Instruction 12-5 The managing partner of an advertising agency believes that his company's sales are related to the industry sales.He uses MicrosoftS1U1P1S1S1P0 Excel's Data Analysis tool to analyse the last four years of quarterly data with the following results: $\begin{array}{l} \text { Regression Statistics }\\ \begin{array}{lll} \text { Multiple R } & 0.802 & \\ \text { R Square } & 0.643 & \\ \text { Adjusted R Square } & 0.618 & \\ \text { Standard Error SYX } & 0.9224 \\ \text { Observations } & 16 \end{array} \end{array}$ ANOVA $\begin{array}{lrcccc} & \text { df } & \text { SS } & \text { MS } & \text { F } & \text { Sig.F } \\ \text { Regression } & 1 & 21.497 & 21.497 & 25.27 & 0.000 \\ \text { Error } & 14 & 11.912 & 0.851 & & \\ \text { Total } & 15 & 33.409 & & & \end{array}$ $\begin{array}{lllll} \text {Predictor} &\text {Coef}&\text {StdError}&\text {t Stat}&\text {P-value}\\ \text { Intercept } & 3.962 & 1.440 & 2.75 & 0.016 \\ \text { Industry } & 0.040451 & 0.008048 & 5.03 & 0.000\\ \\ \text { Durbin-Watson Statistic}&1.59\\ \end{array}$ -Referring to Instruction 12-5,if the Durbin-Watson statistic has a value close to 4,which assumption is violated?

A) Homoscedasticity. B) Normality of the errors. C) Independence of errors. D) None of the above. A) Homoscedasticity. B) Normality of the errors. C) Independence of errors. D) None of the above.

Instruction 12-10 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: \[\begin{array} { | c | c | } \hline \text { Customers } & \text { Sales (Thousands of Dollars) } \\ \hline 907 & 11.20 \\ \hline 926 & 11.05 \\ \hline 713 & 8.21 \\ \hline 741 & 9.21 \\ \hline 780 & 9.42 \\ \hline 698 & 10.08 \\ \hline 510 & 6.73 \\ \hline 529 & 7.02 \\ \hline 460 & 6.12 \\ \hline 872 & 9.52 \\ \hline 650 & 7.53 \\ \hline 603 & 7.25 \\ \hline \end{array}\] -Referring to Instruction 12-10,what are the degrees of freedom of the t test statistic when testing whether the number of customers who make purchases affects weekly sales?

The answer of Instruction 12-10 The management of a chain electronic...

Instruction 12-10 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: \[\begin{array} { | c | c | } \hline \text { Customers } & \text { Sales (Thousands of Dollars) } \\ \hline 907 & 11.20 \\ \hline 926 & 11.05 \\ \hline 713 & 8.21 \\ \hline 741 & 9.21 \\ \hline 780 & 9.42 \\ \hline 698 & 10.08 \\ \hline 510 & 6.73 \\ \hline 529 & 7.02 \\ \hline 460 & 6.12 \\ \hline 872 & 9.52 \\ \hline 650 & 7.53 \\ \hline 603 & 7.25 \\ \hline \end{array}\] -Referring to Instruction 12-10,what is the value of the t test statistic when testing whether the number of customers who make purchases affects weekly sales?

The answer of Instruction 12-10 The management of a chain electronic...

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: ρ = 0 against the two-sided alternative H1: ρ ≠ 0.To test H0: ρ = 0 against the one-sided alternative H1: ρ < 0 at a significance level of 0.2,the p-value is

A) 0.256/2. B) (0.256)2. C) 1 - 0.256. D) 1 - 0.256/2. A) 0.256/2. B) (0.256)2. C) 1 - 0.256. D) 1 - 0.256/2.

Exam 12: Simple Linear Regression

Instruction 12-5 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 analyse the last four 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 SYX 0.9224 Observations 16 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 Statistic 1.59 -Referring to Instruction 12-5,if the Durbin-Watson statistic has a value close to 4,which assumption is violated?

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Question 161

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 (\ ) Sales Toowoomba 1.30 100 Broken Hill 1.60 90 Bendigo 1.80 90 Kalgoorlie 2.00 40 Launceston 2.40 38 Port Augusta 2.90 32 -Referring to Instruction 12-2,what is the standard error of the estimate,S_YX,for the data?

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Question 162

Instruction 12-11 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 Multiple R 0.8691 R Square 0.7554 Adjusted R Square 0.7467 Standard Error 44.4765 Observations 30.0000 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 Emor 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 $Instruction 12-11 A computer software developer would like to use the number of downloads (in thousands)for the trial version of his new shareware to predict the amount of revenue (in thousands of dollars)he can make on the full version of the new shareware.Following is the output from a simple linear regression along with the residual plot and normal probability plot obtained from a data set of 30 different sharewares that he has developed: \begin{array}{lr} \hline {\text { Regression Statistics }} \\ \hline \text { Multiple R } & 0.8691 \\ \hline \text { R Square } & 0.7554 \\ \hline \text { Adjusted R Square } & 0.7467 \\ \text { Standard Error } & 44.4765 \\ \text { Observations } & 30.0000 \\ \hline \end{array} ANOVA \begin{array}{lr|r|r|r|r} \hline & d f & {\text { SS }} &{\text { MS }} & F & \text { Significance } F \\ \hline \text { Regression } & 1 & 171062.9193 & 171062.9193 & 86.4759 & 0.0000 \\ \hline \text { Residual } & 28 & 55388.4309 & 1978.1582 & & \\ \hline \text { Total } & 29 & 226451.3503 & & & \\ \hline \end{array} \begin{array}{lrrrrrrr} \hline & \text { Coefficients } & \text { Standard Emor } & t \text { Stat } & \text { P-value } & \text { Lower 95\% } & \text { Upper 95\% } \\ \hline \text { Intercept } & -95.0614 & 26.9183 & -3.5315 & 0.0015 & -150.2009 & -39.9218 \\ \text { Download } & 3.7297 & 0.4011 & 9.2992 & 0.0000 & 2.9082 & 4.5513 \\ \hline \end{array} -Referring to Instruction 12-11,what is the standard error of estimate?$ $Instruction 12-11 A computer software developer would like to use the number of downloads (in thousands)for the trial version of his new shareware to predict the amount of revenue (in thousands of dollars)he can make on the full version of the new shareware.Following is the output from a simple linear regression along with the residual plot and normal probability plot obtained from a data set of 30 different sharewares that he has developed: \begin{array}{lr} \hline {\text { Regression Statistics }} \\ \hline \text { Multiple R } & 0.8691 \\ \hline \text { R Square } & 0.7554 \\ \hline \text { Adjusted R Square } & 0.7467 \\ \text { Standard Error } & 44.4765 \\ \text { Observations } & 30.0000 \\ \hline \end{array} ANOVA \begin{array}{lr|r|r|r|r} \hline & d f & {\text { SS }} &{\text { MS }} & F & \text { Significance } F \\ \hline \text { Regression } & 1 & 171062.9193 & 171062.9193 & 86.4759 & 0.0000 \\ \hline \text { Residual } & 28 & 55388.4309 & 1978.1582 & & \\ \hline \text { Total } & 29 & 226451.3503 & & & \\ \hline \end{array} \begin{array}{lrrrrrrr} \hline & \text { Coefficients } & \text { Standard Emor } & t \text { Stat } & \text { P-value } & \text { Lower 95\% } & \text { Upper 95\% } \\ \hline \text { Intercept } & -95.0614 & 26.9183 & -3.5315 & 0.0015 & -150.2009 & -39.9218 \\ \text { Download } & 3.7297 & 0.4011 & 9.2992 & 0.0000 & 2.9082 & 4.5513 \\ \hline \end{array} -Referring to Instruction 12-11,what is the standard error of estimate?$ -Referring to Instruction 12-11,what is the standard error of estimate?

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Question 163

Instruction 12-10 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 9.42 698 10.08 510 6.73 529 7.02 460 6.12 872 9.52 650 7.53 603 7.25 -Referring to Instruction 12-10,what are the degrees of freedom of the t test statistic when testing whether the number of customers who make purchases affects weekly sales?

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Question 164

Instruction 12-7 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. Coefficients Standard Error T Stat P-value Intercept 4.866004258 0.35743609 13.61363441 8.7932-13 \& -0.502513506 0.071597152 -7.01862425 2.94942-07 -Referring to Instruction 12-7,which of the following will be a correct conclusion?

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Question 165

If the residuals in a regression analysis of time ordered data are not correlated,the value of the Durbin-Watson D statistic should be near ________.

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Question 166

Instruction 12-11 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 Multiple R 0.8691 R Square 0.7554 Adjusted R Square 0.7467 Standard Error 44.4765 Observations 30.0000 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 $Instruction 12-11 A computer software developer would like to use the number of downloads (in thousands)for the trial version of his new shareware to predict the amount of revenue (in thousands of dollars)he can make on the full version of the new shareware.Following is the output from a simple linear regression along with the residual plot and normal probability plot obtained from a data set of 30 different sharewares that he has developed: \begin{array}{|lr|} \hline{\text { Regression Statistics }} \\ \hline \text { Multiple R } & 0.8691 \\ \hline \text { R Square } & 0.7554 \\ \hline \text { Adjusted R Square } & 0.7467 \\ \text { Standard Error } & 44.4765 \\ \text { Observations } & 30.0000 \\ \hline \end{array} ANOVA \begin{array}{|l|r|r|r|r|r} & {\text { df }} & {\text { SS }} & {\text { MS }} & F & \text { Significance F } \\ \hline \text { Regression } & 1 & 171062.9193 & 171062.9193 & 86.4759 & 0.0000 \\ \hline \text { Residual } & 28 & 55388.4309 & 1978.1582 & & \\ \hline \text { Total } & 29 & 226451.3503 & & & \end{array} \begin{array}{lrrrrrr} \hline & \text { Coefficients } & \text { Standard Error } & t \text { Stat } & \text { P-value } & {\text { Lower 95\% }} & \text { Upper 95\% } \\ \hline \text { Intercept } & -95.0614 & 26.9183 & -3.5315 & 0.0015 & -150.2009 & -39.9218 \\ \hline \text { Download } & 3.7297 & 0.4011 & 9.2992 & 0.0000 & 2.9082 & 4.5513 \end{array} -Referring to Instruction 12-11,the Durbin-Watson statistic is inappropriate for this data set.$ $Instruction 12-11 A computer software developer would like to use the number of downloads (in thousands)for the trial version of his new shareware to predict the amount of revenue (in thousands of dollars)he can make on the full version of the new shareware.Following is the output from a simple linear regression along with the residual plot and normal probability plot obtained from a data set of 30 different sharewares that he has developed: \begin{array}{|lr|} \hline{\text { Regression Statistics }} \\ \hline \text { Multiple R } & 0.8691 \\ \hline \text { R Square } & 0.7554 \\ \hline \text { Adjusted R Square } & 0.7467 \\ \text { Standard Error } & 44.4765 \\ \text { Observations } & 30.0000 \\ \hline \end{array} ANOVA \begin{array}{|l|r|r|r|r|r} & {\text { df }} & {\text { SS }} & {\text { MS }} & F & \text { Significance F } \\ \hline \text { Regression } & 1 & 171062.9193 & 171062.9193 & 86.4759 & 0.0000 \\ \hline \text { Residual } & 28 & 55388.4309 & 1978.1582 & & \\ \hline \text { Total } & 29 & 226451.3503 & & & \end{array} \begin{array}{lrrrrrr} \hline & \text { Coefficients } & \text { Standard Error } & t \text { Stat } & \text { P-value } & {\text { Lower 95\% }} & \text { Upper 95\% } \\ \hline \text { Intercept } & -95.0614 & 26.9183 & -3.5315 & 0.0015 & -150.2009 & -39.9218 \\ \hline \text { Download } & 3.7297 & 0.4011 & 9.2992 & 0.0000 & 2.9082 & 4.5513 \end{array} -Referring to Instruction 12-11,the Durbin-Watson statistic is inappropriate for this data set.$ -Referring to Instruction 12-11,the Durbin-Watson statistic is inappropriate for this data set.

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Question 167

Instruction 12-3 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 CoopJobs JobOffer 1 1 4 2 2 6 3 1 3 4 0 1 -Referring to Instruction 12-3,the regression sum of squares (SSR)is ________.

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Question 168

Instruction 12-9 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 Microsoft Excel output for predicting starting salary (Y)using number of hours spent studying per day (X)for a sample of 51 students.NOTE: Only partial output is shown. Regression Statistics Multiple R 0.8857 R Square 0.7845 Adjusted R Square 0.7801 Standard Error 1.3704 Observations 51 ANOVA df SS MS F Significance F Regression 1 335.0472 335.0473 178.3859 Residual 1.8782 Total 50 427.0798 Coefficients Standard Error t Stat P-value Lower 95\% Upper 95\% Intercept -1.8940 0.4018 -4.7134 2.051-05 -2.7015 -1.0865 Hours 0.9795 0.0733 13.3561 5.944-18 0.8321 1.1269 Note: 2.051E-05 = 2.051 * 10^-0.5 and 5.944E-18 = 5.944 * 10^-18. -Referring to Instruction 12-9,to test the claim that SALARY depends positively on HOURS against the null hypothesis that SALARY does not depend linearly on HOURS,the p-value of the test statistic is

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Question 169

Instruction 12-11 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 Multiple R 0.8691 R Square 0.7554 Adjusted R Square 0.7467 Standard Error 44.4765 Observations 30.0000 ANOVA df SS MS F Significance F Regression 1 171062.9193 171062.9193 86.4759 0.0000 Residuall 28 55388.4309 1978.1582 Total 29 226451.3503 Coefficients Standard Eror 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 $Instruction 12-11 A computer software developer would like to use the number of downloads (in thousands)for the trial version of his new shareware to predict the amount of revenue (in thousands of dollars)he can make on the full version of the new shareware.Following is the output from a simple linear regression along with the residual plot and normal probability plot obtained from a data set of 30 different sharewares that he has developed: \begin{array}{lr} \hline {\text { Regression Statistics }} \\ \hline \text { Multiple R } & 0.8691 \\ \hline \text { R Square } & 0.7554 \\ \hline \text { Adjusted R Square } & 0.7467 \\ \hline \text { Standard Error } & 44.4765 \\ \hline \text { Observations } & 30.0000 \\ \hline \end{array} ANOVA \begin{array}{|l|r|r|r|r|r|} & d f & \text { SS } & {M S} & F & \text { Significance } F \\ \hline \text { Regression } & 1 & 171062.9193 & 171062.9193 & 86.4759 & 0.0000 \\ \hline \text { Residuall } & 28 & 55388.4309 & 1978.1582 & \\ \hline \text { Total } & 29 & 226451.3503 & & \end{array} \begin{array}{lrrrrrrr} \hline & \text { Coefficients } & \text { Standard Eror } &{t \text { Stat }} & \text { P-value } & \text { Lower 95\% } & \text { Upper 95\% } \\ \hline \text { Intercept } & -95.0614 & 26.9183 & -3.5315 & 0.0015 & -150.2009 & -39.9218 \\ \hline \text { Download } & 3.7297 & 0.4011 & 9.2992 & 0.0000 & 2.9082 & 4.5513 \end{array} -Referring to Instruction 12-11,what is the value of the test statistic for testing whether there is a linear relationship between revenue and the number of downloads?$ $Instruction 12-11 A computer software developer would like to use the number of downloads (in thousands)for the trial version of his new shareware to predict the amount of revenue (in thousands of dollars)he can make on the full version of the new shareware.Following is the output from a simple linear regression along with the residual plot and normal probability plot obtained from a data set of 30 different sharewares that he has developed: \begin{array}{lr} \hline {\text { Regression Statistics }} \\ \hline \text { Multiple R } & 0.8691 \\ \hline \text { R Square } & 0.7554 \\ \hline \text { Adjusted R Square } & 0.7467 \\ \hline \text { Standard Error } & 44.4765 \\ \hline \text { Observations } & 30.0000 \\ \hline \end{array} ANOVA \begin{array}{|l|r|r|r|r|r|} & d f & \text { SS } & {M S} & F & \text { Significance } F \\ \hline \text { Regression } & 1 & 171062.9193 & 171062.9193 & 86.4759 & 0.0000 \\ \hline \text { Residuall } & 28 & 55388.4309 & 1978.1582 & \\ \hline \text { Total } & 29 & 226451.3503 & & \end{array} \begin{array}{lrrrrrrr} \hline & \text { Coefficients } & \text { Standard Eror } &{t \text { Stat }} & \text { P-value } & \text { Lower 95\% } & \text { Upper 95\% } \\ \hline \text { Intercept } & -95.0614 & 26.9183 & -3.5315 & 0.0015 & -150.2009 & -39.9218 \\ \hline \text { Download } & 3.7297 & 0.4011 & 9.2992 & 0.0000 & 2.9082 & 4.5513 \end{array} -Referring to Instruction 12-11,what is the value of the test statistic for testing whether there is a linear relationship between revenue and the number of downloads?$ -Referring to Instruction 12-11,what is the value of the test statistic for testing whether there is a linear relationship between revenue and the number of downloads?

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Question 170

Instruction 12-11 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 Multiple R 0.8691 R Square 0.7554 Adjusted R Square 0.7467 Standard Error 44.4765 Observations 30.0000 ANOVA df SS MS F Significance F Regression 1 171062.9193 171062.9193 86.4759 0.0000 Residuall 28 55388.4309 1978.1582 Total 29 226451.3503 Coefficients Standard Eror 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 $Instruction 12-11 A computer software developer would like to use the number of downloads (in thousands)for the trial version of his new shareware to predict the amount of revenue (in thousands of dollars)he can make on the full version of the new shareware.Following is the output from a simple linear regression along with the residual plot and normal probability plot obtained from a data set of 30 different sharewares that he has developed: \begin{array}{lr} \hline {\text { Regression Statistics }} \\ \hline \text { Multiple R } & 0.8691 \\ \hline \text { R Square } & 0.7554 \\ \hline \text { Adjusted R Square } & 0.7467 \\ \hline \text { Standard Error } & 44.4765 \\ \hline \text { Observations } & 30.0000 \\ \hline \end{array} ANOVA \begin{array}{|l|r|r|r|r|r|} & d f & \text { SS } & {M S} & F & \text { Significance } F \\ \hline \text { Regression } & 1 & 171062.9193 & 171062.9193 & 86.4759 & 0.0000 \\ \hline \text { Residuall } & 28 & 55388.4309 & 1978.1582 & \\ \hline \text { Total } & 29 & 226451.3503 & & \end{array} \begin{array}{lrrrrrrr} \hline & \text { Coefficients } & \text { Standard Eror } &{t \text { Stat }} & \text { P-value } & \text { Lower 95\% } & \text { Upper 95\% } \\ \hline \text { Intercept } & -95.0614 & 26.9183 & -3.5315 & 0.0015 & -150.2009 & -39.9218 \\ \hline \text { Download } & 3.7297 & 0.4011 & 9.2992 & 0.0000 & 2.9082 & 4.5513 \end{array} -Referring to Instruction 12-11,the null hypothesis for testing whether there is a linear relationship between revenue and number of dowloads is There is no linear relationship between revenue and number of downloads.$ $Instruction 12-11 A computer software developer would like to use the number of downloads (in thousands)for the trial version of his new shareware to predict the amount of revenue (in thousands of dollars)he can make on the full version of the new shareware.Following is the output from a simple linear regression along with the residual plot and normal probability plot obtained from a data set of 30 different sharewares that he has developed: \begin{array}{lr} \hline {\text { Regression Statistics }} \\ \hline \text { Multiple R } & 0.8691 \\ \hline \text { R Square } & 0.7554 \\ \hline \text { Adjusted R Square } & 0.7467 \\ \hline \text { Standard Error } & 44.4765 \\ \hline \text { Observations } & 30.0000 \\ \hline \end{array} ANOVA \begin{array}{|l|r|r|r|r|r|} & d f & \text { SS } & {M S} & F & \text { Significance } F \\ \hline \text { Regression } & 1 & 171062.9193 & 171062.9193 & 86.4759 & 0.0000 \\ \hline \text { Residuall } & 28 & 55388.4309 & 1978.1582 & \\ \hline \text { Total } & 29 & 226451.3503 & & \end{array} \begin{array}{lrrrrrrr} \hline & \text { Coefficients } & \text { Standard Eror } &{t \text { Stat }} & \text { P-value } & \text { Lower 95\% } & \text { Upper 95\% } \\ \hline \text { Intercept } & -95.0614 & 26.9183 & -3.5315 & 0.0015 & -150.2009 & -39.9218 \\ \hline \text { Download } & 3.7297 & 0.4011 & 9.2992 & 0.0000 & 2.9082 & 4.5513 \end{array} -Referring to Instruction 12-11,the null hypothesis for testing whether there is a linear relationship between revenue and number of dowloads is There is no linear relationship between revenue and number of downloads.$ -Referring to Instruction 12-11,the null hypothesis for testing whether there is a linear relationship between revenue and number of dowloads is "There is no linear relationship between revenue and number of downloads".

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Question 171

Regression analysis is used for prediction,while correlation analysis is used to measure the strength of the association between two numerical variables.

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Question 172

The Regression Sum of Squares (SSR)can never be greater than the Total Sum of Squares (SST).

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Question 173

Instruction 12-10 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 9.42 698 10.08 510 6.73 529 7.02 460 6.12 872 9.52 650 7.53 603 7.25 -Referring to Instruction 12-10,what is the value of the t test statistic when testing whether the number of customers who make purchases affects weekly sales?

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Question 174

Assuming a linear relationship between X and Y,if the coefficient of correlation (r)equals -0.30,

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Question 175

Instruction 12-3 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 CoopJobs JobOffer 1 1 4 2 2 6 3 1 3 4 0 1 -Referring to Instruction 12-3,the director of cooperative education wanted to test the hypothesis that the true slope was equal to 3.0.The value of the test statistic is ________.

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Question 176

Instruction 12-4 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 12-4,the least squares estimate of the Y-intercept is ________.

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Question 177

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: ρ = 0 against the two-sided alternative H₁: ρ ≠ 0.To test H₀: ρ = 0 against the one-sided alternative H₁: ρ < 0 at a significance level of 0.2,the p-value is

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Question 178

Instruction 12-12 The manager of the purchasing department of a large savings and loan organization would like to develop a model to predict the amount of time (measured in hours)it takes to record a loan application.Data are collected from a sample of 30 days,and the number of applications recorded and completion time in hours is recorded.Below is the regression output: Regression Statistics Multiple R 0.9447 R Square 0.8924 Adjusted R 0.8886 Square Standard 0.3342 Error 30 Observations ANOVA df SS 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 t Stat P -value Lower 95\% Upper 95\% Intercept 0.4024 0.1236 3.2559 0.0030 0.1492 0.6555 Applications RECORD 0.0126 0.0008 15.2388 4.3946-15 0.0109 0.0143 Note: 4.3946E-15 is 4.3946 x 10^-15. $Instruction 12-12 The manager of the purchasing department of a large savings and loan organization would like to develop a model to predict the amount of time (measured in hours)it takes to record a loan application.Data are collected from a sample of 30 days,and the number of applications recorded and completion time in hours is recorded.Below is the regression output: \begin{array}{l} \begin{array} { l r } \hline { \text { Regression Statistics } } \\ \hline \text { Multiple R } & 0.9447 \\ \text { R Square } & 0.8924 \\ \text { Adjusted R } & 0.8886 \\ \text { Square } & \\ \text { Standard } & 0.3342 \\ \text { Error } & 30 \\ \text { Observations } & \\ \hline \end{array}\\ \text { ANOVA }\\ \begin{array} { l r r r r r } \hline & d f & { \text { SS } } & { \text { MS } } & F & { \begin{array} { c } \text { Significance } \\ F \end{array} } \\ \hline \text { Regression } & 1 & 25.9438 & 25.9438 & 232.2200 & 4.3946 \mathrm { E } - 15 \\ \text { Residual } & 28 & 3.1282 & 0.1117 & & \\ \text { Total } & 29 & 29.072 & & & \\ \hline \end{array}\\ \begin{array} { l r r r r r r } \hline & \text { Coefficients } & \begin{array} { c } \text { Standard } \\ \text { Error } \end{array} & t \text { Stat } & P \text {-value } & \text { Lower 95\% } & \text { Upper 95\% } \\ \text { Intercept } & 0.4024 & 0.1236 & 3.2559 & 0.0030 & 0.1492 & 0.6555 \\ \text { Applications RECORD } & 0.0126 & 0.0008 & 15.2388 & 4.3946 \mathrm{E}-15 & 0.0109 & 0.0143\\ \hline \end{array} \end{array} Note: 4.3946E-15 is 4.3946 x 10<sup>-15</sup>. -Referring to Instruction 12-12,the value of the measured t test statistic to test whether the amount of time depends linearly on the number of loan applications recorded is$ $Instruction 12-12 The manager of the purchasing department of a large savings and loan organization would like to develop a model to predict the amount of time (measured in hours)it takes to record a loan application.Data are collected from a sample of 30 days,and the number of applications recorded and completion time in hours is recorded.Below is the regression output: \begin{array}{l} \begin{array} { l r } \hline { \text { Regression Statistics } } \\ \hline \text { Multiple R } & 0.9447 \\ \text { R Square } & 0.8924 \\ \text { Adjusted R } & 0.8886 \\ \text { Square } & \\ \text { Standard } & 0.3342 \\ \text { Error } & 30 \\ \text { Observations } & \\ \hline \end{array}\\ \text { ANOVA }\\ \begin{array} { l r r r r r } \hline & d f & { \text { SS } } & { \text { MS } } & F & { \begin{array} { c } \text { Significance } \\ F \end{array} } \\ \hline \text { Regression } & 1 & 25.9438 & 25.9438 & 232.2200 & 4.3946 \mathrm { E } - 15 \\ \text { Residual } & 28 & 3.1282 & 0.1117 & & \\ \text { Total } & 29 & 29.072 & & & \\ \hline \end{array}\\ \begin{array} { l r r r r r r } \hline & \text { Coefficients } & \begin{array} { c } \text { Standard } \\ \text { Error } \end{array} & t \text { Stat } & P \text {-value } & \text { Lower 95\% } & \text { Upper 95\% } \\ \text { Intercept } & 0.4024 & 0.1236 & 3.2559 & 0.0030 & 0.1492 & 0.6555 \\ \text { Applications RECORD } & 0.0126 & 0.0008 & 15.2388 & 4.3946 \mathrm{E}-15 & 0.0109 & 0.0143\\ \hline \end{array} \end{array} Note: 4.3946E-15 is 4.3946 x 10<sup>-15</sup>. -Referring to Instruction 12-12,the value of the measured t test statistic to test whether the amount of time depends linearly on the number of loan applications recorded is$ -Referring to Instruction 12-12,the value of the measured t test statistic to test whether the amount of time depends linearly on the number of loan applications recorded is

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Question 179

Instruction 12-11 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 Multiple R 0.8691 R Square 0.7554 Adjusted R Square 0.7467 Standard Error 44.4765 Observations 30.0000 ANOVA df SS MS F Significance F Regression 1 171062.9193 171062.9193 86.4759 0.0000 Residuall 28 55388.4309 1978.1582 Total 29 226451.3503 Coefficients Standard Eror 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 $Instruction 12-11 A computer software developer would like to use the number of downloads (in thousands)for the trial version of his new shareware to predict the amount of revenue (in thousands of dollars)he can make on the full version of the new shareware.Following is the output from a simple linear regression along with the residual plot and normal probability plot obtained from a data set of 30 different sharewares that he has developed: \begin{array}{lr} \hline {\text { Regression Statistics }} \\ \hline \text { Multiple R } & 0.8691 \\ \hline \text { R Square } & 0.7554 \\ \hline \text { Adjusted R Square } & 0.7467 \\ \hline \text { Standard Error } & 44.4765 \\ \hline \text { Observations } & 30.0000 \\ \hline \end{array} ANOVA \begin{array}{|l|r|r|r|r|r|} & d f & \text { SS } & {M S} & F & \text { Significance } F \\ \hline \text { Regression } & 1 & 171062.9193 & 171062.9193 & 86.4759 & 0.0000 \\ \hline \text { Residuall } & 28 & 55388.4309 & 1978.1582 & \\ \hline \text { Total } & 29 & 226451.3503 & & \end{array} \begin{array}{lrrrrrrr} \hline & \text { Coefficients } & \text { Standard Eror } &{t \text { Stat }} & \text { P-value } & \text { Lower 95\% } & \text { Upper 95\% } \\ \hline \text { Intercept } & -95.0614 & 26.9183 & -3.5315 & 0.0015 & -150.2009 & -39.9218 \\ \hline \text { Download } & 3.7297 & 0.4011 & 9.2992 & 0.0000 & 2.9082 & 4.5513 \end{array} -Referring to Instruction 12-11,there is sufficient evidence that revenue and number of downloads are linearly related at a 5% level of significance.$ $Instruction 12-11 A computer software developer would like to use the number of downloads (in thousands)for the trial version of his new shareware to predict the amount of revenue (in thousands of dollars)he can make on the full version of the new shareware.Following is the output from a simple linear regression along with the residual plot and normal probability plot obtained from a data set of 30 different sharewares that he has developed: \begin{array}{lr} \hline {\text { Regression Statistics }} \\ \hline \text { Multiple R } & 0.8691 \\ \hline \text { R Square } & 0.7554 \\ \hline \text { Adjusted R Square } & 0.7467 \\ \hline \text { Standard Error } & 44.4765 \\ \hline \text { Observations } & 30.0000 \\ \hline \end{array} ANOVA \begin{array}{|l|r|r|r|r|r|} & d f & \text { SS } & {M S} & F & \text { Significance } F \\ \hline \text { Regression } & 1 & 171062.9193 & 171062.9193 & 86.4759 & 0.0000 \\ \hline \text { Residuall } & 28 & 55388.4309 & 1978.1582 & \\ \hline \text { Total } & 29 & 226451.3503 & & \end{array} \begin{array}{lrrrrrrr} \hline & \text { Coefficients } & \text { Standard Eror } &{t \text { Stat }} & \text { P-value } & \text { Lower 95\% } & \text { Upper 95\% } \\ \hline \text { Intercept } & -95.0614 & 26.9183 & -3.5315 & 0.0015 & -150.2009 & -39.9218 \\ \hline \text { Download } & 3.7297 & 0.4011 & 9.2992 & 0.0000 & 2.9082 & 4.5513 \end{array} -Referring to Instruction 12-11,there is sufficient evidence that revenue and number of downloads are linearly related at a 5% level of significance.$ -Referring to Instruction 12-11,there is sufficient evidence that revenue and number of downloads are linearly related at a 5% level of significance.

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Question 180

Showing 161 - 180 of 196

If the residuals in a regression analysis of time ordered data are not correlated,the value of the Durbin-Watson D statistic should be near ________.

Regression analysis is used for prediction,while correlation analysis is used to measure the strength of the association between two numerical variables.

The Regression Sum of Squares (SSR)can never be greater than the Total Sum of Squares (SST).

Assuming a linear relationship between X and Y,if the coefficient of correlation (r)equals -0.30,

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: ρ = 0 against the two-sided alternative H₁: ρ ≠ 0.To test H₀: ρ = 0 against the one-sided alternative H₁: ρ < 0 at a significance level of 0.2,the p-value is

Introduction and Data Collection

Presenting Data in Tables and Charts

Numerical Descriptive Measures

Basic Probability

Some Important Discrete Probability Distributions

The Normal Distribution and Other Continuous Distributions

Sampling Distributions

Confidence Interval Estimation

Fundamentals of Hypothesis Testing: One-Sample Tests

Hypothesis Testing: Two Sample Tests

Analysis of Variance

Introduction to Multiple Regression

Time-Series Forecasting and Index Numbers

Chi-Square Tests

Multiple Regression Model Building

Decision Making

Statistical Applications in Quality and Productivity Management

Further Non-Parametric Tests

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