Section 9.9 (p.577)

1. When null hypothesis is true, the number X of men who prefer type A has a binomial distribution with n=20 and p=0.5. Therefore, Pr(X ³ 15) = 0.0207, which indicates that we should reject H₀ at a₀=0.05.

2. Many things can be done, for example: don’t let these men know the type of blade; randomize the order of blade types to be used etc…

3. The t statistic U=2.039, and U has a t distribution with 14 df. Therefore we should reject H₀ at a₀=0.05 and not at a₀=0.025.

4. (a) signed test: 0.0593; Wilcoxon signed-ranks test: 0.03; t test: between 0.025 and 0.05.

(b) signed test: Independent differences, Pr(differences ³ 0) = Pr(differences £ 0) =0.5;

Wilcoxon signed-ranks test: Independent and symmetric differences

t test: Difference has a normal distribution.

(c) It’s safe to reject H₀ at a₀=0.05. If the tail areas were widely different, we should further consider the assumptions needed for these three test and decide which is the most likely case.

5. (a) X = # of (difference > 0) = 13. And under the null hypothesis, X has a binomial distribution with n=20 and p=0.5. Therefore the tail probability is Pr(X ³ 13) = 0.1316, which indicates that we can’t reject H₀ at a₀=0.1.

(b) We have S_n=181, and Z_n=2.84. The tail probability Pr(Z_n ³ 2.84) = 0.0023, which indicates that we should reject H₀ at a₀=0.01.