TABLE 14-17
the Marketing Manager for a Nationally Franchised

Short Answer

TABLE 14-17
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 (Attitude 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:
$\begin{array}{lccccccrr} & & & & \text { Odds } & \text { 95 \% CI } \\\text {Predictor }& \text {Coef }&\text { SE Coef }&\text { Z }& \text {P} &\text { Ratio} &\text { Lower} & \text {Upper }\\\text {Constant }& -70.49 & 47.22 & -1.49 & 0.135 & & & \\\text {Income }& 0.2868 & 0.1523 & 1.88 & 0.060 & 1.33 & 0.99 & 1.80 \\\text {Lawn Size} & 1.0647 & 0.7472 & 1.42 & 0.154 & 2.90 & 0.67 & 12.54 \\\text {Attitude} & -12.744 & 9.455 & -1.35 & 0.178 & 0.00 & 0.00 & 326.06 \\\text {Teenager} & -0.200 & 1.061 & -0.19 & 0.850 & 0.82 & 0.10 & 6.56 \\\text {Age} & 1.0792 & 0.8783 & 1.23 & 0.219 & 2.94 & 0.53 & 16.45\end{array}$

Log-Likelihood = -4.890
Test that all slopes are zero: G = 31.808, DF = 5, P-Value = 0.000

Goodness-of-Fit Tests
$\begin{array}{lrrr}\text { Method } & \text { Chi-Square } & \text { DF } & \text { P } \\ \text { Pearson } & 9.313 & 24 & 0.997 \\ \text { Deviance } & 9.780 & 24 & 0.995 \\ \text { Hosmer-Lemeshow } & 0.571 & 8 & 1.000\end{array}$



-Referring to Table 14-17, 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?