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 Income makes a significant contribution to the model in the presence of the other independent variables?

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

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 Age makes a significant contribution to the model in the presence of the other independent variables at a 0.05 level of significance.

The p-value for Age is 0.219 which is greater than 0.05, indicating that Age is not significant in the presence of the other independent variables at a 0.05 level of significance.

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 should be the decision ('reject' or 'do not reject')on the null hypothesis when testing whether Age makes a significant contribution to the model in the presence of the other independent variables at a 0.05 level of significance?

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

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 Income makes a significant contribution to the model in the presence of the other independent variables at a 0.05 level of significance.

In the table, the p-value for Income is 0.066, which is greater than the significance level of 0.05. Therefore, we fail to reject the null hypothesis that Income does not make a significant contribution to the model in the presence of the other independent variables. Hence, there is not enough evidence to conclude that Income makes a significant contribution to the model at a 0.05 level of significance.

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 should be the decision ('reject' or 'do not reject')on the null hypothesis when testing whether LawnSize makes a significant contribution to the model in the presence of the other independent variables at a 0.05 level of significance?

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

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 Teenager makes a significant contribution to the model in the presence of the other independent variables?

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

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 should be the decision ('reject' or 'do not reject')on the null hypothesis when testing whether Teenager makes a significant contribution to the model in the presence of the other independent variables at a 0.05 level of significance?

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

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 Attitude makes a significant contribution to the model in the presence of the other independent variables?

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

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 LawnSize makes a significant contribution to the model in the presence of the other independent variables?

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

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 should be the decision ('reject' or 'do not reject')on the null hypothesis when testing whether Attitude makes a significant contribution to the model in the presence of the other independent variables at a 0.05 level of significance?

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

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 LawnSize makes a significant contribution to the model in the presence of the other independent variables at a 0.05 level of significance.

In the "Analysis of Variance" section of the output, the p-value for LawnSize is 0.142, which is greater than the 0.05 level of significance. Therefore, we fail to reject the null hypothesis and conclude that there is not enough evidence to suggest that LawnSize makes a significant contribution to the model in the presence of the other independent variables.

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 should be the decision ('reject' or 'do not reject')on the null hypothesis when testing whether Income makes a significant contribution to the model in the presence of the other independent variables at a 0.05 level of significance?

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

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 Attitude makes a significant contribution to the model in the presence of the other independent variables at a 0.05 level of significance.

The P-value for Attitude is 0.178, which is greater than 0.05. Therefore, we fail to reject the null hypothesis and conclude that Attitude does not make a significant contribution to the model in the presence of the other independent variables at a 0.05 level of significance.

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 Teenager makes a significant contribution to the model in the presence of the other independent variables at a 0.05 level of significance.

The p-value for Teenager is 0.850, which is greater than 0.05, indicating that it is not significant at the 0.05 level of significance. Therefore, we fail to reject the null hypothesis that Teenager does not make a significant contribution to the model in the presence of the other independent variables.

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 Age makes a significant contribution to the model in the presence of the other independent variables?

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

Quiz 14: Introduction to Multiple Regression

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