Deck 12: Simple Linear Regression

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Question
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 standard error of the regression slope estimate, Sb1S _ { b _ { 1 } } ?

A) 0.885
B) 12.650
C) 16.299
D) 0.784
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Question
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,a 95% confidence interval for ?1 is (15,30).Interpret the interval.

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

A) There is sufficient evidence (at the ? = 0.05) to conclude that sales revenue (X) is a useful linear predictor of service charge (Y).
B) Sales revenue (X) is a poor predictor of service charge (Y).
C) There is insufficient evidence (at the ? = 0.10) to conclude that sales revenue (X) is a useful linear predictor of service charge (Y).
D) For every $1 million increase in sales revenue, we expect a service charge to increase $0.034.
Question
Instruction 12.2
A chocolate bar manufacturer is interested in trying to estimate how sales are influenced by the price of their product. To do this, the company randomly chooses six country towns and cities and offers the chocolate bar at different prices. Using chocolate bar sales as the dependent variable, the company will conduct a simple linear regression on the data below:
 City  Price($)  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,to test whether a change in price will have any impact on average sales,what would be the critical values? Use ? = 0.05.

A) ± 2.7765
B) ± 3.1634
C) ± 2.5706
D) ± 3.4954
Question
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 coefficient of correlation for these data?

A) 0.8854
B) 0.7839
C) -0.7839
D) -0.8854
Question
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,if the price of the chocolate bar is set at $2,the predicted sales will be

A) 100.
B) 30.
C) 65.
D) 90.
Question
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  Coop jobs  job Offer 114226313401\begin{array} { | l | l | l | } \hline \text { Student } & \text { Coop jobs } & \text { job Offer } \\\hline 1 & 1 & 4 \\\hline 2 & 2 & 6 \\\hline 3 & 1 & 3 \\\hline 4 & 0 & 1 \\\hline\end{array}

-Referring to Instruction 12.3,the prediction for the number of job offers for a person with two cooperative jobs is____________.
Question
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  Coop jobs  job Offer 114226313401\begin{array} { | l | l | l | } \hline \text { Student } & \text { Coop jobs } & \text { job Offer } \\\hline 1 & 1 & 4 \\\hline 2 & 2 & 6 \\\hline 3 & 1 & 3 \\\hline 4 & 0 & 1 \\\hline\end{array}

-Referring to Instruction 12.3,the total sum of squares (SST)is ____________.
Question
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 estimate of ?0,the Y-intercept of the line.

A) About 95% of the observed service charges fall within $2,700 of the least squares line.
B) There is no practical interpretation since a sales revenue of $0 is a nonsensical value.
C) For every $1 million increase in sales revenue, we expect a service charge to decrease $2,700.
D) All companies will be charged at least $2,700 by the bank.
Question
Instruction 12.2
A chocolate bar manufacturer is interested in trying to estimate how sales are influenced by the price of their product. To do this, the company randomly chooses six country towns and cities and offers the chocolate bar at different prices. Using chocolate bar sales as the dependent variable, the company will conduct a simple linear regression on the data below:
 City  Price($)  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 standard error of the estimate,SYX,for the data?

A) 0.885
B) 0.784
C) 16.299
D) 12.650
Question
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 percentage of the total variation in chocolate bar sales is explained by prices?

A) 78.39%
B) 88.54%
C) 48.19%
D) 100%
Question
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 ?(X * Xˉ\bar { X } )2 for these data?

A) 25.66
B) 0
C) 2.54
D) 1.66
Question
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 percentage of the total variation in chocolate bar sales explained by the regression model?

A) 48.19%
B) 100%
C) 88.54%
D) 78.39%
Question
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  Coop jobs  job Offer 114226313401\begin{array} { | l | l | l | } \hline \text { Student } & \text { Coop jobs } & \text { job Offer } \\\hline 1 & 1 & 4 \\\hline 2 & 2 & 6 \\\hline 3 & 1 & 3 \\\hline 4 & 0 & 1 \\\hline\end{array}

-Referring to Instruction 12.3,the least squares estimate of the Y-intercept is____________.
Question
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?

A) 0.784
B) -3.810
C) 161.386
D) -48.193
Question
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,to test that the regression coefficient,?1 is not equal to 0,what would be the critical values? Use ? = 0.05.

A) ± 2.7765
B) ± 3.1634
C) ± 2.5706
D) ± 3.4954
Question
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  Coop jobs  job Offer 114226313401\begin{array} { | l | l | l | } \hline \text { Student } & \text { Coop jobs } & \text { job Offer } \\\hline 1 & 1 & 4 \\\hline 2 & 2 & 6 \\\hline 3 & 1 & 3 \\\hline 4 & 0 & 1 \\\hline\end{array}

-Referring to Instruction 12.3,the least squares estimate of the slope is_____________.
Question
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 slope parameter for the chocolate bar price and sales data?

A) -48.193
B) -3.810
C) 161.386
D) 0.784
Question
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 estimate of ?,the standard deviation of the random error term (standard error of the estimate)in the model.

A) About 95% of the observed service charges fall within $130 of the least squares line.
B) About 95% of the observed service charges fall within $65 of the least squares line.
C) About 95% of the observed service charges equal their corresponding predicted values.
D) For every $1 million increase in sales revenue, we expect a service charge to increase $65.
Question
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  Coop jobs  job Offer 114226313401\begin{array} { | l | l | l | } \hline \text { Student } & \text { Coop jobs } & \text { job Offer } \\\hline 1 & 1 & 4 \\\hline 2 & 2 & 6 \\\hline 3 & 1 & 3 \\\hline 4 & 0 & 1 \\\hline\end{array}

-Referring to Instruction 12.3,the regression sum of squares (SSR)is____________.
Question
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 127522113734264433555152961534725588365992844103048111731122238\begin{array} { | l | l | l | } \hline \text { Broker } & \text { Clients } & \text { Sales } \\\hline 1 & 27 & 52 \\\hline 2 & 11 & 37 \\\hline 3 & 42 & 64 \\\hline 4 & 33 & 55 \\\hline 5 & 15 & 29 \\\hline 6 & 15 & 34 \\\hline 7 & 25 & 58 \\\hline 8 & 36 & 59 \\\hline 9 & 28 & 44 \\\hline 10 & 30 & 48 \\\hline 11 & 17 & 31 \\\hline 12 & 22 & 38 \\\hline\end{array}

-Referring to Instruction 12.4,the regression sum of squares (SSR)is____________
Question
Instruction 12.9
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) 90711.2092611.057138.217419.217809.4289810.085106.735297.024606.128729.526507.536037.25\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 898 & 10.08 \\\hline 510 & 6.73 \\\hline 529 & 7.02 \\\hline 460 & 6.12 \\\hline 872 & 9.52 \\\hline 650 & 7.53 \\\hline 603 & 7.25 \\\hline\end{array}

-Referring to Instruction 12.9,93.98% of the total variation in weekly sales can be explained by the variation in the number of customers who make purchases.
Question
Instruction 12.8
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 \text { Regression statistics }
 MultipleR 0.8857 R Square 0.7845 Adjusted R  Square 0.7801 Standard Error 1.3704 Observations 51\begin{array}{|l|l|}\hline\text { MultipleR } & 0.8857 \\\hline \text { R Square } & 0.7845 \\\hline \begin{array}{l}\text { Adjusted R } \\\text { Square }\end{array} & 0.7801 \\\hline \text { Standard Error } & 1.3704 \\\hline \text { Observations } & 51\\\hline\end{array}

 ANOVA df55 MS F Significance F Regression 1335.0472335.0473178.3859 Residual 1.8782 Total 50427.0798\begin{array}{l}\text { ANOVA }\\\begin{array}{|l|l|l|l|l|l|l|}\hline & d f & 55 & \text { MS } & F & \begin{array}{l}\text { Significance } \\F\end{array} & \\\hline \text { Regression } & 1 & 335.0472 & 335.0473 & 178.3859 & & \\\hline \text { Residual } & & & 1.8782 & & & \\\hline \text { Total } & 50 & 427.0798 & & & & \\\hline\end{array}\end{array}

 Coefficients  Standard  Error  tStat  p-value  Lower 95%  Upper 95%  Intercept 1.89400.40184.71342.051E052.70151.0865 Hours 0.97950.073313.35615.944E180.83211.1269\begin{array}{|l|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 } & -1.8940 & 0.4018 & -4.7134 & 2.051 \mathrm{E}-05 & -2.7015 & -1.0865 \\\hline \text { Hours } & 0.9795 & 0.0733 & 13.3561 & 5.944 \mathrm{E}-18 & 0.8321 & 1.1269 \\\hline\end{array} Note: 2.051E-05 = 2.051 * 10S1-0.5 and 5.944E-18 = 5.944 * 10S1-18.

-Referring to Instruction 12.8,the estimated average change in salary (in thousands of dollars)as a result of spending an extra hour per day studying is

A) 0.9795.
B) 0.7845.
C) 335.0473.
D) -1.8940.
Question
Instruction 12.9
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) 90711.2092611.057138.217419.217809.4289810.085106.735297.024606.128729.526507.536037.25\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 898 & 10.08 \\\hline 510 & 6.73 \\\hline 529 & 7.02 \\\hline 460 & 6.12 \\\hline 872 & 9.52 \\\hline 650 & 7.53 \\\hline 603 & 7.25 \\\hline\end{array}

-Referring to Instruction 12.9,generate the scatter plot.
Question
Instruction 12.9
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) 90711.2092611.057138.217419.217809.4289810.085106.735297.024606.128729.526507.536037.25\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 898 & 10.08 \\\hline 510 & 6.73 \\\hline 529 & 7.02 \\\hline 460 & 6.12 \\\hline 872 & 9.52 \\\hline 650 & 7.53 \\\hline 603 & 7.25 \\\hline\end{array}

-Referring to Instruction 12.9,what are the values of the estimated intercept and slope?
Question
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 127522113734264433555152961534725588365992844103048111731122238\begin{array} { | l | l | l | } \hline \text { Broker } & \text { Clients } & \text { Sales } \\\hline 1 & 27 & 52 \\\hline 2 & 11 & 37 \\\hline 3 & 42 & 64 \\\hline 4 & 33 & 55 \\\hline 5 & 15 & 29 \\\hline 6 & 15 & 34 \\\hline 7 & 25 & 58 \\\hline 8 & 36 & 59 \\\hline 9 & 28 & 44 \\\hline 10 & 30 & 48 \\\hline 11 & 17 & 31 \\\hline 12 & 22 & 38 \\\hline\end{array}

-Referring to Instruction 12.4,the total sum of squares (SST)is ____________.
Question
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 127522113734264433555152961534725588365992844103048111731122238\begin{array} { | l | l | l | } \hline \text { Broker } & \text { Clients } & \text { Sales } \\\hline 1 & 27 & 52 \\\hline 2 & 11 & 37 \\\hline 3 & 42 & 64 \\\hline 4 & 33 & 55 \\\hline 5 & 15 & 29 \\\hline 6 & 15 & 34 \\\hline 7 & 25 & 58 \\\hline 8 & 36 & 59 \\\hline 9 & 28 & 44 \\\hline 10 & 30 & 48 \\\hline 11 & 17 & 31 \\\hline 12 & 22 & 38 \\\hline\end{array}

-Referring to Instruction 12.4,the prediction for the amount of sales (in $1,000s)for a person who brings 25 new customers into the firm is ____________.
Question
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:
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<sup> </sup> Excel's Data Analysis tool to analyse the last four years of quarterly data with the following results:    -Referring to Instruction 12.5,the prediction for a quarter in which X = 120 is Y = ____________.<div style=padding-top: 35px>

-Referring to Instruction 12.5,the prediction for a quarter in which X = 120 is Y = ____________.
Question
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  Coop jobs  job Offer 114226313401\begin{array} { | l | l | l | } \hline \text { Student } & \text { Coop jobs } & \text { job Offer } \\\hline 1 & 1 & 4 \\\hline 2 & 2 & 6 \\\hline 3 & 1 & 3 \\\hline 4 & 0 & 1 \\\hline\end{array}

-Referring to Instruction 12.3,set up a scatter diagram.
Question
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  Coop jobs  job Offer 114226313401\begin{array} { | l | l | l | } \hline \text { Student } & \text { Coop jobs } & \text { job Offer } \\\hline 1 & 1 & 4 \\\hline 2 & 2 & 6 \\\hline 3 & 1 & 3 \\\hline 4 & 0 & 1 \\\hline\end{array}

-Referring to Instruction 12.3,the coefficient of determination is____________.
Question
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 127522113734264433555152961534725588365992844103048111731122238\begin{array} { | l | l | l | } \hline \text { Broker } & \text { Clients } & \text { Sales } \\\hline 1 & 27 & 52 \\\hline 2 & 11 & 37 \\\hline 3 & 42 & 64 \\\hline 4 & 33 & 55 \\\hline 5 & 15 & 29 \\\hline 6 & 15 & 34 \\\hline 7 & 25 & 58 \\\hline 8 & 36 & 59 \\\hline 9 & 28 & 44 \\\hline 10 & 30 & 48 \\\hline 11 & 17 & 31 \\\hline 12 & 22 & 38 \\\hline\end{array}

-Referring to Instruction 12.4,the least squares estimate of the Y-intercept is ____________.
Question
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 127522113734264433555152961534725588365992844103048111731122238\begin{array} { | l | l | l | } \hline \text { Broker } & \text { Clients } & \text { Sales } \\\hline 1 & 27 & 52 \\\hline 2 & 11 & 37 \\\hline 3 & 42 & 64 \\\hline 4 & 33 & 55 \\\hline 5 & 15 & 29 \\\hline 6 & 15 & 34 \\\hline 7 & 25 & 58 \\\hline 8 & 36 & 59 \\\hline 9 & 28 & 44 \\\hline 10 & 30 & 48 \\\hline 11 & 17 & 31 \\\hline 12 & 22 & 38 \\\hline\end{array}

-Referring to Instruction 12.4,the error or residual sum of squares (SSE)is____________.
Question
Instruction 12.7
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.
<strong>Instruction 12.7 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<sup> </sup> Excel output from regressing average grade on university entrance exam scores using a data set of eight randomly chosen students from a large university.    -Referring to Instruction 12.7,the interpretation of the coefficient of determination in this regression is</strong> A) average grade accounts for 57.74% of the variability of university entrance exam scores. B) university entrance exam scores account for 57.74% of the total fluctuation in average grade. C) 57.74% of the total variation of university entrance exam scores can be explained by average grade. D) None of the above. <div style=padding-top: 35px>

-Referring to Instruction 12.7,the interpretation of the coefficient of determination in this regression is

A) average grade accounts for 57.74% of the variability of university entrance exam scores.
B) university entrance exam scores account for 57.74% of the total fluctuation in average grade.
C) 57.74% of the total variation of university entrance exam scores can be explained by average grade.
D) None of the above.
Question
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 127522113734264433555152961534725588365992844103048111731122238\begin{array} { | l | l | l | } \hline \text { Broker } & \text { Clients } & \text { Sales } \\\hline 1 & 27 & 52 \\\hline 2 & 11 & 37 \\\hline 3 & 42 & 64 \\\hline 4 & 33 & 55 \\\hline 5 & 15 & 29 \\\hline 6 & 15 & 34 \\\hline 7 & 25 & 58 \\\hline 8 & 36 & 59 \\\hline 9 & 28 & 44 \\\hline 10 & 30 & 48 \\\hline 11 & 17 & 31 \\\hline 12 & 22 & 38 \\\hline\end{array}

-Referring to Instruction 12.4,set up a scatter diagram.
Question
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 127522113734264433555152961534725588365992844103048111731122238\begin{array} { | l | l | l | } \hline \text { Broker } & \text { Clients } & \text { Sales } \\\hline 1 & 27 & 52 \\\hline 2 & 11 & 37 \\\hline 3 & 42 & 64 \\\hline 4 & 33 & 55 \\\hline 5 & 15 & 29 \\\hline 6 & 15 & 34 \\\hline 7 & 25 & 58 \\\hline 8 & 36 & 59 \\\hline 9 & 28 & 44 \\\hline 10 & 30 & 48 \\\hline 11 & 17 & 31 \\\hline 12 & 22 & 38 \\\hline\end{array}

-Referring to Instruction 12.4,____% of the total variation in sales generated can be explained by the number of new customers brought in.
Question
Instruction 12.6
The following Microsoft Excel tables are obtained when 'Score received on an exam (measured in percentage points)' (Y) is regressed on 'percentage attendance' (X) for 22 students in a Statistics for Business and Economics course.
 Regression Statistics \text { Regression Statistics }
 MultipleR 0.142620229 R Square 0.02034053 Adjusted R  Square 0.028642444 Standard Error 20.25979924 Observations 22\begin{array}{|l|l|}\hline \text { MultipleR } & 0.142620229 \\\hline \text { R Square } & 0.02034053 \\\hline \begin{array}{l}\text { Adjusted R } \\\text { Square }\end{array} & -0.028642444 \\\hline \text { Standard Error } & 20.25979924 \\\hline \text { Observations } & 22 \\\hline\end{array}

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

-Referring to Instruction 12.6,which of the following statements is true?

A) If attendance increases by 1%, the estimated mean score received will increase by 39.39 percentage points.
B) If attendance increases by 0.341%, the estimated mean score received will increase by 1 percentage point.
C) If the score received increases by 39.39%, the estimated mean attendance will go up by 1%.
D) If attendance increases by 1%, the estimated mean score received will increase by 0.341 percentage points.
Question
Instruction 12.8
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 \text { Regression statistics }
 MultipleR 0.8857 R Square 0.7845 Adjusted R  Square 0.7801 Standard Error 1.3704 Observations 51\begin{array}{|l|l|}\hline\text { MultipleR } & 0.8857 \\\hline \text { R Square } & 0.7845 \\\hline \begin{array}{l}\text { Adjusted R } \\\text { Square }\end{array} & 0.7801 \\\hline \text { Standard Error } & 1.3704 \\\hline \text { Observations } & 51\\\hline\end{array}

 ANOVA df55 MS F Significance F Regression 1335.0472335.0473178.3859 Residual 1.8782 Total 50427.0798\begin{array}{l}\text { ANOVA }\\\begin{array}{|l|l|l|l|l|l|l|}\hline & d f & 55 & \text { MS } & F & \begin{array}{l}\text { Significance } \\F\end{array} & \\\hline \text { Regression } & 1 & 335.0472 & 335.0473 & 178.3859 & & \\\hline \text { Residual } & & & 1.8782 & & & \\\hline \text { Total } & 50 & 427.0798 & & & & \\\hline\end{array}\end{array}

 Coefficients  Standard  Error  tStat  p-value  Lower 95%  Upper 95%  Intercept 1.89400.40184.71342.051E052.70151.0865 Hours 0.97950.073313.35615.944E180.83211.1269\begin{array}{|l|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 } & -1.8940 & 0.4018 & -4.7134 & 2.051 \mathrm{E}-05 & -2.7015 & -1.0865 \\\hline \text { Hours } & 0.9795 & 0.0733 & 13.3561 & 5.944 \mathrm{E}-18 & 0.8321 & 1.1269 \\\hline\end{array} Note: 2.051E-05 = 2.051 * 10S1-0.5 and 5.944E-18 = 5.944 * 10S1-18.

-Referring to Instruction 12.8,the error sum of squares (SSE)of the above regression is

A) 427.079804.
B) 92.0325465.
C) 1.878215.
D) 335.047257.
Question
Instruction 12.7
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.
<strong>Instruction 12.7 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<sup> </sup> Excel output from regressing average grade on university entrance exam scores using a data set of eight randomly chosen students from a large university.    -Referring to Instruction 12.7,what is the predicted average value of average grade when university entrance exam score = 20?</strong> A) 2.80 B) 2.61 C) 2.66 D) 3.12 <div style=padding-top: 35px>

-Referring to Instruction 12.7,what is the predicted average value of average grade when university entrance exam score = 20?

A) 2.80
B) 2.61
C) 2.66
D) 3.12
Question
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  Coop jobs  job Offer 114226313401\begin{array} { | l | l | l | } \hline \text { Student } & \text { Coop jobs } & \text { job Offer } \\\hline 1 & 1 & 4 \\\hline 2 & 2 & 6 \\\hline 3 & 1 & 3 \\\hline 4 & 0 & 1 \\\hline\end{array}

-Referring to Instruction 12.3,the error or residual sum of squares (SSE)is____________.
Question
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:
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<sup> </sup> Excel's Data Analysis tool to analyse the last four years of quarterly data with the following results:    -Referring to Instruction 12.5,the estimates of the Y-intercept and slope are ____________ and ____________,respectively.<div style=padding-top: 35px>

-Referring to Instruction 12.5,the estimates of the Y-intercept and slope are ____________ and ____________,respectively.
Question
When using a regression model to make predictions,the term 'relevant range' refers to____________.
Question
The standard error of the estimate is a measure of

A) total variation of the Y variable.
B) explained variation.
C) the variation of the X variable.
D) the variation around the sample regression line.
Question
What do we mean when we say that a simple linear regression model is 'statistically' useful?

A) The model is 'practically' useful for predicting Y.
B) The model is an excellent predictor of Y.
C) All the statistics computed from the sample make sense.
D) The model is a better predictor of Y than the sample mean, Yˉ\bar { Y } .
Question
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?

A) 13.5
B) 12.0
C) -6.0
D) -2.0
Question
Instruction 12.10
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\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}

 ANOVA \text { ANOVA }
dfSSMSF Significance F Regression 1171062.9193171062.919386.47590.0000 Residual 2855388.43091978.1582 Total 29226451.3503\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}

 Coefficients  Standard  Error  t Stat p-value  Lower 95%  Upper 95%  Intercept 95.061426.91833.53150.0015150.200939.9218 Download 3.72970.40119.29920.00002.90824.5513\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}  <strong>Instruction 12.10 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 12.10,which of the following is the correct interpretation for the slope coefficient?</strong> A) For each increase of 1,000 downloads, the expected revenue is estimated to increase by $3.7297 thousands. B) For each increase of 1,000 dollars in expected revenue, the expected number of downloads is estimated to increase by 3.7297 thousands. C) For each decrease of 1,000 dollars in expected revenue, the expected number of downloads is estimated to increase by 3.7297 thousands. D) For each decrease of 1,000 downloads, the expected revenue is estimated to increase by $3.7297 thousands. <div style=padding-top: 35px>   <strong>Instruction 12.10 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 12.10,which of the following is the correct interpretation for the slope coefficient?</strong> A) For each increase of 1,000 downloads, the expected revenue is estimated to increase by $3.7297 thousands. B) For each increase of 1,000 dollars in expected revenue, the expected number of downloads is estimated to increase by 3.7297 thousands. C) For each decrease of 1,000 dollars in expected revenue, the expected number of downloads is estimated to increase by 3.7297 thousands. D) For each decrease of 1,000 downloads, the expected revenue is estimated to increase by $3.7297 thousands. <div style=padding-top: 35px>

-Referring to Instruction 12.10,which of the following is the correct interpretation for the slope coefficient?

A) For each increase of 1,000 downloads, the expected revenue is estimated to increase by $3.7297 thousands.
B) For each increase of 1,000 dollars in expected revenue, the expected number of downloads is estimated to increase by 3.7297 thousands.
C) For each decrease of 1,000 dollars in expected revenue, the expected number of downloads is estimated to increase by 3.7297 thousands.
D) For each decrease of 1,000 downloads, the expected revenue is estimated to increase by $3.7297 thousands.
Question
The slope (b1)represents

A) the estimated average change in Y per unit change in X.
B) predicted value of Y when X = 0.
C) variation around the line of regression.
D) the predicted value of Y.
Question
The regression sum of squares (SSR)can never be greater than the total sum of squares (SST).
Question
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.
Question
A simple regression has a b0 value of 1.3 and a b1 value of 2.6.What is the predicted Y for an X value of 0?

A) 2.6
B) 3.9
C) 1.3
D) 3.4
Question
Instruction 12.10
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\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}

 ANOVA \text { ANOVA }
dfSSMSF Significance F Regression 1171062.9193171062.919386.47590.0000 Residual 2855388.43091978.1582 Total 29226451.3503\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}

 Coefficients  Standard  Error  t Stat p-value  Lower 95%  Upper 95%  Intercept 95.061426.91833.53150.0015150.200939.9218 Download 3.72970.40119.29920.00002.90824.5513\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}  Instruction 12.10 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 12.10,predict the revenue when the number of downloads is 30,000.<div style=padding-top: 35px>   Instruction 12.10 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 12.10,predict the revenue when the number of downloads is 30,000.<div style=padding-top: 35px>

-Referring to Instruction 12.10,predict the revenue when the number of downloads is 30,000.
Question
The coefficient of determination (r2)tells you

A) the proportion of variation in Y that is explained by the independent variable X in the model.
B) whether r has any significance.
C) that you should not partition the total variation.
D) that the coefficient of correlation (r) is larger than 1.
Question
The least squares method minimises which of the following?

A) SST
B) SSR
C) SSE
D) All of the above.
Question
Instruction 12.12
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 114226313401\begin{array} { | l | l | l | } \hline \text { Student } & \text { Coop jobs } & \text { job Offer } \\\hline 1 & 1 & 4 \\\hline 2 & 2 & 6 \\\hline 3 & 1 & 3 \\\hline 4 & 0 & 1 \\\hline\end{array}

-Referring to Instruction 12.12,the standard error of estimate is ____________.
Question
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 127522113734264433555152961534725588365992844103048111731122238\begin{array} { | l | l | l | } \hline \text { Broker } & \text { Clients } & \text { Sales } \\\hline 1 & 27 & 52 \\\hline 2 & 11 & 37 \\\hline 3 & 42 & 64 \\\hline 4 & 33 & 55 \\\hline 5 & 15 & 29 \\\hline 6 & 15 & 34 \\\hline 7 & 25 & 58 \\\hline 8 & 36 & 59 \\\hline 9 & 28 & 44 \\\hline 10 & 30 & 48 \\\hline 11 & 17 & 31 \\\hline 12 & 22 & 38 \\\hline\end{array}

-Referring to Instruction 12.13,the coefficient of determination is ____________.
Question
The coefficient of determination represents the ratio of SSR to SST.
Question
Instruction 12.11
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 \text { Regression statistics }
 MultipleR 0.9447 R Square 0.8924 Adjusted R  Square 0.8886 Standard Error 0.3342 Observations 30\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}

 ANOVA \text { ANOVA }
dfS5MSF Significance F Regression 125.943825.9438232.22004.3946E15 Residual 283.12820.1117 Total 2929.072\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}

 Coefficients  Standard  Error  tStat  p-value  Lower 95%  Upper 95%  Intercept 0.40240.12363.25590.00300.14920.6555 Applications  Recorded 0.01260.000815.23884.3946E150.01090.0143\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}
Note: 4.3946E-15 is 4.3946 × 10-15.  <strong>Instruction 12.11 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:  \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}  Note: 4.3946E-15 is 4.3946 × 10<sup>-</sup><sup>15</sup>.      -Referring to Instruction 12.11,the estimated mean amount of time it takes to record one additional loan application is</strong> A) 0.4024 more hours. B) 0.0126 more hours. C) 0.0126 fewer hours. D) 0.4024 fewer hours. <div style=padding-top: 35px>   <strong>Instruction 12.11 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:  \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}  Note: 4.3946E-15 is 4.3946 × 10<sup>-</sup><sup>15</sup>.      -Referring to Instruction 12.11,the estimated mean amount of time it takes to record one additional loan application is</strong> A) 0.4024 more hours. B) 0.0126 more hours. C) 0.0126 fewer hours. D) 0.4024 fewer hours. <div style=padding-top: 35px>

-Referring to Instruction 12.11,the estimated mean amount of time it takes to record one additional loan application is

A) 0.4024 more hours.
B) 0.0126 more hours.
C) 0.0126 fewer hours.
D) 0.4024 fewer hours.
Question
The Y-intercept (b0)represents the

A) predicted value of Y.
B) change in estimated average Y per unit change in X.
C) estimated average Y when X = 0.
D) variation around the sample regression line.
Question
Instruction 12.12
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 114226313401\begin{array} { | l | l | l | } \hline \text { Student } & \text { Coop jobs } & \text { job Offer } \\\hline 1 & 1 & 4 \\\hline 2 & 2 & 6 \\\hline 3 & 1 & 3 \\\hline 4 & 0 & 1 \\\hline\end{array}

-Referring to Instruction 12.12,the coefficient of correlation is ____________.
Question
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 127522113734264433555152961534725588365992844103048111731122238\begin{array} { | l | l | l | } \hline \text { Broker } & \text { Clients } & \text { Sales } \\\hline 1 & 27 & 52 \\\hline 2 & 11 & 37 \\\hline 3 & 42 & 64 \\\hline 4 & 33 & 55 \\\hline 5 & 15 & 29 \\\hline 6 & 15 & 34 \\\hline 7 & 25 & 58 \\\hline 8 & 36 & 59 \\\hline 9 & 28 & 44 \\\hline 10 & 30 & 48 \\\hline 11 & 17 & 31 \\\hline 12 & 22 & 38 \\\hline\end{array}

-Referring to Instruction 12.13,the standard error of the estimated slope coefficient is ____________.
Question
Instruction 12.10
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\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}

 ANOVA \text { ANOVA }
dfSSMSF Significance F Regression 1171062.9193171062.919386.47590.0000 Residual 2855388.43091978.1582 Total 29226451.3503\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}

 Coefficients  Standard  Error  t Stat p-value  Lower 95%  Upper 95%  Intercept 95.061426.91833.53150.0015150.200939.9218 Download 3.72970.40119.29920.00002.90824.5513\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}  <strong>Instruction 12.10 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 12.10,which of the following is the correct interpretation for the slope coefficient?</strong> A) For each incsrease of 1 million dollars in box office gross, expected home video units sold are estimated to increase by 4.3331 thousand units. B) For each increase of 1 dollar in box office gross, expected home video units sold are estimated to increase by 4.3331 units. C) For each increase of 1 million dollars in box office gross, expected home video units sold are estimated to increase by 4.3331 units. D) For each increase of 1 dollar in box office gross, expected home video units sold are estimated to increase by 4.3331 thousand units. <div style=padding-top: 35px>   <strong>Instruction 12.10 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 12.10,which of the following is the correct interpretation for the slope coefficient?</strong> A) For each incsrease of 1 million dollars in box office gross, expected home video units sold are estimated to increase by 4.3331 thousand units. B) For each increase of 1 dollar in box office gross, expected home video units sold are estimated to increase by 4.3331 units. C) For each increase of 1 million dollars in box office gross, expected home video units sold are estimated to increase by 4.3331 units. D) For each increase of 1 dollar in box office gross, expected home video units sold are estimated to increase by 4.3331 thousand units. <div style=padding-top: 35px>

-Referring to Instruction 12.10,which of the following is the correct interpretation for the slope coefficient?

A) For each incsrease of 1 million dollars in box office gross, expected home video units sold are estimated to increase by 4.3331 thousand units.
B) For each increase of 1 dollar in box office gross, expected home video units sold are estimated to increase by 4.3331 units.
C) For each increase of 1 million dollars in box office gross, expected home video units sold are estimated to increase by 4.3331 units.
D) For each increase of 1 dollar in box office gross, expected home video units sold are estimated to increase by 4.3331 thousand units.
Question
Instruction 12.16
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) 90711.2092611.057138.217419.217809.4289810.085106.735297.024606.128729.526507.536037.25\begin{array} { | l | 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 898 & 10.08 \\\hline 510 & 6.73 \\\hline 529 & 7.02 \\\hline 460 & 6.12 \\\hline 872 & 9.52 \\\hline 650 & 7.53 \\\hline 603 & 7.25 \\\hline\end{array}

-Referring to Instruction 12.16,what is the value of the standard error of the estimate?
Question
Instruction 12.14
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\begin{array}{|l|l|}\hline \text { Multiple R } & 0.802 \\\hline \text { R Square } & 0.643 \\\hline \begin{array}{l}\text { Adjusted R } \\\text { Square }\end{array} & 0.618 \\\hline \begin{array}{l}\text { Standard Error } \\\text { SYX }\end{array} & 0.9224 \\\hline \text { Observations } & 16 \\\hline\end{array}

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

 Predictor  Coef  StdError  t Stat  p-value  Intercept 3.9621.4402.750.016 Industry 0.0404510.0080485.030.000\begin{array}{|l|l|l|l|l|}\hline\text { Predictor } & \text { Coef } & \text { StdError } & \text { t Stat } & \text { p-value } \\\hline \text { Intercept } & 3.962 & 1.440 & 2.75 & 0.016 \\\hline \text { Industry } & 0.040451 & 0.008048 & 5.03 & 0.000\\\hline\end{array}


 Durbin-Watson 1.59 Statistic \begin{array}{|l|l|}\hline\text { Durbin-Watson } & 1.59 \\\text { Statistic } &\\\hline\end{array}

-Referring to Instruction 12.14,the standard error of the estimated slope coefficient is ____________.
Question
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 127522113734264433555152961534725588365992844103048111731122238\begin{array} { | l | l | l | } \hline \text { Broker } & \text { Clients } & \text { Sales } \\\hline 1 & 27 & 52 \\\hline 2 & 11 & 37 \\\hline 3 & 42 & 64 \\\hline 4 & 33 & 55 \\\hline 5 & 15 & 29 \\\hline 6 & 15 & 34 \\\hline 7 & 25 & 58 \\\hline 8 & 36 & 59 \\\hline 9 & 28 & 44 \\\hline 10 & 30 & 48 \\\hline 11 & 17 & 31 \\\hline 12 & 22 & 38 \\\hline\end{array}

-Referring to Instruction 12.13,the coefficient of correlation is____________.
Question
Instruction 12.18
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 \text { Regression statistics }
 MultipleR 0.9447 R Square 0.8924 Adjusted R  Square 0.8886 Standard Error 0.3342 Observations 30\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}

 ANOVA \text { ANOVA }
dfS5MSF Significance F Regression 125.943825.9438232.22004.3946E15 Residual 283.12820.1117 Total 2929.072\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}

 Coefficients  Standard  Error  tStat  p-value  Lower 95%  Upper 95%  Intercept 0.40240.12363.25590.00300.14920.6555 Applications  Recorded 0.01260.000815.23884.3946E150.01090.0143\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}

Note: 4.3946E-15 is 4.3946 x 10-15.  <strong>Instruction 12.18 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:  \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}   Note: 4.3946E-15 is 4.3946 x 10<sup>-15</sup>.     Note: 4.3946E-15 is 4.3946 × 10<sup>-</sup><sup>15</sup>.  -Referring to Instruction 12.18,the error sum of squares (SSE)of the above regression is</strong> A) 0.1117. B) 29.0720. C) 25.9438. D) 3.1282. <div style=padding-top: 35px>   <strong>Instruction 12.18 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:  \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}   Note: 4.3946E-15 is 4.3946 x 10<sup>-15</sup>.     Note: 4.3946E-15 is 4.3946 × 10<sup>-</sup><sup>15</sup>.  -Referring to Instruction 12.18,the error sum of squares (SSE)of the above regression is</strong> A) 0.1117. B) 29.0720. C) 25.9438. D) 3.1282. <div style=padding-top: 35px>  Note: 4.3946E-15 is 4.3946 × 10-15.

-Referring to Instruction 12.18,the error sum of squares (SSE)of the above regression is

A) 0.1117.
B) 29.0720.
C) 25.9438.
D) 3.1282.
Question
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\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}

 ANOVA \text { ANOVA }
dfSSMSF Significance F Regression 1171062.9193171062.919386.47590.0000 Residual 2855388.43091978.1582 Total 29226451.3503\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}

 Coefficients  Standard  Error  t Stat p-value  Lower 95%  Upper 95%  Intercept 95.061426.91833.53150.0015150.200939.9218 Download 3.72970.40119.29920.00002.90824.5513\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}
 <strong>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?</strong> A) 72.8% of the variation in the video unit sales can be explained by the variation in the box office gross. B) 71.8% of the variation in the video unit sales can be explained by the variation in the box office gross. C) 71.8% of the variation in the box office gross can be explained by the variation in the video unit sales. D) 72.8% of the variation in the box office gross can be explained by the variation in the video unit sales. <div style=padding-top: 35px>   <strong>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?</strong> A) 72.8% of the variation in the video unit sales can be explained by the variation in the box office gross. B) 71.8% of the variation in the video unit sales can be explained by the variation in the box office gross. C) 71.8% of the variation in the box office gross can be explained by the variation in the video unit sales. D) 72.8% of the variation in the box office gross can be explained by the variation in the video unit sales. <div style=padding-top: 35px>

-Referring to Instruction 12.17,which of the following is the correct interpretation for the coefficient of determination?

A) 72.8% of the variation in the video unit sales can be explained by the variation in the box office gross.
B) 71.8% of the variation in the video unit sales can be explained by the variation in the box office gross.
C) 71.8% of the variation in the box office gross can be explained by the variation in the video unit sales.
D) 72.8% of the variation in the box office gross can be explained by the variation in the video unit sales.
Question
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 127522113734264433555152961534725588365992844103048111731122238\begin{array} { | l | l | l | } \hline \text { Broker } & \text { Clients } & \text { Sales } \\\hline 1 & 27 & 52 \\\hline 2 & 11 & 37 \\\hline 3 & 42 & 64 \\\hline 4 & 33 & 55 \\\hline 5 & 15 & 29 \\\hline 6 & 15 & 34 \\\hline 7 & 25 & 58 \\\hline 8 & 36 & 59 \\\hline 9 & 28 & 44 \\\hline 10 & 30 & 48 \\\hline 11 & 17 & 31 \\\hline 12 & 22 & 38 \\\hline\end{array}

-Referring to Instruction 12.13,the standard error of estimate is ____________.
Question
If the plot of the residuals is fan shaped,which assumption is violated?

A) Homoscedasticity
B) Independence of errors
C) Normality
D) No assumptions are violated, the graph should resemble a fan
Question
Instruction 12.14
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\begin{array}{|l|l|}\hline \text { Multiple R } & 0.802 \\\hline \text { R Square } & 0.643 \\\hline \begin{array}{l}\text { Adjusted R } \\\text { Square }\end{array} & 0.618 \\\hline \begin{array}{l}\text { Standard Error } \\\text { SYX }\end{array} & 0.9224 \\\hline \text { Observations } & 16 \\\hline\end{array}

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

 Predictor  Coef  StdError  t Stat  p-value  Intercept 3.9621.4402.750.016 Industry 0.0404510.0080485.030.000\begin{array}{|l|l|l|l|l|}\hline\text { Predictor } & \text { Coef } & \text { StdError } & \text { t Stat } & \text { p-value } \\\hline \text { Intercept } & 3.962 & 1.440 & 2.75 & 0.016 \\\hline \text { Industry } & 0.040451 & 0.008048 & 5.03 & 0.000\\\hline\end{array}


 Durbin-Watson 1.59 Statistic \begin{array}{|l|l|}\hline\text { Durbin-Watson } & 1.59 \\\text { Statistic } &\\\hline\end{array}

-Referring to Instruction 12.14,the correlation coefficient is ____________.
Question
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\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}

 ANOVA \text { ANOVA }
dfSSMSF Significance F Regression 1171062.9193171062.919386.47590.0000 Residual 2855388.43091978.1582 Total 29226451.3503\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}

 Coefficients  Standard  Error  t Stat p-value  Lower 95%  Upper 95%  Intercept 95.061426.91833.53150.0015150.200939.9218 Download 3.72970.40119.29920.00002.90824.5513\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}
 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?<div style=padding-top: 35px>   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?<div style=padding-top: 35px>

-Referring to Instruction 12.17,what is the standard deviation around the regression line?
Question
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\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}

 ANOVA \text { ANOVA }
dfSSMSF Significance F Regression 1171062.9193171062.919386.47590.0000 Residual 2855388.43091978.1582 Total 29226451.3503\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}

 Coefficients  Standard  Error  t Stat p-value  Lower 95%  Upper 95%  Intercept 95.061426.91833.53150.0015150.200939.9218 Download 3.72970.40119.29920.00002.90824.5513\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}
 <strong>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?</strong> A) 74.67% of the variation in revenue can be explained by the variation in the number of downloads. B) 75.54% of the variation in revenue can be explained by the variation in the number of downloads. C) 75.54% of the variation in the number of downloads can be explained by the variation in revenue. D) 74.67% of the variation in the number of downloads can be explained by the variation in revenue. <div style=padding-top: 35px>   <strong>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?</strong> A) 74.67% of the variation in revenue can be explained by the variation in the number of downloads. B) 75.54% of the variation in revenue can be explained by the variation in the number of downloads. C) 75.54% of the variation in the number of downloads can be explained by the variation in revenue. D) 74.67% of the variation in the number of downloads can be explained by the variation in revenue. <div style=padding-top: 35px>

-Referring to Instruction 12.17,which of the following is the correct interpretation for the coefficient of determination?

A) 74.67% of the variation in revenue can be explained by the variation in the number of downloads.
B) 75.54% of the variation in revenue can be explained by the variation in the number of downloads.
C) 75.54% of the variation in the number of downloads can be explained by the variation in revenue.
D) 74.67% of the variation in the number of downloads can be explained by the variation in revenue.
Question
Which of the following assumptions concerning the probability distribution of the random error term is stated incorrectly?

A) The variance of the distribution increases as X increases.
B) The mean of the distribution is 0.
C) The errors are independent.
D) The distribution is normal.
Question
Instruction 12.15
The following Microsoft Excel tables are obtained when 'Score received on an exam (measured in percentage points)' (Y) is regressed on 'percentage attendance' (X) for 22 students in a Statistics for Business and Economics course.
 Regression Statistics \text { Regression Statistics }
 MultipleR 0.142620229 R Square 0.02034053 Adjusted R  Square 0.028642444 Standard Error 20.25979924 Observations 22\begin{array}{|l|l|}\hline\text { MultipleR } & 0.142620229 \\\hline \text { R Square } & 0.02034053 \\\hline \begin{array}{l}\text { Adjusted R } \\\text { Square }\end{array} & -0.028642444 \\\hline \text { Standard Error } & 20.25979924 \\\hline \text { Observations } & 22\\\hline\end{array}

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



-Referring to Instruction 12.15,which of the following statements is true?

A) 14.2% of the total variability in percentage attendance can be explained by score received.
B) 2% of the total variability in score received can be explained by percentage attendance.
C) 2% of the total variability in percentage attendance can be explained by score received.
D) 14.26% of the total variability in score received can be explained by percentage attendance.
Question
Instruction 12.14
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\begin{array}{|l|l|}\hline \text { Multiple R } & 0.802 \\\hline \text { R Square } & 0.643 \\\hline \begin{array}{l}\text { Adjusted R } \\\text { Square }\end{array} & 0.618 \\\hline \begin{array}{l}\text { Standard Error } \\\text { SYX }\end{array} & 0.9224 \\\hline \text { Observations } & 16 \\\hline\end{array}

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

 Predictor  Coef  StdError  t Stat  p-value  Intercept 3.9621.4402.750.016 Industry 0.0404510.0080485.030.000\begin{array}{|l|l|l|l|l|}\hline\text { Predictor } & \text { Coef } & \text { StdError } & \text { t Stat } & \text { p-value } \\\hline \text { Intercept } & 3.962 & 1.440 & 2.75 & 0.016 \\\hline \text { Industry } & 0.040451 & 0.008048 & 5.03 & 0.000\\\hline\end{array}


 Durbin-Watson 1.59 Statistic \begin{array}{|l|l|}\hline\text { Durbin-Watson } & 1.59 \\\text { Statistic } &\\\hline\end{array}

-Referring to Instruction 12.14,the standard error of the estimate is____________.
Question
Instruction 12.14
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\begin{array}{|l|l|}\hline \text { Multiple R } & 0.802 \\\hline \text { R Square } & 0.643 \\\hline \begin{array}{l}\text { Adjusted R } \\\text { Square }\end{array} & 0.618 \\\hline \begin{array}{l}\text { Standard Error } \\\text { SYX }\end{array} & 0.9224 \\\hline \text { Observations } & 16 \\\hline\end{array}

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

 Predictor  Coef  StdError  t Stat  p-value  Intercept 3.9621.4402.750.016 Industry 0.0404510.0080485.030.000\begin{array}{|l|l|l|l|l|}\hline\text { Predictor } & \text { Coef } & \text { StdError } & \text { t Stat } & \text { p-value } \\\hline \text { Intercept } & 3.962 & 1.440 & 2.75 & 0.016 \\\hline \text { Industry } & 0.040451 & 0.008048 & 5.03 & 0.000\\\hline\end{array}


 Durbin-Watson 1.59 Statistic \begin{array}{|l|l|}\hline\text { Durbin-Watson } & 1.59 \\\text { Statistic } &\\\hline\end{array}

-Referring to Instruction 12.14,the coefficient of determination is ____________.
Question
The residuals represent

A) the square root of the slope.
B) the difference between the actual Y values and the predicted Y values.
C) the difference between the actual Y values and the mean of Y.
D) the predicted value of Y for the average X value.
Question
Based on the residual plot below,you will conclude that there might be a violation of which of the following assumptions? <strong>Based on the residual plot below,you will conclude that there might be a violation of which of the following assumptions?  </strong> A) Homoscedasticity B) Independence of errors C) Linearity of the relationship D) Normality of errors <div style=padding-top: 35px>

A) Homoscedasticity
B) Independence of errors
C) Linearity of the relationship
D) Normality of errors
Question
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\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}

 ANOVA \text { ANOVA }
dfSSMSF Significance F Regression 1171062.9193171062.919386.47590.0000 Residual 2855388.43091978.1582 Total 29226451.3503\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}

 Coefficients  Standard  Error  t Stat p-value  Lower 95%  Upper 95%  Intercept 95.061426.91833.53150.0015150.200939.9218 Download 3.72970.40119.29920.00002.90824.5513\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}
 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 error of estimate?<div style=padding-top: 35px>   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 error of estimate?<div style=padding-top: 35px>

-Referring to Instruction 12.17,what is the standard error of estimate?
Question
In performing a regression analysis involving two numerical variables,you are assuming

A) the variances of X and Y are equal.
B) that X and Y are independent.
C) the variation around the line of regression is the same for each X value.
D) All of the above.
Question
Instruction 12.16
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) 90711.2092611.057138.217419.217809.4289810.085106.735297.024606.128729.526507.536037.25\begin{array} { | l | 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 898 & 10.08 \\\hline 510 & 6.73 \\\hline 529 & 7.02 \\\hline 460 & 6.12 \\\hline 872 & 9.52 \\\hline 650 & 7.53 \\\hline 603 & 7.25 \\\hline\end{array}

-Referring to Instruction 12.16,what is the value of the coefficient of correlation?
Question
Instruction 12.16
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) 90711.2092611.057138.217419.217809.4289810.085106.735297.024606.128729.526507.536037.25\begin{array} { | l | 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 898 & 10.08 \\\hline 510 & 6.73 \\\hline 529 & 7.02 \\\hline 460 & 6.12 \\\hline 872 & 9.52 \\\hline 650 & 7.53 \\\hline 603 & 7.25 \\\hline\end{array}

-Referring to Instruction 12.16,what is the value of the coefficient of determination?
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Deck 12: Simple Linear Regression
1
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 standard error of the regression slope estimate, Sb1S _ { b _ { 1 } } ?

A) 0.885
B) 12.650
C) 16.299
D) 0.784
12.650
2
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,a 95% confidence interval for ?1 is (15,30).Interpret the interval.

A) At the ? = 0.05 level, there is no evidence of a linear relationship between service charge (Y) and sales revenue (X).
B) We are 95% confident that average service charge (Y) will increase between $15 and $30 for every $1 million increase in sales revenue (X).
C) We are 95% confident that the mean service charge will fall between $15 and $30 per month.
D) We are 95% confident that the sales revenue (X) will increase between $15 and $30 million for every $1 increase in service charge (Y).
We are 95% confident that average service charge (Y) will increase between $15 and $30 for every $1 million increase in sales revenue (X).
3
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.

A) There is sufficient evidence (at the ? = 0.05) to conclude that sales revenue (X) is a useful linear predictor of service charge (Y).
B) Sales revenue (X) is a poor predictor of service charge (Y).
C) There is insufficient evidence (at the ? = 0.10) to conclude that sales revenue (X) is a useful linear predictor of service charge (Y).
D) For every $1 million increase in sales revenue, we expect a service charge to increase $0.034.
There is sufficient evidence (at the ? = 0.05) to conclude that sales revenue (X) is a useful linear predictor of service charge (Y).
4
Instruction 12.2
A chocolate bar manufacturer is interested in trying to estimate how sales are influenced by the price of their product. To do this, the company randomly chooses six country towns and cities and offers the chocolate bar at different prices. Using chocolate bar sales as the dependent variable, the company will conduct a simple linear regression on the data below:
 City  Price($)  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,to test whether a change in price will have any impact on average sales,what would be the critical values? Use ? = 0.05.

A) ± 2.7765
B) ± 3.1634
C) ± 2.5706
D) ± 3.4954
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5
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 coefficient of correlation for these data?

A) 0.8854
B) 0.7839
C) -0.7839
D) -0.8854
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6
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,if the price of the chocolate bar is set at $2,the predicted sales will be

A) 100.
B) 30.
C) 65.
D) 90.
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7
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  Coop jobs  job Offer 114226313401\begin{array} { | l | l | l | } \hline \text { Student } & \text { Coop jobs } & \text { job Offer } \\\hline 1 & 1 & 4 \\\hline 2 & 2 & 6 \\\hline 3 & 1 & 3 \\\hline 4 & 0 & 1 \\\hline\end{array}

-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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8
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  Coop jobs  job Offer 114226313401\begin{array} { | l | l | l | } \hline \text { Student } & \text { Coop jobs } & \text { job Offer } \\\hline 1 & 1 & 4 \\\hline 2 & 2 & 6 \\\hline 3 & 1 & 3 \\\hline 4 & 0 & 1 \\\hline\end{array}

-Referring to Instruction 12.3,the total sum of squares (SST)is ____________.
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9
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 estimate of ?0,the Y-intercept of the line.

A) About 95% of the observed service charges fall within $2,700 of the least squares line.
B) There is no practical interpretation since a sales revenue of $0 is a nonsensical value.
C) For every $1 million increase in sales revenue, we expect a service charge to decrease $2,700.
D) All companies will be charged at least $2,700 by the bank.
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10
Instruction 12.2
A chocolate bar manufacturer is interested in trying to estimate how sales are influenced by the price of their product. To do this, the company randomly chooses six country towns and cities and offers the chocolate bar at different prices. Using chocolate bar sales as the dependent variable, the company will conduct a simple linear regression on the data below:
 City  Price($)  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 standard error of the estimate,SYX,for the data?

A) 0.885
B) 0.784
C) 16.299
D) 12.650
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11
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 percentage of the total variation in chocolate bar sales is explained by prices?

A) 78.39%
B) 88.54%
C) 48.19%
D) 100%
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12
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 ?(X * Xˉ\bar { X } )2 for these data?

A) 25.66
B) 0
C) 2.54
D) 1.66
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13
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 percentage of the total variation in chocolate bar sales explained by the regression model?

A) 48.19%
B) 100%
C) 88.54%
D) 78.39%
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14
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  Coop jobs  job Offer 114226313401\begin{array} { | l | l | l | } \hline \text { Student } & \text { Coop jobs } & \text { job Offer } \\\hline 1 & 1 & 4 \\\hline 2 & 2 & 6 \\\hline 3 & 1 & 3 \\\hline 4 & 0 & 1 \\\hline\end{array}

-Referring to Instruction 12.3,the least squares estimate of the Y-intercept is____________.
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15
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?

A) 0.784
B) -3.810
C) 161.386
D) -48.193
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16
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,to test that the regression coefficient,?1 is not equal to 0,what would be the critical values? Use ? = 0.05.

A) ± 2.7765
B) ± 3.1634
C) ± 2.5706
D) ± 3.4954
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17
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  Coop jobs  job Offer 114226313401\begin{array} { | l | l | l | } \hline \text { Student } & \text { Coop jobs } & \text { job Offer } \\\hline 1 & 1 & 4 \\\hline 2 & 2 & 6 \\\hline 3 & 1 & 3 \\\hline 4 & 0 & 1 \\\hline\end{array}

-Referring to Instruction 12.3,the least squares estimate of the slope is_____________.
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18
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 slope parameter for the chocolate bar price and sales data?

A) -48.193
B) -3.810
C) 161.386
D) 0.784
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19
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 estimate of ?,the standard deviation of the random error term (standard error of the estimate)in the model.

A) About 95% of the observed service charges fall within $130 of the least squares line.
B) About 95% of the observed service charges fall within $65 of the least squares line.
C) About 95% of the observed service charges equal their corresponding predicted values.
D) For every $1 million increase in sales revenue, we expect a service charge to increase $65.
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20
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  Coop jobs  job Offer 114226313401\begin{array} { | l | l | l | } \hline \text { Student } & \text { Coop jobs } & \text { job Offer } \\\hline 1 & 1 & 4 \\\hline 2 & 2 & 6 \\\hline 3 & 1 & 3 \\\hline 4 & 0 & 1 \\\hline\end{array}

-Referring to Instruction 12.3,the regression sum of squares (SSR)is____________.
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21
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 127522113734264433555152961534725588365992844103048111731122238\begin{array} { | l | l | l | } \hline \text { Broker } & \text { Clients } & \text { Sales } \\\hline 1 & 27 & 52 \\\hline 2 & 11 & 37 \\\hline 3 & 42 & 64 \\\hline 4 & 33 & 55 \\\hline 5 & 15 & 29 \\\hline 6 & 15 & 34 \\\hline 7 & 25 & 58 \\\hline 8 & 36 & 59 \\\hline 9 & 28 & 44 \\\hline 10 & 30 & 48 \\\hline 11 & 17 & 31 \\\hline 12 & 22 & 38 \\\hline\end{array}

-Referring to Instruction 12.4,the regression sum of squares (SSR)is____________
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22
Instruction 12.9
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) 90711.2092611.057138.217419.217809.4289810.085106.735297.024606.128729.526507.536037.25\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 898 & 10.08 \\\hline 510 & 6.73 \\\hline 529 & 7.02 \\\hline 460 & 6.12 \\\hline 872 & 9.52 \\\hline 650 & 7.53 \\\hline 603 & 7.25 \\\hline\end{array}

-Referring to Instruction 12.9,93.98% of the total variation in weekly sales can be explained by the variation in the number of customers who make purchases.
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23
Instruction 12.8
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 \text { Regression statistics }
 MultipleR 0.8857 R Square 0.7845 Adjusted R  Square 0.7801 Standard Error 1.3704 Observations 51\begin{array}{|l|l|}\hline\text { MultipleR } & 0.8857 \\\hline \text { R Square } & 0.7845 \\\hline \begin{array}{l}\text { Adjusted R } \\\text { Square }\end{array} & 0.7801 \\\hline \text { Standard Error } & 1.3704 \\\hline \text { Observations } & 51\\\hline\end{array}

 ANOVA df55 MS F Significance F Regression 1335.0472335.0473178.3859 Residual 1.8782 Total 50427.0798\begin{array}{l}\text { ANOVA }\\\begin{array}{|l|l|l|l|l|l|l|}\hline & d f & 55 & \text { MS } & F & \begin{array}{l}\text { Significance } \\F\end{array} & \\\hline \text { Regression } & 1 & 335.0472 & 335.0473 & 178.3859 & & \\\hline \text { Residual } & & & 1.8782 & & & \\\hline \text { Total } & 50 & 427.0798 & & & & \\\hline\end{array}\end{array}

 Coefficients  Standard  Error  tStat  p-value  Lower 95%  Upper 95%  Intercept 1.89400.40184.71342.051E052.70151.0865 Hours 0.97950.073313.35615.944E180.83211.1269\begin{array}{|l|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 } & -1.8940 & 0.4018 & -4.7134 & 2.051 \mathrm{E}-05 & -2.7015 & -1.0865 \\\hline \text { Hours } & 0.9795 & 0.0733 & 13.3561 & 5.944 \mathrm{E}-18 & 0.8321 & 1.1269 \\\hline\end{array} Note: 2.051E-05 = 2.051 * 10S1-0.5 and 5.944E-18 = 5.944 * 10S1-18.

-Referring to Instruction 12.8,the estimated average change in salary (in thousands of dollars)as a result of spending an extra hour per day studying is

A) 0.9795.
B) 0.7845.
C) 335.0473.
D) -1.8940.
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24
Instruction 12.9
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) 90711.2092611.057138.217419.217809.4289810.085106.735297.024606.128729.526507.536037.25\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 898 & 10.08 \\\hline 510 & 6.73 \\\hline 529 & 7.02 \\\hline 460 & 6.12 \\\hline 872 & 9.52 \\\hline 650 & 7.53 \\\hline 603 & 7.25 \\\hline\end{array}

-Referring to Instruction 12.9,generate the scatter plot.
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25
Instruction 12.9
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) 90711.2092611.057138.217419.217809.4289810.085106.735297.024606.128729.526507.536037.25\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 898 & 10.08 \\\hline 510 & 6.73 \\\hline 529 & 7.02 \\\hline 460 & 6.12 \\\hline 872 & 9.52 \\\hline 650 & 7.53 \\\hline 603 & 7.25 \\\hline\end{array}

-Referring to Instruction 12.9,what are the values of the estimated intercept and slope?
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26
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 127522113734264433555152961534725588365992844103048111731122238\begin{array} { | l | l | l | } \hline \text { Broker } & \text { Clients } & \text { Sales } \\\hline 1 & 27 & 52 \\\hline 2 & 11 & 37 \\\hline 3 & 42 & 64 \\\hline 4 & 33 & 55 \\\hline 5 & 15 & 29 \\\hline 6 & 15 & 34 \\\hline 7 & 25 & 58 \\\hline 8 & 36 & 59 \\\hline 9 & 28 & 44 \\\hline 10 & 30 & 48 \\\hline 11 & 17 & 31 \\\hline 12 & 22 & 38 \\\hline\end{array}

-Referring to Instruction 12.4,the total sum of squares (SST)is ____________.
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27
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 127522113734264433555152961534725588365992844103048111731122238\begin{array} { | l | l | l | } \hline \text { Broker } & \text { Clients } & \text { Sales } \\\hline 1 & 27 & 52 \\\hline 2 & 11 & 37 \\\hline 3 & 42 & 64 \\\hline 4 & 33 & 55 \\\hline 5 & 15 & 29 \\\hline 6 & 15 & 34 \\\hline 7 & 25 & 58 \\\hline 8 & 36 & 59 \\\hline 9 & 28 & 44 \\\hline 10 & 30 & 48 \\\hline 11 & 17 & 31 \\\hline 12 & 22 & 38 \\\hline\end{array}

-Referring to Instruction 12.4,the prediction for the amount of sales (in $1,000s)for a person who brings 25 new customers into the firm is ____________.
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28
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:
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<sup> </sup> Excel's Data Analysis tool to analyse the last four years of quarterly data with the following results:    -Referring to Instruction 12.5,the prediction for a quarter in which X = 120 is Y = ____________.

-Referring to Instruction 12.5,the prediction for a quarter in which X = 120 is Y = ____________.
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29
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  Coop jobs  job Offer 114226313401\begin{array} { | l | l | l | } \hline \text { Student } & \text { Coop jobs } & \text { job Offer } \\\hline 1 & 1 & 4 \\\hline 2 & 2 & 6 \\\hline 3 & 1 & 3 \\\hline 4 & 0 & 1 \\\hline\end{array}

-Referring to Instruction 12.3,set up a scatter diagram.
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30
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  Coop jobs  job Offer 114226313401\begin{array} { | l | l | l | } \hline \text { Student } & \text { Coop jobs } & \text { job Offer } \\\hline 1 & 1 & 4 \\\hline 2 & 2 & 6 \\\hline 3 & 1 & 3 \\\hline 4 & 0 & 1 \\\hline\end{array}

-Referring to Instruction 12.3,the coefficient of determination is____________.
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31
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 127522113734264433555152961534725588365992844103048111731122238\begin{array} { | l | l | l | } \hline \text { Broker } & \text { Clients } & \text { Sales } \\\hline 1 & 27 & 52 \\\hline 2 & 11 & 37 \\\hline 3 & 42 & 64 \\\hline 4 & 33 & 55 \\\hline 5 & 15 & 29 \\\hline 6 & 15 & 34 \\\hline 7 & 25 & 58 \\\hline 8 & 36 & 59 \\\hline 9 & 28 & 44 \\\hline 10 & 30 & 48 \\\hline 11 & 17 & 31 \\\hline 12 & 22 & 38 \\\hline\end{array}

-Referring to Instruction 12.4,the least squares estimate of the Y-intercept is ____________.
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32
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 127522113734264433555152961534725588365992844103048111731122238\begin{array} { | l | l | l | } \hline \text { Broker } & \text { Clients } & \text { Sales } \\\hline 1 & 27 & 52 \\\hline 2 & 11 & 37 \\\hline 3 & 42 & 64 \\\hline 4 & 33 & 55 \\\hline 5 & 15 & 29 \\\hline 6 & 15 & 34 \\\hline 7 & 25 & 58 \\\hline 8 & 36 & 59 \\\hline 9 & 28 & 44 \\\hline 10 & 30 & 48 \\\hline 11 & 17 & 31 \\\hline 12 & 22 & 38 \\\hline\end{array}

-Referring to Instruction 12.4,the error or residual sum of squares (SSE)is____________.
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33
Instruction 12.7
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.
<strong>Instruction 12.7 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<sup> </sup> Excel output from regressing average grade on university entrance exam scores using a data set of eight randomly chosen students from a large university.    -Referring to Instruction 12.7,the interpretation of the coefficient of determination in this regression is</strong> A) average grade accounts for 57.74% of the variability of university entrance exam scores. B) university entrance exam scores account for 57.74% of the total fluctuation in average grade. C) 57.74% of the total variation of university entrance exam scores can be explained by average grade. D) None of the above.

-Referring to Instruction 12.7,the interpretation of the coefficient of determination in this regression is

A) average grade accounts for 57.74% of the variability of university entrance exam scores.
B) university entrance exam scores account for 57.74% of the total fluctuation in average grade.
C) 57.74% of the total variation of university entrance exam scores can be explained by average grade.
D) None of the above.
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34
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 127522113734264433555152961534725588365992844103048111731122238\begin{array} { | l | l | l | } \hline \text { Broker } & \text { Clients } & \text { Sales } \\\hline 1 & 27 & 52 \\\hline 2 & 11 & 37 \\\hline 3 & 42 & 64 \\\hline 4 & 33 & 55 \\\hline 5 & 15 & 29 \\\hline 6 & 15 & 34 \\\hline 7 & 25 & 58 \\\hline 8 & 36 & 59 \\\hline 9 & 28 & 44 \\\hline 10 & 30 & 48 \\\hline 11 & 17 & 31 \\\hline 12 & 22 & 38 \\\hline\end{array}

-Referring to Instruction 12.4,set up a scatter diagram.
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35
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 127522113734264433555152961534725588365992844103048111731122238\begin{array} { | l | l | l | } \hline \text { Broker } & \text { Clients } & \text { Sales } \\\hline 1 & 27 & 52 \\\hline 2 & 11 & 37 \\\hline 3 & 42 & 64 \\\hline 4 & 33 & 55 \\\hline 5 & 15 & 29 \\\hline 6 & 15 & 34 \\\hline 7 & 25 & 58 \\\hline 8 & 36 & 59 \\\hline 9 & 28 & 44 \\\hline 10 & 30 & 48 \\\hline 11 & 17 & 31 \\\hline 12 & 22 & 38 \\\hline\end{array}

-Referring to Instruction 12.4,____% of the total variation in sales generated can be explained by the number of new customers brought in.
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36
Instruction 12.6
The following Microsoft Excel tables are obtained when 'Score received on an exam (measured in percentage points)' (Y) is regressed on 'percentage attendance' (X) for 22 students in a Statistics for Business and Economics course.
 Regression Statistics \text { Regression Statistics }
 MultipleR 0.142620229 R Square 0.02034053 Adjusted R  Square 0.028642444 Standard Error 20.25979924 Observations 22\begin{array}{|l|l|}\hline \text { MultipleR } & 0.142620229 \\\hline \text { R Square } & 0.02034053 \\\hline \begin{array}{l}\text { Adjusted R } \\\text { Square }\end{array} & -0.028642444 \\\hline \text { Standard Error } & 20.25979924 \\\hline \text { Observations } & 22 \\\hline\end{array}

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

-Referring to Instruction 12.6,which of the following statements is true?

A) If attendance increases by 1%, the estimated mean score received will increase by 39.39 percentage points.
B) If attendance increases by 0.341%, the estimated mean score received will increase by 1 percentage point.
C) If the score received increases by 39.39%, the estimated mean attendance will go up by 1%.
D) If attendance increases by 1%, the estimated mean score received will increase by 0.341 percentage points.
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37
Instruction 12.8
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 \text { Regression statistics }
 MultipleR 0.8857 R Square 0.7845 Adjusted R  Square 0.7801 Standard Error 1.3704 Observations 51\begin{array}{|l|l|}\hline\text { MultipleR } & 0.8857 \\\hline \text { R Square } & 0.7845 \\\hline \begin{array}{l}\text { Adjusted R } \\\text { Square }\end{array} & 0.7801 \\\hline \text { Standard Error } & 1.3704 \\\hline \text { Observations } & 51\\\hline\end{array}

 ANOVA df55 MS F Significance F Regression 1335.0472335.0473178.3859 Residual 1.8782 Total 50427.0798\begin{array}{l}\text { ANOVA }\\\begin{array}{|l|l|l|l|l|l|l|}\hline & d f & 55 & \text { MS } & F & \begin{array}{l}\text { Significance } \\F\end{array} & \\\hline \text { Regression } & 1 & 335.0472 & 335.0473 & 178.3859 & & \\\hline \text { Residual } & & & 1.8782 & & & \\\hline \text { Total } & 50 & 427.0798 & & & & \\\hline\end{array}\end{array}

 Coefficients  Standard  Error  tStat  p-value  Lower 95%  Upper 95%  Intercept 1.89400.40184.71342.051E052.70151.0865 Hours 0.97950.073313.35615.944E180.83211.1269\begin{array}{|l|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 } & -1.8940 & 0.4018 & -4.7134 & 2.051 \mathrm{E}-05 & -2.7015 & -1.0865 \\\hline \text { Hours } & 0.9795 & 0.0733 & 13.3561 & 5.944 \mathrm{E}-18 & 0.8321 & 1.1269 \\\hline\end{array} Note: 2.051E-05 = 2.051 * 10S1-0.5 and 5.944E-18 = 5.944 * 10S1-18.

-Referring to Instruction 12.8,the error sum of squares (SSE)of the above regression is

A) 427.079804.
B) 92.0325465.
C) 1.878215.
D) 335.047257.
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38
Instruction 12.7
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.
<strong>Instruction 12.7 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<sup> </sup> Excel output from regressing average grade on university entrance exam scores using a data set of eight randomly chosen students from a large university.    -Referring to Instruction 12.7,what is the predicted average value of average grade when university entrance exam score = 20?</strong> A) 2.80 B) 2.61 C) 2.66 D) 3.12

-Referring to Instruction 12.7,what is the predicted average value of average grade when university entrance exam score = 20?

A) 2.80
B) 2.61
C) 2.66
D) 3.12
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39
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  Coop jobs  job Offer 114226313401\begin{array} { | l | l | l | } \hline \text { Student } & \text { Coop jobs } & \text { job Offer } \\\hline 1 & 1 & 4 \\\hline 2 & 2 & 6 \\\hline 3 & 1 & 3 \\\hline 4 & 0 & 1 \\\hline\end{array}

-Referring to Instruction 12.3,the error or residual sum of squares (SSE)is____________.
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40
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:
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<sup> </sup> Excel's Data Analysis tool to analyse the last four years of quarterly data with the following results:    -Referring to Instruction 12.5,the estimates of the Y-intercept and slope are ____________ and ____________,respectively.

-Referring to Instruction 12.5,the estimates of the Y-intercept and slope are ____________ and ____________,respectively.
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41
When using a regression model to make predictions,the term 'relevant range' refers to____________.
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42
The standard error of the estimate is a measure of

A) total variation of the Y variable.
B) explained variation.
C) the variation of the X variable.
D) the variation around the sample regression line.
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43
What do we mean when we say that a simple linear regression model is 'statistically' useful?

A) The model is 'practically' useful for predicting Y.
B) The model is an excellent predictor of Y.
C) All the statistics computed from the sample make sense.
D) The model is a better predictor of Y than the sample mean, Yˉ\bar { Y } .
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44
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?

A) 13.5
B) 12.0
C) -6.0
D) -2.0
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45
Instruction 12.10
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\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}

 ANOVA \text { ANOVA }
dfSSMSF Significance F Regression 1171062.9193171062.919386.47590.0000 Residual 2855388.43091978.1582 Total 29226451.3503\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}

 Coefficients  Standard  Error  t Stat p-value  Lower 95%  Upper 95%  Intercept 95.061426.91833.53150.0015150.200939.9218 Download 3.72970.40119.29920.00002.90824.5513\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}  <strong>Instruction 12.10 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 12.10,which of the following is the correct interpretation for the slope coefficient?</strong> A) For each increase of 1,000 downloads, the expected revenue is estimated to increase by $3.7297 thousands. B) For each increase of 1,000 dollars in expected revenue, the expected number of downloads is estimated to increase by 3.7297 thousands. C) For each decrease of 1,000 dollars in expected revenue, the expected number of downloads is estimated to increase by 3.7297 thousands. D) For each decrease of 1,000 downloads, the expected revenue is estimated to increase by $3.7297 thousands.   <strong>Instruction 12.10 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 12.10,which of the following is the correct interpretation for the slope coefficient?</strong> A) For each increase of 1,000 downloads, the expected revenue is estimated to increase by $3.7297 thousands. B) For each increase of 1,000 dollars in expected revenue, the expected number of downloads is estimated to increase by 3.7297 thousands. C) For each decrease of 1,000 dollars in expected revenue, the expected number of downloads is estimated to increase by 3.7297 thousands. D) For each decrease of 1,000 downloads, the expected revenue is estimated to increase by $3.7297 thousands.

-Referring to Instruction 12.10,which of the following is the correct interpretation for the slope coefficient?

A) For each increase of 1,000 downloads, the expected revenue is estimated to increase by $3.7297 thousands.
B) For each increase of 1,000 dollars in expected revenue, the expected number of downloads is estimated to increase by 3.7297 thousands.
C) For each decrease of 1,000 dollars in expected revenue, the expected number of downloads is estimated to increase by 3.7297 thousands.
D) For each decrease of 1,000 downloads, the expected revenue is estimated to increase by $3.7297 thousands.
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46
The slope (b1)represents

A) the estimated average change in Y per unit change in X.
B) predicted value of Y when X = 0.
C) variation around the line of regression.
D) the predicted value of Y.
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47
The regression sum of squares (SSR)can never be greater than the total sum of squares (SST).
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48
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.
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49
A simple regression has a b0 value of 1.3 and a b1 value of 2.6.What is the predicted Y for an X value of 0?

A) 2.6
B) 3.9
C) 1.3
D) 3.4
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50
Instruction 12.10
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\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}

 ANOVA \text { ANOVA }
dfSSMSF Significance F Regression 1171062.9193171062.919386.47590.0000 Residual 2855388.43091978.1582 Total 29226451.3503\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}

 Coefficients  Standard  Error  t Stat p-value  Lower 95%  Upper 95%  Intercept 95.061426.91833.53150.0015150.200939.9218 Download 3.72970.40119.29920.00002.90824.5513\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}  Instruction 12.10 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 12.10,predict the revenue when the number of downloads is 30,000.  Instruction 12.10 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 12.10,predict the revenue when the number of downloads is 30,000.

-Referring to Instruction 12.10,predict the revenue when the number of downloads is 30,000.
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51
The coefficient of determination (r2)tells you

A) the proportion of variation in Y that is explained by the independent variable X in the model.
B) whether r has any significance.
C) that you should not partition the total variation.
D) that the coefficient of correlation (r) is larger than 1.
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52
The least squares method minimises which of the following?

A) SST
B) SSR
C) SSE
D) All of the above.
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53
Instruction 12.12
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 114226313401\begin{array} { | l | l | l | } \hline \text { Student } & \text { Coop jobs } & \text { job Offer } \\\hline 1 & 1 & 4 \\\hline 2 & 2 & 6 \\\hline 3 & 1 & 3 \\\hline 4 & 0 & 1 \\\hline\end{array}

-Referring to Instruction 12.12,the standard error of estimate is ____________.
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54
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 127522113734264433555152961534725588365992844103048111731122238\begin{array} { | l | l | l | } \hline \text { Broker } & \text { Clients } & \text { Sales } \\\hline 1 & 27 & 52 \\\hline 2 & 11 & 37 \\\hline 3 & 42 & 64 \\\hline 4 & 33 & 55 \\\hline 5 & 15 & 29 \\\hline 6 & 15 & 34 \\\hline 7 & 25 & 58 \\\hline 8 & 36 & 59 \\\hline 9 & 28 & 44 \\\hline 10 & 30 & 48 \\\hline 11 & 17 & 31 \\\hline 12 & 22 & 38 \\\hline\end{array}

-Referring to Instruction 12.13,the coefficient of determination is ____________.
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55
The coefficient of determination represents the ratio of SSR to SST.
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56
Instruction 12.11
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 \text { Regression statistics }
 MultipleR 0.9447 R Square 0.8924 Adjusted R  Square 0.8886 Standard Error 0.3342 Observations 30\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}

 ANOVA \text { ANOVA }
dfS5MSF Significance F Regression 125.943825.9438232.22004.3946E15 Residual 283.12820.1117 Total 2929.072\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}

 Coefficients  Standard  Error  tStat  p-value  Lower 95%  Upper 95%  Intercept 0.40240.12363.25590.00300.14920.6555 Applications  Recorded 0.01260.000815.23884.3946E150.01090.0143\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}
Note: 4.3946E-15 is 4.3946 × 10-15.  <strong>Instruction 12.11 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:  \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}  Note: 4.3946E-15 is 4.3946 × 10<sup>-</sup><sup>15</sup>.      -Referring to Instruction 12.11,the estimated mean amount of time it takes to record one additional loan application is</strong> A) 0.4024 more hours. B) 0.0126 more hours. C) 0.0126 fewer hours. D) 0.4024 fewer hours.   <strong>Instruction 12.11 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:  \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}  Note: 4.3946E-15 is 4.3946 × 10<sup>-</sup><sup>15</sup>.      -Referring to Instruction 12.11,the estimated mean amount of time it takes to record one additional loan application is</strong> A) 0.4024 more hours. B) 0.0126 more hours. C) 0.0126 fewer hours. D) 0.4024 fewer hours.

-Referring to Instruction 12.11,the estimated mean amount of time it takes to record one additional loan application is

A) 0.4024 more hours.
B) 0.0126 more hours.
C) 0.0126 fewer hours.
D) 0.4024 fewer hours.
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57
The Y-intercept (b0)represents the

A) predicted value of Y.
B) change in estimated average Y per unit change in X.
C) estimated average Y when X = 0.
D) variation around the sample regression line.
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58
Instruction 12.12
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 114226313401\begin{array} { | l | l | l | } \hline \text { Student } & \text { Coop jobs } & \text { job Offer } \\\hline 1 & 1 & 4 \\\hline 2 & 2 & 6 \\\hline 3 & 1 & 3 \\\hline 4 & 0 & 1 \\\hline\end{array}

-Referring to Instruction 12.12,the coefficient of correlation is ____________.
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59
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 127522113734264433555152961534725588365992844103048111731122238\begin{array} { | l | l | l | } \hline \text { Broker } & \text { Clients } & \text { Sales } \\\hline 1 & 27 & 52 \\\hline 2 & 11 & 37 \\\hline 3 & 42 & 64 \\\hline 4 & 33 & 55 \\\hline 5 & 15 & 29 \\\hline 6 & 15 & 34 \\\hline 7 & 25 & 58 \\\hline 8 & 36 & 59 \\\hline 9 & 28 & 44 \\\hline 10 & 30 & 48 \\\hline 11 & 17 & 31 \\\hline 12 & 22 & 38 \\\hline\end{array}

-Referring to Instruction 12.13,the standard error of the estimated slope coefficient is ____________.
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60
Instruction 12.10
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\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}

 ANOVA \text { ANOVA }
dfSSMSF Significance F Regression 1171062.9193171062.919386.47590.0000 Residual 2855388.43091978.1582 Total 29226451.3503\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}

 Coefficients  Standard  Error  t Stat p-value  Lower 95%  Upper 95%  Intercept 95.061426.91833.53150.0015150.200939.9218 Download 3.72970.40119.29920.00002.90824.5513\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}  <strong>Instruction 12.10 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 12.10,which of the following is the correct interpretation for the slope coefficient?</strong> A) For each incsrease of 1 million dollars in box office gross, expected home video units sold are estimated to increase by 4.3331 thousand units. B) For each increase of 1 dollar in box office gross, expected home video units sold are estimated to increase by 4.3331 units. C) For each increase of 1 million dollars in box office gross, expected home video units sold are estimated to increase by 4.3331 units. D) For each increase of 1 dollar in box office gross, expected home video units sold are estimated to increase by 4.3331 thousand units.   <strong>Instruction 12.10 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 12.10,which of the following is the correct interpretation for the slope coefficient?</strong> A) For each incsrease of 1 million dollars in box office gross, expected home video units sold are estimated to increase by 4.3331 thousand units. B) For each increase of 1 dollar in box office gross, expected home video units sold are estimated to increase by 4.3331 units. C) For each increase of 1 million dollars in box office gross, expected home video units sold are estimated to increase by 4.3331 units. D) For each increase of 1 dollar in box office gross, expected home video units sold are estimated to increase by 4.3331 thousand units.

-Referring to Instruction 12.10,which of the following is the correct interpretation for the slope coefficient?

A) For each incsrease of 1 million dollars in box office gross, expected home video units sold are estimated to increase by 4.3331 thousand units.
B) For each increase of 1 dollar in box office gross, expected home video units sold are estimated to increase by 4.3331 units.
C) For each increase of 1 million dollars in box office gross, expected home video units sold are estimated to increase by 4.3331 units.
D) For each increase of 1 dollar in box office gross, expected home video units sold are estimated to increase by 4.3331 thousand units.
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61
Instruction 12.16
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) 90711.2092611.057138.217419.217809.4289810.085106.735297.024606.128729.526507.536037.25\begin{array} { | l | 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 898 & 10.08 \\\hline 510 & 6.73 \\\hline 529 & 7.02 \\\hline 460 & 6.12 \\\hline 872 & 9.52 \\\hline 650 & 7.53 \\\hline 603 & 7.25 \\\hline\end{array}

-Referring to Instruction 12.16,what is the value of the standard error of the estimate?
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62
Instruction 12.14
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\begin{array}{|l|l|}\hline \text { Multiple R } & 0.802 \\\hline \text { R Square } & 0.643 \\\hline \begin{array}{l}\text { Adjusted R } \\\text { Square }\end{array} & 0.618 \\\hline \begin{array}{l}\text { Standard Error } \\\text { SYX }\end{array} & 0.9224 \\\hline \text { Observations } & 16 \\\hline\end{array}

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

 Predictor  Coef  StdError  t Stat  p-value  Intercept 3.9621.4402.750.016 Industry 0.0404510.0080485.030.000\begin{array}{|l|l|l|l|l|}\hline\text { Predictor } & \text { Coef } & \text { StdError } & \text { t Stat } & \text { p-value } \\\hline \text { Intercept } & 3.962 & 1.440 & 2.75 & 0.016 \\\hline \text { Industry } & 0.040451 & 0.008048 & 5.03 & 0.000\\\hline\end{array}


 Durbin-Watson 1.59 Statistic \begin{array}{|l|l|}\hline\text { Durbin-Watson } & 1.59 \\\text { Statistic } &\\\hline\end{array}

-Referring to Instruction 12.14,the standard error of the estimated slope coefficient is ____________.
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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 127522113734264433555152961534725588365992844103048111731122238\begin{array} { | l | l | l | } \hline \text { Broker } & \text { Clients } & \text { Sales } \\\hline 1 & 27 & 52 \\\hline 2 & 11 & 37 \\\hline 3 & 42 & 64 \\\hline 4 & 33 & 55 \\\hline 5 & 15 & 29 \\\hline 6 & 15 & 34 \\\hline 7 & 25 & 58 \\\hline 8 & 36 & 59 \\\hline 9 & 28 & 44 \\\hline 10 & 30 & 48 \\\hline 11 & 17 & 31 \\\hline 12 & 22 & 38 \\\hline\end{array}

-Referring to Instruction 12.13,the coefficient of correlation is____________.
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Instruction 12.18
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 \text { Regression statistics }
 MultipleR 0.9447 R Square 0.8924 Adjusted R  Square 0.8886 Standard Error 0.3342 Observations 30\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}

 ANOVA \text { ANOVA }
dfS5MSF Significance F Regression 125.943825.9438232.22004.3946E15 Residual 283.12820.1117 Total 2929.072\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}

 Coefficients  Standard  Error  tStat  p-value  Lower 95%  Upper 95%  Intercept 0.40240.12363.25590.00300.14920.6555 Applications  Recorded 0.01260.000815.23884.3946E150.01090.0143\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}

Note: 4.3946E-15 is 4.3946 x 10-15.  <strong>Instruction 12.18 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:  \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}   Note: 4.3946E-15 is 4.3946 x 10<sup>-15</sup>.     Note: 4.3946E-15 is 4.3946 × 10<sup>-</sup><sup>15</sup>.  -Referring to Instruction 12.18,the error sum of squares (SSE)of the above regression is</strong> A) 0.1117. B) 29.0720. C) 25.9438. D) 3.1282.   <strong>Instruction 12.18 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:  \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}   Note: 4.3946E-15 is 4.3946 x 10<sup>-15</sup>.     Note: 4.3946E-15 is 4.3946 × 10<sup>-</sup><sup>15</sup>.  -Referring to Instruction 12.18,the error sum of squares (SSE)of the above regression is</strong> A) 0.1117. B) 29.0720. C) 25.9438. D) 3.1282.  Note: 4.3946E-15 is 4.3946 × 10-15.

-Referring to Instruction 12.18,the error sum of squares (SSE)of the above regression is

A) 0.1117.
B) 29.0720.
C) 25.9438.
D) 3.1282.
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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\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}

 ANOVA \text { ANOVA }
dfSSMSF Significance F Regression 1171062.9193171062.919386.47590.0000 Residual 2855388.43091978.1582 Total 29226451.3503\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}

 Coefficients  Standard  Error  t Stat p-value  Lower 95%  Upper 95%  Intercept 95.061426.91833.53150.0015150.200939.9218 Download 3.72970.40119.29920.00002.90824.5513\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}
 <strong>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?</strong> A) 72.8% of the variation in the video unit sales can be explained by the variation in the box office gross. B) 71.8% of the variation in the video unit sales can be explained by the variation in the box office gross. C) 71.8% of the variation in the box office gross can be explained by the variation in the video unit sales. D) 72.8% of the variation in the box office gross can be explained by the variation in the video unit sales.   <strong>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?</strong> A) 72.8% of the variation in the video unit sales can be explained by the variation in the box office gross. B) 71.8% of the variation in the video unit sales can be explained by the variation in the box office gross. C) 71.8% of the variation in the box office gross can be explained by the variation in the video unit sales. D) 72.8% of the variation in the box office gross can be explained by the variation in the video unit sales.

-Referring to Instruction 12.17,which of the following is the correct interpretation for the coefficient of determination?

A) 72.8% of the variation in the video unit sales can be explained by the variation in the box office gross.
B) 71.8% of the variation in the video unit sales can be explained by the variation in the box office gross.
C) 71.8% of the variation in the box office gross can be explained by the variation in the video unit sales.
D) 72.8% of the variation in the box office gross can be explained by the variation in the video unit sales.
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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 127522113734264433555152961534725588365992844103048111731122238\begin{array} { | l | l | l | } \hline \text { Broker } & \text { Clients } & \text { Sales } \\\hline 1 & 27 & 52 \\\hline 2 & 11 & 37 \\\hline 3 & 42 & 64 \\\hline 4 & 33 & 55 \\\hline 5 & 15 & 29 \\\hline 6 & 15 & 34 \\\hline 7 & 25 & 58 \\\hline 8 & 36 & 59 \\\hline 9 & 28 & 44 \\\hline 10 & 30 & 48 \\\hline 11 & 17 & 31 \\\hline 12 & 22 & 38 \\\hline\end{array}

-Referring to Instruction 12.13,the standard error of estimate is ____________.
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If the plot of the residuals is fan shaped,which assumption is violated?

A) Homoscedasticity
B) Independence of errors
C) Normality
D) No assumptions are violated, the graph should resemble a fan
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Instruction 12.14
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\begin{array}{|l|l|}\hline \text { Multiple R } & 0.802 \\\hline \text { R Square } & 0.643 \\\hline \begin{array}{l}\text { Adjusted R } \\\text { Square }\end{array} & 0.618 \\\hline \begin{array}{l}\text { Standard Error } \\\text { SYX }\end{array} & 0.9224 \\\hline \text { Observations } & 16 \\\hline\end{array}

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

 Predictor  Coef  StdError  t Stat  p-value  Intercept 3.9621.4402.750.016 Industry 0.0404510.0080485.030.000\begin{array}{|l|l|l|l|l|}\hline\text { Predictor } & \text { Coef } & \text { StdError } & \text { t Stat } & \text { p-value } \\\hline \text { Intercept } & 3.962 & 1.440 & 2.75 & 0.016 \\\hline \text { Industry } & 0.040451 & 0.008048 & 5.03 & 0.000\\\hline\end{array}


 Durbin-Watson 1.59 Statistic \begin{array}{|l|l|}\hline\text { Durbin-Watson } & 1.59 \\\text { Statistic } &\\\hline\end{array}

-Referring to Instruction 12.14,the correlation coefficient is ____________.
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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\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}

 ANOVA \text { ANOVA }
dfSSMSF Significance F Regression 1171062.9193171062.919386.47590.0000 Residual 2855388.43091978.1582 Total 29226451.3503\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}

 Coefficients  Standard  Error  t Stat p-value  Lower 95%  Upper 95%  Intercept 95.061426.91833.53150.0015150.200939.9218 Download 3.72970.40119.29920.00002.90824.5513\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}
 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?
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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\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}

 ANOVA \text { ANOVA }
dfSSMSF Significance F Regression 1171062.9193171062.919386.47590.0000 Residual 2855388.43091978.1582 Total 29226451.3503\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}

 Coefficients  Standard  Error  t Stat p-value  Lower 95%  Upper 95%  Intercept 95.061426.91833.53150.0015150.200939.9218 Download 3.72970.40119.29920.00002.90824.5513\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}
 <strong>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?</strong> A) 74.67% of the variation in revenue can be explained by the variation in the number of downloads. B) 75.54% of the variation in revenue can be explained by the variation in the number of downloads. C) 75.54% of the variation in the number of downloads can be explained by the variation in revenue. D) 74.67% of the variation in the number of downloads can be explained by the variation in revenue.   <strong>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?</strong> A) 74.67% of the variation in revenue can be explained by the variation in the number of downloads. B) 75.54% of the variation in revenue can be explained by the variation in the number of downloads. C) 75.54% of the variation in the number of downloads can be explained by the variation in revenue. D) 74.67% of the variation in the number of downloads can be explained by the variation in revenue.

-Referring to Instruction 12.17,which of the following is the correct interpretation for the coefficient of determination?

A) 74.67% of the variation in revenue can be explained by the variation in the number of downloads.
B) 75.54% of the variation in revenue can be explained by the variation in the number of downloads.
C) 75.54% of the variation in the number of downloads can be explained by the variation in revenue.
D) 74.67% of the variation in the number of downloads can be explained by the variation in revenue.
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71
Which of the following assumptions concerning the probability distribution of the random error term is stated incorrectly?

A) The variance of the distribution increases as X increases.
B) The mean of the distribution is 0.
C) The errors are independent.
D) The distribution is normal.
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72
Instruction 12.15
The following Microsoft Excel tables are obtained when 'Score received on an exam (measured in percentage points)' (Y) is regressed on 'percentage attendance' (X) for 22 students in a Statistics for Business and Economics course.
 Regression Statistics \text { Regression Statistics }
 MultipleR 0.142620229 R Square 0.02034053 Adjusted R  Square 0.028642444 Standard Error 20.25979924 Observations 22\begin{array}{|l|l|}\hline\text { MultipleR } & 0.142620229 \\\hline \text { R Square } & 0.02034053 \\\hline \begin{array}{l}\text { Adjusted R } \\\text { Square }\end{array} & -0.028642444 \\\hline \text { Standard Error } & 20.25979924 \\\hline \text { Observations } & 22\\\hline\end{array}

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



-Referring to Instruction 12.15,which of the following statements is true?

A) 14.2% of the total variability in percentage attendance can be explained by score received.
B) 2% of the total variability in score received can be explained by percentage attendance.
C) 2% of the total variability in percentage attendance can be explained by score received.
D) 14.26% of the total variability in score received can be explained by percentage attendance.
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73
Instruction 12.14
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\begin{array}{|l|l|}\hline \text { Multiple R } & 0.802 \\\hline \text { R Square } & 0.643 \\\hline \begin{array}{l}\text { Adjusted R } \\\text { Square }\end{array} & 0.618 \\\hline \begin{array}{l}\text { Standard Error } \\\text { SYX }\end{array} & 0.9224 \\\hline \text { Observations } & 16 \\\hline\end{array}

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

 Predictor  Coef  StdError  t Stat  p-value  Intercept 3.9621.4402.750.016 Industry 0.0404510.0080485.030.000\begin{array}{|l|l|l|l|l|}\hline\text { Predictor } & \text { Coef } & \text { StdError } & \text { t Stat } & \text { p-value } \\\hline \text { Intercept } & 3.962 & 1.440 & 2.75 & 0.016 \\\hline \text { Industry } & 0.040451 & 0.008048 & 5.03 & 0.000\\\hline\end{array}


 Durbin-Watson 1.59 Statistic \begin{array}{|l|l|}\hline\text { Durbin-Watson } & 1.59 \\\text { Statistic } &\\\hline\end{array}

-Referring to Instruction 12.14,the standard error of the estimate is____________.
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74
Instruction 12.14
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\begin{array}{|l|l|}\hline \text { Multiple R } & 0.802 \\\hline \text { R Square } & 0.643 \\\hline \begin{array}{l}\text { Adjusted R } \\\text { Square }\end{array} & 0.618 \\\hline \begin{array}{l}\text { Standard Error } \\\text { SYX }\end{array} & 0.9224 \\\hline \text { Observations } & 16 \\\hline\end{array}

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

 Predictor  Coef  StdError  t Stat  p-value  Intercept 3.9621.4402.750.016 Industry 0.0404510.0080485.030.000\begin{array}{|l|l|l|l|l|}\hline\text { Predictor } & \text { Coef } & \text { StdError } & \text { t Stat } & \text { p-value } \\\hline \text { Intercept } & 3.962 & 1.440 & 2.75 & 0.016 \\\hline \text { Industry } & 0.040451 & 0.008048 & 5.03 & 0.000\\\hline\end{array}


 Durbin-Watson 1.59 Statistic \begin{array}{|l|l|}\hline\text { Durbin-Watson } & 1.59 \\\text { Statistic } &\\\hline\end{array}

-Referring to Instruction 12.14,the coefficient of determination is ____________.
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75
The residuals represent

A) the square root of the slope.
B) the difference between the actual Y values and the predicted Y values.
C) the difference between the actual Y values and the mean of Y.
D) the predicted value of Y for the average X value.
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76
Based on the residual plot below,you will conclude that there might be a violation of which of the following assumptions? <strong>Based on the residual plot below,you will conclude that there might be a violation of which of the following assumptions?  </strong> A) Homoscedasticity B) Independence of errors C) Linearity of the relationship D) Normality of errors

A) Homoscedasticity
B) Independence of errors
C) Linearity of the relationship
D) Normality of errors
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77
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\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}

 ANOVA \text { ANOVA }
dfSSMSF Significance F Regression 1171062.9193171062.919386.47590.0000 Residual 2855388.43091978.1582 Total 29226451.3503\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}

 Coefficients  Standard  Error  t Stat p-value  Lower 95%  Upper 95%  Intercept 95.061426.91833.53150.0015150.200939.9218 Download 3.72970.40119.29920.00002.90824.5513\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}
 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 error of estimate?  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 error of estimate?

-Referring to Instruction 12.17,what is the standard error of estimate?
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78
In performing a regression analysis involving two numerical variables,you are assuming

A) the variances of X and Y are equal.
B) that X and Y are independent.
C) the variation around the line of regression is the same for each X value.
D) All of the above.
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79
Instruction 12.16
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) 90711.2092611.057138.217419.217809.4289810.085106.735297.024606.128729.526507.536037.25\begin{array} { | l | 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 898 & 10.08 \\\hline 510 & 6.73 \\\hline 529 & 7.02 \\\hline 460 & 6.12 \\\hline 872 & 9.52 \\\hline 650 & 7.53 \\\hline 603 & 7.25 \\\hline\end{array}

-Referring to Instruction 12.16,what is the value of the coefficient of correlation?
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80
Instruction 12.16
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) 90711.2092611.057138.217419.217809.4289810.085106.735297.024606.128729.526507.536037.25\begin{array} { | l | 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 898 & 10.08 \\\hline 510 & 6.73 \\\hline 529 & 7.02 \\\hline 460 & 6.12 \\\hline 872 & 9.52 \\\hline 650 & 7.53 \\\hline 603 & 7.25 \\\hline\end{array}

-Referring to Instruction 12.16,what is the value of the coefficient of determination?
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