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TABLE 14-12
a Weight-Loss Clinic Wants to Use Regression Analysis Y=β0+β1X1+β2X2+β3X3+β4X1X2+β5X1X3+ε Y=\beta_{0}+\beta_{1} X_{1}+\beta_{2} X_{2}+\beta_{3} X_{3}+\beta_{4} X_{1} X_{2}+\beta_{5} X_{1} X_{3}+\varepsilon

Question 64

Multiple Choice

TABLE 14-12
A weight-loss clinic wants to use regression analysis to build a model for weight-loss of a client (measured in pounds) . Two variables thought to affect weight-loss are client's length of time on the weight loss program and time of session. These variables are described below:
Y = Weight- loss (in pounds)
X1 = Length of time in weight- loss program (in months)
X
2 = 1 if morning session, 0 if not
X3 = 1 if afternoon session, 0 if not (Base level = evening session)
Data for 12 clients on a weight- loss program at the clinic were collected and used to fit the interaction model:
Y=β0+β1X1+β2X2+β3X3+β4X1X2+β5X1X3+ε Y=\beta_{0}+\beta_{1} X_{1}+\beta_{2} X_{2}+\beta_{3} X_{3}+\beta_{4} X_{1} X_{2}+\beta_{5} X_{1} X_{3}+\varepsilon
Partial output from Microsoft Excel follows:
 Regression Statistics  Multiple R 0.73514 R Square 0.540438 Adjusted R Square 0.157469 Standard Error 12.4147 Observations 12\begin{array}{l}\text { Regression Statistics }\\\begin{array} { l c } \hline \text { Multiple R } & 0.73514 \\\text { R Square } & 0.540438 \\\text { Adjusted R Square } & 0.157469 \\\text { Standard Error } & 12.4147 \\\text { Observations } & 12 \\\hline\end{array}\end{array}

ANOVA
F=5.41118 Significance F=0.040201F = 5.41118 \quad\text { Significance } F = 0.040201

Coefficients  Standard Error  t Stat  p -valueIntercept 0.08974414.1270.00600.9951Length (X1) 6.225382.434732.549560.0479Morn Ses (X2) 2.21727222.14160.1001410.9235Aft Ses (X3) 11.82333.15453.5589010.0165Length*Morn Ses0.770583.5620.2163340.8359Length * Aft Ses0.541473.359880.1611580.8773\begin{array}{lcccr}\hline & \text {Coefficients }& \text { Standard Error }& \text { t Stat }& \text { p -value} \\\hline \text {Intercept }& 0.089744 & 14.127 & 0.0060 & 0.9951 \\ \text {Length (X1) }& 6.22538 & 2.43473 & 2.54956 & 0.0479 \\ \text {Morn Ses (X2) }& 2.217272 & 22.1416 & 0.100141 & 0.9235 \\ \text {Aft Ses (X3) } & 11.8233 & 3.1545 & 3.558901 & 0.0165 \\ \text {Length*Morn Ses} & 0.77058 & 3.562 & 0.216334 & 0.8359 \\ \text {Length * Aft Ses} & -0.54147 & 3.35988 & -0.161158 & 0.8773 \\\hline\end{array}


TABLE 14-17
 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}

Regression Staxistics
 Multiple R 0.73514 R Square 0.540438 Adjusted R Square 0.157469 Standard E rror 12.4147 Observations12\begin{array}{lc}\hline \text { Multiple R } & 0.73514 \\\text { R Square } & 0.540438 \\\text { Adjusted R Square } & 0.157469 \\\text { Standard E rror } & 12.4147\\\\\text { Observations}& 12\end{array}

ANOVA
F=5.41118F = 5.41118 \quad Significance F=0.040201F = 0.040201

 Coefficients  Standard Error tStat p-value  Intercept 0.08974414.1270.00600.9951 Length (X1) 6.225382.434732.549560.0479 Morn Ses (X2) 2.21727222.14160.1001410.9235 Aft Ses (X3) 11.82333.15453.5589010.0165 Length  Morn S es 0.770583.5620.2163340.8359 Length  Aft Ses 0.541473.359880.1611580.8773\begin{array}{lllll}\hline\text { Coefficients }&\text { Standard Error}&\text { tStat}\text { p-value }\\\hline\text { Intercept } & 0.089744 & 14.127 & 0.0060 & 0.9951 \\\text { Length }\left(X_{1}\right) & 6.22538 & 2.43473 & 2.54956 & 0.0479 \\\text { Morn Ses }\left(X_{2}\right) & 2.217272 & 22.1416 & 0.100141 & 0.9235 \\\text { Aft Ses }\left(X_{3}\right) & 11.8233 & 3.1545 & 3.558901 & 0.0165 \\\text { Length }{ }^{*} \text { Morn S es }& 0.77058 & 3.562 & 0.216334 & 0.8359 \\\text { Length }{ }^{*} \text { Aft Ses }-0.54147 & 3.35988 & -0.161158 & 0.8773 &\end{array}
-Referring to Table 14-12, in terms of the þ's in the model, give the average change in weight-loss (Y) for every 1 month increase in time in the program (X1) when attending the evening session.


A) þ4 + þ5
B) þ1 + þ4
C) þ1 + þ5
D) þ1

Correct Answer:

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