Exam 12: Simple Linear Regression

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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 percentage of the total variation in chocolate bar sales explained by the regression model?

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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 managers of the brokerage firm wanted to test the hypothesis that the true slope was equal to 0.At a level of significance of 0.01,the null hypothesis should be ________ (rejected or not rejected).

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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 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,the standard error of the estimated slope coefficient is ________.

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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 prediction for the number of job offers for a person with two cooperative jobs is ________.

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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 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.At a level of significance of 0.01,the decision that should be made implies that the number of new customers brought in ________ (had or did not have)a positive impact on the amount of sales generated.

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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 90% confidence interval for the mean change in the amount of time needed as a result of recording one additional loan application 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 90% confidence interval for the mean change in the amount of time needed as a result of recording one additional loan application is -Referring to Instruction 12-12,the 90% confidence interval for the mean change in the amount of time needed as a result of recording one additional loan application is

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A simple regression has a b0 value of 5 and a b1 value of 3.5.What is the predicted Y for an X value of -2?

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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 managers of the brokerage firm wanted to test the hypothesis that the true slope was equal to 0.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 ________.

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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,construct a 95% prediction interval for the weekly sales of a store that has 600 purchasing customers.

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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,to test whether the prison stocks portfolio is negatively related to the All Ordinaries index,the appropriate null and alternative hypotheses are,________ respectively,

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Instruction 12-8 It is believed that average grade (based on a four -point scale)should have a positive linear relationship with university entrance exam scores.Given below is the Microsoft Excel output from regressing average grade on university entrance exam scores using a data set of eight randomly chosen students from a large university. Regressing average grade on university entrance exam score }\\ Regression Statistics Multiple R 0.7598 R Square 0.5774 Adjusted R Square 0.5069 Standard Error 0.2691 Observations 8 ANOVA df SS MS F Significance F Regression 1 0.5940 0.5940 8.1986 0.0286 Residual 6 0.4347 0.0724 Total 7 1.0287 Coefficients Standard Error t Stat P-value Lower 95\% Upper 95\% Intercept 0.5681 0.9284 0.6119 0.5630 -1.7036 2.8398 University entrance 0.1895 exam score 0.1021 0.0356 2.8633 0.0286 0.0148 0.1895 -Referring to Instruction 12-8,the interpretation of the coefficient of determination in this regression is

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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 p-value of the F test statistic when testing whether the number of customers who make purchases is a good predictor for weekly sales?

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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 If 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 -Referring to Instruction 12-5,the value of the quantity that the least squares regression line minimises is ________.

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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,generate the scatter plot.

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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 significanceF 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}{l|r} \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|} \hline &df&SS&MS&F&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 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 \\ \text { Download } & 3.7297 & 0.4011 & 9.2992 & 0.0000 & 2.9082 & 4.5513 \\ \hline \end{array}      -Referring to Instruction 12-11,which of the following is the correct interpretation for the slope coefficient?  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}{l|r} \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|} \hline &df&SS&MS&F&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 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 \\ \text { Download } & 3.7297 & 0.4011 & 9.2992 & 0.0000 & 2.9082 & 4.5513 \\ \hline \end{array}      -Referring to Instruction 12-11,which of the following is the correct interpretation for the slope coefficient? -Referring to Instruction 12-11,which of the following is the correct interpretation for the slope coefficient?

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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 are the lower and upper limits of the 95% confidence interval estimate for population slope?  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 are the lower and upper limits of the 95% confidence interval estimate for population slope? -Referring to Instruction 12-11,what are the lower and upper limits of the 95% confidence interval estimate for population slope?

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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 total sum of squares (SST)is ________.

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When using a regression model to make predictions,the term "relevant range" refers to ________.

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Instruction 12-8 It is believed that average grade (based on a four -point scale)should have a positive linear relationship with university entrance exam scores.Given below is the Microsoft Excel output from regressing average grade on university entrance exam scores using a data set of eight randomly chosen students from a large university. Regressing average grade on university entrance exam score Regression Statistics Multiple R 0.7598 R Square 0.5774 Adjusted R Square 0.5069 Standard Error 0.2691 Observations 8 ANOVA df SS MS F Significance F Regression 1 0.5940 0.5940 8.1986 0.0286 Residual 6 0.4347 0.0724 Total 7 1.0287 Coefficients Standard Error t Stat P-value Lower 95\% Upper 95\% Intercept 0.5681 0.9284 0.6119 0.5630 -1.7036 2.8398 University entrance exam score 0.1021 0.0356 2.8633 0.0286 0.0148 0.1895 -Referring to Instruction 12-8,the value of the measured (observed)test statistic of the F test for H0: ?1 = 0 versus H1: ?1 ? 0

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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,to test the claim that the mean amount of time depends positively on the number of loan applications recorded against the null hypothesis that the mean amount of time does not depend linearly on the number of invoices processed,the p-value of the test statistic 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,to test the claim that the mean amount of time depends positively on the number of loan applications recorded against the null hypothesis that the mean amount of time does not depend linearly on the number of invoices processed,the p-value of the test statistic is -Referring to Instruction 12-12,to test the claim that the mean amount of time depends positively on the number of loan applications recorded against the null hypothesis that the mean amount of time does not depend linearly on the number of invoices processed,the p-value of the test statistic is

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