Question 1

An academic advisor wants to predict the typical starting salary of a graduate at a top business school using the GMAT score of the school as a predictor variable. A simple linear regression of SALARY versus GMAT using 25 data points is shown below. $$\hat { \beta } _ { 0 } = - 92040 \hat { \beta } 1 = 228 s = 3213 r ^ { 2 } = .66 r = .81 \mathrm { df } = 23 \quad t = 6.67$$ A 95% prediction interval for SALARY when GMAT = 600 is approximately ($37,915, $51,984). Interpret this interval.
A) We are 95% confident that the SALARY of a top business school graduate with a GMAT of 600 will fall between $37,915 and $51,984.
B) We are 95% confident that the mean SALARY of all top business school graduates with GMATs of 600 will fall between $37,915 and $51,984.
C) We are 95% confident that the increase in SALARY for a 600-point increase in GMAT will fall between $37,915 and $51,984.
D) We are 95% confident that the SALARY of a top business school graduate will fall between $37,915 and $51,984.

Accepted Answer

A prediction interval provides a range where we expect the salary of an individual graduate with a GMAT score of 600 to fall, with 95% confidence.

Question 2

To investigate the relationship between yield of potatoes, y, and level of fertilizer application, x, a researcher divides a field into eight plots of equal size and applies differing amounts of fertilizer to each. The yield of potatoes (in pounds) and the fertilizer application (in pounds) are recorded for each plot. The data are as follows: $\begin{array}{l|cccccccc}
\hline x & 1 & 1.5 & 2 & 2.5 & 3 & 3.5 & 4 & 4.5 \
\hline y & 25 & 31 & 27 & 28 & 36 & 35 & 32 & 34 \
\hline
\end{array}$

Summary statistics yield $S S _ { x x } = 10.5 , S S _ { y y } = 112 , S S _ { x y } = 25$, and $S S E = 52.476$. Calculate the coefficient of determination.

Accepted Answer

The answer of To investigate the relationship between yield of...

Question 3

The dean of the Business School at a small Florida college wishes to determine whether the grade -point average (GPA) of a graduating student can be used to predict the graduate's starting salary. More specifically, the dean wants to know whether higher GPAs lead to higher starting salaries. Records for 23 of last year's Business School graduates are selected at random, and data on GPA (x) and starting salary (y, in $thousands) for each graduate were used to fit the model $$E ( y ) = \beta _ { 0 } + \beta _ { 1 } x$$
The results of the simple linear regression are provided below.
$$\begin{array} { l l } 
\hat{y} = 4.25 + 2.75 x , \quad & S S x y = 5.15 , S S x x = 1.87 \
& S S y y = 15.17 , S S E = 1.0075 \
\text { Range of the } x \text {-values: } & 2.23 - 3.85 \
\text { Range of the } y \text {-values: } & 9.3 - 15.6
\end{array}$$ Suppose a 95% prediction interval for y when x = 3.00 is (16, 21). Interpret the interval.
A) We are 95% confident that the starting salary of a Business School graduate with a GPA of 3.00 will fall between $16,000 and $21,000.
B) We are 95% confident that the mean starting salary of all Business School graduates with GPAs of 3.00 will fall between $16,000 and $21,000.
C) We are 95% confident that the starting salary of a Business School graduate will increase between $16,000 and $21,000 for every 3-point increase in GPA.
D) We are 95% confident that the starting salary of a Business School graduate will fall between $16,000 and $21,000.

Accepted Answer

A prediction interval provides a range in which we expect the starting salary of an individual graduate with a GPA of 3.00 to fall, with 95% confidence.

Question 4

A study of the top 75 MBA programs attempted to predict the average starting salary (in $1000's) of graduates of the program based on the amount of tuition (in $1000's) charged by the program. The results of a simple linear regression analysis are shown below: Least Squares Linear Regression of Salary Predictor
$\begin{array} { l r c c l } \text { Variables } & \text { Coefficient } & \text { Std Error } & \text { T } & \text { P } \ \text { Constant } & 18.1849 & 10.3336 & 1.76 & 0.0826 \ \text { Size } & 1.47494 & 0.14017 & 10.52 & 0.0000 \end{array}$

R-Squared $\quad 0.6027 \quad$ Resid. Mean Square (MSE) $532.986$
Adjusted R-Squared $0.5972$ Standard Deviation $23.0865$ The model was then used to create 95% confidence and prediction intervals for y and for E(Y) when the tuition charged by the MBA program was $75,000. The results are shown here: 95% confidence interval for E(Y): ($123,390, $134,220) 95% prediction interval for Y: ($82,476, $175,130) Which of the following interpretations is correct if you want to use the model to estimate E(Y) for all MBA programs?
A) We are 95% confident that the average starting salary for graduates of a single MBA program that charges $75,000 in tuition will fall between $123,390 and $134,220.
B) We are 95% confident that the average starting salary for graduates of a single MBA program that charges $75,000 in tuition will fall between $82,476 and $175,130.
C) We are 95% confident that the average of all starting salaries for graduates of all MBA programs that charge $75,000 in tuition will fall between $123,390 and $134,220.
D) We are 95% confident that the average of all starting salaries for graduates of all MBA programs that charge $75,000 in tuition will fall between $82,476 and $175,130.

Accepted Answer

The answer of A study of the top 75 MBA...

Question 5

The least squares model provides very good estimates of y for values of x far outside the range of x values contained in the sample.

Accepted Answer

The least squares model is only valid within the range of x values used to estimate the model. Extrapolation outside of this range can be unreliable and should be done with caution.

Question 6

What is the relationship between diamond price and carat size? 307 diamonds were sampled and a straight-line relationship was hypothesized between y = diamond price (in dollars) and x = size of the diamond (in carats). The simple linear regression for the analysis is shown below: Least Squares Linear Regression of PRICE $$\begin{array} { l c c c c c } 
\text { Predictor } & & & & \
\text { Variables } & \text { Coefficient } & \text { Std Error } & \text { T } & \text { P } & \
\text { Constant } & - 2298.36 & 158.531 & - 14.50 & 0.0000 & \
\text { Size } & 11598.9 & 230.111 & 50.41 & 0.0000 & \
& & & & & \
\text { R-Squared } & 0.8925 & \text { Resid. Mean Square (MSE) } & 1248950 \
\text { Adjusted R-Squared } & 0.8922 & { \text { Standard Deviation } } & 1117.56
\end{array}$$ Interpret the coefficient of determination for the regression model.
A) We can explain 89.25% of the variation in the sampled diamond prices around their mean using the size of the diamond in a linear model.
B) There is sufficient evidence to indicate that the size of the diamond is a useful predictor of the price of a diamond when testing at alpha = 0.05.
C) We expect most of the sampled diamond prices to fall within $2235.12 of their least squares predicted values.
D) For every 1-carat increase in the size of a diamond, we estimate that the price of the diamond will increase by $1117.56.
11.6 Using the Model for Estimation and Prediction
1 Calculate and Compare Confidence Intervals for Mean of y

Accepted Answer

The R-squared value of 0.8925 indicates that 89.25% of the variation in diamond prices can be explained by the size of the diamond in the linear regression model.

Question 7

An academic advisor wants to predict the typical starting salary of a graduate at a top business school using the GMAT score of the school as a predictor variable. A simple linear regression of SALARY versus GMAT using 25 data points is shown below. $$\hat { \beta } _ { 0 } = - 92040 \hat { \beta } 1 = 228 \mathrm {~s} = 3213 \mathrm { r } ^ { 2 } = .66 \quad r = .81 \mathrm { df } = 23 \quad t = 6.67$$
Give a practical interpretation of $r ^ { 2 } = .66$.
A) $66 \%$ of the sample variation in SALARY can be explained by using GMAT in a straight -line model.
B) We expect to predict SALARY to within $2 [ \sqrt { .66 } ]$ of its true value using GMAT in a straight-line model.
C) We estimate SALARY to increase $\$ .66$ for every 1 -point increase in GMAT.
D) We can predict SALARY correctly $66 \%$ of the time using GMAT in a straight-line model.

Accepted Answer

The answer of An academic advisor wants to predict the...

Question 8

Consider the following pairs of observations: $$\begin{array} { | l | l | l | l | l | l | } 
\hline x & 2 & 0 & 3 & 3 & 5 \
\hline y & 1 & 3 & 4 & 6 & 7 \
\hline
\end{array}$$ a. Construct a scattergram for the data. b. Find the least squares line, and plot it on your scattergram. c. Find a 99% confidence interval for the mean value of y when x = 1. d. Find a 99% prediction interval for a new value of y when x = 1.

Accepted Answer

The answer of Consider the following pairs of observations: \[\begin{array}...

Question 9

Probabilistic models are commonly used to estimate both the mean value of y and a new individual value of y for a particular value of x.

Accepted Answer

Probabilistic models allow for the estimation of the mean value of y for a given x, and also provide a probability distribution that can be used to estimate the likelihood of a new individual value of y for that same value of x.

Question 10

A study of the top 75 MBA programs attempted to predict the average starting salary (in $1000's) of graduates of the program based on the amount of tuition (in $1000's) charged by the program. The results of a simple linear regression analysis are shown below: Least Squares Linear Regression of Salary Predictor
$\begin{array} { l r r l l } \text { Variables } & \text { Coefficient } & \text { Std Error } & \text { T } & \text { P } \ \text { Constant } & 18.1849 & 10.3336 & 1.76 & 0.0826 \ \text { Size } & 1.47494 & 0.14017 & 10.52 & 0.0000 \end{array}$

$\begin{array}{lccc}
\text { R-Squared } & 0.6027 & \text { Resid. Mean Square (MSE) } & 532.986 \
\text { Adjusted R-Squared } &0.5972 & \text { Standard Deviation } & 23.0865
\end{array}$
 The model was then used to create 95% confidence and prediction intervals for y and for E(Y) when the tuition charged by the MBA program was $75,000. The results are shown here: 95% confidence interval for E(Y): ($123,390, $134,220) 95% prediction interval for Y: ($82,476, $175,130) Which of the following interpretations is correct if you want to use the model to predict Y for a single MBA programs?
A) We are 95% confident that the average starting salary for graduates of a single MBA program that charges $75,000 in tuition will fall between $123,390 and $134,220.
B) We are 95% confident that the average starting salary for graduates of a single MBA program that charges $75,000 in tuition will fall between $82,476 and $175,130.
C) We are 95% confident that the average of all starting salaries for graduates of all MBA programs that charge $75,000 in tuition will fall between $123,390 and $134,220.
D) We are 95% confident that the average of all starting salaries for graduates of all MBA programs that charge $75,000 in tuition will fall between $82,476 and $175,130.

Accepted Answer

The prediction interval is used to predict the salary for a single MBA program, which is wider due to more uncertainty compared to the confidence interval for the mean.

Question 11

What is the relationship between diamond price and carat size? 307 diamonds were sampled and a straight-line relationship was hypothesized between y = diamond price (in dollars) and x = size of the diamond (in carats). The simple linear regression for the analysis is shown below: Least Squares Linear Regression of PRICE $$\begin{array}{l}
\text { Predictor }\
\begin{array} { l c c c l l } 
\text { Variables } & \text { Coefficient } & \text { Std Error } & \text { T } & \text { P } \
\text { Constant } & - 2298.36 & 158.531 & - 14.50 & 0.0000 \
\text { Size } & 11598.9 & 230.111 & 50.41 & 0.0000
\end{array}\\
\begin{array} { l c c c } 
\text { R-Squared } & 0.8925 & \text { Resid. Mean Square (MSE) } & 1248950 \
\text { Adjusted R-Squared } & 0.8922 & \text { Standard Deviation } & 1117.56
\end{array}
\end{array}$$ The model was then used to create 95% confidence and prediction intervals for y and for E(Y) when the carat size of the diamond was 1 carat. The results are shown here: 95% confidence interval for E(Y): ($9091.60, $9509.40) 95% prediction interval for Y: ($7091.50, $11,510.00) Which of the following interpretations is correct if you want to use the model to estimate E(Y) for all 1-carat diamonds?
A) We are 95% confident that the price of a 1-carat diamond will fall between $7091.50 and $11,510.00.
B) We are 95% confident that the average price of all 1-carat diamonds will fall between $7091.50 and $11,510.00.
C) We are 95% confident that the price of a 1-carat diamond will fall between $9091.60 and $9509.40.
D) We are 95% confident that the average price of all 1-carat diamonds will fall between $9091.60 and $9509.40.

Accepted Answer

The 95% confidence interval for E(Y) is used to estimate the average price of all 1-carat diamonds, which is between $9091.60 and $9509.40.

Question 12

Consider the following pairs of observations: $$\begin{array} { | c | c | c | c | c | c | } 
\hline x & 2 & 3 & 5 & 5 & 6 \
\hline y & 1.3 & 1.6 & 2.1 & 2.2 & 2.7 \
\hline
\end{array}$$ Find and interpret the value of the coefficient of determination.

Accepted Answer

The answer of Consider the following pairs of observations: \[\begin{array}...

Question 13

The dean of the Business School at a small Florida college wishes to determine whether the grade -point average (GPA) of a graduating student can be used to predict the graduate's starting salary. More specifically, the dean wants to know whether higher GPAs lead to higher starting salaries. Records for 23 of last year's Business School graduates are selected at random, and data on GPA (x) and starting salary (y, in $thousands) for each graduate were used to fit the model $$E ( y ) = \beta _ { 0 } + \beta _ { 1 } x$$
The results of the simple linear regression are provided below.
$\begin{array}{ll}
\hat{y}=4.25+2.75 x, & S S_{x y}=5.15, S S_{x x}=1.87 \
S S y y=15.17, S S E=1.0075
\end{array}$

Calculate the value of $r ^ { 2 }$, the coefficient of determination.
A) 0.934
B) 0.661
C) 0.872
D) 0.339

Accepted Answer

The coefficient of determination is calculated as $r ^ { 2 } = \frac { S S _ { x y } ^ { 2 } } { S S _ { x x } S S _ { y y } } = \frac { 5.15 ^ { 2 } } { 1.87 \times 15.17 } = 0.934$.

Question 14

What is the relationship between diamond price and carat size? 307 diamonds were sampled and a straight-line relationship was hypothesized between y = diamond price (in dollars) and x = size of the diamond (in carats). The simple linear regression for the analysis is shown below: Least Squares Linear Regression of PRICE  The model was then used to create 95% confidence and prediction intervals for y and for E(Y) when the carat size of the diamond was 1 carat. The results are shown here: 95% confidence interval for E(Y): ($9091.60, $9509.40) 95% prediction interval for Y: ($7091.50, $11,510.00) Which of the following interpretations is correct if you want to use the model to determine the price of a single 1-carat diamond?
A) We are 95% confident that the price of a 1-carat diamond will fall between $7091.50 and $11,510.00.
B) We are 95% confident that the average price of all 1-carat diamonds will fall between $7091.50 and $11,510.00.
C) We are 95% confident that the price of a 1-carat diamond will fall between $9091.60 and $9509.40.
D) We are 95% confident that the average price of all 1-carat diamonds will fall between $9091.60 and $9509.40.

Accepted Answer

The answer of What is the relationship between diamond price...

Question 15

A company keeps extensive records on its new salespeople on the premise that sales should increase with experience. A random sample of seven new salespeople produced the data on experience and sales shown in the table. $\begin{array}{c|c}
\text { Months on Job } & \begin{array}{c}
\text { Monthly Sales } \
\text { (\$ thousands) }
\end{array} \
\hline 2 & 2.4 \
4 & 7.0 \
8 & 11.3 \
12 & 15.0 \
1 & .8 \
5 & 3.7 \
9 & 12.0
\end{array}$

Stumary statistics yield $S S _ { x x } = 94.8571 , S S _ { x y } = 124.7571 , S S _ { y y } = 176.5171 , \bar { x } = 5.8571$, and $\bar { y } = 7.4571$. Using $S S E = 12.435$, find and interpret the coefficient of determination.

Accepted Answer

The answer of A company keeps extensive records on its...

Question 16

A breeder of Thoroughbred horses wishes to model the relationship between the gestation period and the length of life of a horse. The breeder believes that the two variables may follow a linear trend. The information in the table was supplied to the breeder from various thoroughbred stables across the state. $\begin{array}{cccccc}
\begin{array}{cccc}
\text { Horse }
\end{array} & \begin{array}{c}
\text { Gestation } \
\text { period }\
x \text { (days) }  
\end{array} & \begin{array}{c}
\text { Life } \
\text { Length }\
 y \text { (years) }  
\end{array} & \text { Horse } & \begin{array}{c}
\text { Gestation } \
\text { period } \
x \text { (days) }  
\end{array} & \begin{array}{c}
\text { Life } \
\text { Length }\
 y \text { (years) }  
\end{array} \
1 & 416 & 24 & 5 & 356 & 22 \
2 & 279 & 25.5 & 6 & 403 & 23.5 \
3 & 298 & 20 & 7 & 265 & 21 \
4 & 307 & 21.5 & & &
\end{array}$

Summary statistics yield $S S _ { x x } = 21,752 , S S _ { x y } = 236.5 , S S _ { y y } = 22 , \bar { x } = 332$, and $\bar { y } = 22.5$. Find a $95 \%$ prediction interval for the length of life of a horse that had a gestation period of 300 days. Use $s = 2$ as an estimate of $\sigma$ and use $y = 18.89 + .01087 x$

Accepted Answer

The answer of A breeder of Thoroughbred horses wishes to...

Question 17

A manufacturer of boiler drums wants to use regression to predict the number of man-hours needed to erect drums in the future. The manufacturer collected a random sample of 35 boilers and measured the following two variables: MANHRS: $\quad y =$ Number of man-hours required to erect the drum
PRESSURE: $\quad x =$ Boiler design pressure (pounds per square inch, i.e., $\mathrm { psi }$ )
The simple linear model $E ( y ) = \beta _ { 1 } + \beta _ { 1 } x$ was fit to the data. A printout for the analysis appears below:
UNWEIGHTED LEAST SQUARES LINEAR REGRESSION OF MANHRS
$\begin{array}{c|c|c|c|c}
\text { PREDICTOR } & & & & \
\text { VARIABLES } & \text { COEFFICIENT } & \text { STD ERROR } & \text { STUDENT'S T } & \text { P } \
\hline \text { CONSTANT } & 1.88059 & 0.58380 & 3.22 & 0.0028 \
\text { PRESSURE } & 0.00321 & 0.00163 & 2.17 & 0.0300
\end{array}$

$\begin{array} { l l l l } \text { R-SQUARED } & 0.4342 & \text { RESID. MEAN SQUARE (MSE) } & 4.25460 \ \text { ADJUSTED R-SQUARED } & 0.4176 & \text { STANDARD DEVIATION } & 2.06267 \end{array}$

$\begin{array}{l|r|c|c|c|c}
\text { SOURCE } & \text { DF } & {\text { SS }} & \text { MS } & \mathrm{F} & \mathrm{P} \
\hline \text { REGRESSION } & 1 & 111.008 & 111.008 & 5.19 & 0.0300 \
\text { RESIDUAL } & 34 & 144.656 & 4.25160 & & \
\text { TOTAL } & 35 & 255.665 & & &
\end{array}$

Give a practical interpretation of the coefficient of determination, $r ^ { 2 }$.
A) About 43% of the sample variation in number of man-hours can be explained by the simple linear model.
B) We are 43% confident that the design pressure will be a useful predictor of number of man-hours required to build a steam drum.
C) Approximately 95% of the actual man-hours required to build a drum will fall within 43 hours of their predicted values.
D) About 2.06% of the sample variation in number of man-hours can be explained by the simple linear model.

Accepted Answer

About 43% of the sample variation in number of man-hours can be explained by the simple linear model.

Question 18

In a comprehensive road test on new car models, one variable measured is the time it takes the car to accelerate from 0 to 60 miles per hour. To model acceleration time, a regression analysis is conducted on a random sample of 129 new cars. TIME60: $\quad y =$ Elapsed time (in seconds) from $0 \mathrm { mph }$ to $60 \mathrm { mph }$
MAX $x =$ Maximum speed attained (miles per hour)
The simple linear model $E ( y ) = \beta _ { 0 } + \beta _ { 1 } x$ was fit to the data. Computer printouts for the analysis are given below:

NWEIGHTED LEAST SQUARES LINEAR REGRESSION OF TIME 60

$\begin{array}{l|c|c|c|c}
\text { PREDICTOR } & & & & \
\text { VARIABLES } & \text { COEFFICIENT } & \text { STD ERROR } & \text { STUDENT S T } & \text { P } \
\hline \text { CONSTANT } & 18.7171 & 0.63708 & 29.38 & 0.0000 \
\text { MAX } & -0.08365 & 0.00491 & -17.05 & 0.0000
\end{array}$

$\begin{array}{llll}
\text { R-SQUARED } & 0.6960 & \text { RESID. MEAN SQUARE (MSE) } & 1.28695 \
\text { ADJUSED R-SQUARED } & 0.6937 & \text { STANDARD DEVIATION } & 1.13444
\end{array}$

$\begin{array}{l|r|r|c|c|c}
\text { SOURCE } & \text { DF } & {\text { SS }} & \text { MS } & \text { F } & \text { P } \
\hline \text { REGRESSION } & 1 & 374.285 & 374.285 & 290.83 & 0.0000 \
\text { RESIDUAL } & 127 & 163.443 & 1.28695 & & \
\text { TOTAL } & 128 & 537.728 & & &
\end{array}$

CASES INCLUDED 129 MISSING CASES 0 Approximately what percentage of the sample variation in acceleration time can be explained by the simple linear model?
A) 70%
B) 0%
C) -17%
D) 8%

Accepted Answer

The R-squared value is 0.6960, which indicates that approximately 69.60% of the variation in acceleration time can be explained by the model. This is closest to 70%.

Question 19

A company keeps extensive records on its new salespeople on the premise that sales should increase with experience. A random sample of seven new salespeople produced the data on experience and sales shown in the table. $\begin{array}{c|c}
\text { Months on Job } & \begin{array}{c}
\text { Monthly Sales } \
\text { (\$ thousands) }
\end{array} \
\hline 2 & 2.4 \
4 & 7.0 \
8 & 11.3 \
12 & 15.0 \
1 & .8 \
5 & 3.7 \
9 & 12.0
\end{array}$

Summary statistics yield $S S _ { x x } = 94.8571 , S S _ { x y } = 124.7571 , S S _ { y y } = 176.5171 , \bar { x } = 5.8571$, and $\bar { y } = 7.4571$. Calculate a $90 \%$ confidence interval for $E ( y )$ when $x = 5$ months. Assume $s = 1.577$ and the prediction equation is $\hat { y } = - .25 + 1.315 x$.

Accepted Answer

The answer of A company keeps extensive records on its...

Question 20

A realtor collected the following data for a random sample of ten homes that recently sold in her area. $$\begin{array} { | c | c | c | } 
\hline \text { House } & \text { Asking Price } & \text { Days on Market } \
\hline \text { A } & \$ 114,500 & 29 \
\hline \text { B } & \$ 149,900 & 16 \
\hline \text { C } & \$ 154,700 & 59 \
\hline \text { D } & \$ 159,900 & 42 \
\hline \text { E } & \$ 160,000 & 72 \
\hline \text { F } & \$ 165,900 & 45 \
\hline \text { G } & \$ 169,700 & 12 \
\hline \text { H } & \$ 171,900 & 39 \
\hline \text { I } & \$ 175,000 & 81 \
\hline \text { J } & \$ 289,900 & 121 \
\hline
\end{array}$$ a. Find a 90% confidence interval for the mean number of days on the market for all houses listed at $150,000. b. Suppose a house has just been listed at $150,000. Find a 90% prediction interval for the number of days the house will be on the market before it sells.

Accepted Answer

The answer of A realtor collected the following data for...

Quiz 11: Simple Linear Regression

The least squares model provides very good estimates of y for values of x far outside the range of x values contained in the sample.

Probabilistic models are commonly used to estimate both the mean value of y and a new individual value of y for a particular value of x.

Consider the following pairs of observations: $\begin{array} { | c | c | c | c | c | c | } \hline x & 2 & 3 & 5 & 5 & 6 \\\hline y & 1.3 & 1.6 & 2.1 & 2.2 & 2.7 \\\hline\end{array}$ Find and interpret the value of the coefficient of determination.