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TABLE 14-17
the Marketing Manager for a Nationally

question 164

True/False

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:
 Odds  95 % CI Predictor Coef  SE Coef  Z P Ratio LowerUpper Constant 70.4947.221.490.135Income 0.28680.15231.880.0601.330.991.80Lawn Size1.06470.74721.420.1542.900.6712.54Attitude12.7449.4551.350.1780.000.00326.06Teenager0.2001.0610.190.8500.820.106.56Age1.07920.87831.230.2192.940.5316.45\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
 Method  Chi-Square  DF  P  Pearson 9.313240.997 Deviance 9.780240.995 Hosmer-Lemeshow 0.57181.000 \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, there is not enough evidence to conclude that the model is not a good-fitting model at a 0.05 level of significance.


Definitions:

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