**NAME: **

- 1.
- Ray Bob says he can't decide between two MLR models for
the same set of data. He says that he likes large
*R*^{2}_{a}and small MSE, but that model 1 has a larger*R*^{2}_{a}than model 2, but also a larger MSE. What do you say to Ray Bob?**ANS:**This can't be true, because (see book, p. 474). Since*S*is the same regardless of model, a model with larger^{2}*R*^{2}_{a}must have smaller MSE. - 2.
- An experiment was conducted to determine the effect of
miles driven and temperature on tire wear. Thirty-two tires of the
same model were randomly assigned to temperature and distance
settings. They were then subjected to simulated driving at the
assigned temperature and distance. At the end of the driving period,
the tire wear was recorded. The response is a measure of tire wear.
Figures 1-3 contain SAS/INSIGHT output for three fits to these data. CTEMP and CMILES are the centered predictors temperature and miles driven.

- (a)
- Compare the three
models based on all relevant measures and
graphs. Which do you think provides the best
fit? Why?
Model Measure 1 2 3 Residual versus Under-predicts Good Good fitted plot center values *R*^{2}0.4759 0.7449 0.7499 *R*^{2}_{a}0.4397 0.7273 0.7231 MSE 0.0184 0.0090 0.0091 **ANS:**Based on the above measures, model 2 is best, closely followed by model 3. Model 1 is very bad. - (b)
- Write out the equation for
the
**best fitting**model. Interpret this equation in terms even Professor P. can understand.*The intercept, 3.0105, is the estimated mean wear when temperature and miles are at their means.**The estimated change in mean wear per unit increase in temperature is .**The estimated change in mean wear per unit increase in miles is .* - (c)
- What proportion of the total
variation in the tire wear measurement is explained by using the
**best fitting**model?**ANS:**0.7449 - c.
- You are going on a trip of 400
miles this weekend, and you think the temperature will be 58
^{o}F. Use the**best fitting**model to predict your tire wear. The mean of the miles driven variable in the data set is 192.875, and that of the temperature variable is 71.975.**ANS:**, and , so the prediction is

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