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.