Regression Revisited

Solutions to Practice Problems

1. Before any analysis, make a scatter plot of the data.  Here,  with horse power as the explanatory (x) variable and miles per gallon as the response (y) variable. The scatter plot looks reasonably linear, so a linear model should be appropriate here.

For the hypothesis test, our parameter is beta, the true slope for predicting gas mileage (mpg) from horse power across all motor vehicles. The hypotheses are H0: beta = 0 versus Ha: beta <> 0.

Make a residual plot and a normal quantile plot of the residuals. While not perfect, the plots do not suggest a severe violation of linearity, constant variance, or normality. We will proceed with the test.

Using our computer, we find the sample slope is b = -0.06940, the standard error of the slope is sb = 0.02329, the test statistic is t = (-0.06940 - 0)/0.02329 = -2.98 with df = n-2 = 15-2 = 13, and the p-value is 0.011. Since this is less than the standard alpha = 0.05, we reject the null hypothesis.

At the 5% level, we conclude that horse power is a useful linear predictor of gas mileage (mpg).

2. At 95% confidence, the t critical value at 13 df is 2.16, and so a 95% confidence interval for beta is b +/- 2.16*sb = -0.06940 +/- 2.16*0.02329 = (-.1197,-.0191). Hence, we are 95% confident that a positive difference of 1 horse power corresponds, on average, to a negative difference of between .0191mpg and .1197mpg.

3. We should again check that the requisite assumptions are met. A scatter plot shows a moderate positive relationship between price and horse power. However, a careful examination of the residuals reveals two problems: (1) the residual plot clearly shows non-constant variance, and (2) a normal quantile plot shows the residuals might be non-normal, athough the latter problem is less pronounced. Hence, a blind calculation based on the standard t formula would not be valid here.