D.31 Answers: Correlation

Answers to exercises in Sect. 34.8.

Answer to Exercise 34.1: Many correct answers.
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.