Comparing Two Populations: Hypothesis Testing for Proportions, Study notes of Mathematical Statistics

Download Comparing Two Populations: Hypothesis Testing for Proportions and more Study notes Mathematical Statistics in PDF only on Docsity! Lecture 19. Comparing two proportions. Math 283, Fall 2004 Suggested problems [not to be turned in] (answers to odd-numbered problems are in the back) 8.33, 8.37, 8.41, 8.47, 8.63 (Answers might not coincide completely with yours, since the book uses Wilson approach) Goal: compare two population proportions p1 and p2 H0 : equal, i.e. p1 = p2 HA : p1 ≠ p2 (two-tailed) , or p1 > p2 (right-tailed) , or p1 < p2 (left-tailed) Classical approach Use pi = Xi/ni, i = 1,2 (Xi is number of “successes”) SE(p1 − p2) = Sqrt[p1(1−p1)/n1 + p2(1−p2)/n2] Based on normal approximation, leads to the C.I. Assumptions: all npi and n(1 – pi) are larger than 10 Example Binge drinking (Example 8.8 p. 589) ni Xi pi Men 7180 1630 Women 9916 1684 Wilson approach (used in the Book) Use 1 ~p = (X1+1)/(n1+2) etc. when ni are small (we will not use this). Note: when ni are large, both approaches give similar results. p1 − p2 ± z* SE(p1 − p2)