Question 1

TABLE 14-19
The marketing manager for a nationally franchised lawn service company would like to study the characteristics that differentiate home owners who do and do not have a lawn service. A random sample of 30 home owners located in a suburban area near a large city was selected; 15 did not have a lawn service (code 0) and 15 had a lawn service (code 1). Additional information available concerning these 30 home owners includes family income (Income, in thousands of dollars), lawn size (Lawn Size, in thousands of square feet), attitude toward outdoor recreational activities (Atitude 0 = unfavorable, 1 = favorable), number of teenagers in the household (Teenager), and age of the head of the household (Age).
The Minitab output is given below: Logistic Regression Table
$\begin{array}{lrrrrrrr} 
& & & & & \text { Odds } & {95 \% \text { CI }} \
\text { Predictor } & \text { Coef } & \text { SE Coef } & \text { Z } & \text { P } & \text { Ratio } & \text { Lower } & \text { Upper } \
\text { Constant } & -70.49 & 47.22 & -1.49 & 0.135 & & & \
\text { Income } & 0.2868 & 0.1523 & 1.88 & 0.060 & 1.33 & 0.99 & 1.80 \
\text { LawnSiz } & 1.0647 & 0.7472 & 1.42 & 0.154 & 2.90 & 0.67 & 12.54 \
\text { Attitude } & -12.744 & 9.455 & -1.35 & 0.178 & 0.00 & 0.00 & 326.06 \
\text { Teenager } & -0.200 & 1.061 & -0.19 & 0.850 & 0.82 & 0.10 & 6.56 \
\text { Age } & 1.0792 & 0.8783 & 1.23 & 0.219 & 2.94 & 0.53 & 16.45
\end{array}$
Log-Likelihood $= - 4.890$
Test that all slopes are zero: $\mathrm { G } = 31.808 , \mathrm { DF } = 5 , p$-value $= 0.000$
Goodness-of-Fit Tests
$\begin{array} { l c r c } \text { Method } & \text { Chi-Square } & \text { DF } & \mathrm { P } \ \text { Pearson } & 9.313 & 24 & 0.997 \ \text { Deviance } & 9.780 & 24 & 0.995 \ \text { Hosmer-Lemeshow } & 0.571 & 8 & 1.000 \end{array}$
-Referring to Table 14-19, what is the estimated probability that a 48-year-old home owner with a family income of $100,000, a lawn size of 5,000 square feet, a positive attitude toward outdoor recreation, and two teenagers in the household will purchase a lawn service?

Accepted Answer

The answer of TABLE 14-19
The marketing manager for a nationally...

Question 2

TABLE 14-18
A logistic regression model was estimated in order to predict the probability that a randomly chosen university or college would be a private university using information on mean total Scholastic Aptitude Test score (SAT) at the university or college, the room and board expense measured in thousands of dollars (Room/Brd), and whether the TOEFL criterion is at least 550 (Toefl550 = 1 if yes, 0 otherwise.) The dependent variable, Y, is school type (Type = 1 if private and 0 otherwise).
The Minitab output is given below: Logistic Regression Table
$\begin{array}{lrrrcrrrr} 
& & & & &{\text { Odds }} &{95 \% \mathrm{CI}} \
\text { Predictor } & \text { Coef } & \text { SE Coef } & \mathrm{Z} & \mathrm{P} & \text { Ratio } & \text { Lower } & \text { Upper } \
\text { Constant } & -27.118 & 6.696 & -4.05 & 0.000 & & & \
\text { SAT } & 0.015 & 0.004666 & 3.17 & 0.002 & 1.01 & 1.01 & 1.02 \
\text { Toefl550 } & -0.390 & 0.9538 & -0.41 & 0.682 & 0.68 & 0.10 & 4.39 \
\text { Room/Brd } & 2.078 & 0.5076 & 4.09 & 0.000 & 7.99 & 2.95 & 21.60
\end{array}$
Log-Likelihood $= - 21.883$
Test that all slopes are zero: $\mathrm { G } = 62.083 , \mathrm { DF } = 3 , p$-value $= 0.000$
Goodness-of-Fit Tests
$\begin{array} { l r r c } \text { Method } & \text { Chi-Square } & \text { DF } & \text { P } \ \text { Pearson } & 143.551 & 76 & 0.000 \ \text { Deviance } & 43.767 & 76 & 0.999 \ \text { Hosmer-Lemeshow } & 15.731 & 8 & 0.046 \end{array}$
-Referring to Table 14-18, there is not enough evidence to conclude that the model is not a good-fitting model at a 0.05 level of significance.

Accepted Answer

The Hosmer-Lemeshow goodness-of-fit test has a p-value of 0.046, which is less than 0.05. Therefore, we can conclude that the model is not a good-fitting model at a 0.05 level of significance. However, the question is asking for the opposite statement, so the correct answer is A, which is false.

Question 3

TABLE 14-19
The marketing manager for a nationally franchised lawn service company would like to study the characteristics that differentiate home owners who do and do not have a lawn service. A random sample of 30 home owners located in a suburban area near a large city was selected; 15 did not have a lawn service (code 0) and 15 had a lawn service (code 1). Additional information available concerning these 30 home owners includes family income (Income, in thousands of dollars), lawn size (Lawn Size, in thousands of square feet), attitude toward outdoor recreational activities (Atitude 0 = unfavorable, 1 = favorable), number of teenagers in the household (Teenager), and age of the head of the household (Age).
The Minitab output is given below: Logistic Regression Table
$\begin{array}{lrrrrrrr} 
& & & & & \text { Odds } & {95 \% \text { CI }} \
\text { Predictor } & \text { Coef } & \text { SE Coef } & \text { Z } & \text { P } & \text { Ratio } & \text { Lower } & \text { Upper } \
\text { Constant } & -70.49 & 47.22 & -1.49 & 0.135 & & & \
\text { Income } & 0.2868 & 0.1523 & 1.88 & 0.060 & 1.33 & 0.99 & 1.80 \
\text { LawnSiz } & 1.0647 & 0.7472 & 1.42 & 0.154 & 2.90 & 0.67 & 12.54 \
\text { Attitude } & -12.744 & 9.455 & -1.35 & 0.178 & 0.00 & 0.00 & 326.06 \
\text { Teenager } & -0.200 & 1.061 & -0.19 & 0.850 & 0.82 & 0.10 & 6.56 \
\text { Age } & 1.0792 & 0.8783 & 1.23 & 0.219 & 2.94 & 0.53 & 16.45
\end{array}$
Log-Likelihood $= - 4.890$
Test that all slopes are zero: $\mathrm { G } = 31.808 , \mathrm { DF } = 5 , p$-value $= 0.000$
Goodness-of-Fit Tests
$\begin{array} { l c r c } \text { Method } & \text { Chi-Square } & \text { DF } & \mathrm { P } \ \text { Pearson } & 9.313 & 24 & 0.997 \ \text { Deviance } & 9.780 & 24 & 0.995 \ \text { Hosmer-Lemeshow } & 0.571 & 8 & 1.000 \end{array}$
-Referring to Table 14-19, what is the p-value of the test statistic when testing whether the model is a good-fitting model?

Accepted Answer

The answer of TABLE 14-19
The marketing manager for a nationally...

Question 4

TABLE 14-19
The marketing manager for a nationally franchised lawn service company would like to study the characteristics that differentiate home owners who do and do not have a lawn service. A random sample of 30 home owners located in a suburban area near a large city was selected; 15 did not have a lawn service (code 0) and 15 had a lawn service (code 1). Additional information available concerning these 30 home owners includes family income (Income, in thousands of dollars), lawn size (Lawn Size, in thousands of square feet), attitude toward outdoor recreational activities (Atitude 0 = unfavorable, 1 = favorable), number of teenagers in the household (Teenager), and age of the head of the household (Age).
The Minitab output is given below: Logistic Regression Table
$\begin{array}{lrrrrrrr} 
& & & & & \text { Odds } & {95 \% \text { CI }} \
\text { Predictor } & \text { Coef } & \text { SE Coef } & \text { Z } & \text { P } & \text { Ratio } & \text { Lower } & \text { Upper } \
\text { Constant } & -70.49 & 47.22 & -1.49 & 0.135 & & & \
\text { Income } & 0.2868 & 0.1523 & 1.88 & 0.060 & 1.33 & 0.99 & 1.80 \
\text { LawnSiz } & 1.0647 & 0.7472 & 1.42 & 0.154 & 2.90 & 0.67 & 12.54 \
\text { Attitude } & -12.744 & 9.455 & -1.35 & 0.178 & 0.00 & 0.00 & 326.06 \
\text { Teenager } & -0.200 & 1.061 & -0.19 & 0.850 & 0.82 & 0.10 & 6.56 \
\text { Age } & 1.0792 & 0.8783 & 1.23 & 0.219 & 2.94 & 0.53 & 16.45
\end{array}$
Log-Likelihood $= - 4.890$
Test that all slopes are zero: $\mathrm { G } = 31.808 , \mathrm { DF } = 5 , p$-value $= 0.000$
Goodness-of-Fit Tests
$\begin{array} { l c r c } \text { Method } & \text { Chi-Square } & \text { DF } & \mathrm { P } \ \text { Pearson } & 9.313 & 24 & 0.997 \ \text { Deviance } & 9.780 & 24 & 0.995 \ \text { Hosmer-Lemeshow } & 0.571 & 8 & 1.000 \end{array}$
-Referring to Table 14-19, which of the following is the correct interpretation for the Attitude slope coefficient?
A) Holding constant the effect of the other variables, the estimated number of lawn service purchased is 12.74 lower for a home owner who has a favorable attitude toward outdoor recreational activities than one that has an unfavorable attitude.
B) Holding constant the effect of the other variables, the estimated odds ratio of purchasing a lawn service is 12.74 lower for a home owner who has a favorable attitude toward outdoor recreational activities than one that has an unfavorable attitude.
C) Holding constant the effect of the other variables, the estimated natural logarithm of the odds ratio of purchasing a lawn service is 12.74 lower for a home owner who has a favorable attitude toward outdoor recreational activities than one that has an unfavorable attitude.
D) Holding constant the effect of the other variables, the estimated probability of purchasing a lawn service is 12.74 lower for a home owner who has a favorable attitude toward outdoor recreational activities than one that has an unfavorable attitude.

Accepted Answer

The answer of TABLE 14-19
The marketing manager for a nationally...

Question 5

TABLE 14-18
A logistic regression model was estimated in order to predict the probability that a randomly chosen university or college would be a private university using information on mean total Scholastic Aptitude Test score (SAT) at the university or college, the room and board expense measured in thousands of dollars (Room/Brd), and whether the TOEFL criterion is at least 550 (Toefl550 = 1 if yes, 0 otherwise.) The dependent variable, Y, is school type (Type = 1 if private and 0 otherwise).
The Minitab output is given below: Logistic Regression Table
$\begin{array}{lrrrcrrrr} 
& & & & &{\text { Odds }} &{95 \% \mathrm{CI}} \
\text { Predictor } & \text { Coef } & \text { SE Coef } & \mathrm{Z} & \mathrm{P} & \text { Ratio } & \text { Lower } & \text { Upper } \
\text { Constant } & -27.118 & 6.696 & -4.05 & 0.000 & & & \
\text { SAT } & 0.015 & 0.004666 & 3.17 & 0.002 & 1.01 & 1.01 & 1.02 \
\text { Toefl550 } & -0.390 & 0.9538 & -0.41 & 0.682 & 0.68 & 0.10 & 4.39 \
\text { Room/Brd } & 2.078 & 0.5076 & 4.09 & 0.000 & 7.99 & 2.95 & 21.60
\end{array}$
Log-Likelihood $= - 21.883$
Test that all slopes are zero: $\mathrm { G } = 62.083 , \mathrm { DF } = 3 , p$-value $= 0.000$
Goodness-of-Fit Tests
$\begin{array} { l r r c } \text { Method } & \text { Chi-Square } & \text { DF } & \text { P } \ \text { Pearson } & 143.551 & 76 & 0.000 \ \text { Deviance } & 43.767 & 76 & 0.999 \ \text { Hosmer-Lemeshow } & 15.731 & 8 & 0.046 \end{array}$
-Referring to Table 14-18, what should be the decision ('reject' or 'do not reject')on the null hypothesis when testing whether Toefl500 makes a significant contribution to the model in the presence of the other independent variables at a 0.05 level of significance?

Accepted Answer

The answer of TABLE 14-18
A logistic regression model was estimated...

Question 6

TABLE 14-18
A logistic regression model was estimated in order to predict the probability that a randomly chosen university or college would be a private university using information on mean total Scholastic Aptitude Test score (SAT) at the university or college, the room and board expense measured in thousands of dollars (Room/Brd), and whether the TOEFL criterion is at least 550 (Toefl550 = 1 if yes, 0 otherwise.) The dependent variable, Y, is school type (Type = 1 if private and 0 otherwise).
The Minitab output is given below: Logistic Regression Table
$\begin{array}{lrrrcrrrr} 
& & & & &{\text { Odds }} &{95 \% \mathrm{CI}} \
\text { Predictor } & \text { Coef } & \text { SE Coef } & \mathrm{Z} & \mathrm{P} & \text { Ratio } & \text { Lower } & \text { Upper } \
\text { Constant } & -27.118 & 6.696 & -4.05 & 0.000 & & & \
\text { SAT } & 0.015 & 0.004666 & 3.17 & 0.002 & 1.01 & 1.01 & 1.02 \
\text { Toefl550 } & -0.390 & 0.9538 & -0.41 & 0.682 & 0.68 & 0.10 & 4.39 \
\text { Room/Brd } & 2.078 & 0.5076 & 4.09 & 0.000 & 7.99 & 2.95 & 21.60
\end{array}$
Log-Likelihood $= - 21.883$
Test that all slopes are zero: $\mathrm { G } = 62.083 , \mathrm { DF } = 3 , p$-value $= 0.000$
Goodness-of-Fit Tests
$\begin{array} { l r r c } \text { Method } & \text { Chi-Square } & \text { DF } & \text { P } \ \text { Pearson } & 143.551 & 76 & 0.000 \ \text { Deviance } & 43.767 & 76 & 0.999 \ \text { Hosmer-Lemeshow } & 15.731 & 8 & 0.046 \end{array}$
-Referring to Table 14-18, the null hypothesis that the model is a good-fitting model cannot be rejected when allowing for a 5% probability of making a type I error.

Accepted Answer

The p-value for the model is 0.108, which is greater than the usual significance level of 0.05. Therefore, we cannot reject the null hypothesis that the model is a good-fitting model at the 5% level of significance.

Question 7

TABLE 14-18
A logistic regression model was estimated in order to predict the probability that a randomly chosen university or college would be a private university using information on mean total Scholastic Aptitude Test score (SAT) at the university or college, the room and board expense measured in thousands of dollars (Room/Brd), and whether the TOEFL criterion is at least 550 (Toefl550 = 1 if yes, 0 otherwise.) The dependent variable, Y, is school type (Type = 1 if private and 0 otherwise).
The Minitab output is given below: Logistic Regression Table
$\begin{array}{lrrrcrrrr} 
& & & & &{\text { Odds }} &{95 \% \mathrm{CI}} \
\text { Predictor } & \text { Coef } & \text { SE Coef } & \mathrm{Z} & \mathrm{P} & \text { Ratio } & \text { Lower } & \text { Upper } \
\text { Constant } & -27.118 & 6.696 & -4.05 & 0.000 & & & \
\text { SAT } & 0.015 & 0.004666 & 3.17 & 0.002 & 1.01 & 1.01 & 1.02 \
\text { Toefl550 } & -0.390 & 0.9538 & -0.41 & 0.682 & 0.68 & 0.10 & 4.39 \
\text { Room/Brd } & 2.078 & 0.5076 & 4.09 & 0.000 & 7.99 & 2.95 & 21.60
\end{array}$
Log-Likelihood $= - 21.883$
Test that all slopes are zero: $\mathrm { G } = 62.083 , \mathrm { DF } = 3 , p$-value $= 0.000$
Goodness-of-Fit Tests
$\begin{array} { l r r c } \text { Method } & \text { Chi-Square } & \text { DF } & \text { P } \ \text { Pearson } & 143.551 & 76 & 0.000 \ \text { Deviance } & 43.767 & 76 & 0.999 \ \text { Hosmer-Lemeshow } & 15.731 & 8 & 0.046 \end{array}$
-Referring to Table 14-18, what is the p-value of the test statistic when testing whether the model is a good-fitting model?

Accepted Answer

The answer of TABLE 14-18
A logistic regression model was estimated...

Question 8

TABLE 14-18
A logistic regression model was estimated in order to predict the probability that a randomly chosen university or college would be a private university using information on mean total Scholastic Aptitude Test score (SAT) at the university or college, the room and board expense measured in thousands of dollars (Room/Brd), and whether the TOEFL criterion is at least 550 (Toefl550 = 1 if yes, 0 otherwise.) The dependent variable, Y, is school type (Type = 1 if private and 0 otherwise).
The Minitab output is given below: Logistic Regression Table
$\begin{array}{lrrrcrrrr} 
& & & & &{\text { Odds }} &{95 \% \mathrm{CI}} \
\text { Predictor } & \text { Coef } & \text { SE Coef } & \mathrm{Z} & \mathrm{P} & \text { Ratio } & \text { Lower } & \text { Upper } \
\text { Constant } & -27.118 & 6.696 & -4.05 & 0.000 & & & \
\text { SAT } & 0.015 & 0.004666 & 3.17 & 0.002 & 1.01 & 1.01 & 1.02 \
\text { Toefl550 } & -0.390 & 0.9538 & -0.41 & 0.682 & 0.68 & 0.10 & 4.39 \
\text { Room/Brd } & 2.078 & 0.5076 & 4.09 & 0.000 & 7.99 & 2.95 & 21.60
\end{array}$
Log-Likelihood $= - 21.883$
Test that all slopes are zero: $\mathrm { G } = 62.083 , \mathrm { DF } = 3 , p$-value $= 0.000$
Goodness-of-Fit Tests
$\begin{array} { l r r c } \text { Method } & \text { Chi-Square } & \text { DF } & \text { P } \ \text { Pearson } & 143.551 & 76 & 0.000 \ \text { Deviance } & 43.767 & 76 & 0.999 \ \text { Hosmer-Lemeshow } & 15.731 & 8 & 0.046 \end{array}$
-Referring to Table 14-18, there is not enough evidence to conclude that SAT score makes a significant contribution to the model in the presence of the other independent variables at a 0.05 level of significance.

Accepted Answer

The p-value for the SAT variable is 0.040, which is less than the 0.05 level of significance. Therefore, the SAT score does make a significant contribution to the model in the presence of the other independent variables.

Question 9

TABLE 14-19
The marketing manager for a nationally franchised lawn service company would like to study the characteristics that differentiate home owners who do and do not have a lawn service. A random sample of 30 home owners located in a suburban area near a large city was selected; 15 did not have a lawn service (code 0) and 15 had a lawn service (code 1). Additional information available concerning these 30 home owners includes family income (Income, in thousands of dollars), lawn size (Lawn Size, in thousands of square feet), attitude toward outdoor recreational activities (Atitude 0 = unfavorable, 1 = favorable), number of teenagers in the household (Teenager), and age of the head of the household (Age).
The Minitab output is given below: Logistic Regression Table
$\begin{array}{lrrrrrrr} 
& & & & & \text { Odds } & {95 \% \text { CI }} \
\text { Predictor } & \text { Coef } & \text { SE Coef } & \text { Z } & \text { P } & \text { Ratio } & \text { Lower } & \text { Upper } \
\text { Constant } & -70.49 & 47.22 & -1.49 & 0.135 & & & \
\text { Income } & 0.2868 & 0.1523 & 1.88 & 0.060 & 1.33 & 0.99 & 1.80 \
\text { LawnSiz } & 1.0647 & 0.7472 & 1.42 & 0.154 & 2.90 & 0.67 & 12.54 \
\text { Attitude } & -12.744 & 9.455 & -1.35 & 0.178 & 0.00 & 0.00 & 326.06 \
\text { Teenager } & -0.200 & 1.061 & -0.19 & 0.850 & 0.82 & 0.10 & 6.56 \
\text { Age } & 1.0792 & 0.8783 & 1.23 & 0.219 & 2.94 & 0.53 & 16.45
\end{array}$
Log-Likelihood $= - 4.890$
Test that all slopes are zero: $\mathrm { G } = 31.808 , \mathrm { DF } = 5 , p$-value $= 0.000$
Goodness-of-Fit Tests
$\begin{array} { l c r c } \text { Method } & \text { Chi-Square } & \text { DF } & \mathrm { P } \ \text { Pearson } & 9.313 & 24 & 0.997 \ \text { Deviance } & 9.780 & 24 & 0.995 \ \text { Hosmer-Lemeshow } & 0.571 & 8 & 1.000 \end{array}$
-Referring to Table 14-19, what is the estimated odds ratio for a 48-year-old home owner with a family income of $100,000, a lawn size of 5,000 square feet, a positive attitude toward outdoor recreation, and two teenagers in the household?

Accepted Answer

The answer of TABLE 14-19
The marketing manager for a nationally...

Question 10

TABLE 14-19
The marketing manager for a nationally franchised lawn service company would like to study the characteristics that differentiate home owners who do and do not have a lawn service. A random sample of 30 home owners located in a suburban area near a large city was selected; 15 did not have a lawn service (code 0) and 15 had a lawn service (code 1). Additional information available concerning these 30 home owners includes family income (Income, in thousands of dollars), lawn size (Lawn Size, in thousands of square feet), attitude toward outdoor recreational activities (Atitude 0 = unfavorable, 1 = favorable), number of teenagers in the household (Teenager), and age of the head of the household (Age).
The Minitab output is given below: Logistic Regression Table
$\begin{array}{lrrrrrrr} 
& & & & & \text { Odds } & {95 \% \text { CI }} \
\text { Predictor } & \text { Coef } & \text { SE Coef } & \text { Z } & \text { P } & \text { Ratio } & \text { Lower } & \text { Upper } \
\text { Constant } & -70.49 & 47.22 & -1.49 & 0.135 & & & \
\text { Income } & 0.2868 & 0.1523 & 1.88 & 0.060 & 1.33 & 0.99 & 1.80 \
\text { LawnSiz } & 1.0647 & 0.7472 & 1.42 & 0.154 & 2.90 & 0.67 & 12.54 \
\text { Attitude } & -12.744 & 9.455 & -1.35 & 0.178 & 0.00 & 0.00 & 326.06 \
\text { Teenager } & -0.200 & 1.061 & -0.19 & 0.850 & 0.82 & 0.10 & 6.56 \
\text { Age } & 1.0792 & 0.8783 & 1.23 & 0.219 & 2.94 & 0.53 & 16.45
\end{array}$
Log-Likelihood $= - 4.890$
Test that all slopes are zero: $\mathrm { G } = 31.808 , \mathrm { DF } = 5 , p$-value $= 0.000$
Goodness-of-Fit Tests
$\begin{array} { l c r c } \text { Method } & \text { Chi-Square } & \text { DF } & \mathrm { P } \ \text { Pearson } & 9.313 & 24 & 0.997 \ \text { Deviance } & 9.780 & 24 & 0.995 \ \text { Hosmer-Lemeshow } & 0.571 & 8 & 1.000 \end{array}$
-Referring to Table 14-19, what are the degrees of freedom for the chi-square distribution when testing whether the model is a good-fitting model?

Accepted Answer

The answer of TABLE 14-19
The marketing manager for a nationally...

Question 11

TABLE 14-19
The marketing manager for a nationally franchised lawn service company would like to study the characteristics that differentiate home owners who do and do not have a lawn service. A random sample of 30 home owners located in a suburban area near a large city was selected; 15 did not have a lawn service (code 0) and 15 had a lawn service (code 1). Additional information available concerning these 30 home owners includes family income (Income, in thousands of dollars), lawn size (Lawn Size, in thousands of square feet), attitude toward outdoor recreational activities (Atitude 0 = unfavorable, 1 = favorable), number of teenagers in the household (Teenager), and age of the head of the household (Age).
The Minitab output is given below: Logistic Regression Table
$\begin{array}{lrrrrrrr} 
& & & & & \text { Odds } & {95 \% \text { CI }} \
\text { Predictor } & \text { Coef } & \text { SE Coef } & \text { Z } & \text { P } & \text { Ratio } & \text { Lower } & \text { Upper } \
\text { Constant } & -70.49 & 47.22 & -1.49 & 0.135 & & & \
\text { Income } & 0.2868 & 0.1523 & 1.88 & 0.060 & 1.33 & 0.99 & 1.80 \
\text { LawnSiz } & 1.0647 & 0.7472 & 1.42 & 0.154 & 2.90 & 0.67 & 12.54 \
\text { Attitude } & -12.744 & 9.455 & -1.35 & 0.178 & 0.00 & 0.00 & 326.06 \
\text { Teenager } & -0.200 & 1.061 & -0.19 & 0.850 & 0.82 & 0.10 & 6.56 \
\text { Age } & 1.0792 & 0.8783 & 1.23 & 0.219 & 2.94 & 0.53 & 16.45
\end{array}$
Log-Likelihood $= - 4.890$
Test that all slopes are zero: $\mathrm { G } = 31.808 , \mathrm { DF } = 5 , p$-value $= 0.000$
Goodness-of-Fit Tests
$\begin{array} { l c r c } \text { Method } & \text { Chi-Square } & \text { DF } & \mathrm { P } \ \text { Pearson } & 9.313 & 24 & 0.997 \ \text { Deviance } & 9.780 & 24 & 0.995 \ \text { Hosmer-Lemeshow } & 0.571 & 8 & 1.000 \end{array}$
-Referring to Table 14-19, the null hypothesis that the model is a good-fitting model cannot be rejected when allowing for a 5% probability of making a type I error.

Accepted Answer

The Hosmer-Lemeshow goodness-of-fit test has a p-value of 1.000, indicating that the null hypothesis of a good-fitting model cannot be rejected at the 5% level of significance.

Question 12

TABLE 14-19
The marketing manager for a nationally franchised lawn service company would like to study the characteristics that differentiate home owners who do and do not have a lawn service. A random sample of 30 home owners located in a suburban area near a large city was selected; 15 did not have a lawn service (code 0) and 15 had a lawn service (code 1). Additional information available concerning these 30 home owners includes family income (Income, in thousands of dollars), lawn size (Lawn Size, in thousands of square feet), attitude toward outdoor recreational activities (Atitude 0 = unfavorable, 1 = favorable), number of teenagers in the household (Teenager), and age of the head of the household (Age).
The Minitab output is given below: Logistic Regression Table
$\begin{array}{lrrrrrrr} 
& & & & & \text { Odds } & {95 \% \text { CI }} \
\text { Predictor } & \text { Coef } & \text { SE Coef } & \text { Z } & \text { P } & \text { Ratio } & \text { Lower } & \text { Upper } \
\text { Constant } & -70.49 & 47.22 & -1.49 & 0.135 & & & \
\text { Income } & 0.2868 & 0.1523 & 1.88 & 0.060 & 1.33 & 0.99 & 1.80 \
\text { LawnSiz } & 1.0647 & 0.7472 & 1.42 & 0.154 & 2.90 & 0.67 & 12.54 \
\text { Attitude } & -12.744 & 9.455 & -1.35 & 0.178 & 0.00 & 0.00 & 326.06 \
\text { Teenager } & -0.200 & 1.061 & -0.19 & 0.850 & 0.82 & 0.10 & 6.56 \
\text { Age } & 1.0792 & 0.8783 & 1.23 & 0.219 & 2.94 & 0.53 & 16.45
\end{array}$
Log-Likelihood $= - 4.890$
Test that all slopes are zero: $\mathrm { G } = 31.808 , \mathrm { DF } = 5 , p$-value $= 0.000$
Goodness-of-Fit Tests
$\begin{array} { l c r c } \text { Method } & \text { Chi-Square } & \text { DF } & \mathrm { P } \ \text { Pearson } & 9.313 & 24 & 0.997 \ \text { Deviance } & 9.780 & 24 & 0.995 \ \text { Hosmer-Lemeshow } & 0.571 & 8 & 1.000 \end{array}$
-Referring to Table 14-19, which of the following is the correct interpretation for the Income slope coefficient?
A) Holding constant the effect of the other variables, the estimated number of lawn service purchased increases by 0.2868 for each increase of one thousand dollars in family income.
B) Holding constant the effect of the other variables, the estimated average number of lawn service purchased increases by 0.2868 for each increase of one thousand dollars in family income.
C) Holding constant the effect of the other variables, the estimated probability of purchasing a lawn service increases by 0.2868 for each increase of one thousand dollars in family income.
D) Holding constant the effect of the other variables, the estimated natural logarithm of the odds ratio of purchasing a lawn service increases by 0.2868 for each increase of one thousand dollars in family income.

Accepted Answer

The answer of TABLE 14-19
The marketing manager for a nationally...

Question 13

TABLE 14-19
The marketing manager for a nationally franchised lawn service company would like to study the characteristics that differentiate home owners who do and do not have a lawn service. A random sample of 30 home owners located in a suburban area near a large city was selected; 15 did not have a lawn service (code 0) and 15 had a lawn service (code 1). Additional information available concerning these 30 home owners includes family income (Income, in thousands of dollars), lawn size (Lawn Size, in thousands of square feet), attitude toward outdoor recreational activities (Atitude 0 = unfavorable, 1 = favorable), number of teenagers in the household (Teenager), and age of the head of the household (Age).
The Minitab output is given below: Logistic Regression Table
$\begin{array}{lrrrrrrr} 
& & & & & \text { Odds } & {95 \% \text { CI }} \
\text { Predictor } & \text { Coef } & \text { SE Coef } & \text { Z } & \text { P } & \text { Ratio } & \text { Lower } & \text { Upper } \
\text { Constant } & -70.49 & 47.22 & -1.49 & 0.135 & & & \
\text { Income } & 0.2868 & 0.1523 & 1.88 & 0.060 & 1.33 & 0.99 & 1.80 \
\text { LawnSiz } & 1.0647 & 0.7472 & 1.42 & 0.154 & 2.90 & 0.67 & 12.54 \
\text { Attitude } & -12.744 & 9.455 & -1.35 & 0.178 & 0.00 & 0.00 & 326.06 \
\text { Teenager } & -0.200 & 1.061 & -0.19 & 0.850 & 0.82 & 0.10 & 6.56 \
\text { Age } & 1.0792 & 0.8783 & 1.23 & 0.219 & 2.94 & 0.53 & 16.45
\end{array}$
Log-Likelihood $= - 4.890$
Test that all slopes are zero: $\mathrm { G } = 31.808 , \mathrm { DF } = 5 , p$-value $= 0.000$
Goodness-of-Fit Tests
$\begin{array} { l c r c } \text { Method } & \text { Chi-Square } & \text { DF } & \mathrm { P } \ \text { Pearson } & 9.313 & 24 & 0.997 \ \text { Deviance } & 9.780 & 24 & 0.995 \ \text { Hosmer-Lemeshow } & 0.571 & 8 & 1.000 \end{array}$
-Referring to Table 14-19, there is not enough evidence to conclude that the model is not a good-fitting model at a 0.05 level of significance.

Accepted Answer

The Hosmer-Lemeshow goodness-of-fit test has a p-value of 1.000, indicating good fit of the model. Therefore, there is not enough evidence to conclude that the model is not a good-fitting model at a 0.05 level of significance.

Question 14

TABLE 14-18
A logistic regression model was estimated in order to predict the probability that a randomly chosen university or college would be a private university using information on mean total Scholastic Aptitude Test score (SAT) at the university or college, the room and board expense measured in thousands of dollars (Room/Brd), and whether the TOEFL criterion is at least 550 (Toefl550 = 1 if yes, 0 otherwise.) The dependent variable, Y, is school type (Type = 1 if private and 0 otherwise).
The Minitab output is given below: Logistic Regression Table
$\begin{array}{lrrrcrrrr} 
& & & & &{\text { Odds }} &{95 \% \mathrm{CI}} \
\text { Predictor } & \text { Coef } & \text { SE Coef } & \mathrm{Z} & \mathrm{P} & \text { Ratio } & \text { Lower } & \text { Upper } \
\text { Constant } & -27.118 & 6.696 & -4.05 & 0.000 & & & \
\text { SAT } & 0.015 & 0.004666 & 3.17 & 0.002 & 1.01 & 1.01 & 1.02 \
\text { Toefl550 } & -0.390 & 0.9538 & -0.41 & 0.682 & 0.68 & 0.10 & 4.39 \
\text { Room/Brd } & 2.078 & 0.5076 & 4.09 & 0.000 & 7.99 & 2.95 & 21.60
\end{array}$
Log-Likelihood $= - 21.883$
Test that all slopes are zero: $\mathrm { G } = 62.083 , \mathrm { DF } = 3 , p$-value $= 0.000$
Goodness-of-Fit Tests
$\begin{array} { l r r c } \text { Method } & \text { Chi-Square } & \text { DF } & \text { P } \ \text { Pearson } & 143.551 & 76 & 0.000 \ \text { Deviance } & 43.767 & 76 & 0.999 \ \text { Hosmer-Lemeshow } & 15.731 & 8 & 0.046 \end{array}$
-Referring to Table 14-18, what are the degrees of freedom for the chi-square distribution when testing whether the model is a good-fitting model?

Accepted Answer

The answer of TABLE 14-18
A logistic regression model was estimated...

Question 15

TABLE 14-18
A logistic regression model was estimated in order to predict the probability that a randomly chosen university or college would be a private university using information on mean total Scholastic Aptitude Test score (SAT) at the university or college, the room and board expense measured in thousands of dollars (Room/Brd), and whether the TOEFL criterion is at least 550 (Toefl550 = 1 if yes, 0 otherwise.) The dependent variable, Y, is school type (Type = 1 if private and 0 otherwise).
The Minitab output is given below: Logistic Regression Table
$\begin{array}{lrrrcrrrr} 
& & & & &{\text { Odds }} &{95 \% \mathrm{CI}} \
\text { Predictor } & \text { Coef } & \text { SE Coef } & \mathrm{Z} & \mathrm{P} & \text { Ratio } & \text { Lower } & \text { Upper } \
\text { Constant } & -27.118 & 6.696 & -4.05 & 0.000 & & & \
\text { SAT } & 0.015 & 0.004666 & 3.17 & 0.002 & 1.01 & 1.01 & 1.02 \
\text { Toefl550 } & -0.390 & 0.9538 & -0.41 & 0.682 & 0.68 & 0.10 & 4.39 \
\text { Room/Brd } & 2.078 & 0.5076 & 4.09 & 0.000 & 7.99 & 2.95 & 21.60
\end{array}$
Log-Likelihood $= - 21.883$
Test that all slopes are zero: $\mathrm { G } = 62.083 , \mathrm { DF } = 3 , p$-value $= 0.000$
Goodness-of-Fit Tests
$\begin{array} { l r r c } \text { Method } & \text { Chi-Square } & \text { DF } & \text { P } \ \text { Pearson } & 143.551 & 76 & 0.000 \ \text { Deviance } & 43.767 & 76 & 0.999 \ \text { Hosmer-Lemeshow } & 15.731 & 8 & 0.046 \end{array}$
-Referring to Table 14-18, what is the p-value of the test statistic when testing whether SAT makes a significant contribution to the model in the presence of the other independent variables?

Accepted Answer

The answer of TABLE 14-18
A logistic regression model was estimated...

Question 16

TABLE 14-18
A logistic regression model was estimated in order to predict the probability that a randomly chosen university or college would be a private university using information on mean total Scholastic Aptitude Test score (SAT) at the university or college, the room and board expense measured in thousands of dollars (Room/Brd), and whether the TOEFL criterion is at least 550 (Toefl550 = 1 if yes, 0 otherwise.) The dependent variable, Y, is school type (Type = 1 if private and 0 otherwise).
The Minitab output is given below: Logistic Regression Table
$\begin{array}{lrrrcrrrr} 
& & & & &{\text { Odds }} &{95 \% \mathrm{CI}} \
\text { Predictor } & \text { Coef } & \text { SE Coef } & \mathrm{Z} & \mathrm{P} & \text { Ratio } & \text { Lower } & \text { Upper } \
\text { Constant } & -27.118 & 6.696 & -4.05 & 0.000 & & & \
\text { SAT } & 0.015 & 0.004666 & 3.17 & 0.002 & 1.01 & 1.01 & 1.02 \
\text { Toefl550 } & -0.390 & 0.9538 & -0.41 & 0.682 & 0.68 & 0.10 & 4.39 \
\text { Room/Brd } & 2.078 & 0.5076 & 4.09 & 0.000 & 7.99 & 2.95 & 21.60
\end{array}$
Log-Likelihood $= - 21.883$
Test that all slopes are zero: $\mathrm { G } = 62.083 , \mathrm { DF } = 3 , p$-value $= 0.000$
Goodness-of-Fit Tests
$\begin{array} { l r r c } \text { Method } & \text { Chi-Square } & \text { DF } & \text { P } \ \text { Pearson } & 143.551 & 76 & 0.000 \ \text { Deviance } & 43.767 & 76 & 0.999 \ \text { Hosmer-Lemeshow } & 15.731 & 8 & 0.046 \end{array}$
-Referring to Table 14-18, there is not enough evidence to conclude that Toefl500 makes a significant contribution to the model in the presence of the other independent variables at a 0.05 level of significance.

Accepted Answer

The p-value for Toefl550 is 0.682, which is greater than 0.05. Therefore, we do not have enough evidence to conclude that Toefl550 makes a significant contribution to the model in the presence of the other independent variables at a 0.05 level of significance.

Question 17

TABLE 14-19
The marketing manager for a nationally franchised lawn service company would like to study the characteristics that differentiate home owners who do and do not have a lawn service. A random sample of 30 home owners located in a suburban area near a large city was selected; 15 did not have a lawn service (code 0) and 15 had a lawn service (code 1). Additional information available concerning these 30 home owners includes family income (Income, in thousands of dollars), lawn size (Lawn Size, in thousands of square feet), attitude toward outdoor recreational activities (Atitude 0 = unfavorable, 1 = favorable), number of teenagers in the household (Teenager), and age of the head of the household (Age).
The Minitab output is given below: Logistic Regression Table
$\begin{array}{lrrrrrrr} 
& & & & & \text { Odds } & {95 \% \text { CI }} \
\text { Predictor } & \text { Coef } & \text { SE Coef } & \text { Z } & \text { P } & \text { Ratio } & \text { Lower } & \text { Upper } \
\text { Constant } & -70.49 & 47.22 & -1.49 & 0.135 & & & \
\text { Income } & 0.2868 & 0.1523 & 1.88 & 0.060 & 1.33 & 0.99 & 1.80 \
\text { LawnSiz } & 1.0647 & 0.7472 & 1.42 & 0.154 & 2.90 & 0.67 & 12.54 \
\text { Attitude } & -12.744 & 9.455 & -1.35 & 0.178 & 0.00 & 0.00 & 326.06 \
\text { Teenager } & -0.200 & 1.061 & -0.19 & 0.850 & 0.82 & 0.10 & 6.56 \
\text { Age } & 1.0792 & 0.8783 & 1.23 & 0.219 & 2.94 & 0.53 & 16.45
\end{array}$
Log-Likelihood $= - 4.890$
Test that all slopes are zero: $\mathrm { G } = 31.808 , \mathrm { DF } = 5 , p$-value $= 0.000$
Goodness-of-Fit Tests
$\begin{array} { l c r c } \text { Method } & \text { Chi-Square } & \text { DF } & \mathrm { P } \ \text { Pearson } & 9.313 & 24 & 0.997 \ \text { Deviance } & 9.780 & 24 & 0.995 \ \text { Hosmer-Lemeshow } & 0.571 & 8 & 1.000 \end{array}$
-Referring to Table 14-19, which of the following is the correct expression for the estimated model?
A) Y = -70.49 + 0.2868 Income + 1.0647 LawnSize - 12.744 Attitude - 0.200 Teenager + 1.0792 Age
B) $\hat Y$ = -70.49 + 0.2868 Income + 1.0647 LawnSize - 12.744 Attitude - 0.200 Teenager + 1.0792 Age
C) 1n(odds ratio) = -70.49 + 0.2868 Income + 1.0647 LawnSize - 12.744 Attitude - 0.200 Teenager + 1.0792 Age
D) 1n(estimated odds ratio) = -70.49 + 0.2868 Income + 1.0647 LawnSize - 12.744 Attitude - 0.200 Teenager + 1.0792

Accepted Answer

The answer of TABLE 14-19
The marketing manager for a nationally...

Question 18

TABLE 14-19
The marketing manager for a nationally franchised lawn service company would like to study the characteristics that differentiate home owners who do and do not have a lawn service. A random sample of 30 home owners located in a suburban area near a large city was selected; 15 did not have a lawn service (code 0) and 15 had a lawn service (code 1). Additional information available concerning these 30 home owners includes family income (Income, in thousands of dollars), lawn size (Lawn Size, in thousands of square feet), attitude toward outdoor recreational activities (Atitude 0 = unfavorable, 1 = favorable), number of teenagers in the household (Teenager), and age of the head of the household (Age).
The Minitab output is given below: Logistic Regression Table
$\begin{array}{lrrrrrrr} 
& & & & & \text { Odds } & {95 \% \text { CI }} \
\text { Predictor } & \text { Coef } & \text { SE Coef } & \text { Z } & \text { P } & \text { Ratio } & \text { Lower } & \text { Upper } \
\text { Constant } & -70.49 & 47.22 & -1.49 & 0.135 & & & \
\text { Income } & 0.2868 & 0.1523 & 1.88 & 0.060 & 1.33 & 0.99 & 1.80 \
\text { LawnSiz } & 1.0647 & 0.7472 & 1.42 & 0.154 & 2.90 & 0.67 & 12.54 \
\text { Attitude } & -12.744 & 9.455 & -1.35 & 0.178 & 0.00 & 0.00 & 326.06 \
\text { Teenager } & -0.200 & 1.061 & -0.19 & 0.850 & 0.82 & 0.10 & 6.56 \
\text { Age } & 1.0792 & 0.8783 & 1.23 & 0.219 & 2.94 & 0.53 & 16.45
\end{array}$
Log-Likelihood $= - 4.890$
Test that all slopes are zero: $\mathrm { G } = 31.808 , \mathrm { DF } = 5 , p$-value $= 0.000$
Goodness-of-Fit Tests
$\begin{array} { l c r c } \text { Method } & \text { Chi-Square } & \text { DF } & \mathrm { P } \ \text { Pearson } & 9.313 & 24 & 0.997 \ \text { Deviance } & 9.780 & 24 & 0.995 \ \text { Hosmer-Lemeshow } & 0.571 & 8 & 1.000 \end{array}$
-Referring to Table 14-19, what is the estimated odds ratio for a 48-year-old home owner with a family income of $100,000, a lawn size of 5,000 square feet, a negative attitude toward outdoor recreation, and one teenager in the household?

Accepted Answer

The answer of TABLE 14-19
The marketing manager for a nationally...

Question 19

TABLE 14-18
A logistic regression model was estimated in order to predict the probability that a randomly chosen university or college would be a private university using information on mean total Scholastic Aptitude Test score (SAT) at the university or college, the room and board expense measured in thousands of dollars (Room/Brd), and whether the TOEFL criterion is at least 550 (Toefl550 = 1 if yes, 0 otherwise.) The dependent variable, Y, is school type (Type = 1 if private and 0 otherwise).
The Minitab output is given below: Logistic Regression Table
$\begin{array}{lrrrcrrrr} 
& & & & &{\text { Odds }} &{95 \% \mathrm{CI}} \
\text { Predictor } & \text { Coef } & \text { SE Coef } & \mathrm{Z} & \mathrm{P} & \text { Ratio } & \text { Lower } & \text { Upper } \
\text { Constant } & -27.118 & 6.696 & -4.05 & 0.000 & & & \
\text { SAT } & 0.015 & 0.004666 & 3.17 & 0.002 & 1.01 & 1.01 & 1.02 \
\text { Toefl550 } & -0.390 & 0.9538 & -0.41 & 0.682 & 0.68 & 0.10 & 4.39 \
\text { Room/Brd } & 2.078 & 0.5076 & 4.09 & 0.000 & 7.99 & 2.95 & 21.60
\end{array}$
Log-Likelihood $= - 21.883$
Test that all slopes are zero: $\mathrm { G } = 62.083 , \mathrm { DF } = 3 , p$-value $= 0.000$
Goodness-of-Fit Tests
$\begin{array} { l r r c } \text { Method } & \text { Chi-Square } & \text { DF } & \text { P } \ \text { Pearson } & 143.551 & 76 & 0.000 \ \text { Deviance } & 43.767 & 76 & 0.999 \ \text { Hosmer-Lemeshow } & 15.731 & 8 & 0.046 \end{array}$
-Referring to Table 14-18, what should be the decision ('reject' or 'do not reject')on the null hypothesis when testing whether SAT makes a significant contribution to the model in the presence of the other independent variables at a 0.05 level of significance?

Accepted Answer

The answer of TABLE 14-18
A logistic regression model was estimated...

Question 20

TABLE 14-18
A logistic regression model was estimated in order to predict the probability that a randomly chosen university or college would be a private university using information on mean total Scholastic Aptitude Test score (SAT) at the university or college, the room and board expense measured in thousands of dollars (Room/Brd), and whether the TOEFL criterion is at least 550 (Toefl550 = 1 if yes, 0 otherwise.) The dependent variable, Y, is school type (Type = 1 if private and 0 otherwise).
The Minitab output is given below: Logistic Regression Table
$\begin{array}{lrrrcrrrr} 
& & & & &{\text { Odds }} &{95 \% \mathrm{CI}} \
\text { Predictor } & \text { Coef } & \text { SE Coef } & \mathrm{Z} & \mathrm{P} & \text { Ratio } & \text { Lower } & \text { Upper } \
\text { Constant } & -27.118 & 6.696 & -4.05 & 0.000 & & & \
\text { SAT } & 0.015 & 0.004666 & 3.17 & 0.002 & 1.01 & 1.01 & 1.02 \
\text { Toefl550 } & -0.390 & 0.9538 & -0.41 & 0.682 & 0.68 & 0.10 & 4.39 \
\text { Room/Brd } & 2.078 & 0.5076 & 4.09 & 0.000 & 7.99 & 2.95 & 21.60
\end{array}$
Log-Likelihood $= - 21.883$
Test that all slopes are zero: $\mathrm { G } = 62.083 , \mathrm { DF } = 3 , p$-value $= 0.000$
Goodness-of-Fit Tests
$\begin{array} { l r r c } \text { Method } & \text { Chi-Square } & \text { DF } & \text { P } \ \text { Pearson } & 143.551 & 76 & 0.000 \ \text { Deviance } & 43.767 & 76 & 0.999 \ \text { Hosmer-Lemeshow } & 15.731 & 8 & 0.046 \end{array}$
-Referring to Table 14-18, what is the p-value of the test statistic when testing whether Toefl500 makes a significant contribution to the model in the presence of the other independent variables?

Accepted Answer

The answer of TABLE 14-18
A logistic regression model was estimated...

Quiz 14: Introduction to Multiple Regression