Solution Found!
Forecasting daily admission of a water park (cont’d).
Chapter 12, Problem 162SE(choose chapter or problem)
Forecasting daily admission of a water park (cont’d). Refer to Exercise 12.161. The owners of the water adventure park are advised that the prediction model could probably be improved if interaction terms were added. In Particular, it is thought that the rate at which mean attendance increases as predicted high temperature increases will be greater on weekends than on weekdays. The following model is therefore proposed:
\(E(y)=\beta_{0}+\beta_{1} x_{1}+\beta_{2} x_{2}+\beta_{3} x_{3}+\beta_{4} x_{1} x_{3}\)
The same 30 days of data used in Exercise 12.161 areagain used to obtain the least squares model
\(\hat{y}=250-700 x_{1}+100 x_{2}+5 x_{3}+15 x_{1} x_{3}\)
with
\(s_{\hat{\beta}_{4}}=3.0\ \ \ \quad R^{2}=.96\)
a. Graph the predicted day’s attendance, y, against the day’s predicted high temperature, \(x_3\), for a sunny weekday and for a sunny weekend day. Plot both on the same graph for \(x_3\) between \(70^{\circ}F\) and \(100^{\circ}F\). Note the increase in slope for the weekend day. Interpret this.
b. Do the data indicate that the interaction term is a useful addition to the model? Use \(\alpha\ =\ .05\).
c. Use this model to predict the attendance for a sunny weekday with a predicted high temperature of \(95^{\circ}F\).
d. Suppose the 90% prediction interval for part c is (800, 850). Compare this result with the prediction interval for the model without interaction in Exercise 12.161,part e. Do the relative widths of the confidence intervals support or refute your conclusion about the utility of the interaction term (part b)?
e. The owners, noting that the coefficient \(\hat{\beta}_1=-700\), conclude the model is ridiculous because it seems to imply that the mean attendance will be 700 less on weekends than on weekdays. Explain why this is not the case.
Questions & Answers
QUESTION:
Forecasting daily admission of a water park (cont’d). Refer to Exercise 12.161. The owners of the water adventure park are advised that the prediction model could probably be improved if interaction terms were added. In Particular, it is thought that the rate at which mean attendance increases as predicted high temperature increases will be greater on weekends than on weekdays. The following model is therefore proposed:
\(E(y)=\beta_{0}+\beta_{1} x_{1}+\beta_{2} x_{2}+\beta_{3} x_{3}+\beta_{4} x_{1} x_{3}\)
The same 30 days of data used in Exercise 12.161 areagain used to obtain the least squares model
\(\hat{y}=250-700 x_{1}+100 x_{2}+5 x_{3}+15 x_{1} x_{3}\)
with
\(s_{\hat{\beta}_{4}}=3.0\ \ \ \quad R^{2}=.96\)
a. Graph the predicted day’s attendance, y, against the day’s predicted high temperature, \(x_3\), for a sunny weekday and for a sunny weekend day. Plot both on the same graph for \(x_3\) between \(70^{\circ}F\) and \(100^{\circ}F\). Note the increase in slope for the weekend day. Interpret this.
b. Do the data indicate that the interaction term is a useful addition to the model? Use \(\alpha\ =\ .05\).
c. Use this model to predict the attendance for a sunny weekday with a predicted high temperature of \(95^{\circ}F\).
d. Suppose the 90% prediction interval for part c is (800, 850). Compare this result with the prediction interval for the model without interaction in Exercise 12.161,part e. Do the relative widths of the confidence intervals support or refute your conclusion about the utility of the interaction term (part b)?
e. The owners, noting that the coefficient \(\hat{\beta}_1=-700\), conclude the model is ridiculous because it seems to imply that the mean attendance will be 700 less on weekends than on weekdays. Explain why this is not the case.
ANSWER:Step 1 of 9
(a)
The least squares model is given below:
where y = Predicted day’s attendance
For a sunny weekday, 
Table (1)