TABLE 14-19
the Marketing Manager for a Nationally Franchised Lawn

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