } } \\ \hline \text { Multiple R } & 0.9947 \\ \text { R Square } & 0.8924 \\ \text { Adjusted R Square } & 0.8886 \\ \text { Standard Error } & 0.3342 \\ \text { Observations } & 30 \\ \hline \end{array}   \text{ANOVA}    \begin{array}{llcc} &\text { df} &\text {SS } &\text {M S}&\text { F}&\text { Significance F}\\ \text { Regression } &1&25.9438&25 .9438&232 .2200&4 .3946 \mathrm{E}-15\\ \text {Residual }&28 &3.12820 .1117 \\ \text { Total } & 29 & 29.072 \end{array}         -Referring to Table 13-12, the degrees of freedom for the t test on whether the number of invoices processed affects the amount of time are</strong> A) 1. B) 28. C) 30. D) 29.
 <strong>TABLE 13-12 The manager of the purchasing department of a large banking organization would like to develop a model to predict the amount of time (measured in hours) it takes to process invoices. Data are collected from a sample of 30 days, and the number of invoices processed and completion time in hours is recorded. Below is the regression output:  \begin{array} { l c } { \text { Regression } \text { Statistics } } \\ \hline \text { Multiple R } & 0.9947 \\ \text { R Square } & 0.8924 \\ \text { Adjusted R Square } & 0.8886 \\ \text { Standard Error } & 0.3342 \\ \text { Observations } & 30 \\ \hline \end{array}   \text{ANOVA}    \begin{array}{llcc} &\text { df} &\text {SS } &\text {M S}&\text { F}&\text { Significance F}\\ \text { Regression } &1&25.9438&25 .9438&232 .2200&4 .3946 \mathrm{E}-15\\ \text {Residual }&28 &3.12820 .1117 \\ \text { Total } & 29 & 29.072 \end{array}         -Referring to Table 13-12, the degrees of freedom for the t test on whether the number of invoices processed affects the amount of time are</strong> A) 1. B) 28. C) 30. D) 29.   <strong>TABLE 13-12 The manager of the purchasing department of a large banking organization would like to develop a model to predict the amount of time (measured in hours) it takes to process invoices. Data are collected from a sample of 30 days, and the number of invoices processed and completion time in hours is recorded. Below is the regression output:  \begin{array} { l c } { \text { Regression } \text { Statistics } } \\ \hline \text { Multiple R } & 0.9947 \\ \text { R Square } & 0.8924 \\ \text { Adjusted R Square } & 0.8886 \\ \text { Standard Error } & 0.3342 \\ \text { Observations } & 30 \\ \hline \end{array}   \text{ANOVA}    \begin{array}{llcc} &\text { df} &\text {SS } &\text {M S}&\text { F}&\text { Significance F}\\ \text { Regression } &1&25.9438&25 .9438&232 .2200&4 .3946 \mathrm{E}-15\\ \text {Residual }&28 &3.12820 .1117 \\ \text { Total } & 29 & 29.072 \end{array}         -Referring to Table 13-12, the degrees of freedom for the t test on whether the number of invoices processed affects the amount of time are</strong> A) 1. B) 28. C) 30. D) 29.

-Referring to Table 13-12, the degrees of freedom for the t test on whether the number of invoices processed affects the amount of time are

A) 1.
B) 28.
C) 30.
D) 29.
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20
The least squares method minimizes which of the following?

A) SSR
B) SST
C) SSE
D) all of the above
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21
TABLE 13-5
The managing partner of an advertising agency believes that his company's sales are related to the industry sales. He uses Microsoft Excel's Data Analysis tool to analyze the last 4 years of quarterly data with the following results:
 Regression Statistics  Multiple R 0.802 R Square 0.643 Adjusted R Square 0.618 Standard Error SYX0.9224 Observations 16\begin{array}{l}\text { Regression Statistics }\\\begin{array} { l c } \hline \text { Multiple R } & 0.802 \\\text { R Square } & 0.643 \\\text { Adjusted R Square } & 0.618 \\\text { Standard Error } S Y X & 0.9224 \\\text { Observations } & 16 \\\hline\end{array}\end{array}  ANOVA df SS  MS F Sig. F Regression 121.49721.49725.270.000 Error 1411.9120.851 Total 1533.409\begin{array}{l}\text { ANOVA }\\\begin{array} { l r l r l r } \hline & d f & \text { SS } & \text { MS } & F & \text { Sig. } F \\\hline \text { Regression } & 1 & 21.497 & 21.497 & 25.27 & 0.000 \\\text { Error } & 14 & 11.912 & 0.851 & & \\\text { Total } & 15 & 33.409 & & &\end{array}\end{array}  Predictor  Coefficients Standard Error t Stat p-value  Intercept 3.9621.4402.750.016 Industry 0.0404510.0080485.030.000\begin{array}{l}\text { Predictor }&\text { Coefficients}&\text { Standard Error }& t \text { Stat } &p \text {-value }\\\text { Intercept } & 3.962 & 1.440 & 2.75 & 0.016\\\text { Industry } & 0.040451 & 0.008048 & 5.03 & 0.000\end{array} Durbin- Watson Statistic 1.59

-Referring to Table 13-5, the partner wants to test for autocorrelation using the Durbin-Watson statistic. Using a level of significance of 0.05, the decision he should make is

A) there is no evidence of autocorrelation.
B) there is not enough information to perform the test.
C) there is evidence of autocorrelation.
D) the test is unable to make a definite conclusion.
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22
TABLE 13-2
A candy bar manufacturer is interested in trying to estimate how sales are influenced by the price of their product. To do this, the company randomly chooses 6 small cities and offers the candy bar at different prices. Using candy bar sales as the dependent variable, the company will conduct a simple linear regression on the data below:
 City  Price ($) Sales  River Falls 1.30100 Hudson 1.6090 Ellsworth 1.8090 Prescott 2.0040 Rock Elm 2.4038 Stillwater 2.9032\begin{array} { l l c } \hline \text { City } & \text { Price } ( \$ ) & \text { Sales } \\\hline \text { River Falls } & 1.30 & 100 \\\text { Hudson } & 1.60 & 90 \\\text { Ellsworth } & 1.80 & 90 \\\text { Prescott } & 2.00 & 40 \\\text { Rock Elm } & 2.40 & 38 \\\text { Stillwater } & 2.90 & 32 \\\hline\end{array}

-Referring to Table 13-2, to test whether a change in price will have any impact on average sales, what would be the critical values? Use ? = 0.05.

A) ±2.7765
B) ±2.5706
C) ±3.1634
D) ±3.4954
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23
If the correlation coefficient (r) = 1.00, then

A) there is no explained variation.
B) there is no unexplained variation.
C) the Y-intercept (b0) must equal 0.
D) the explained variation equals the unexplained variation.
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24
If the plot of the residuals is fan shaped, which assumption is violated?

A) independence of errors
B) normality
C) homoscedasticity
D) No assumptions are violated; the graph should resemble a fan.
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25
TABLE 13-12
The manager of the purchasing department of a large banking organization would like to develop a model to predict the amount of time (measured in hours) it takes to process invoices. Data are collected from a sample of 30 days, and the number of invoices processed and completion time in hours is recorded. Below is the regression output:
 Regression Statistics  Multiple R 0.9947 R Square 0.8924 Adjusted R Square 0.8886 Standard Error 0.3342 ations 30\begin{array}{l}\text { Regression Statistics }\\\begin{array} { l c } \hline \text { Multiple R } & 0.9947 \\\text { R Square } & 0.8924 \\\text { Adjusted R Square } & 0.8886 \\\text { Standard Error } & 0.3342 \\\text { ations } & 30 \\\hline\end{array}\end{array}

 d f  SS MS F  Significance F Regression125.943825.9438232.22004.3946E15Residual 283.12820.1117Total 2929.072\begin{array}{lrrccc}\hline & \text { d f } & \text { SS } & \text {MS} & \text { F } & \text { Significance F } \\\hline \text {Regression} & 1 & 25.9438 & 25.9438 & 232.2200 & 4.3946 \mathrm{E}-15 \\ \text {Residual }& 28 & 3.1282 & 0.1117 & & \\ \text {Total }& 29 & 29.072 & & & \\\hline\end{array}

 Coefficients  Standard Error  t Stat  p -valueLower 95%Upper 95% Invoices 0.40240.12363.25590.00300.14920.6555Processed 0.01260.000815.23884.3946E150.01090.0143\begin{array}{lrrrrrr}\hline & \text { Coefficients }& \text { Standard Error }& \text { t Stat }& \text { p -value}& \text {Lower 95\%} & \text {Upper 95\%} \\\hline \text { Invoices }& 0.4024 & 0.1236 & 3.2559 & 0.0030 & 0.1492 & 0.6555 \\ \text {Processed }& 0.0126 & 0.0008 & 15.2388 & 4.3946 \mathrm{E}-15 & 0.0109 & 0.0143 \\\hline\end{array}

 <strong>TABLE 13-12 The manager of the purchasing department of a large banking organization would like to develop a model to predict the amount of time (measured in hours) it takes to process invoices. Data are collected from a sample of 30 days, and the number of invoices processed and completion time in hours is recorded. Below is the regression output:  \begin{array}{l} \text { Regression Statistics }\\ \begin{array} { l c } \hline \text { Multiple R } & 0.9947 \\ \text { R Square } & 0.8924 \\ \text { Adjusted R Square } & 0.8886 \\ \text { Standard Error } & 0.3342 \\ \text { ations } & 30 \\ \hline \end{array} \end{array}    \begin{array}{lrrccc} \hline & \text { d f } &  \text { SS } &  \text {MS} &  \text { F } & \text { Significance  F } \\ \hline  \text {Regression} & 1 & 25.9438 & 25.9438 & 232.2200 & 4.3946 \mathrm{E}-15  \\ \text {Residual }& 28 & 3.1282 & 0.1117 & & \\ \text {Total }& 29 & 29.072 & & & \\ \hline \end{array}    \begin{array}{lrrrrrr} \hline & \text { Coefficients }& \text { Standard Error }& \text { t Stat }&  \text { p -value}& \text {Lower 95\%} &  \text {Upper 95\%} \\ \hline \text { Invoices }& 0.4024 & 0.1236 & 3.2559 & 0.0030 & 0.1492 & 0.6555 \\ \text {Processed }& 0.0126 & 0.0008 & 15.2388 &  4.3946 \mathrm{E}-15  & 0.0109 & 0.0143 \\ \hline \end{array}          -Referring to Table 13-12, the p-value of the measured t-test statistic to test whether the number of invoices processed affects the amount of time is</strong> A) (0.0030)/2. B) 0.0030. C) 4.3946E-15. D) (4.3946E-15)/2.

 <strong>TABLE 13-12 The manager of the purchasing department of a large banking organization would like to develop a model to predict the amount of time (measured in hours) it takes to process invoices. Data are collected from a sample of 30 days, and the number of invoices processed and completion time in hours is recorded. Below is the regression output:  \begin{array}{l} \text { Regression Statistics }\\ \begin{array} { l c } \hline \text { Multiple R } & 0.9947 \\ \text { R Square } & 0.8924 \\ \text { Adjusted R Square } & 0.8886 \\ \text { Standard Error } & 0.3342 \\ \text { ations } & 30 \\ \hline \end{array} \end{array}    \begin{array}{lrrccc} \hline & \text { d f } &  \text { SS } &  \text {MS} &  \text { F } & \text { Significance  F } \\ \hline  \text {Regression} & 1 & 25.9438 & 25.9438 & 232.2200 & 4.3946 \mathrm{E}-15  \\ \text {Residual }& 28 & 3.1282 & 0.1117 & & \\ \text {Total }& 29 & 29.072 & & & \\ \hline \end{array}    \begin{array}{lrrrrrr} \hline & \text { Coefficients }& \text { Standard Error }& \text { t Stat }&  \text { p -value}& \text {Lower 95\%} &  \text {Upper 95\%} \\ \hline \text { Invoices }& 0.4024 & 0.1236 & 3.2559 & 0.0030 & 0.1492 & 0.6555 \\ \text {Processed }& 0.0126 & 0.0008 & 15.2388 &  4.3946 \mathrm{E}-15  & 0.0109 & 0.0143 \\ \hline \end{array}          -Referring to Table 13-12, the p-value of the measured t-test statistic to test whether the number of invoices processed affects the amount of time is</strong> A) (0.0030)/2. B) 0.0030. C) 4.3946E-15. D) (4.3946E-15)/2.


-Referring to Table 13-12, the p-value of the measured t-test statistic to test whether the number of invoices processed affects the amount of time is

A) (0.0030)/2.
B) 0.0030.
C) 4.3946E-15.
D) (4.3946E-15)/2.
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26
Assuming a linear relationship between X and Y, if the coefficient of correlation (r) equals - 0.30,

A) the slope (b1) is negative.
B) there is no correlation.
C) variable X is larger than variable Y.
D) the variance of X is negative.
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27
TABLE 13-01
A large national bank charges local companies for using their services. A bank official reported the results of a regression analysis designed to predict the bank's charges (Y) -- measured in dollars per month -- for services rendered to local companies. One independent variable used to predict service charge to a company is the company's sales revenue (X) -- measured in millions of dollars. Data for 21 companies who use the bank's services were used to fit the model:
E(Y)=β0+β1XE(Y)=\beta_{0}+\beta_{1} X
The results of the simple linear regression are provided below.
Y^=2,700+20X, SYX=65, two-tailed p value =0.034( for testing β1)\hat{Y}=-2,700+20 X, \mathrm{~S}_{Y X}=65, \text { two-tailed } p \text { value }=0.034\left(\text { for testing } \beta_{1}\right)


-Referring to Table 13-1, interpret the p-value for testing whether þ1 exceeds 0.

A) There is sufficient evidence (at the ? = 0.05) to conclude that sales revenue (X) is a useful linear predictor of service charge (Y).
B) For every $1 million increase in sales revenue, we expect a service charge to increase $0.034.
C) There is insufficient evidence (at the ? = 0.10) to conclude that sales revenue (X) is a useful linear predictor of service charge (Y).
D) Sales revenue (X) is a poor predictor of service charge (Y).
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28
TABLE 13-2
A candy bar manufacturer is interested in trying to estimate how sales are influenced by the price of their product. To do this, the company randomly chooses 6 small cities and offers the candy bar at different prices. Using candy bar sales as the dependent variable, the company will conduct a simple linear regression on the data below:
 City  Price ($) Sales  River Falls 1.30100 Hudson 1.6090 Ellsworth 1.8090 Prescott 2.0040 Rock Elm 2.4038 Stillwater 2.9032\begin{array} { l l c } \hline \text { City } & \text { Price } ( \$ ) & \text { Sales } \\\hline \text { River Falls } & 1.30 & 100 \\\text { Hudson } & 1.60 & 90 \\\text { Ellsworth } & 1.80 & 90 \\\text { Prescott } & 2.00 & 40 \\\text { Rock Elm } & 2.40 & 38 \\\text { Stillwater } & 2.90 & 32 \\\hline\end{array}

-Referring to Table 13-2, what percentage of the total variation in candy bar sales is explained by prices?

A) 100%
B) 78.39%
C) 88.54%
D) 48.19%
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29
The Y-intercept (b0) represents the

A) change in estimated average Y per unit change in X.
B) variation around the sample regression line.
C) predicted value of Y.
D) estimated average Y when X = 0.
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30
TABLE 13- 11
A company that has the distribution rights to home video sales of previously released movies would like to use the box office gross (in millions of dollars) to estimate the number of units (in thousands of units) that it can expect to sell. Following is the output from a simple linear regression along with the residual plot and normal probability plot obtained from a data set of 30 different movie titles:
 Regression Statistics  Multiple R 0.8531 RSquare 0.7278 Adjusted R Square 0.7180 Standard Error 47.8668 Observations 30\begin{array}{l}\text { Regression Statistics }\\\begin{array} { l c } \hline \text { Multiple R } & 0.8531 \\\text { RSquare } & 0.7278 \\\text { Adjusted R Square } & 0.7180 \\\text { Standard Error } & 47.8668 \\\text { Observations } & 30\end{array}\end{array}
ANOVA
 d f SS  MS Significance FRegression 1171499.78171499.7874.85052.1259E09Residual2864154.422291.23Total29235654.20\begin{array}{lrrrrr}\hline &\text { d f}& \text { SS } & \text { MS } & \text {F }& \text {Significance F} \\\hline \text {Regression }& 1 & 171499.78 & 171499.78 & 74.8505 & 2.1259E-09 \\\text {Residual} & 28 & 64154.42 & 2291.23 & & \\\text {Total} & 29 & 235654.20 & & & \\\hline\end{array}

Coefficients  Standard Error t Stat  p -value Lower 95% Upper 95%  Intercept 76.535111.83186.46865.24E0752.2987100.7716Gross4.33310.50088.65162.13E093.30725.3590\begin{array}{lrrrrrr}\hline & \text {Coefficients }& \text { Standard Error}& \text { t Stat }& \text { p -value }& \text {Lower 95\% }& \text {Upper 95\% }\\\hline \text { Intercept }& 76.5351 & 11.8318 & 6.4686 & 5.24 \mathrm{E}-07& 52.2987 & 100.7716 \\ \text {Gross} & 4.3331 & 0.5008 & 8.6516 & 2.13 \mathrm{E}-09 & 3.3072 & 5.3590 \\\hline\end{array}

 <strong>TABLE 13- 11 A company that has the distribution rights to home video sales of previously released movies would like to use the box office gross (in millions of dollars) to estimate the number of units (in thousands of units) that it can expect to sell. Following is the output from a simple linear regression along with the residual plot and normal probability plot obtained from a data set of 30 different movie titles:  \begin{array}{l} \text { Regression Statistics }\\ \begin{array} { l c } \hline \text { Multiple R } & 0.8531 \\ \text { RSquare } & 0.7278 \\ \text { Adjusted R Square } & 0.7180 \\ \text { Standard Error } & 47.8668 \\ \text { Observations } & 30 \end{array} \end{array}  ANOVA  \begin{array}{lrrrrr} \hline &\text { d f}& \text { SS } & \text { MS } & \text {F }& \text {Significance F}  \\ \hline \text {Regression }& 1 & 171499.78 & 171499.78 & 74.8505 & 2.1259E-09 \\ \text {Residual} & 28 & 64154.42 & 2291.23 & & \\ \text {Total} & 29 & 235654.20 & & & \\ \hline\end{array}    \begin{array}{lrrrrrr} \hline &  \text {Coefficients }& \text { Standard Error}& \text { t  Stat }&  \text { p -value }&  \text {Lower 95\% }& \text {Upper 95\% }\\ \hline \text { Intercept }& 76.5351 & 11.8318 & 6.4686 & 5.24 \mathrm{E}-07& 52.2987 & 100.7716 \\ \text {Gross} & 4.3331 & 0.5008 & 8.6516 & 2.13 \mathrm{E}-09 & 3.3072 & 5.3590 \\ \hline \end{array}        -Referring to Table 13-11, which of the following is the correct interpretation for the slope coefficient?</strong> A) For each increase of 1 dollar in box office gross, expected home video units sold are estimated to increase by 4.3331 thousand units. B) For each increase of 1 million dollars in box office gross, expected home video units sold are estimated to increase by 4.3331 units. C) For each increase of 1 dollar in box office gross, expected home video units sold are estimated to increase by 4.3331 units. D) For each increase of 1 million dollars in box office gross, expected home video units sold are estimated to increase by 4.3331 thousand units.   <strong>TABLE 13- 11 A company that has the distribution rights to home video sales of previously released movies would like to use the box office gross (in millions of dollars) to estimate the number of units (in thousands of units) that it can expect to sell. Following is the output from a simple linear regression along with the residual plot and normal probability plot obtained from a data set of 30 different movie titles:  \begin{array}{l} \text { Regression Statistics }\\ \begin{array} { l c } \hline \text { Multiple R } & 0.8531 \\ \text { RSquare } & 0.7278 \\ \text { Adjusted R Square } & 0.7180 \\ \text { Standard Error } & 47.8668 \\ \text { Observations } & 30 \end{array} \end{array}  ANOVA  \begin{array}{lrrrrr} \hline &\text { d f}& \text { SS } & \text { MS } & \text {F }& \text {Significance F}  \\ \hline \text {Regression }& 1 & 171499.78 & 171499.78 & 74.8505 & 2.1259E-09 \\ \text {Residual} & 28 & 64154.42 & 2291.23 & & \\ \text {Total} & 29 & 235654.20 & & & \\ \hline\end{array}    \begin{array}{lrrrrrr} \hline &  \text {Coefficients }& \text { Standard Error}& \text { t  Stat }&  \text { p -value }&  \text {Lower 95\% }& \text {Upper 95\% }\\ \hline \text { Intercept }& 76.5351 & 11.8318 & 6.4686 & 5.24 \mathrm{E}-07& 52.2987 & 100.7716 \\ \text {Gross} & 4.3331 & 0.5008 & 8.6516 & 2.13 \mathrm{E}-09 & 3.3072 & 5.3590 \\ \hline \end{array}        -Referring to Table 13-11, which of the following is the correct interpretation for the slope coefficient?</strong> A) For each increase of 1 dollar in box office gross, expected home video units sold are estimated to increase by 4.3331 thousand units. B) For each increase of 1 million dollars in box office gross, expected home video units sold are estimated to increase by 4.3331 units. C) For each increase of 1 dollar in box office gross, expected home video units sold are estimated to increase by 4.3331 units. D) For each increase of 1 million dollars in box office gross, expected home video units sold are estimated to increase by 4.3331 thousand units.

-Referring to Table 13-11, which of the following is the correct interpretation for the slope coefficient?

A) For each increase of 1 dollar in box office gross, expected home video units sold are estimated to increase by 4.3331 thousand units.
B) For each increase of 1 million dollars in box office gross, expected home video units sold are estimated to increase by 4.3331 units.
C) For each increase of 1 dollar in box office gross, expected home video units sold are estimated to increase by 4.3331 units.
D) For each increase of 1 million dollars in box office gross, expected home video units sold are estimated to increase by 4.3331 thousand units.
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31
TABLE 13-01
A large national bank charges local companies for using their services. A bank official reported the results of a regression analysis designed to predict the bank's charges
E(Y)=β0+β1XE(Y)=\beta_{0}+\beta_{1} X
The results of the simple linear regression are provided below.
Y^=2,700+20X, SYX=65, two-tailed p value =0.034( for testing β1)\hat{Y}=-2,700+20 X, \mathrm{~S}_{Y X}=65, \text { two-tailed } p \text { value }=0.034\left(\text { for testing } \beta_{1}\right)

(Y) -- measured in dollars per month -- for services rendered to local companies. One independent variable used to predict service charge to a company is the company's sales revenue (X) -- measured in millions of dollars. Data for 21 companies who use the bank's services were used to fit the model:



-Referring to Table 13-1, interpret the estimate of a, the standard deviation of the random error term (standard error of the estimate) in the model.

A) About 95% of the observed service charges fall within $130 of the least squares line.
B) For every $1 million increase in sales revenue, we expect a service charge to increase $65.
C) About 95% of the observed service charges fall within $65 of the least squares line.
D) About 95% of the observed service charges equal their corresponding predicted values.
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32
TABLE 13-01
A large national bank charges local companies for using their services. A bank official reported the results of a regression analysis designed to predict the bank's charges (Y) -- measured in dollars per month -- for services rendered to local companies. One independent variable used to predict service charge to a company is the company's sales revenue (X) -- measured in millions of dollars. Data for 21 companies who use the bank's services were used to fit the model:
E(Y) =?0 + ?1X
The results of the simple linear regression are provided below.

Y^=2,700+20X,SYX=65 \hat{Y}=-2,700+20 X, S_{Y X}=65 , two-tailed p p value =0.034( =0.034\left(\right. for testing β1) \left.\beta_{1}\right)


-Referring to Table 13-1, a 95% confidence interval for ?1 is (15, 30). Interpret the interval.

A) At the ? = 0.05 level, there is no evidence of a linear relationship between service charge (Y) and sales revenue (X).
B) We are 95% confident that the mean service charge will fall between $15 and $30 per month.
C) We are 95% confident that average service charge (Y) will increase between $15 and $30 for every $1 million increase in sales revenue (X).
D) We are 95% confident that the sales revenue (X) will increase between $15 and $30 million for every $1 increase in service charge (Y).
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33
The slope (b1) represents

A) the estimated average change in Y per unit change in X.
B) predicted value of Y when X = 0.
C) the predicted value of Y.
D) variation around the line of regression.
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34
TABLE 13-7
An investment specialist claims that if one holds a portfolio that moves in the opposite direction to the market index like the S&P 500, then it is possible to reduce the variability of the portfolio's return. In other words, one can create a portfolio with positive returns but less exposure to risk. A sample of 26 years of S&P 500 index and a portfolio consisting of stocks of private prisons, which are believed to be negatively related to the S&P 500 index, is collected. A regression analysis was performed by regressing the returns of the prison stocks portfolio (Y) on the returns of S&P 500 index (X) to prove that the prison stocks portfolio is negatively related to the S&P 500 index at a 5% level of significance. The results are given in the following EXCEL output.
 Coefficients  Standard Enor  T Stat p-value  Intercept 4.8660042580.3574360913.613634418.7932E13 S&P 0.5025135060.0715971527.018624252.94942E07\begin{array} { l c c c c } \hline & \text { Coefficients } & \text { Standard Enor } & \text { T Stat } & p \text {-value } \\\hline \text { Intercept } & 4.866004258 & 0.35743609 & 13.61363441 & 8.7932 \mathrm { E } - 13 \\\text { S\&P } & - 0.502513506 & 0.071597152 -& 7.01862425 & 2.94942 \mathrm { E } - 07 \\\hline\end{array}

-Referring to Table 13-7, which of the following will be a correct conclusion?

A) We can reject the null hypothesis and, therefore, conclude that there is sufficient evidence to show that the prisons stock portfolio and S&P 500 index are negatively related.
B) We can reject the null hypothesis and conclude that there is not sufficient evidence to show that the prisons stock portfolio and S&P 500 index are negatively related.
C) We cannot reject the null hypothesis and, therefore, conclude that there is not sufficient evidence to show that the prisons stock portfolio and S&P 500 index are negatively related.
D) We cannot reject the null hypothesis and, therefore, conclude that there is sufficient evidence to show that the prisons stock portfolio and S&P 500 index are negatively related.
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35
 TABLE 13-10 \text { TABLE 13-10 } The management of a chain electronic store would like to develop a model for predicting the weekly sales (in thousand of dollars) for individual stores based on the number of customers who made purchases. A random sample of 12 stores yields the following results:
 Cugtomera  5aleg (Thougardaof Dollara) 90711.2092611.057138.217419.217809.4289810.085106.735297.024606.128729.526507.536037.25\begin{array} { | c | c | } \hline \text { Cugtomera } & \text { 5aleg (Thougardaof Dollara) } \\\hline \mathbf { 9 0 7 } & 11.20 \\\hline \mathbf { 9 2 6 } & 11.05 \\\hline 713 & \mathbf { 8 . 2 1 } \\\hline 741 & \mathbf { 9 . 2 1 } \\\hline 780 & \mathbf { 9 . 4 2 } \\\hline \mathbf { 8 9 8 } & \mathbf { 1 0 . 0 8 } \\\hline 510 & \mathbf { 6 . 7 3 } \\\hline 529 & 7.02 \\\hline 460 & \mathbf { 6 . 1 2 } \\\hline \mathbf { 8 7 2 } & \mathbf { 9 . 5 2 } \\\hline \mathbf { 6 5 0 } & 7.53 \\\hline \mathbf { 6 0 3 } & 7.25 \\\hline\end{array}

-Referring to Table 13-10, the residual plot indicates possible violation of which assumptions?

A) normality
B) autocorrelation
C) linearity of the relationship
D) homoscedasticity
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36
In a simple linear regression problem, r and b1

A) must have the same sign.
B) may have opposite signs.
C) must have opposite signs.
D) are equal.
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37
The standard error of the estimate is a measure of

A) the variation of the X variable.
B) total variation of the Y variable.
C) the variation around the sample regression line.
D) explained variation.
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38
TABLE 13-9
It is believed that, the average numbers of hours spent studying per day (HOURS) during undergraduate education should have a positive linear relationship with the starting salary (SALARY, measured in thousands of dollars per month) after graduation. Given below is the Excel output from regressing starting salary on number of hours spent studying per day for a sample of 51 students. NOTE: Some of the numbers in the output are purposely erased.
 Regression Stedistics  Multiple R 0.8857 RSquare 0.7845 Adjusted R Square 0.7801 Standard Error 1.3704\begin{array}{ll}\text { Regression Stedistics } \\\hline \text { Multiple R } & 0.8857 \\\text { RSquare } & 0.7845 \\\text { Adjusted R Square } & 0.7801 \\\text { Standard Error } & 1.3704\end{array}

 Observations 51\text { Observations } \quad 51

ANOVAANOVA
dfSSMSF Significance F Regression 1335.0472335.0473178.3859 Residual 1.8782 Total 50427.0798\begin{array}{lccrcc}\hline & d f & S S &{M S} & F & \text { Significance } F \\\hline \text { Regression } & 1 & 335.0472 & 335.0473 & 178.3859 \\\text { Residual } & & 1.8782 & \\\text { Total } & 50 & 427.0798 & & \\\hline\end{array}


 Coeffcients  Standard Error t Stat p-value  Lower 95% Upper 95%  Intercept 1.89400.40184.71342.051E052.70151.0865 Hours 0.97950.073313.35615.944E180.83211.1269\begin{array}{lccccrr}\hline & \text { Coeffcients } & \text { Standard Error }& t \text { Stat } & p \text {-value } & \text { Lower 95\%} & \text { Upper 95\% } \\\hline \text { Intercept } & -1.8940 & 0.4018 & -4.7134 & 2.051 \mathrm{E}-05 & -2.7015 & -1.0865 \\\text { Hours } & 0.9795 & 0.0733 & 13.3561 & 5.944 \mathrm{E}-18 & 0.8321 & 1.1269 \\\hline\end{array}



-Referring to Table 13-9, to test the claim that average SALARY depends positively on HOURS against the null hypothesis that average SALARY does not depend linearly on HOURS, the p-value of the test statistic is

A) (2.051E-05)/2.
B) 5.944E-18.
C) (5.944E-18)/2.
D) 2.051E-05.
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39
The strength of the linear relationship between two numerical variables may be measured by the

A) slope.
B) Y-intercept.
C) coefficient of correlation.
D) scatter diagram.
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40
TABLE 13-8
It is believed that GPA (grade point average, based on a four point scale) should have a positive linear relationship with ACT scores. Given below is the Excel output from regressing GPA on ACT scores using a data set of 8 randomly chosen students from a Big-Ten university.
 Regressing GPA on ACT Regression Statistics  Multiple R 0.7598 R Square 0.5774 Adjusted R Square 0.5069 Standard E rror 0.2691 Observations 8\begin{array}{l}\text { Regressing GPA on } \mathrm { ACT }\\\begin{array} { l c } \hline & \text { Regression Statistics } \\\hline \text { Multiple R } & 0.7598 \\\text { R Square } & 0.5774 \\\text { Adjusted R Square } & 0.5069 \\\text { Standard E rror } & 0.2691 \\\text { Observations } & 8\end{array}\end{array}

ANOVA
d fSS MS  F  Significance FRegression 10.59400.59408.19860.0286Residual 60.43470.0724Total 71.0287\begin{array}{lcllcc}\hline & \text {d f} & \text {SS} & \text { MS }& \text { F } & \text { Significance F} \\\hline \text {Regression }&1 & 0.5940 & 0.5940 & 8.1986 & 0.0286 \\ \text {Residual }& 6 & 0.4347 & 0.0724 & & \\ \text {Total }& 7 & 1.0287 & & & \\\hline\end{array}

Coefficients Standard Error T Stat  p -value Lower 95% Upper 95%  Intercept0.56810.92840.61190.56301.70362.8398ACT 0.10210.03562.86330.02860.01480.1895\begin{array}{lrrrrrr}\hline & \text {Coefficients }& \text {Standard Error }& \text {T Stat }& \text { p -value }& \text {Lower 95\%}& \text { Upper 95\% } \\\hline \text { Intercept} & 0.5681 & 0.9284 & 0.6119 & 0.5630 & -1.7036 & 2.8398 \\ \text {ACT }& 0.1021 & 0.0356 & 2.8633 & 0.0286 & 0.0148 & 0.1895 \\\hline\end{array}


-Referring to Table 13-8, the value of the measured (observed) test statistic of the F-test for H0 : ?1 = 0 versus H1 : ?1 ? 0

A) is always positive.
B) has the same sign as the corresponding t test statistic.
C) may be negative.
D) is always negative.
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41
The coefficient of determination (r2) tells us

A) the proportion of total variation that is explained.
B) that we should not partition the total variation.
C) whether r has any significance.
D) that the coefficient of correlation (r) is larger than 1.
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42
TABLE 13-9
It is believed that, the average numbers of hours spent studying per day (HOURS) during undergraduate education should have a positive linear relationship with the starting salary (SALARY, measured in thousands of dollars per month) after graduation. Given below is the Excel output from regressing starting salary on number of hours spent studying per day for a sample of 51 students. NOTE: Some of the numbers in the output are purposely erased.
 Regression Stedistics  Multiple R 0.8857 R Square 0.7845 Adjusted R Square 0.7801 Standard Error 1.3704 Observations 51\begin{array}{l}\text { Regression Stedistics }\\\begin{array} { l c } \hline \text { Multiple R } & 0.8857 \\\text { R Square } & 0.7845 \\ \text { Adjusted R Square } & 0.7801 \\\text { Standard Error } & 1.3704 \\\text { Observations } & 51\end{array}\end{array}  ANOVA \text { ANOVA }
dfSSMSF Significance F Regression 1335.0472335.0473178.3859 Residual 1.8782 Total 50427.0798\begin{array}{lccrcc}\hline & d f & S S &{M S} & F & \text { Significance } F \\\hline \text { Regression } & 1 & 335.0472 & 335.0473 & 178.3859 \\\text { Residual } & & & 1.8782 & \\\text { Total } & 50 & 427.0798 & &\end{array}  Coefficients  Standard Enror t Stat p-value  Lower 95%  Upper 95%  Intercept 1.89400.40184.71342.051E052.70151.0865 Hours 0.97950.073313.35615.944E180.83211.1269\begin{array} { l r c c c r r } & \text { Coefficients } & \text { Standard Enror } & t \text { Stat } & { p \text {-value } } & \text { Lower 95\% } & \text { Upper 95\% } \\\hline \text { Intercept } & - 1.8940 & 0.4018 & - 4.7134 & 2.051 \mathrm { E } - 05 & - 2.7015 & - 1.0865 \\\text { Hours } & 0.9795 & 0.0733 & 13.3561 & 5.944 \mathrm { E } - 18 & 0.8321 & 1.1269 \\\hline\end{array} \begin{array} { l r r r r r r } \end{array}

-Referring to Table 13-9, the error sum of squares (SSE) of the above regression is

A) 1.878215.
B) 335.047257.
C) 92.0325465.
D) 427.079804.
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43
TABLE 13-7
An investment specialist claims that if one holds a portfolio that moves in the opposite direction to the market index like the S&P 500, then it is possible to reduce the variability of the portfolio's return. In other words, one can create a portfolio with positive returns but less exposure to risk. A sample of 26 years of S&P 500 index and a portfolio consisting of stocks of private prisons, which are believed to be negatively related to the S&P 500 index, is collected. A regression analysis was performed by regressing the returns of the prison stocks portfolio (Y) on the returns of S&P 500 index (X) to prove that the prison stocks portfolio is negatively related to the S&P 500 index at a 5% level of significance. The results are given in the following EXCEL output.
 Coefficients  Standard Error  T Stat p-value  Intercept 4.8660042580.3574360913.613634418.7932E13 S& P 0.5025135060.0715971527.01862425294942E07\begin{array} { l c c c c } \hline & \text { Coefficients } & \text { Standard Error } & \text { T Stat } & { p \text {-value } } \\\hline \text { Intercept } & 4.866004258 & 0.35743609 & 13.61363441 & 8.7932 \mathrm { E } - 13 \\\text { S\& P } & - 0.502513506 & 0.071597152 &- 7.01862425 & 294942 \mathrm { E } - 07 \\\hline\end{array}

-Referring to Table 13-7, to test whether the prison stocks portfolio is negatively related to the S&P 500 index, the p-value of the associated test statistic is

A) 8.7932E- 13.
B) 2.94942E- 07.
C) (2.94942E- 07)2.
D) 2.94942E- 07/2.
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44
TABLE 13-2
A candy bar manufacturer is interested in trying to estimate how sales are influenced by the price of their product. To do this, the company randomly chooses 6 small cities and offers the candy bar at different prices. Using candy bar sales as the dependent variable, the company will conduct a simple linear regression on the data below:
 City  Price ($) Sales  River Falls 1.30100 Hudson 1.6090 Ellsworth 1.8090 Prescott 2.0040 Rock Elm 2.4038 Stillwater 2.9032\begin{array} { l l c } \hline \text { City } & \text { Price } ( \$ ) & \text { Sales } \\\hline \text { River Falls } & 1.30 & 100 \\\text { Hudson } & 1.60 & 90 \\\text { Ellsworth } & 1.80 & 90 \\\text { Prescott } & 2.00 & 40 \\\text { Rock Elm } & 2.40 & 38 \\\text { Stillwater } & 2.90 & 32 \\\hline\end{array}

-Referring to Table 13-2, what is the standard error of the regression slope estimate, Sb1?

A) 12.650
B) 0.784
C) 0.885
D) 16.299
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45
TABLE 13-2
A candy bar manufacturer is interested in trying to estimate how sales are influenced by the price of their product. To do this, the company randomly chooses 6 small cities and offers the candy bar at different prices. Using candy bar sales as the dependent variable, the company will conduct a simple linear regression on the data below:
 City  Price ($) Sales  River Falls 1.30100 Hudson 1.6090 Ellsworth 1.8090 Prescott 2.0040 Rock Elm 2.4038 Stillwater 2.9032\begin{array} { l l c } \hline \text { City } & \text { Price } ( \$ ) & \text { Sales } \\\hline \text { River Falls } & 1.30 & 100 \\\text { Hudson } & 1.60 & 90 \\\text { Ellsworth } & 1.80 & 90 \\\text { Prescott } & 2.00 & 40 \\\text { Rock Elm } & 2.40 & 38 \\\text { Stillwater } & 2.90 & 32 \\\hline\end{array}

-Referring to Table 13-2, what is the standard error of the estimate, SYX, for the data?

A) 0.885
B) 16.299
C) 0.784
D) 12.650
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46
The sample correlation coefficient between X and Y is 0.375. It has been found out that the p-value is 0.256 when testing H0 : ? = 0 against the two-sided alternative H0 :? ? 0. To test H0 : ? = 0 against the one-sided alternative H0 : ? < 0 at a significance level of 0.2, the p-value is

A) 1 - 0.256/2.
B) (0.256)2.
C) 1 - 0.256.
D) 0.256/2.
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47
The width of the prediction interval for the predicted value of Y is dependent on

A) the standard error of the estimate.
B) the sample size.
C) the value of X for which the prediction is being made.
D) all of the above
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48
The sample correlation coefficient between X and Y is 0.375. It has been found out that the p- value is 0.256 when testing H0 : ? = 0 against the two- sided alternative H0: ?? 0. To test H0 :? = 0 against the one- sided alternative H0 : ? > 0 at a significance level of 0.2, the p- value is

A) (0.256)2.
B) 1 - 0.256/2.
C) 0.256/2.
D) 1 - 0.256.
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49
Based on the residual plot below, you will conclude that there might be a violation of which of the following assumptions?
<strong>Based on the residual plot below, you will conclude that there might be a violation of which of the following assumptions?  </strong> A) Homoscedasticity B) Independence of errors C) Normality of errors D) Linearity of the relationship

A) Homoscedasticity
B) Independence of errors
C) Normality of errors
D) Linearity of the relationship
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50
TABLE 13-12
The manager of the purchasing department of a large banking organization would like to develop a model to predict the amount of time (measured in hours) it takes to process invoices. Data are collected from a sample of 30 days, and the number of invoices processed and completion time in hours is recorded. Below is the regression output:
 Regression Statistics  Multiple R 0.9947 R Square 0.8924 Adjusted R Square 0.8886 Standard Error 0.3342 ations 30\begin{array}{l}\text { Regression Statistics }\\\begin{array} { l c } \hline \text { Multiple R } & 0.9947 \\\text { R Square } & 0.8924 \\\text { Adjusted R Square } & 0.8886 \\\text { Standard Error } & 0.3342 \\\text { ations } & 30 \\\hline\end{array}\end{array}

 d f  SS MS F  Significance F Regression125.943825.9438232.22004.3946E15Residual 283.12820.1117Total 2929.072\begin{array}{lrrccc}\hline & \text { d f } & \text { SS } & \text {MS} & \text { F } & \text { Significance F } \\\hline \text {Regression} & 1 & 25.9438 & 25.9438 & 232.2200 & 4.3946 \mathrm{E}-15 \\ \text {Residual }& 28 & 3.1282 & 0.1117 & & \\ \text {Total }& 29 & 29.072 & & & \\\hline\end{array}

 Coefficients  Standard Error  t Stat  p -valueLower 95%Upper 95% Invoices 0.40240.12363.25590.00300.14920.6555Processed 0.01260.000815.23884.3946E150.01090.0143\begin{array}{lrrrrrr}\hline & \text { Coefficients }& \text { Standard Error }& \text { t Stat }& \text { p -value}& \text {Lower 95\%} & \text {Upper 95\%} \\\hline \text { Invoices }& 0.4024 & 0.1236 & 3.2559 & 0.0030 & 0.1492 & 0.6555 \\ \text {Processed }& 0.0126 & 0.0008 & 15.2388 & 4.3946 \mathrm{E}-15 & 0.0109 & 0.0143 \\\hline\end{array}

 <strong>TABLE 13-12 The manager of the purchasing department of a large banking organization would like to develop a model to predict the amount of time (measured in hours) it takes to process invoices. Data are collected from a sample of 30 days, and the number of invoices processed and completion time in hours is recorded. Below is the regression output:  \begin{array}{l} \text { Regression Statistics }\\ \begin{array} { l c } \hline \text { Multiple R } & 0.9947 \\ \text { R Square } & 0.8924 \\ \text { Adjusted R Square } & 0.8886 \\ \text { Standard Error } & 0.3342 \\ \text { ations } & 30 \\ \hline \end{array} \end{array}    \begin{array}{lrrccc} \hline & \text { d f } &  \text { SS } &  \text {MS} &  \text { F } & \text { Significance  F } \\ \hline  \text {Regression} & 1 & 25.9438 & 25.9438 & 232.2200 & 4.3946 \mathrm{E}-15  \\ \text {Residual }& 28 & 3.1282 & 0.1117 & & \\ \text {Total }& 29 & 29.072 & & & \\ \hline \end{array}    \begin{array}{lrrrrrr} \hline & \text { Coefficients }& \text { Standard Error }& \text { t Stat }&  \text { p -value}& \text {Lower 95\%} &  \text {Upper 95\%} \\ \hline \text { Invoices }& 0.4024 & 0.1236 & 3.2559 & 0.0030 & 0.1492 & 0.6555 \\ \text {Processed }& 0.0126 & 0.0008 & 15.2388 &  4.3946 \mathrm{E}-15  & 0.0109 & 0.0143 \\ \hline \end{array}          -Referring to Table 13-12, the p-value of the measured F-test statistic to test whether the number of invoices processed affects the amount of time is</strong> A) (4.3946E-15)/2. B) (0.0030)/2. C) 0.0030. D) 4.3946E-15.

 <strong>TABLE 13-12 The manager of the purchasing department of a large banking organization would like to develop a model to predict the amount of time (measured in hours) it takes to process invoices. Data are collected from a sample of 30 days, and the number of invoices processed and completion time in hours is recorded. Below is the regression output:  \begin{array}{l} \text { Regression Statistics }\\ \begin{array} { l c } \hline \text { Multiple R } & 0.9947 \\ \text { R Square } & 0.8924 \\ \text { Adjusted R Square } & 0.8886 \\ \text { Standard Error } & 0.3342 \\ \text { ations } & 30 \\ \hline \end{array} \end{array}    \begin{array}{lrrccc} \hline & \text { d f } &  \text { SS } &  \text {MS} &  \text { F } & \text { Significance  F } \\ \hline  \text {Regression} & 1 & 25.9438 & 25.9438 & 232.2200 & 4.3946 \mathrm{E}-15  \\ \text {Residual }& 28 & 3.1282 & 0.1117 & & \\ \text {Total }& 29 & 29.072 & & & \\ \hline \end{array}    \begin{array}{lrrrrrr} \hline & \text { Coefficients }& \text { Standard Error }& \text { t Stat }&  \text { p -value}& \text {Lower 95\%} &  \text {Upper 95\%} \\ \hline \text { Invoices }& 0.4024 & 0.1236 & 3.2559 & 0.0030 & 0.1492 & 0.6555 \\ \text {Processed }& 0.0126 & 0.0008 & 15.2388 &  4.3946 \mathrm{E}-15  & 0.0109 & 0.0143 \\ \hline \end{array}          -Referring to Table 13-12, the p-value of the measured F-test statistic to test whether the number of invoices processed affects the amount of time is</strong> A) (4.3946E-15)/2. B) (0.0030)/2. C) 0.0030. D) 4.3946E-15.


-Referring to Table 13-12, the p-value of the measured F-test statistic to test whether the number of invoices processed affects the amount of time is

A) (4.3946E-15)/2.
B) (0.0030)/2.
C) 0.0030.
D) 4.3946E-15.
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51
If the correlation coefficient (r) = 1.00, then

A) all the data points must fall exactly on a straight line with a slope that equals 1.00.
B) all the data points must fall exactly on a straight line with a negative slope.
C) all the data points must fall exactly on a horizontal straight line with a zero slope.
D) all the data points must fall exactly on a straight line with a positive slope.
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52
TABLE 13- 11
A company that has the distribution rights to home video sales of previously released movies would like to use the box office gross (in millions of dollars) to estimate the number of units (in thousands of units) that it can expect to sell. Following is the output from a simple linear regression along with the residual plot and normal probability plot obtained from a data set of 30 different movie titles:
 Regression Statistics  Multiple R 0.8531 RSquare 0.7278 Adjusted R Square 0.7180 Standard Error 47.8668 Observations 30\begin{array}{l}\text { Regression Statistics }\\\begin{array} { l c } \hline \text { Multiple R } & 0.8531 \\\text { RSquare } & 0.7278 \\\text { Adjusted R Square } & 0.7180 \\\text { Standard Error } & 47.8668 \\\text { Observations } & 30\end{array}\end{array}
ANOVA
 d f SS  MS Significance FRegression 1171499.78171499.7874.85052.1259E09Residual2864154.422291.23Total29235654.20\begin{array}{lrrrrr}\hline &\text { d f}& \text { SS } & \text { MS } & \text {F }& \text {Significance F} \\\hline \text {Regression }& 1 & 171499.78 & 171499.78 & 74.8505 & 2.1259E-09 \\\text {Residual} & 28 & 64154.42 & 2291.23 & & \\\text {Total} & 29 & 235654.20 & & & \\\hline\end{array}

Coefficients  Standard Error t Stat  p -value Lower 95% Upper 95%  Intercept 76.535111.83186.46865.24E0752.2987100.7716Gross4.33310.50088.65162.13E093.30725.3590\begin{array}{lrrrrrr}\hline & \text {Coefficients }& \text { Standard Error}& \text { t Stat }& \text { p -value }& \text {Lower 95\% }& \text {Upper 95\% }\\\hline \text { Intercept }& 76.5351 & 11.8318 & 6.4686 & 5.24 \mathrm{E}-07& 52.2987 & 100.7716 \\ \text {Gross} & 4.3331 & 0.5008 & 8.6516 & 2.13 \mathrm{E}-09 & 3.3072 & 5.3590 \\\hline\end{array}

 <strong>TABLE 13- 11 A company that has the distribution rights to home video sales of previously released movies would like to use the box office gross (in millions of dollars) to estimate the number of units (in thousands of units) that it can expect to sell. Following is the output from a simple linear regression along with the residual plot and normal probability plot obtained from a data set of 30 different movie titles:  \begin{array}{l} \text { Regression Statistics }\\ \begin{array} { l c } \hline \text { Multiple R } & 0.8531 \\ \text { RSquare } & 0.7278 \\ \text { Adjusted R Square } & 0.7180 \\ \text { Standard Error } & 47.8668 \\ \text { Observations } & 30 \end{array} \end{array}  ANOVA  \begin{array}{lrrrrr} \hline &\text { d f}& \text { SS } & \text { MS } & \text {F }& \text {Significance F}  \\ \hline \text {Regression }& 1 & 171499.78 & 171499.78 & 74.8505 & 2.1259E-09 \\ \text {Residual} & 28 & 64154.42 & 2291.23 & & \\ \text {Total} & 29 & 235654.20 & & & \\ \hline\end{array}    \begin{array}{lrrrrrr} \hline &  \text {Coefficients }& \text { Standard Error}& \text { t  Stat }&  \text { p -value }&  \text {Lower 95\% }& \text {Upper 95\% }\\ \hline \text { Intercept }& 76.5351 & 11.8318 & 6.4686 & 5.24 \mathrm{E}-07& 52.2987 & 100.7716 \\ \text {Gross} & 4.3331 & 0.5008 & 8.6516 & 2.13 \mathrm{E}-09 & 3.3072 & 5.3590 \\ \hline \end{array}        -Referring to Table 13-11, which of the following assumptions appears to have been violated?</strong> A) homoscedasticity B) independence of errors C) normality of error D) none of the above   <strong>TABLE 13- 11 A company that has the distribution rights to home video sales of previously released movies would like to use the box office gross (in millions of dollars) to estimate the number of units (in thousands of units) that it can expect to sell. Following is the output from a simple linear regression along with the residual plot and normal probability plot obtained from a data set of 30 different movie titles:  \begin{array}{l} \text { Regression Statistics }\\ \begin{array} { l c } \hline \text { Multiple R } & 0.8531 \\ \text { RSquare } & 0.7278 \\ \text { Adjusted R Square } & 0.7180 \\ \text { Standard Error } & 47.8668 \\ \text { Observations } & 30 \end{array} \end{array}  ANOVA  \begin{array}{lrrrrr} \hline &\text { d f}& \text { SS } & \text { MS } & \text {F }& \text {Significance F}  \\ \hline \text {Regression }& 1 & 171499.78 & 171499.78 & 74.8505 & 2.1259E-09 \\ \text {Residual} & 28 & 64154.42 & 2291.23 & & \\ \text {Total} & 29 & 235654.20 & & & \\ \hline\end{array}    \begin{array}{lrrrrrr} \hline &  \text {Coefficients }& \text { Standard Error}& \text { t  Stat }&  \text { p -value }&  \text {Lower 95\% }& \text {Upper 95\% }\\ \hline \text { Intercept }& 76.5351 & 11.8318 & 6.4686 & 5.24 \mathrm{E}-07& 52.2987 & 100.7716 \\ \text {Gross} & 4.3331 & 0.5008 & 8.6516 & 2.13 \mathrm{E}-09 & 3.3072 & 5.3590 \\ \hline \end{array}        -Referring to Table 13-11, which of the following assumptions appears to have been violated?</strong> A) homoscedasticity B) independence of errors C) normality of error D) none of the above

-Referring to Table 13-11, which of the following assumptions appears to have been violated?

A) homoscedasticity
B) independence of errors
C) normality of error
D) none of the above
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53
TABLE 13-8
It is believed that GPA (grade point average, based on a four point scale) should have a positive linear relationship with ACT scores. Given below is the Excel output from regressing GPA on ACT scores using a data set of 8 randomly chosen students from a Big-Ten university.
Regressing GPA on ACT\text {Regressing GPA on \(\mathrm { ACT }\)}
 Regressing GPA on ACT Regression Statistics  Multiple R 0.7598 R Square 0.5774 Adjusted R Square 0.5069 Standard E rror 0.2691 Observations 8\begin{array}{l}\text { Regressing GPA on } \mathrm { ACT }\\\begin{array} { l c } \hline & \text { Regression Statistics } \\\hline \text { Multiple R } & 0.7598 \\\text { R Square } & 0.5774 \\\text { Adjusted R Square } & 0.5069 \\\text { Standard E rror } & 0.2691 \\\text { Observations } & 8\end{array}\end{array}

ANOVA
d fSS MS  F  Significance FRegression 10.59400.59408.19860.0286Residual 60.43470.0724Total 71.0287\begin{array}{lcllcc}\hline & \text {d f} & \text {SS} & \text { MS }& \text { F } & \text { Significance F} \\\hline \text {Regression }&1 & 0.5940 & 0.5940 & 8.1986 & 0.0286 \\ \text {Residual }& 6 & 0.4347 & 0.0724 & & \\ \text {Total }& 7 & 1.0287 & & & \\\hline\end{array}

Coefficients Standard Error T Stat  p -value Lower 95% Upper 95%  Intercept0.56810.92840.61190.56301.70362.8398ACT 0.10210.03562.86330.02860.01480.1895\begin{array}{lrrrrrr}\hline & \text {Coefficients }& \text {Standard Error }& \text {T Stat }& \text { p -value }& \text {Lower 95\%}& \text { Upper 95\% } \\\hline \text { Intercept} & 0.5681 & 0.9284 & 0.6119 & 0.5630 & -1.7036 & 2.8398 \\ \text {ACT }& 0.1021 & 0.0356 & 2.8633 & 0.0286 & 0.0148 & 0.1895 \\\hline\end{array}



-Referring to Table 13-8, the interpretation of the coefficient of determination in this regression is

A) ACT scores account for 57.74% of the total fluctuation in GPA.
B) 57.74% of the total variation of ACT scores can be explained by GPA.
C) GPA accounts for 57.74% of the variability of ACT scores.
D) none of the above
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54
TABLE 13-6
The following EXCEL tables are obtained when "Score received on an exam (measured in percentage points)" (Y) is regressed on "percentage attendance" (X) for 22 students in a Statistics for Business and Economics course.
 Regression Statistics  Multiple R 0.142620229 R Square 0.02034053 Adjusted R Square 0.028642444 Standard Error20.25979924 Observations 22\begin{array}{lc}\text { Regression Statistics } \\\text { Multiple R } & 0.142620229 \\\text { R Square } & 0.02034053 \\\text { Adjusted R Square } & -0.028642444 \\\text { Standard Error} &20.25979924 \\\text { Observations } & 22 \\\hline\end{array}

 Coefficients  Standard Enor  T Stat p-value  Intercept 39.3902730937.243476591.0576422160.302826622 Attendance 0.3405835730.528524520.6444044890.526635689\begin{array} { l c r c c } \hline & \text { Coefficients } & \text { Standard Enor } & \text { T Stat } & p \text {-value } \\\hline \text { Intercept } & 39.39027309 & 37.24347659 & 1.057642216 & 0.302826622 \\\text { Attendance } & 0.340583573 & 0.52852452 & 0.644404489 & 0.526635689 \\\hline\end{array}

-Referring to Table 13-6, which of the following statements is true?

A) If attendance increases by 1%, the estimated average score received will increase by 39.39 percentage points.
B) If the score received increases by 39.39%, the estimated average attendance will go up by 1%.
C) If attendance increases by 1%, the estimated average score received will increase by 0.341 percentage points.
D) If attendance increases by 0.341%, the estimated average score received will increase by 1 percentage point.
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55
TABLE 13-12
The manager of the purchasing department of a large banking organization would like to develop a model to predict the amount of time (measured in hours) it takes to process invoices. Data are collected from a sample of 30 days, and the number of invoices processed and completion time in hours is recorded. Below is the regression output:
 Regression Statistics  Multiple R 0.9947 R Square 0.8924 Adjusted R Square 0.8886 Standard Error 0.3342 ations 30\begin{array}{l}\text { Regression Statistics }\\\begin{array} { l c } \hline \text { Multiple R } & 0.9947 \\\text { R Square } & 0.8924 \\\text { Adjusted R Square } & 0.8886 \\\text { Standard Error } & 0.3342 \\\text { ations } & 30 \\\hline\end{array}\end{array}

 d f  SS MS F  Significance F Regression125.943825.9438232.22004.3946E15Residual 283.12820.1117Total 2929.072\begin{array}{lrrccc}\hline & \text { d f } & \text { SS } & \text {MS} & \text { F } & \text { Significance F } \\\hline \text {Regression} & 1 & 25.9438 & 25.9438 & 232.2200 & 4.3946 \mathrm{E}-15 \\ \text {Residual }& 28 & 3.1282 & 0.1117 & & \\ \text {Total }& 29 & 29.072 & & & \\\hline\end{array}

 Coefficients  Standard Error  t Stat  p -valueLower 95%Upper 95% Invoices 0.40240.12363.25590.00300.14920.6555Processed 0.01260.000815.23884.3946E150.01090.0143\begin{array}{lrrrrrr}\hline & \text { Coefficients }& \text { Standard Error }& \text { t Stat }& \text { p -value}& \text {Lower 95\%} & \text {Upper 95\%} \\\hline \text { Invoices }& 0.4024 & 0.1236 & 3.2559 & 0.0030 & 0.1492 & 0.6555 \\ \text {Processed }& 0.0126 & 0.0008 & 15.2388 & 4.3946 \mathrm{E}-15 & 0.0109 & 0.0143 \\\hline\end{array}

 <strong>TABLE 13-12 The manager of the purchasing department of a large banking organization would like to develop a model to predict the amount of time (measured in hours) it takes to process invoices. Data are collected from a sample of 30 days, and the number of invoices processed and completion time in hours is recorded. Below is the regression output:  \begin{array}{l} \text { Regression Statistics }\\ \begin{array} { l c } \hline \text { Multiple R } & 0.9947 \\ \text { R Square } & 0.8924 \\ \text { Adjusted R Square } & 0.8886 \\ \text { Standard Error } & 0.3342 \\ \text { ations } & 30 \\ \hline \end{array} \end{array}    \begin{array}{lrrccc} \hline & \text { d f } &  \text { SS } &  \text {MS} &  \text { F } & \text { Significance  F } \\ \hline  \text {Regression} & 1 & 25.9438 & 25.9438 & 232.2200 & 4.3946 \mathrm{E}-15  \\ \text {Residual }& 28 & 3.1282 & 0.1117 & & \\ \text {Total }& 29 & 29.072 & & & \\ \hline \end{array}    \begin{array}{lrrrrrr} \hline & \text { Coefficients }& \text { Standard Error }& \text { t Stat }&  \text { p -value}& \text {Lower 95\%} &  \text {Upper 95\%} \\ \hline \text { Invoices }& 0.4024 & 0.1236 & 3.2559 & 0.0030 & 0.1492 & 0.6555 \\ \text {Processed }& 0.0126 & 0.0008 & 15.2388 &  4.3946 \mathrm{E}-15  & 0.0109 & 0.0143 \\ \hline \end{array}          -Referring to Table 13-12, the error sum of squares (SSE) of the above regression is</strong> A) 0.1117. B) 29.0720. C) 25.9438. D) 3.1282.

 <strong>TABLE 13-12 The manager of the purchasing department of a large banking organization would like to develop a model to predict the amount of time (measured in hours) it takes to process invoices. Data are collected from a sample of 30 days, and the number of invoices processed and completion time in hours is recorded. Below is the regression output:  \begin{array}{l} \text { Regression Statistics }\\ \begin{array} { l c } \hline \text { Multiple R } & 0.9947 \\ \text { R Square } & 0.8924 \\ \text { Adjusted R Square } & 0.8886 \\ \text { Standard Error } & 0.3342 \\ \text { ations } & 30 \\ \hline \end{array} \end{array}    \begin{array}{lrrccc} \hline & \text { d f } &  \text { SS } &  \text {MS} &  \text { F } & \text { Significance  F } \\ \hline  \text {Regression} & 1 & 25.9438 & 25.9438 & 232.2200 & 4.3946 \mathrm{E}-15  \\ \text {Residual }& 28 & 3.1282 & 0.1117 & & \\ \text {Total }& 29 & 29.072 & & & \\ \hline \end{array}    \begin{array}{lrrrrrr} \hline & \text { Coefficients }& \text { Standard Error }& \text { t Stat }&  \text { p -value}& \text {Lower 95\%} &  \text {Upper 95\%} \\ \hline \text { Invoices }& 0.4024 & 0.1236 & 3.2559 & 0.0030 & 0.1492 & 0.6555 \\ \text {Processed }& 0.0126 & 0.0008 & 15.2388 &  4.3946 \mathrm{E}-15  & 0.0109 & 0.0143 \\ \hline \end{array}          -Referring to Table 13-12, the error sum of squares (SSE) of the above regression is</strong> A) 0.1117. B) 29.0720. C) 25.9438. D) 3.1282.


-Referring to Table 13-12, the error sum of squares (SSE) of the above regression is

A) 0.1117.
B) 29.0720.
C) 25.9438.
D) 3.1282.
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56
In performing a regression analysis involving two numerical variables, we are assuming

A) the variation around the line of regression is the same for each X value.
B) that X and Y are independent.
C) the variances of X and Y are equal.
D) all of the above
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57
TABLE 13- 11
A company that has the distribution rights to home video sales of previously released movies would like to use the box office gross (in millions of dollars) to estimate the number of units (in thousands of units) that it can expect to sell. Following is the output from a simple linear regression along with the residual plot and normal probability plot obtained from a data set of 30 different movie titles:
 Regression Statistics  Multiple R 0.8531 R Square 0.7278 Adjusted R Square 0.7180 Standard Error 47.8668 Observations 30\begin{array} { l c } & \text { Regression Statistics } \\\hline \text { Multiple R } & 0.8531 \\\text { R Square } & 0.7278 \\\text { Adjusted R Square } & 0.7180 \\\text { Standard Error } & 47.8668 \\\text { Observations } & 30\end{array} ANOVA
 d f SS  MS Significance FRegression 1171499.78171499.7874.85052.1259E09Residual2864154.422291.23Total29235654.20\begin{array}{lrrrrr}\hline &\text { d f}& \text { SS } & \text { MS } & \text {F }& \text {Significance F} \\\hline \text {Regression }& 1 & 171499.78 & 171499.78 & 74.8505 & 2.1259E-09 \\\text {Residual} & 28 & 64154.42 & 2291.23 & & \\\text {Total} & 29 & 235654.20 & & & \\\hline\end{array}

Coefficients  Standard Error t Stat  p -value Lower 95% Upper 95%  Intercept 76.535111.83186.46865.24E0752.2987100.7716Gross4.33310.50088.65162.13E093.30725.3590\begin{array}{lrrrrrr}\hline & \text {Coefficients }& \text { Standard Error}& \text { t Stat }& \text { p -value }& \text {Lower 95\% }& \text {Upper 95\% }\\\hline \text { Intercept }& 76.5351 & 11.8318 & 6.4686 & 5.24 \mathrm{E}-07& 52.2987 & 100.7716 \\ \text {Gross} & 4.3331 & 0.5008 & 8.6516 & 2.13 \mathrm{E}-09 & 3.3072 & 5.3590 \\\hline\end{array}
 <strong>TABLE 13- 11 A company that has the distribution rights to home video sales of previously released movies would like to use the box office gross (in millions of dollars) to estimate the number of units (in thousands of units) that it can expect to sell. Following is the output from a simple linear regression along with the residual plot and normal probability plot obtained from a data set of 30 different movie titles:  \begin{array} { l c } & \text { Regression Statistics } \\ \hline \text { Multiple R } & 0.8531 \\ \text { R Square } & 0.7278 \\ \text { Adjusted R Square } & 0.7180 \\ \text { Standard Error } & 47.8668 \\ \text { Observations } & 30 \end{array}  ANOVA  \begin{array}{lrrrrr} \hline &\text { d f}& \text { SS } & \text { MS } & \text {F }& \text {Significance F}  \\ \hline \text {Regression }& 1 & 171499.78 & 171499.78 & 74.8505 & 2.1259E-09 \\ \text {Residual} & 28 & 64154.42 & 2291.23 & & \\ \text {Total} & 29 & 235654.20 & & & \\ \hline\end{array}    \begin{array}{lrrrrrr} \hline &  \text {Coefficients }& \text { Standard Error}& \text { t  Stat }&  \text { p -value }&  \text {Lower 95\% }& \text {Upper 95\% }\\ \hline \text { Intercept }& 76.5351 & 11.8318 & 6.4686 & 5.24 \mathrm{E}-07& 52.2987 & 100.7716 \\ \text {Gross} & 4.3331 & 0.5008 & 8.6516 & 2.13 \mathrm{E}-09 & 3.3072 & 5.3590 \\ \hline \end{array}        -Referring to Table 13-11, which of the following is the correct alternative hypothesis for testing whether there is a linear relationship between box office gross and home video unit sales?</strong> A)   H_{1}: \beta_{1}=0   B)   H_{1}: b_{1}=0   C)   H_{1}: \beta_{1} \neq 0   D)   H_{1}: b_{1} \neq 0
 <strong>TABLE 13- 11 A company that has the distribution rights to home video sales of previously released movies would like to use the box office gross (in millions of dollars) to estimate the number of units (in thousands of units) that it can expect to sell. Following is the output from a simple linear regression along with the residual plot and normal probability plot obtained from a data set of 30 different movie titles:  \begin{array} { l c } & \text { Regression Statistics } \\ \hline \text { Multiple R } & 0.8531 \\ \text { R Square } & 0.7278 \\ \text { Adjusted R Square } & 0.7180 \\ \text { Standard Error } & 47.8668 \\ \text { Observations } & 30 \end{array}  ANOVA  \begin{array}{lrrrrr} \hline &\text { d f}& \text { SS } & \text { MS } & \text {F }& \text {Significance F}  \\ \hline \text {Regression }& 1 & 171499.78 & 171499.78 & 74.8505 & 2.1259E-09 \\ \text {Residual} & 28 & 64154.42 & 2291.23 & & \\ \text {Total} & 29 & 235654.20 & & & \\ \hline\end{array}    \begin{array}{lrrrrrr} \hline &  \text {Coefficients }& \text { Standard Error}& \text { t  Stat }&  \text { p -value }&  \text {Lower 95\% }& \text {Upper 95\% }\\ \hline \text { Intercept }& 76.5351 & 11.8318 & 6.4686 & 5.24 \mathrm{E}-07& 52.2987 & 100.7716 \\ \text {Gross} & 4.3331 & 0.5008 & 8.6516 & 2.13 \mathrm{E}-09 & 3.3072 & 5.3590 \\ \hline \end{array}        -Referring to Table 13-11, which of the following is the correct alternative hypothesis for testing whether there is a linear relationship between box office gross and home video unit sales?</strong> A)   H_{1}: \beta_{1}=0   B)   H_{1}: b_{1}=0   C)   H_{1}: \beta_{1} \neq 0   D)   H_{1}: b_{1} \neq 0


-Referring to Table 13-11, which of the following is the correct alternative hypothesis for testing whether there is a linear relationship between box office gross and home video unit sales?

A) H1:β1=0 H_{1}: \beta_{1}=0
B) H1:b1=0 H_{1}: b_{1}=0
C) H1:β10 H_{1}: \beta_{1} \neq 0
D) H1:b10 H_{1}: b_{1} \neq 0
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58
If the Durbin-Watson statistic has a value close to 4, which assumption is violated?

A) normality of the errors
B) homoscedasticity
C) independence of errors
D) none of the above
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59
TABLE 13-2
A candy bar manufacturer is interested in trying to estimate how sales are influenced by the price of their product. To do this, the company randomly chooses 6 small cities and offers the candy bar at different prices. Using candy bar sales as the dependent variable, the company will conduct a simple linear regression on the data below:
 City  Price ($) Sales  River Falls 1.30100 Hudson 1.6090 Ellsworth 1.8090 Prescott 2.0040 Rock Elm 2.4038 Stillwater 2.9032\begin{array} { l l c } \hline \text { City } & \text { Price } ( \$ ) & \text { Sales } \\\hline \text { River Falls } & 1.30 & 100 \\\text { Hudson } & 1.60 & 90 \\\text { Ellsworth } & 1.80 & 90 \\\text { Prescott } & 2.00 & 40 \\\text { Rock Elm } & 2.40 & 38 \\\text { Stillwater } & 2.90 & 32 \\\hline\end{array}

-Referring to Table 13-2, what is the estimated slope parameter for the candy bar price and sales data?

A) 161.386
B) - 48.193
C) 0.784
D) - 3.810
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60
TABLE 13-8
It is believed that GPA (grade point average, based on a four point scale) should have a positive linear relationship with ACT scores. Given below is the Excel output from regressing GPA on ACT scores using a data set of 8 randomly chosen students from a Big-Ten university.
Regressing GPA on ACT\text {Regressing GPA on \(\mathrm { ACT }\)}
 Regressing GPA on ACT Regression Statistics  Multiple R 0.7598 R Square 0.5774 Adjusted R Square 0.5069 Standard E rror 0.2691 Observations 8\begin{array}{l}\text { Regressing GPA on } \mathrm { ACT }\\\begin{array} { l c } \hline & \text { Regression Statistics } \\\hline \text { Multiple R } & 0.7598 \\\text { R Square } & 0.5774 \\\text { Adjusted R Square } & 0.5069 \\\text { Standard E rror } & 0.2691 \\\text { Observations } & 8\end{array}\end{array}

ANOVA
d fSS MS  F  Significance FRegression 10.59400.59408.19860.0286Residual 60.43470.0724Total 71.0287\begin{array}{lcllcc}\hline & \text {d f} & \text {SS} & \text { MS }& \text { F } & \text { Significance F} \\\hline \text {Regression }&1 & 0.5940 & 0.5940 & 8.1986 & 0.0286 \\ \text {Residual }& 6 & 0.4347 & 0.0724 & & \\ \text {Total }& 7 & 1.0287 & & & \\\hline\end{array}

Coefficients Standard Error T Stat  p -value Lower 95% Upper 95%  Intercept0.56810.92840.61190.56301.70362.8398ACT 0.10210.03562.86330.02860.01480.1895\begin{array}{lrrrrrr}\hline & \text {Coefficients }& \text {Standard Error }& \text {T Stat }& \text { p -value }& \text {Lower 95\%}& \text { Upper 95\% } \\\hline \text { Intercept} & 0.5681 & 0.9284 & 0.6119 & 0.5630 & -1.7036 & 2.8398 \\ \text {ACT }& 0.1021 & 0.0356 & 2.8633 & 0.0286 & 0.0148 & 0.1895 \\\hline\end{array}



-Referring to Table 13-8, what is the predicted average value of GPA when ACT = 20?

A) 2.66
B) 2.61
C) 3.12
D) 2.80
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61
TABLE 13-8
It is believed that GPA (grade point average, based on a four point scale) should have a positive linear relationship with ACT scores. Given below is the Excel output from regressing GPA on ACT scores using a data set of 8 randomly chosen students from a Big-Ten university.
 Regressing GPA on ACT Regression Statistics  Multiple R 0.7598 R Square 0.5774 Adjusted R Square 0.5069 Standard E rror 0.2691 Observations 8\begin{array}{l}\text { Regressing GPA on } \mathrm { ACT }\\\begin{array} { l c } \hline & \text { Regression Statistics } \\\hline \text { Multiple R } & 0.7598 \\\text { R Square } & 0.5774 \\\text { Adjusted R Square } & 0.5069 \\\text { Standard E rror } & 0.2691 \\\text { Observations } & 8\end{array}\end{array}

ANOVA
d fSS MS  F  Significance FRegression 10.59400.59408.19860.0286Residual 60.43470.0724Total 71.0287\begin{array}{lcllcc}\hline & \text {d f} & \text {SS} & \text { MS }& \text { F } & \text { Significance F} \\\hline \text {Regression }&1 & 0.5940 & 0.5940 & 8.1986 & 0.0286 \\ \text {Residual }& 6 & 0.4347 & 0.0724 & & \\ \text {Total }& 7 & 1.0287 & & & \\\hline\end{array}

Coefficients Standard Error T Stat  p -value Lower 95% Upper 95%  Intercept0.56810.92840.61190.56301.70362.8398ACT 0.10210.03562.86330.02860.01480.1895\begin{array}{lrrrrrr}\hline & \text {Coefficients }& \text {Standard Error }& \text {T Stat }& \text { p -value }& \text {Lower 95\%}& \text { Upper 95\% } \\\hline \text { Intercept} & 0.5681 & 0.9284 & 0.6119 & 0.5630 & -1.7036 & 2.8398 \\ \text {ACT }& 0.1021 & 0.0356 & 2.8633 & 0.0286 & 0.0148 & 0.1895 \\\hline\end{array}


-Referring to Table 13-8, what are the decision and conclusion on testing whether there is any linear relationship at 1% level of significance between GPA and ACT scores?

A) Do not reject the null hypothesis; hence there is sufficient evidence to show that ACT scores and GPA are linearly related.
B) Reject the null hypothesis; hence there is not sufficient evidence to show that ACT scores and GPA are linearly related.
C) Reject the null hypothesis; hence there is sufficient evidence to show that ACT scores and GPA are linearly related.
D) Do not reject the null hypothesis; hence there is not sufficient evidence to show that ACT scores and GPA are linearly related.
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62
TABLE 13-2
A candy bar manufacturer is interested in trying to estimate how sales are influenced by the price of their product. To do this, the company randomly chooses 6 small cities and offers the candy bar at different prices. Using candy bar sales as the dependent variable, the company will conduct a simple linear regression on the data below:
 City  Price ($) Sales  River Falls 1.30100 Hudson 1.6090 Ellsworth 1.8090 Prescott 2.0040 Rock Elm 2.4038 Stillwater 2.9032\begin{array} { l l c } \hline \text { City } & \text { Price } ( \$ ) & \text { Sales } \\\hline \text { River Falls } & 1.30 & 100 \\\text { Hudson } & 1.60 & 90 \\\text { Ellsworth } & 1.80 & 90 \\\text { Prescott } & 2.00 & 40 \\\text { Rock Elm } & 2.40 & 38 \\\text { Stillwater } & 2.90 & 32\end{array}

-Referring to Table 13-2, to test that the regression coefficient, þ1, is not equal to 0, what would be the critical values? Use ? = 0.05.

A) ±3.1634
B) ±2.5706
C) ±2.7764
D) ±3.4954
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63
TABLE 13-2
A candy bar manufacturer is interested in trying to estimate how sales are influenced by the price of their product. To do this, the company randomly chooses 6 small cities and offers the candy bar at different prices. Using candy bar sales as the dependent variable, the company will conduct a simple linear regression on the data below:
 City  Price ($) Sales  River Falls 1.30100 Hudson 1.6090 Ellsworth 1.8090 Prescott 2.0040 Rock Elm 2.4038 Stillwater 2.9032\begin{array} { l l c } \hline \text { City } & \text { Price } ( \$ ) & \text { Sales } \\\hline \text { River Falls } & 1.30 & 100 \\\text { Hudson } & 1.60 & 90 \\\text { Ellsworth } & 1.80 & 90 \\\text { Prescott } & 2.00 & 40 \\\text { Rock Elm } & 2.40 & 38 \\\text { Stillwater } & 2.90 & 32\end{array}

-Referring to Table 13-2, if the price of the candy bar is set at $2, the predicted sales will be

A) 90.
B) 65.
C) 100.
D) 30.
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64
The Y-intercept (b0) represents the

A) change in estimated average Y per unit change in X.
B) predicted value of Y.
C) predicted value of Y when X = 0.
D) variation around the sample regression line.
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65
The residuals represent

A) the difference between the actual Y values and the predicted Y values.
B) the square root of the slope.
C) the predicted value of Y for the average X value.
D) the difference between the actual Y values and the mean of Y.
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66
TABLE 13-2
A candy bar manufacturer is interested in trying to estimate how sales are influenced by the price of their product. To do this, the company randomly chooses 6 small cities and offers the candy bar at different prices. Using candy bar sales as the dependent variable, the company will conduct a simple linear regression on the data below:
 City  Price ($) Sales  River Falls 1.30100 Hudson 1.6090 Ellsworth 1.8090 Prescott 2.0040 Rock Elm 2.4038 Stillwater 2.9032\begin{array} { l l c } \hline \text { City } & \text { Price } ( \$ ) & \text { Sales } \\\hline \text { River Falls } & 1.30 & 100 \\\text { Hudson } & 1.60 & 90 \\\text { Ellsworth } & 1.80 & 90 \\\text { Prescott } & 2.00 & 40 \\\text { Rock Elm } & 2.40 & 38 \\\text { Stillwater } & 2.90 & 32 \\\hline\end{array}

-Referring to Table 13-2, what is the estimated average change in the sales of the candy bar if price goes up by $1.00?

A) - 3.810
B) - 48.193
C) 161.386
D) 0.784
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67
TABLE 13-12
The manager of the purchasing department of a large banking organization would like to develop a model to predict the amount of time (measured in hours) it takes to process invoices. Data are collected from a sample of 30 days, and the number of invoices processed and completion time in hours is recorded. Below is the regression output:
 Regression Statistics  Multiple R 0.9947 R Square 0.8924 Adjusted R Square 0.8886 Standard Error 0.3342 ations 30\begin{array}{l}\text { Regression Statistics }\\\begin{array} { l c } \hline \text { Multiple R } & 0.9947 \\\text { R Square } & 0.8924 \\\text { Adjusted R Square } & 0.8886 \\\text { Standard Error } & 0.3342 \\\text { ations } & 30 \\\hline\end{array}\end{array}

 d f  SS MS F  Significance F Regression125.943825.9438232.22004.3946E15Residual 283.12820.1117Total 2929.072\begin{array}{lrrccc}\hline & \text { d f } & \text { SS } & \text {MS} & \text { F } & \text { Significance F } \\\hline \text {Regression} & 1 & 25.9438 & 25.9438 & 232.2200 & 4.3946 \mathrm{E}-15 \\ \text {Residual }& 28 & 3.1282 & 0.1117 & & \\ \text {Total }& 29 & 29.072 & & & \\\hline\end{array}

 Coefficients  Standard Error  t Stat  p -valueLower 95%Upper 95% Invoices 0.40240.12363.25590.00300.14920.6555Processed 0.01260.000815.23884.3946E150.01090.0143\begin{array}{lrrrrrr}\hline & \text { Coefficients }& \text { Standard Error }& \text { t Stat }& \text { p -value}& \text {Lower 95\%} & \text {Upper 95\%} \\\hline \text { Invoices }& 0.4024 & 0.1236 & 3.2559 & 0.0030 & 0.1492 & 0.6555 \\ \text {Processed }& 0.0126 & 0.0008 & 15.2388 & 4.3946 \mathrm{E}-15 & 0.0109 & 0.0143 \\\hline\end{array}

 <strong>TABLE 13-12 The manager of the purchasing department of a large banking organization would like to develop a model to predict the amount of time (measured in hours) it takes to process invoices. Data are collected from a sample of 30 days, and the number of invoices processed and completion time in hours is recorded. Below is the regression output:  \begin{array}{l} \text { Regression Statistics }\\ \begin{array} { l c } \hline \text { Multiple R } & 0.9947 \\ \text { R Square } & 0.8924 \\ \text { Adjusted R Square } & 0.8886 \\ \text { Standard Error } & 0.3342 \\ \text { ations } & 30 \\ \hline \end{array} \end{array}    \begin{array}{lrrccc} \hline & \text { d f } &  \text { SS } &  \text {MS} &  \text { F } & \text { Significance  F } \\ \hline  \text {Regression} & 1 & 25.9438 & 25.9438 & 232.2200 & 4.3946 \mathrm{E}-15  \\ \text {Residual }& 28 & 3.1282 & 0.1117 & & \\ \text {Total }& 29 & 29.072 & & & \\ \hline \end{array}    \begin{array}{lrrrrrr} \hline & \text { Coefficients }& \text { Standard Error }& \text { t Stat }&  \text { p -value}& \text {Lower 95\%} &  \text {Upper 95\%} \\ \hline \text { Invoices }& 0.4024 & 0.1236 & 3.2559 & 0.0030 & 0.1492 & 0.6555 \\ \text {Processed }& 0.0126 & 0.0008 & 15.2388 &  4.3946 \mathrm{E}-15  & 0.0109 & 0.0143 \\ \hline \end{array}          -Referring to Table 13-12, the estimated average amount of time it takes to process one additional invoice is</strong> A) 0.0126 more hours. B) 0.0126 fewer hours. C) 0.4024 more hours. D) 0.4024 fewer hours.

 <strong>TABLE 13-12 The manager of the purchasing department of a large banking organization would like to develop a model to predict the amount of time (measured in hours) it takes to process invoices. Data are collected from a sample of 30 days, and the number of invoices processed and completion time in hours is recorded. Below is the regression output:  \begin{array}{l} \text { Regression Statistics }\\ \begin{array} { l c } \hline \text { Multiple R } & 0.9947 \\ \text { R Square } & 0.8924 \\ \text { Adjusted R Square } & 0.8886 \\ \text { Standard Error } & 0.3342 \\ \text { ations } & 30 \\ \hline \end{array} \end{array}    \begin{array}{lrrccc} \hline & \text { d f } &  \text { SS } &  \text {MS} &  \text { F } & \text { Significance  F } \\ \hline  \text {Regression} & 1 & 25.9438 & 25.9438 & 232.2200 & 4.3946 \mathrm{E}-15  \\ \text {Residual }& 28 & 3.1282 & 0.1117 & & \\ \text {Total }& 29 & 29.072 & & & \\ \hline \end{array}    \begin{array}{lrrrrrr} \hline & \text { Coefficients }& \text { Standard Error }& \text { t Stat }&  \text { p -value}& \text {Lower 95\%} &  \text {Upper 95\%} \\ \hline \text { Invoices }& 0.4024 & 0.1236 & 3.2559 & 0.0030 & 0.1492 & 0.6555 \\ \text {Processed }& 0.0126 & 0.0008 & 15.2388 &  4.3946 \mathrm{E}-15  & 0.0109 & 0.0143 \\ \hline \end{array}          -Referring to Table 13-12, the estimated average amount of time it takes to process one additional invoice is</strong> A) 0.0126 more hours. B) 0.0126 fewer hours. C) 0.4024 more hours. D) 0.4024 fewer hours.


-Referring to Table 13-12, the estimated average amount of time it takes to process one additional invoice is

A) 0.0126 more hours.
B) 0.0126 fewer hours.
C) 0.4024 more hours.
D) 0.4024 fewer hours.
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68
TABLE 13- 11
A company that has the distribution rights to home video sales of previously released movies would like to use the box office gross (in millions of dollars) to estimate the number of units (in thousands of units) that it can expect to sell. Following is the output from a simple linear regression along with the residual plot and normal probability plot obtained from a data set of 30 different movie titles:
 Regression Stadistics  Multiple R 0.8531 R Square 0.7278 Adjusted R Square 0.7180 Standard Error 47.8668 Observations 30\begin{array} { l c } \hline & \text { Regression Stadistics } \\\hline \text { Multiple R } & 0.8531 \\\text { R Square } & 0.7278 \\\text { Adjusted R Square } & 0.7180 \\\text { Standard Error } & 47.8668 \\\text { Observations } & 30\end{array} ANOVA
 d f SS  MS Significance FRegression 1171499.78171499.7874.85052.1259E09Residual2864154.422291.23Total29235654.20\begin{array}{lrrrrr}\hline &\text { d f}& \text { SS } & \text { MS } & \text {F }& \text {Significance F} \\\hline \text {Regression }& 1 & 171499.78 & 171499.78 & 74.8505 & 2.1259E-09 \\\text {Residual} & 28 & 64154.42 & 2291.23 & & \\\text {Total} & 29 & 235654.20 & & & \\\hline\end{array}

Coefficients  Standard Error t Stat  p -value Lower 95% Upper 95%  Intercept 76.535111.83186.46865.24E0752.2987100.7716Gross4.33310.50088.65162.13E093.30725.3590\begin{array}{lrrrrrr}\hline & \text {Coefficients }& \text { Standard Error}& \text { t Stat }& \text { p -value }& \text {Lower 95\% }& \text {Upper 95\% }\\\hline \text { Intercept }& 76.5351 & 11.8318 & 6.4686 & 5.24 \mathrm{E}-07& 52.2987 & 100.7716 \\ \text {Gross} & 4.3331 & 0.5008 & 8.6516 & 2.13 \mathrm{E}-09 & 3.3072 & 5.3590 \\\hline\end{array}

 TABLE 13- 11 A company that has the distribution rights to home video sales of previously released movies would like to use the box office gross (in millions of dollars) to estimate the number of units (in thousands of units) that it can expect to sell. Following is the output from a simple linear regression along with the residual plot and normal probability plot obtained from a data set of 30 different movie titles:  \begin{array} { l c } \hline & \text { Regression Stadistics } \\ \hline \text { Multiple R } & 0.8531 \\ \text { R Square } & 0.7278 \\ \text { Adjusted R Square } & 0.7180 \\ \text { Standard Error } & 47.8668 \\ \text { Observations } & 30 \end{array}  ANOVA  \begin{array}{lrrrrr} \hline &\text { d f}& \text { SS } & \text { MS } & \text {F }& \text {Significance F}  \\ \hline \text {Regression }& 1 & 171499.78 & 171499.78 & 74.8505 & 2.1259E-09 \\ \text {Residual} & 28 & 64154.42 & 2291.23 & & \\ \text {Total} & 29 & 235654.20 & & & \\ \hline\end{array}    \begin{array}{lrrrrrr} \hline &  \text {Coefficients }& \text { Standard Error}& \text { t  Stat }&  \text { p -value }&  \text {Lower 95\% }& \text {Upper 95\% }\\ \hline \text { Intercept }& 76.5351 & 11.8318 & 6.4686 & 5.24 \mathrm{E}-07& 52.2987 & 100.7716 \\ \text {Gross} & 4.3331 & 0.5008 & 8.6516 & 2.13 \mathrm{E}-09 & 3.3072 & 5.3590 \\ \hline \end{array}        -Referring to Table 13-11, what is the standard deviation around the regression line?  TABLE 13- 11 A company that has the distribution rights to home video sales of previously released movies would like to use the box office gross (in millions of dollars) to estimate the number of units (in thousands of units) that it can expect to sell. Following is the output from a simple linear regression along with the residual plot and normal probability plot obtained from a data set of 30 different movie titles:  \begin{array} { l c } \hline & \text { Regression Stadistics } \\ \hline \text { Multiple R } & 0.8531 \\ \text { R Square } & 0.7278 \\ \text { Adjusted R Square } & 0.7180 \\ \text { Standard Error } & 47.8668 \\ \text { Observations } & 30 \end{array}  ANOVA  \begin{array}{lrrrrr} \hline &\text { d f}& \text { SS } & \text { MS } & \text {F }& \text {Significance F}  \\ \hline \text {Regression }& 1 & 171499.78 & 171499.78 & 74.8505 & 2.1259E-09 \\ \text {Residual} & 28 & 64154.42 & 2291.23 & & \\ \text {Total} & 29 & 235654.20 & & & \\ \hline\end{array}    \begin{array}{lrrrrrr} \hline &  \text {Coefficients }& \text { Standard Error}& \text { t  Stat }&  \text { p -value }&  \text {Lower 95\% }& \text {Upper 95\% }\\ \hline \text { Intercept }& 76.5351 & 11.8318 & 6.4686 & 5.24 \mathrm{E}-07& 52.2987 & 100.7716 \\ \text {Gross} & 4.3331 & 0.5008 & 8.6516 & 2.13 \mathrm{E}-09 & 3.3072 & 5.3590 \\ \hline \end{array}        -Referring to Table 13-11, what is the standard deviation around the regression line?

-Referring to Table 13-11, what is the standard deviation around the regression line?
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69
TABLE 13-6
The following EXCEL tables are obtained when "Score received on an exam (measured in percentage points)" (Y) is regressed on "percentage attendance" (X) for 22 students in a Statistics for Business and Economics course.
Regression Statistics  Multiple R0.142620229R Square0.02034053Adjusted R Square0.028642444Standard Error20.25979924Observations22\begin{array}{ll}\hline \text {Regression Statistics } \\\hline \text { Multiple R} & 0.142620229 \\ \text {R Square} & 0.02034053 \\ \text {Adjusted R Square} & -0.028642444 \\ \text {Standard Error} & 20.25979924 \\ \text {Observations} & 22 \\\hline\end{array}

 Coefficients  Standard Enor  T Stat p-value  Intercept 39.3902730937.243476591.0576422160.302826622 Attendance 0.3405835730.528524520.6444044890.526635689\begin{array}{lcrrc} & \text { Coefficients } & \text { Standard Enor } & {\text { T Stat }} & p \text {-value } \\\hline \text { Intercept } & 39.39027309 & 37.24347659 & 1.057642216 & 0.302826622 \\\text { Attendance } & 0.340583573 & 0.52852452 & 0.644404489 & 0.526635689 \\\hline\end{array}


-Referring to Table 13-6, which of the following statements is true?

A) 2% of the total variability in percentage attendance can be explained by score received.
B) -2.86% of the total variability in percentage attendance can be explained by score received.
C) -2.86% of the total variability in score received can be explained by percentage attendance.
D) 2% of the total variability in score received can be explained by percentage attendance.
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70
TABLE 13-9
It is believed that, the average numbers of hours spent studying per day (HOURS) during undergraduate education should have a positive linear relationship with the starting salary (SALARY, measured in thousands of dollars per month) after graduation. Given below is the Excel output from regressing starting salary on number of hours spent studying per day for a sample of 51 students. NOTE: Some of the numbers in the output are purposely erased.
Regression Stedistics
 Multiple R 0.8857 RSquare 0.7845 Adjusted R Square 0.7801 Standard Error 1.3704\begin{array} { l l } \text { Multiple R } & 0.8857 \\ \text { RSquare } & 0.7845 \\ \text { Adjusted R Square } & 0.7801 \\ \text { Standard Error } & 1.3704 \end{array}
Observations 51

ANOVA

dfSSMSF Significance F Regression 1335.0472335.0473178.3859 Residual 1.8782 Total 50427.0798\begin{array}{lccrcc}\hline & d f & S S &{M S} & F & \text { Significance } F \\\hline \text { Regression } & 1 & 335.0472 & 335.0473 & 178.3859 \\\text { Residual } & & & 1.8782 & \\\text { Total } & 50 & 427.0798 & &\end{array}

 Coeffcients  Standard Error t Stat p-value  Lower 95%  Upper 95%  Intercept 1.89400.40184.71342.051E052.70151.0865 Hours 0.97950.073313.35615.944E180.83211.1269\begin{array}{lrccccr} & \text { Coeffcients } & \text { Standard Error } &t \text { Stat } & p-{\text {-value }} & \text { Lower 95\% } & \text { Upper 95\% } \\\hline \text { Intercept } & -1.8940 & 0.4018 & -4.7134 & 2.051 \mathrm{E}-05 & -2.7015 & -1.0865 \\\text { Hours } & 0.9795 & 0.0733 & 13.3561 & 5.944 \mathrm{E}-18 & 0.8321 & 1.1269 \\\hline\end{array}


-Referring to Table 13-9, the p-value of the measured F-test statistic to test whether HOURS affects SALARY is

A) 2.051E-05.
B) (2.051E-05)/2.
C) 5.944E-18.
D) (5.944E-18)/2.
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71
TABLE 13-3
The director of cooperative education at a state college wants to examine the effect of cooperative education job experience on marketability in the work place. She takes a random sample of 4 students. For these 4, she finds out how many times each had a cooperative education job and how many job offers they received upon graduation. These data are presented in the table below.
 Student  CoopJobs  JobOffer 114226313401\begin{array} { l l l } \hline \text { Student } & \text { CoopJobs } & \text { JobOffer } \\\hline 1 & 1 & 4 \\2 & 2 & 6 \\3 & 1 & 3 \\4 & 0 & 1\end{array}

-Referring to Table 13-3, the total sum of squares (SST) is _____.
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72
TABLE 13-01
A large national bank charges local companies for using their services. A bank official reported the results of a regression analysis designed to predict the bank's charges (Y) -- measured in dollars per month -- for services rendered to local companies. One independent variable used to predict service charge to a company is the company's sales revenue (X) -- measured in millions of dollars. Data for 21 companies who use the bank's services were used to fit the model:
E(Y)=β0+β1XE(Y)=\beta_{0}+\beta_{1} X
The results of the simple linear regression are provided below.
Y^=2,700+20X, SYX=65, two-tailed p value =0.034( for testing β1)\hat{Y}=-2,700+20 X, \mathrm{~S}_{Y X}=65, \text { two-tailed } p \text { value }=0.034\left(\text { for testing } \beta_{1}\right)


-Referring to Table 13-1, interpret the estimate of þ0, the Y-intercept of the line.

A) About 95% of the observed service charges fall within $2,700 of the least squares line.
B) For every $1 million increase in sales revenue, we expect a service charge to decrease $2,700.
C) All companies will be charged at least $2,700 by the bank.
D) There is no practical interpretation since a sales revenue of $0 is a nonsensical value.
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73
TABLE 13- 11
A company that has the distribution rights to home video sales of previously released movies would like to use the box office gross (in millions of dollars) to estimate the number of units (in thousands of units) that it can expect to sell. Following is the output from a simple linear regression along with the residual plot and normal probability plot obtained from a data set of 30 different movie titles:
 Regression Statistics  Multiple R 0.8531 RSquare 0.7278 Adjusted R Square 0.7180 Standard Error 47.8668 Observations 30\begin{array}{l}\text { Regression Statistics }\\\begin{array} { l c } \hline \text { Multiple R } & 0.8531 \\\text { RSquare } & 0.7278 \\\text { Adjusted R Square } & 0.7180 \\\text { Standard Error } & 47.8668 \\\text { Observations } & 30\end{array}\end{array}
ANOVA
 d f SS  MS Significance FRegression 1171499.78171499.7874.85052.1259E09Residual2864154.422291.23Total29235654.20\begin{array}{lrrrrr}\hline &\text { d f}& \text { SS } & \text { MS } & \text {F }& \text {Significance F} \\\hline \text {Regression }& 1 & 171499.78 & 171499.78 & 74.8505 & 2.1259E-09 \\\text {Residual} & 28 & 64154.42 & 2291.23 & & \\\text {Total} & 29 & 235654.20 & & & \\\hline\end{array}

Coefficients  Standard Error t Stat  p -value Lower 95% Upper 95%  Intercept 76.535111.83186.46865.24E0752.2987100.7716Gross4.33310.50088.65162.13E093.30725.3590\begin{array}{lrrrrrr}\hline & \text {Coefficients }& \text { Standard Error}& \text { t Stat }& \text { p -value }& \text {Lower 95\% }& \text {Upper 95\% }\\\hline \text { Intercept }& 76.5351 & 11.8318 & 6.4686 & 5.24 \mathrm{E}-07& 52.2987 & 100.7716 \\ \text {Gross} & 4.3331 & 0.5008 & 8.6516 & 2.13 \mathrm{E}-09 & 3.3072 & 5.3590 \\\hline\end{array}

 <strong>TABLE 13- 11 A company that has the distribution rights to home video sales of previously released movies would like to use the box office gross (in millions of dollars) to estimate the number of units (in thousands of units) that it can expect to sell. Following is the output from a simple linear regression along with the residual plot and normal probability plot obtained from a data set of 30 different movie titles:  \begin{array}{l} \text { Regression Statistics }\\ \begin{array} { l c } \hline \text { Multiple R } & 0.8531 \\ \text { RSquare } & 0.7278 \\ \text { Adjusted R Square } & 0.7180 \\ \text { Standard Error } & 47.8668 \\ \text { Observations } & 30 \end{array} \end{array}  ANOVA  \begin{array}{lrrrrr} \hline &\text { d f}& \text { SS } & \text { MS } & \text {F }& \text {Significance F}  \\ \hline \text {Regression }& 1 & 171499.78 & 171499.78 & 74.8505 & 2.1259E-09 \\ \text {Residual} & 28 & 64154.42 & 2291.23 & & \\ \text {Total} & 29 & 235654.20 & & & \\ \hline\end{array}    \begin{array}{lrrrrrr} \hline &  \text {Coefficients }& \text { Standard Error}& \text { t  Stat }&  \text { p -value }&  \text {Lower 95\% }& \text {Upper 95\% }\\ \hline \text { Intercept }& 76.5351 & 11.8318 & 6.4686 & 5.24 \mathrm{E}-07& 52.2987 & 100.7716 \\ \text {Gross} & 4.3331 & 0.5008 & 8.6516 & 2.13 \mathrm{E}-09 & 3.3072 & 5.3590 \\ \hline \end{array}        -Referring to Table 13-11, which of the following is the correct interpretation for the coefficient of determination?</strong> A) 72.8% of the variation in the box office gross can be explained by the variation in the video unit sales. B) 71.8% of the variation in the video unit sales can be explained by the variation in the box office gross. C) 71.8% of the variation in the box office gross can be explained by the variation in the video unit sales. D) 72.8% of the variation in the video unit sales can be explained by the variation in the box office gross.   <strong>TABLE 13- 11 A company that has the distribution rights to home video sales of previously released movies would like to use the box office gross (in millions of dollars) to estimate the number of units (in thousands of units) that it can expect to sell. Following is the output from a simple linear regression along with the residual plot and normal probability plot obtained from a data set of 30 different movie titles:  \begin{array}{l} \text { Regression Statistics }\\ \begin{array} { l c } \hline \text { Multiple R } & 0.8531 \\ \text { RSquare } & 0.7278 \\ \text { Adjusted R Square } & 0.7180 \\ \text { Standard Error } & 47.8668 \\ \text { Observations } & 30 \end{array} \end{array}  ANOVA  \begin{array}{lrrrrr} \hline &\text { d f}& \text { SS } & \text { MS } & \text {F }& \text {Significance F}  \\ \hline \text {Regression }& 1 & 171499.78 & 171499.78 & 74.8505 & 2.1259E-09 \\ \text {Residual} & 28 & 64154.42 & 2291.23 & & \\ \text {Total} & 29 & 235654.20 & & & \\ \hline\end{array}    \begin{array}{lrrrrrr} \hline &  \text {Coefficients }& \text { Standard Error}& \text { t  Stat }&  \text { p -value }&  \text {Lower 95\% }& \text {Upper 95\% }\\ \hline \text { Intercept }& 76.5351 & 11.8318 & 6.4686 & 5.24 \mathrm{E}-07& 52.2987 & 100.7716 \\ \text {Gross} & 4.3331 & 0.5008 & 8.6516 & 2.13 \mathrm{E}-09 & 3.3072 & 5.3590 \\ \hline \end{array}        -Referring to Table 13-11, which of the following is the correct interpretation for the coefficient of determination?</strong> A) 72.8% of the variation in the box office gross can be explained by the variation in the video unit sales. B) 71.8% of the variation in the video unit sales can be explained by the variation in the box office gross. C) 71.8% of the variation in the box office gross can be explained by the variation in the video unit sales. D) 72.8% of the variation in the video unit sales can be explained by the variation in the box office gross.

-Referring to Table 13-11, which of the following is the correct interpretation for the coefficient of determination?

A) 72.8% of the variation in the box office gross can be explained by the variation in the video unit sales.
B) 71.8% of the variation in the video unit sales can be explained by the variation in the box office gross.
C) 71.8% of the variation in the box office gross can be explained by the variation in the video unit sales.
D) 72.8% of the variation in the video unit sales can be explained by the variation in the box office gross.
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74
TABLE 13-8
It is believed that GPA (grade point average, based on a four point scale) should have a positive linear relationship with ACT scores. Given below is the Excel output from regressing GPA on ACT scores using a data set of 8 randomly chosen students from a Big-Ten university.
Regressing GPA on ACTA C T
 Regression Statistics  Multiple R 0.7598 R Square 0.5774 Adjusted R Square 0.5069 Standard Error 0.2691 Observations 8\begin{array}{lc}\text { Regression Statistics } \\\hline \text { Multiple R } & 0.7598 \\\text { R Square } & 0.5774 \\\text { Adjusted R Square } & 0.5069 \\\text { Standard Error } & 0.2691 \\\text { Observations } & 8\end{array}
ANOVA
dfSSMSF Significance F Regression 1 0.59400.59408.19860.0286 Residual 60.43470.0724 Total 71.0287\begin{array}{lccccc} & d f & S S & M S & F & \text { Significance } F \\\hline \text { Regression 1 } & 0.5940 & 0.5940 & 8.1986 & 0.0286 \\\text { Residual } & 6 & 0.4347 & 0.0724 & & \\\text { Total } & 7 & 1.0287 & & & \\\hline\end{array}

Coefficients Standard Error T Stat  p -value Lower 95% Upper 95%  Intercept0.56810.92840.61190.56301.70362.8398ACT 0.10210.03562.86330.02860.01480.1895\begin{array}{lrrrrrr}\hline & \text {Coefficients }& \text {Standard Error }& \text {T Stat }& \text { p -value }& \text {Lower 95\%}& \text { Upper 95\% } \\\hline \text { Intercept} & 0.5681 & 0.9284 & 0.6119 & 0.5630 & -1.7036 & 2.8398 \\ \text {ACT }& 0.1021 & 0.0356 & 2.8633 & 0.0286 & 0.0148 & 0.1895 \\\hline\end{array}



-Referring to Table 13-8, the value of the measured test statistic to test whether there is any linear relationship between GPA and ACT is

A) 0.7598.
B) 0.0356.
C) 0.1021.
D) 2.8633.
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75
TABLE 13-10
The management of a chain electronic store would like to develop a model for predicting the weekly sales (in thousand of dollars) for individual stores based on the number of customers who made purchases. A random sample of 12 stores yields the following results:
 Customers  Sales (Thousands of Dollars) 90711.2092611.057138.217419.217809.4289810.085106.735297.024606.128729.526507.536037.25\begin{array} { | l | l | } \hline \text { Customers } & \text { Sales (Thousands of Dollars) } \\\hline 907 & 11.20 \\\hline 926 & 11.05 \\\hline 713 & 8.21 \\\hline 741 & 9.21 \\\hline 780 & 9.42 \\\hline 898 & 10.08 \\\hline 510 & 6.73 \\\hline 529 & 7.02 \\\hline 460 & 6.12 \\\hline 872 & 9.52 \\\hline 650 & 7.53 \\\hline 603 & 7.25 \\\hline\end{array}

-Referring to Table 13-10, what is the value of the coefficient of correlation?
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76
TABLE 13-12
The manager of the purchasing department of a large banking organization would like to develop a model to predict the amount of time (measured in hours) it takes to process invoices. Data are collected from a sample of 30 days, and the number of invoices processed and completion time in hours is recorded. Below is the regression output:
 Regression Statistics  Multiple R 0.9947 R Square 0.8924 Adjusted R Square 0.8886 Standard Error 0.3342 ations 30\begin{array}{l}\text { Regression Statistics }\\\begin{array} { l c } \hline \text { Multiple R } & 0.9947 \\\text { R Square } & 0.8924 \\\text { Adjusted R Square } & 0.8886 \\\text { Standard Error } & 0.3342 \\\text { ations } & 30 \\\hline\end{array}\end{array}

 d f  SS MS F  Significance F Regression125.943825.9438232.22004.3946E15Residual 283.12820.1117Total 2929.072\begin{array}{lrrccc}\hline & \text { d f } & \text { SS } & \text {MS} & \text { F } & \text { Significance F } \\\hline \text {Regression} & 1 & 25.9438 & 25.9438 & 232.2200 & 4.3946 \mathrm{E}-15 \\ \text {Residual }& 28 & 3.1282 & 0.1117 & & \\ \text {Total }& 29 & 29.072 & & & \\\hline\end{array}

 Coefficients  Standard Error  t Stat  p -valueLower 95%Upper 95% Invoices 0.40240.12363.25590.00300.14920.6555Processed 0.01260.000815.23884.3946E150.01090.0143\begin{array}{lrrrrrr}\hline & \text { Coefficients }& \text { Standard Error }& \text { t Stat }& \text { p -value}& \text {Lower 95\%} & \text {Upper 95\%} \\\hline \text { Invoices }& 0.4024 & 0.1236 & 3.2559 & 0.0030 & 0.1492 & 0.6555 \\ \text {Processed }& 0.0126 & 0.0008 & 15.2388 & 4.3946 \mathrm{E}-15 & 0.0109 & 0.0143 \\\hline\end{array}

 <strong>TABLE 13-12 The manager of the purchasing department of a large banking organization would like to develop a model to predict the amount of time (measured in hours) it takes to process invoices. Data are collected from a sample of 30 days, and the number of invoices processed and completion time in hours is recorded. Below is the regression output:  \begin{array}{l} \text { Regression Statistics }\\ \begin{array} { l c } \hline \text { Multiple R } & 0.9947 \\ \text { R Square } & 0.8924 \\ \text { Adjusted R Square } & 0.8886 \\ \text { Standard Error } & 0.3342 \\ \text { ations } & 30 \\ \hline \end{array} \end{array}    \begin{array}{lrrccc} \hline & \text { d f } &  \text { SS } &  \text {MS} &  \text { F } & \text { Significance  F } \\ \hline  \text {Regression} & 1 & 25.9438 & 25.9438 & 232.2200 & 4.3946 \mathrm{E}-15  \\ \text {Residual }& 28 & 3.1282 & 0.1117 & & \\ \text {Total }& 29 & 29.072 & & & \\ \hline \end{array}    \begin{array}{lrrrrrr} \hline & \text { Coefficients }& \text { Standard Error }& \text { t Stat }&  \text { p -value}& \text {Lower 95\%} &  \text {Upper 95\%} \\ \hline \text { Invoices }& 0.4024 & 0.1236 & 3.2559 & 0.0030 & 0.1492 & 0.6555 \\ \text {Processed }& 0.0126 & 0.0008 & 15.2388 &  4.3946 \mathrm{E}-15  & 0.0109 & 0.0143 \\ \hline \end{array}          -Referring to Table 13-12, the value of the measured t-test statistic to test whether the amount of time depends linearly on the number of invoices processed is</strong> A) 0.8924. B) 15.2388. C) 232.2200. D) 3.2559.

 <strong>TABLE 13-12 The manager of the purchasing department of a large banking organization would like to develop a model to predict the amount of time (measured in hours) it takes to process invoices. Data are collected from a sample of 30 days, and the number of invoices processed and completion time in hours is recorded. Below is the regression output:  \begin{array}{l} \text { Regression Statistics }\\ \begin{array} { l c } \hline \text { Multiple R } & 0.9947 \\ \text { R Square } & 0.8924 \\ \text { Adjusted R Square } & 0.8886 \\ \text { Standard Error } & 0.3342 \\ \text { ations } & 30 \\ \hline \end{array} \end{array}    \begin{array}{lrrccc} \hline & \text { d f } &  \text { SS } &  \text {MS} &  \text { F } & \text { Significance  F } \\ \hline  \text {Regression} & 1 & 25.9438 & 25.9438 & 232.2200 & 4.3946 \mathrm{E}-15  \\ \text {Residual }& 28 & 3.1282 & 0.1117 & & \\ \text {Total }& 29 & 29.072 & & & \\ \hline \end{array}    \begin{array}{lrrrrrr} \hline & \text { Coefficients }& \text { Standard Error }& \text { t Stat }&  \text { p -value}& \text {Lower 95\%} &  \text {Upper 95\%} \\ \hline \text { Invoices }& 0.4024 & 0.1236 & 3.2559 & 0.0030 & 0.1492 & 0.6555 \\ \text {Processed }& 0.0126 & 0.0008 & 15.2388 &  4.3946 \mathrm{E}-15  & 0.0109 & 0.0143 \\ \hline \end{array}          -Referring to Table 13-12, the value of the measured t-test statistic to test whether the amount of time depends linearly on the number of invoices processed is</strong> A) 0.8924. B) 15.2388. C) 232.2200. D) 3.2559.


-Referring to Table 13-12, the value of the measured t-test statistic to test whether the amount of time depends linearly on the number of invoices processed is

A) 0.8924.
B) 15.2388.
C) 232.2200.
D) 3.2559.
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TABLE 13-3
The director of cooperative education at a state college wants to examine the effect of cooperative education job experience on marketability in the work place. She takes a random sample of 4 students. For these 4, she finds out how many times each had a cooperative education job and how many job offers they received upon graduation. These data are presented in the table below.
 Student  CoopJobs  JobOffer 114226313401\begin{array} { l l l } \hline \text { Student } & \text { CoopJobs } & \text { JobOffer } \\\hline 1 & 1 & 4 \\2 & 2 & 6 \\3 & 1 & 3 \\4 & 0 & 1\end{array}

-Referring to Table 13-3, suppose the director of cooperative education wants to obtain a 95% confidence-interval estimate for the mean number of job offers received by people who have had exactly one cooperative education job. The t critical value she would use is_______
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78
If the Durbin-Watson statistic has a value close to 0, which assumption is violated?

A) homoscedasticity
B) normality of the errors
C) independence of errors
D) none of the above
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79
TABLE 13-7
An investment specialist claims that if one holds a portfolio that moves in the opposite direction to the market index like the S&P 500, then it is possible to reduce the variability of the portfolio's return. In other words, one can create a portfolio with positive returns but less exposure to risk. A sample of 26 years of S&P 500 index and a portfolio consisting of stocks of private prisons, which are believed to be negatively related to the S&P 500 index, is collected. A regression analysis was performed by regressing the returns of the prison stocks portfolio (Y) on the returns of S&P 500 index (X) to prove that the prison stocks portfolio is negatively related to the S&P 500 index at a 5% level of significance. The results are given in the following EXCEL output.
 Coefficients  Standard Error  T Stat p-value  Intercept 4.8660042580.3574360913.613634418.7932E13 S& P 0.5025135060.0715971527.01862425294942E07\begin{array} { l c c c c } \hline & \text { Coefficients } & \text { Standard Error } & \text { T Stat } & { p \text {-value } } \\\hline \text { Intercept } & 4.866004258 & 0.35743609 & 13.61363441 & 8.7932 \mathrm { E } - 13 \\\text { S\& P } & - 0.502513506 & 0.071597152 &- 7.01862425 & 294942 \mathrm { E } - 07 \\\hline\end{array}

-Referring to Table 13-7, to test whether the prison stocks portfolio is negatively related to the S&P 500 index, the appropriate null and alternative hypotheses are, respectively,

A) H0:r0 H_{0}: r \leq 0 versus H1:r>0 H_{1}: r>0 .
B) H0:r0 H_{0}: r \geq 0 versus H1:r<0 H_{1}: r<0 .
C) H0:ρ0 H_{0}: \rho \geq 0 versus H1:ρ<0 H_{1}: \rho<0 .
D) H0:ρ0 H_{0}: \rho \leq 0 versus H1:ρ>0 H_{1}: \rho>0 .
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80
TABLE 13-5
The managing partner of an advertising agency believes that his company's sales are related to the industry sales. He uses Microsoft Excel's Data Analysis tool to analyze the last 4 years of quarterly data with the following results:
 Regression Sttuistics  Multiple R 0.802 R Square 0.643 Adjusted RSquare 0.618 Standard Error SYX 0.9224\begin{array}{ll}\text { Regression Sttuistics } \\\hline \text { Multiple R } & 0.802 \\\text { R Square } & 0.643 \\\text { Adjusted RSquare } & 0.618 \\\text { Standard Error SYX } & 0.9224\end{array}

Observations 16
ANOVA

df SS  MS F Sig.F  Regression 121.49721.49725.270.000 Error 1411.9120.851 Total 1533.409\begin{array}{lrlrlr} & d f & \text { SS } & \text { MS } & F & \text { Sig.F } \\\hline \text { Regression } & 1 & 21.497 & 21.497 & 25.27 & 0.000 \\\text { Error } & 14 & 11.912 & 0.851 & & \\\text { Total } & 15 & 33.409 & & &\end{array}

 Predictor Coeffcients  Standard Error  tStat  p-value  Intercept 3.9621.4402.750.016 Industry 0.0404510.0080485.030.000 Durbin- Watson Statistic 1.59\begin{array}{l}\begin{array}{lllrl}\hline \text { Predictor}& \text { Coeffcients } & \text { Standard Error } & \text { tStat } & \text { p-value } \\\text { Intercept } & 3.962 & 1.440 & 2.75 & 0.016 \\\text { Industry } & 0.040451 & 0.008048 & 5.03 & 0.000 \\\hline\end{array}\\\text { Durbin- Watson Statistic } 1.59\end{array}


-Referring to Table 13-5, the standard error of the estimated slope coefficient is________ .
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