Answer to Exercise 34.2: 1. $$R^2 = 0.881^2 = 77.6$$%. About 77.6% of the variation in punting distance can be explained by the variation in right-leg strength. 2. $$H_0$$: $$\rho=0$$ and $$H_1$$: $$\rho\ne0$$. $$t=6.16$$; $$P$$-value very small. Very strong evidence of a correlation in the population.
Answer to Exercise 34.3: The plot looks linear; $$n=25$$; variation doesn’t seem constant.
Answer to Exercise 34.4: 1. Very close to $$-1$$. 2. $$r = -\sqrt{0.9929} = -0.9964$$. ($$r$$ must be negative!) 3. Very small. This is a very large value for $$r$$ on a reasonable sized sample. 4. Yes.
Answer to Exercise 35.5: 1. $$b_0$$: When someone spends no time on sunscreen application, an average of 0.27g has been applied; nonsense. $$b_1$$: Each extra minute spent on application adds an average of 2.21g of sunscreen: sensible. 2. The value of $$\beta_0$$ could be zero… which would make sense. 3. $$\hat{y} = 0.27 + (2.21\times 8) = 17.95$$; an average of about 18g. 4. About 64% of the variation in sunscreen amount applied can be explained by the variation in the time spent on application. 5. $$r = \sqrt{0.64} = 0.8$$, and need a positive value of $$r$$. A strong and positive correlation between the variables.