18.05 Lecture Notes - Lecture 7: Mit Opencourseware, Null Distribution
Document Summary
Problem 1. (10 pts. ) (a) h0: = 0. 5. Ha: one-sided > 0. 5, two-sided = 0. Test statistic: x = number of heads in 250 spins. One-sided data at least as extreme: x 140. Using r we compute the one-sided p-value is. 5. p = p (x 140|h0) = 1 - pbinom(139, 250, 0. 5) = 0. 03321. The one-sided p-value computes the probability in the one-sided tail. Because our null distribution (binomial(250, 0. 5)) is symmetric, each tail in the rejection region will have probability /2. In this case the two-sided p-value is computed by doubling the smaller of the one sided values. We computed the right tail p-value just above. This is the smaller of the two p-values so our two-sided p-value is 2 0. 03321 = 0. 06642. This rounds to 0. 07, so the gure of 7% is the two-sided p-value. Note: we could use the normal approximation binomial(250, 0. 5) n(125, 250/4)
