Exam 12: Simple Linear Regression

arrow
  • Select Tags
search iconSearch Question
flashcardsStudy Flashcards
  • Select Tags

onfident that the mean amount of time needed to record one additional loan application is somewhere  Regression statistics \text { Regression statistics } MultipleR 0.9447 R Square 0.8924 Adjusted R Square 0.8886 Standard Error 0.3342 Observations 30  ANOVA \text { ANOVA } df S5 MS F Significance F Regression 1 25.9438 25.9438 232.2200 4.3946-15 Residual 28 3.1282 0.1117 Total 29 29.072 Coefficients Standard Error tStat p-value Lower 95\% Upper 95\% Intercept 0.4024 0.1236 3.2559 0.0030 0.1492 0.6555 Applications Recorded 0.0126 0.0008 15.2388 4.3946- 15 0.0109 0.0143  Note: 4.3946E15 is 4.3946×1015\text { Note: } 4.3946 \mathrm { E } - 15 \text { is } 4.3946 \times 10 ^ { - 15 } \text {. }  onfident that the mean amount of time needed to record one additional loan application is somewhere  \text { Regression statistics }   \begin{array}{|l|l|} \hline\text { MultipleR } & 0.9447 \\ \hline \text { R Square } & 0.8924 \\ \hline \begin{array}{l} \text { Adjusted R } \\ \text { Square } \end{array} & 0.8886 \\ \hline \text { Standard Error } & 0.3342 \\ \hline \text { Observations } & 30 \\ \hline \end{array}    \text { ANOVA }   \begin{array}{|l|l|l|l|l|l|} \hline & d f & S 5 & M S & F & \begin{array}{l} \text { Significance } \\ F \end{array} \\ \hline \text { Regression } & 1 & 25.9438 & 25.9438 & 232.2200 & 4.3946 \mathrm{E}-15 \\ \hline \text { Residual } & 28 & 3.1282 & 0.1117 & & \\ \hline \text { Total } & 29 & 29.072 & & & \\ \hline \end{array}    \begin{array}{|l|l|l|l|l|l|l|} \hline & \text { Coefficients } & \begin{array}{l} \text { Standard } \\ \text { Error } \end{array} & \text { tStat } & \text { p-value } & \text { Lower 95\% } & \text { Upper 95\% } \\ \hline \text { Intercept } & 0.4024 & 0.1236 & 3.2559 & 0.0030 & 0.1492 & 0.6555 \\ \hline \begin{array}{l} \text { Applications } \\ \text { Recorded } \end{array} & 0.0126 & 0.0008 & 15.2388 & \begin{array}{l} 4.3946 \mathrm{E}- \\ 15 \end{array} & 0.0109 & 0.0143 \\ \hline \end{array}     \text { Note: } 4.3946 \mathrm { E } - 15 \text { is } 4.3946 \times 10 ^ { - 15 } \text {. }    -Referring to Instruction 12.36,the p-value of the measured t test statistic to test whether the number of loan applications recorded affects the amount of time is -Referring to Instruction 12.36,the p-value of the measured t test statistic to test whether the number of loan applications recorded affects the amount of time is

Free
(Multiple Choice)
5.0/5
(36)
Correct Answer:
Verified

A

Instruction 12.17 A computer software developer would like to use the number of downloads (in thousands) for the trial version of his new shareware to predict the amount of revenue (in thousands of dollars) he can make on the full version of the new shareware. 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 sharewares that he has developed:  Regression statistics \text { Regression statistics } MultipleR 0.8691 R Square 0.7554 Adjusted R Square 0.7467 Standard Error 44.4765 Observations 30.0000  ANOVA \text { ANOVA } df SS MS F Significance F Regression 1 171062.9193 171062.9193 86.4759 0.0000 Residual 28 55388.4309 1978.1582 Total 29 226451.3503 Coefficients Standard Error t Stat p -value Lower 95\% Upper 95\% Intercept -95.0614 26.9183 -3.5315 0.0015 -150.2009 -39.9218 Download 3.7297 0.4011 9.2992 0.0000 2.9082 4.5513  Instruction 12.17 A computer software developer would like to use the number of downloads (in thousands) for the trial version of his new shareware to predict the amount of revenue (in thousands of dollars) he can make on the full version of the new shareware. 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 sharewares that he has developed:   \text { Regression statistics }   \begin{array}{|l|l|} \hline \text { MultipleR } & 0.8691 \\ \hline \text { R Square } & 0.7554 \\ \hline \begin{array}{l} \text { Adjusted R } \\ \text { Square } \end{array} & 0.7467 \\ \hline \text { Standard Error } & 44.4765 \\ \hline \text { Observations } & 30.0000\\ \hline  \end{array}    \text { ANOVA }   \begin{array}{|l|l|l|l|l|l|} \hline & d f & S S & M S & F & \begin{array}{l} \text { Significance } \\ F \end{array} \\ \hline \text { Regression } & 1 & 171062.9193 & 171062.9193 & 86.4759 & 0.0000 \\ \hline \text { Residual } & 28 & 55388.4309 & 1978.1582 & & \\ \hline \text { Total } & 29 & 226451.3503 & & & \\ \hline \end{array}    \begin{array}{|l|l|l|l|l|l|l|} \hline & \text { Coefficients } & \begin{array}{l} \text { Standard } \\ \text { Error } \end{array} & \text { t Stat } & p \text {-value } & \text { Lower 95\% } & \text { Upper 95\% } \\ \hline \text { Intercept } & -95.0614 & 26.9183 & -3.5315 & 0.0015 & -150.2009 & -39.9218 \\ \hline \text { Download } & 3.7297 & 0.4011 & 9.2992 & 0.0000 & 2.9082 & 4.5513 \\ \hline \end{array}       -Referring to Instruction 12.17,what is the standard deviation around the regression line?  Instruction 12.17 A computer software developer would like to use the number of downloads (in thousands) for the trial version of his new shareware to predict the amount of revenue (in thousands of dollars) he can make on the full version of the new shareware. 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 sharewares that he has developed:   \text { Regression statistics }   \begin{array}{|l|l|} \hline \text { MultipleR } & 0.8691 \\ \hline \text { R Square } & 0.7554 \\ \hline \begin{array}{l} \text { Adjusted R } \\ \text { Square } \end{array} & 0.7467 \\ \hline \text { Standard Error } & 44.4765 \\ \hline \text { Observations } & 30.0000\\ \hline  \end{array}    \text { ANOVA }   \begin{array}{|l|l|l|l|l|l|} \hline & d f & S S & M S & F & \begin{array}{l} \text { Significance } \\ F \end{array} \\ \hline \text { Regression } & 1 & 171062.9193 & 171062.9193 & 86.4759 & 0.0000 \\ \hline \text { Residual } & 28 & 55388.4309 & 1978.1582 & & \\ \hline \text { Total } & 29 & 226451.3503 & & & \\ \hline \end{array}    \begin{array}{|l|l|l|l|l|l|l|} \hline & \text { Coefficients } & \begin{array}{l} \text { Standard } \\ \text { Error } \end{array} & \text { t Stat } & p \text {-value } & \text { Lower 95\% } & \text { Upper 95\% } \\ \hline \text { Intercept } & -95.0614 & 26.9183 & -3.5315 & 0.0015 & -150.2009 & -39.9218 \\ \hline \text { Download } & 3.7297 & 0.4011 & 9.2992 & 0.0000 & 2.9082 & 4.5513 \\ \hline \end{array}       -Referring to Instruction 12.17,what is the standard deviation around the regression line? -Referring to Instruction 12.17,what is the standard deviation around the regression line?

Free
(Short Answer)
4.7/5
(36)
Correct Answer:
Verified

$44.4765 thousands

The residuals represent

Free
(Multiple Choice)
4.8/5
(31)
Correct Answer:
Verified

B

Instruction 12.30 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 analyse the last four years of quarterly data with the following results:  Regression statistics \text { Regression statistics } Multiple R 0.802 R Square 0.643 Adjusted R Square 0.618 Standard Error SYX 0.9224 Observations 16  ANOVA \text { ANOVA } df SS MS F Sig.F Regression 1 21.497 21.497 25.27 0.000 Error 14 11.912 0.851 Total 15 33.409 Predictor Coef StdError t Stat p-value Intercept 3.962 1.440 2.75 0.016 Industry 0.040451 0.008048 5.03 0.000 Durbin-Watson 1.59 Statistic -Referring to Instruction 12.30,the value of the quantity that the least squares regression line minimises is____________.

(Short Answer)
4.8/5
(41)

The value of r is always positive.

(True/False)
4.8/5
(35)

Instruction 12.2 A chocolate 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 six country towns and cities and offers the chocolate bar at different prices. Using chocolate bar sales as the dependent variable, the company will conduct a simple linear regression on the data below:  City  Price($)  Syles  Toowoomba 1.30100 Broken Hill 1.6050 Bendigo 1.8050 Kalgoorlie 2.0040 Launceston 2.4038 Port Augusta 2.5032\begin{array} { | l | l | l | } \hline \text { City } & \text { Price(\$) } & \text { Syles } \\\hline \text { Toowoomba } & 1.30 & 100 \\\hline \text { Broken Hill } & 1.60 & 50 \\\hline \text { Bendigo } & 1.80 & 50 \\\hline \text { Kalgoorlie } & 2.00 & 40 \\\hline \text { Launceston } & 2.40 & 38 \\\hline \text { Port Augusta } & 2.50 & 32 \\\hline\end{array} -Referring to Instruction 12.2,what is the estimated average change in the sales of the chocolate bar if price goes up by $1.00?

(Multiple Choice)
4.8/5
(46)

Instruction 12.31 An investment specialist claims that if one holds a portfolio that moves in opposite direction to the market index like the All Ordinaries Index, 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 the All Ordinaries index and a portfolio consisting of stocks of private prisons, which are believed to be negatively related to the All Ordinaries index, is collected. A regression analysis was performed by regressing the returns of the prison stocks portfolio (Y) on the returns of All Ordinaries index (X) to prove that the prison stocks portfolio is negatively related to the All Ordinaries index at a 5% level of significance. The results are given in the following Microsoft Excel output. Coefflelents Standard Error tStat p -vahse Intercept 4.866004258 0.35743609 13.61363441 8.7932-13 S\&P -0.502513506 0.071597152 -7.01862425 2.94942-07 -Referring to Instruction 12.31,to test whether the prison stocks portfolio is negatively related to the All Ordinaries index,the measured value of the test statistic is

(Multiple Choice)
4.9/5
(40)

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: Y1=b0+b1x1+εjY _ { 1 } = b _ { 0 } + b _ { 1 } x _ { 1 } + \varepsilon _ { j } The results of the simple linear regression are provided below: Y^=2,700+20X,SYX=65, two-tailed p-value =0.034 (for testing θ1 ) \hat { Y } = - 2,700 + 20 X , S _ { Y X } = 65 \text {, two-tailed } p \text {-value } = 0.034 \text { (for testing } \theta _ { 1 } \text { ) } -Referring to Instruction 12.1,interpret the p-value for testing whether ?1 exceeds 0.

(Multiple Choice)
4.9/5
(31)

Instruction 12.13 The managers of a brokerage firm are interested in finding out if the number of new customers a broker brings into the firm affects the sales generated by the broker. They sample 12 brokers and determine the number of new customers they have enrolled in the last year and their sales amounts in thousands of dollars. These data are presented in the table that follows. Broker Clients Sales 1 27 52 2 11 37 3 42 64 4 33 55 5 15 29 6 15 34 7 25 58 8 36 59 9 28 44 10 30 48 11 17 31 12 22 38 -Referring to Instruction 12.13,the coefficient of correlation is____________.

(Short Answer)
4.9/5
(40)

Instruction 12.29 The managers of a brokerage firm are interested in finding out if the number of new customers a broker brings into the firm affects the sales generated by the broker. They sample 12 brokers and determine the number of new customers they have enrolled in the last year and their sales amounts in thousands of dollars. These data are presented in the table that follows. Broker Clients Sales 1 27 52 2 11 37 3 42 64 4 33 55 5 15 29 6 15 34 7 25 58 8 36 59 9 28 44 10 30 48 11 17 31 12 22 38 -Referring to Instruction 12.29,the managers of the brokerage firm wanted to test the hypothesis that the number of new customers brought in had a positive impact on the amount of sales generated.For a test with a level of significance of 0.01,the null hypothesis should be rejected if the value of the test statistic is____________.

(Short Answer)
4.7/5
(43)

Instruction 12.35 A computer software developer would like to use the number of downloads (in thousands) for the trial version of his new shareware to predict the amount of revenue (in thousands of dollars) he can make on the full version of the new shareware. 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 sharewares that he has developed:  Regression statistics \text { Regression statistics } MultipleR 0.8691 R Square 0.7554 Adjusted R Square 0.7467 Standard Error 44.4765 Observations 30.0000  ANOVA \text { ANOVA } df SS MS F Significance F Regression 1 171062.9193 171062.9193 86.4759 0.0000 Residual 28 55388.4309 1978.1582 Total 29 226451.3503 Coefficients Standard Error t Stat p -value Lower 95\% Upper 95\% Intercept -95.0614 26.9183 -3.5315 0.0015 -150.2009 -39.9218 Download 3.7297 0.4011 9.2992 0.0000 2.9082 4.5513  Instruction 12.35 A computer software developer would like to use the number of downloads (in thousands) for the trial version of his new shareware to predict the amount of revenue (in thousands of dollars) he can make on the full version of the new shareware. 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 sharewares that he has developed:   \text { Regression statistics }   \begin{array}{|l|l|} \hline \text { MultipleR } & 0.8691 \\ \hline \text { R Square } & 0.7554 \\ \hline \begin{array}{l} \text { Adjusted R } \\ \text { Square } \end{array} & 0.7467 \\ \hline \text { Standard Error } & 44.4765 \\ \hline \text { Observations } & 30.0000\\ \hline  \end{array}    \text { ANOVA }   \begin{array}{|l|l|l|l|l|l|} \hline & d f & S S & M S & F & \begin{array}{l} \text { Significance } \\ F \end{array} \\ \hline \text { Regression } & 1 & 171062.9193 & 171062.9193 & 86.4759 & 0.0000 \\ \hline \text { Residual } & 28 & 55388.4309 & 1978.1582 & & \\ \hline \text { Total } & 29 & 226451.3503 & & & \\ \hline \end{array}    \begin{array}{|l|l|l|l|l|l|l|} \hline & \text { Coefficients } & \begin{array}{l} \text { Standard } \\ \text { Error } \end{array} & \text { t Stat } & p \text {-value } & \text { Lower 95\% } & \text { Upper 95\% } \\ \hline \text { Intercept } & -95.0614 & 26.9183 & -3.5315 & 0.0015 & -150.2009 & -39.9218 \\ \hline \text { Download } & 3.7297 & 0.4011 & 9.2992 & 0.0000 & 2.9082 & 4.5513 \\ \hline \end{array}         -Referring to Instruction 12.35,the null hypothesis for testing whether there is a linear relationship between revenue and number of downloads is 'There is no linear relationship between revenue and number of downloads'.  Instruction 12.35 A computer software developer would like to use the number of downloads (in thousands) for the trial version of his new shareware to predict the amount of revenue (in thousands of dollars) he can make on the full version of the new shareware. 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 sharewares that he has developed:   \text { Regression statistics }   \begin{array}{|l|l|} \hline \text { MultipleR } & 0.8691 \\ \hline \text { R Square } & 0.7554 \\ \hline \begin{array}{l} \text { Adjusted R } \\ \text { Square } \end{array} & 0.7467 \\ \hline \text { Standard Error } & 44.4765 \\ \hline \text { Observations } & 30.0000\\ \hline  \end{array}    \text { ANOVA }   \begin{array}{|l|l|l|l|l|l|} \hline & d f & S S & M S & F & \begin{array}{l} \text { Significance } \\ F \end{array} \\ \hline \text { Regression } & 1 & 171062.9193 & 171062.9193 & 86.4759 & 0.0000 \\ \hline \text { Residual } & 28 & 55388.4309 & 1978.1582 & & \\ \hline \text { Total } & 29 & 226451.3503 & & & \\ \hline \end{array}    \begin{array}{|l|l|l|l|l|l|l|} \hline & \text { Coefficients } & \begin{array}{l} \text { Standard } \\ \text { Error } \end{array} & \text { t Stat } & p \text {-value } & \text { Lower 95\% } & \text { Upper 95\% } \\ \hline \text { Intercept } & -95.0614 & 26.9183 & -3.5315 & 0.0015 & -150.2009 & -39.9218 \\ \hline \text { Download } & 3.7297 & 0.4011 & 9.2992 & 0.0000 & 2.9082 & 4.5513 \\ \hline \end{array}         -Referring to Instruction 12.35,the null hypothesis for testing whether there is a linear relationship between revenue and number of downloads is 'There is no linear relationship between revenue and number of downloads'. -Referring to Instruction 12.35,the null hypothesis for testing whether there is a linear relationship between revenue and number of downloads is 'There is no linear relationship between revenue and number of downloads'.

(True/False)
4.8/5
(43)

Testing for the existence of correlation is equivalent to

(Multiple Choice)
4.9/5
(34)

Instruction 12.34 The management of a chain electronic store would like to develop a model for predicting the weekly sales (in thousands 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) 907 11.20 926 11.05 713 8.21 741 9.21 780 9.42 898 10.08 510 6.73 529 7.02 460 6.12 872 9.52 650 7.53 603 7.25 -Referring to Instruction 12.34,the null hypothesis for testing whether the number of customers who make purchase effects weekly sales cannot be rejected if 1% probability of committing a Type I error is desired.

(True/False)
4.8/5
(30)

Instruction 12.17 A computer software developer would like to use the number of downloads (in thousands) for the trial version of his new shareware to predict the amount of revenue (in thousands of dollars) he can make on the full version of the new shareware. 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 sharewares that he has developed:  Regression statistics \text { Regression statistics } MultipleR 0.8691 R Square 0.7554 Adjusted R Square 0.7467 Standard Error 44.4765 Observations 30.0000  ANOVA \text { ANOVA } df SS MS F Significance F Regression 1 171062.9193 171062.9193 86.4759 0.0000 Residual 28 55388.4309 1978.1582 Total 29 226451.3503 Coefficients Standard Error t Stat p -value Lower 95\% Upper 95\% Intercept -95.0614 26.9183 -3.5315 0.0015 -150.2009 -39.9218 Download 3.7297 0.4011 9.2992 0.0000 2.9082 4.5513  Instruction 12.17 A computer software developer would like to use the number of downloads (in thousands) for the trial version of his new shareware to predict the amount of revenue (in thousands of dollars) he can make on the full version of the new shareware. 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 sharewares that he has developed:   \text { Regression statistics }   \begin{array}{|l|l|} \hline \text { MultipleR } & 0.8691 \\ \hline \text { R Square } & 0.7554 \\ \hline \begin{array}{l} \text { Adjusted R } \\ \text { Square } \end{array} & 0.7467 \\ \hline \text { Standard Error } & 44.4765 \\ \hline \text { Observations } & 30.0000\\ \hline  \end{array}    \text { ANOVA }   \begin{array}{|l|l|l|l|l|l|} \hline & d f & S S & M S & F & \begin{array}{l} \text { Significance } \\ F \end{array} \\ \hline \text { Regression } & 1 & 171062.9193 & 171062.9193 & 86.4759 & 0.0000 \\ \hline \text { Residual } & 28 & 55388.4309 & 1978.1582 & & \\ \hline \text { Total } & 29 & 226451.3503 & & & \\ \hline \end{array}    \begin{array}{|l|l|l|l|l|l|l|} \hline & \text { Coefficients } & \begin{array}{l} \text { Standard } \\ \text { Error } \end{array} & \text { t Stat } & p \text {-value } & \text { Lower 95\% } & \text { Upper 95\% } \\ \hline \text { Intercept } & -95.0614 & 26.9183 & -3.5315 & 0.0015 & -150.2009 & -39.9218 \\ \hline \text { Download } & 3.7297 & 0.4011 & 9.2992 & 0.0000 & 2.9082 & 4.5513 \\ \hline \end{array}       -Referring to Instruction 12.17,which of the following is the correct interpretation for the coefficient of determination?  Instruction 12.17 A computer software developer would like to use the number of downloads (in thousands) for the trial version of his new shareware to predict the amount of revenue (in thousands of dollars) he can make on the full version of the new shareware. 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 sharewares that he has developed:   \text { Regression statistics }   \begin{array}{|l|l|} \hline \text { MultipleR } & 0.8691 \\ \hline \text { R Square } & 0.7554 \\ \hline \begin{array}{l} \text { Adjusted R } \\ \text { Square } \end{array} & 0.7467 \\ \hline \text { Standard Error } & 44.4765 \\ \hline \text { Observations } & 30.0000\\ \hline  \end{array}    \text { ANOVA }   \begin{array}{|l|l|l|l|l|l|} \hline & d f & S S & M S & F & \begin{array}{l} \text { Significance } \\ F \end{array} \\ \hline \text { Regression } & 1 & 171062.9193 & 171062.9193 & 86.4759 & 0.0000 \\ \hline \text { Residual } & 28 & 55388.4309 & 1978.1582 & & \\ \hline \text { Total } & 29 & 226451.3503 & & & \\ \hline \end{array}    \begin{array}{|l|l|l|l|l|l|l|} \hline & \text { Coefficients } & \begin{array}{l} \text { Standard } \\ \text { Error } \end{array} & \text { t Stat } & p \text {-value } & \text { Lower 95\% } & \text { Upper 95\% } \\ \hline \text { Intercept } & -95.0614 & 26.9183 & -3.5315 & 0.0015 & -150.2009 & -39.9218 \\ \hline \text { Download } & 3.7297 & 0.4011 & 9.2992 & 0.0000 & 2.9082 & 4.5513 \\ \hline \end{array}       -Referring to Instruction 12.17,which of the following is the correct interpretation for the coefficient of determination? -Referring to Instruction 12.17,which of the following is the correct interpretation for the coefficient of determination?

(Multiple Choice)
4.8/5
(36)

When using a regression model to make predictions,you should not predict Y for values of X larger or smaller than the values used to develop the model.

(True/False)
4.9/5
(36)

onfident that the mean amount of time needed to record one additional loan application is somewhere  Regression statistics \text { Regression statistics } MultipleR 0.9447 R Square 0.8924 Adjusted R Square 0.8886 Standard Error 0.3342 Observations 30  ANOVA \text { ANOVA } df S5 MS F Significance F Regression 1 25.9438 25.9438 232.2200 4.3946-15 Residual 28 3.1282 0.1117 Total 29 29.072 Coefficients Standard Error tStat p-value Lower 95\% Upper 95\% Intercept 0.4024 0.1236 3.2559 0.0030 0.1492 0.6555 Applications Recorded 0.0126 0.0008 15.2388 4.3946- 15 0.0109 0.0143  Note: 4.3946E15 is 4.3946×1015\text { Note: } 4.3946 \mathrm { E } - 15 \text { is } 4.3946 \times 10 ^ { - 15 } \text {. }  onfident that the mean amount of time needed to record one additional loan application is somewhere  \text { Regression statistics }   \begin{array}{|l|l|} \hline\text { MultipleR } & 0.9447 \\ \hline \text { R Square } & 0.8924 \\ \hline \begin{array}{l} \text { Adjusted R } \\ \text { Square } \end{array} & 0.8886 \\ \hline \text { Standard Error } & 0.3342 \\ \hline \text { Observations } & 30 \\ \hline \end{array}    \text { ANOVA }   \begin{array}{|l|l|l|l|l|l|} \hline & d f & S 5 & M S & F & \begin{array}{l} \text { Significance } \\ F \end{array} \\ \hline \text { Regression } & 1 & 25.9438 & 25.9438 & 232.2200 & 4.3946 \mathrm{E}-15 \\ \hline \text { Residual } & 28 & 3.1282 & 0.1117 & & \\ \hline \text { Total } & 29 & 29.072 & & & \\ \hline \end{array}    \begin{array}{|l|l|l|l|l|l|l|} \hline & \text { Coefficients } & \begin{array}{l} \text { Standard } \\ \text { Error } \end{array} & \text { tStat } & \text { p-value } & \text { Lower 95\% } & \text { Upper 95\% } \\ \hline \text { Intercept } & 0.4024 & 0.1236 & 3.2559 & 0.0030 & 0.1492 & 0.6555 \\ \hline \begin{array}{l} \text { Applications } \\ \text { Recorded } \end{array} & 0.0126 & 0.0008 & 15.2388 & \begin{array}{l} 4.3946 \mathrm{E}- \\ 15 \end{array} & 0.0109 & 0.0143 \\ \hline \end{array}     \text { Note: } 4.3946 \mathrm { E } - 15 \text { is } 4.3946 \times 10 ^ { - 15 } \text {. }    -Referring to Instruction 12.36,you can be 95% confident that the mean amount of time needed to record one additional loan application is somewhere between 0.0109 and 0.0143 hours. -Referring to Instruction 12.36,you can be 95% confident that the mean amount of time needed to record one additional loan application is somewhere between 0.0109 and 0.0143 hours.

(True/False)
4.7/5
(33)

Instruction 12.28 The managers of a brokerage firm are interested in finding out if the number of new customers a broker brings into the firm affects the sales generated by the broker. They sample 12 brokers and determine the number of new customers they have enrolled in the last year and their sales amounts in thousands of dollars. These data are presented in the table that follows. Broker Clients 5les 1 27 52 2 11 37 3 42 64 4 33 55 5 15 29 6 15 34 7 25 58 8 36 59 9 28 44 10 30 48 11 17 31 12 22 38 -Referring to Instruction 12.28,the managers of the brokerage firm wanted to test the hypothesis that the true slope was equal to 0.The value of the test statistic is ____________.

(Short Answer)
4.9/5
(32)

The coefficient of determination (r2)tells you

(Multiple Choice)
4.9/5
(38)

Instruction 12.28 The managers of a brokerage firm are interested in finding out if the number of new customers a broker brings into the firm affects the sales generated by the broker. They sample 12 brokers and determine the number of new customers they have enrolled in the last year and their sales amounts in thousands of dollars. These data are presented in the table that follows. Broker Clients 5les 1 27 52 2 11 37 3 42 64 4 33 55 5 15 29 6 15 34 7 25 58 8 36 59 9 28 44 10 30 48 11 17 31 12 22 38 -Referring to Instruction 12.28,the managers of the brokerage firm wanted to test the hypothesis that the true slope was equal to 0.The p-value of the test is ____________.

(Essay)
4.8/5
(36)

Instruction 12.27 The director of cooperative education at a university wants to examine the effect of cooperative education job experience on marketability in the workplace. She takes a random sample of four students. For these four, 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 Coop jobs job Offer 1 1 4 2 2 6 3 1 3 4 0 1 -Referring to Instruction 12.27,the director of cooperative education wanted to test the hypothesis that the true slope was equal to 0.The p-value of the test is between____________ and ____________.

(Essay)
4.8/5
(32)
Showing 1 - 20 of 207
close modal

Filters

  • Essay(0)
  • Multiple Choice(0)
  • Short Answer(0)
  • True False(0)
  • Matching(0)