This preview shows pages 1–2. Sign up to view the full content.
This preview has intentionally blurred sections. Sign up to view the full version.
View Full Document
Unformatted text preview: Section 9.4 Exercise 9.19 . The hypotheses are H : & 1 & & 2 = 0 versus H a : & 1 & & 2 < : With n 1 = 36, n 2 = 45, x 1 = 1 : 24, x 2 = 1 : 31, s 2 1 = 0 : 0560, s 2 2 = 0 : 0540, the test statistic is z = 1 : 24 & 1 : 31 q : 0560 36 + : 0540 45 = & 1 : 334 : Hence, the pvalue is pvalue = P f z < & 1 : 334 g : 0918 : Since the pvalue > : 05, H cannot be rejected and there is no su&cient evidence to indicate that the mean for population 1 is smaller than the mean for population 2. Exercise 9.21 . ( a ) The hypotheses are H : & 1 & & 2 = 0 versus H a : & 1 & & 2 > : This is a righttailed test. ( b ) With n 1 = 35, n 2 = 35, x 1 = 6 : 9, x 2 = 5 : 8, s 1 = 2 : 9, s 2 = 1 : 2, the test statistic is z = 6 : 9 & 5 : 8 r (2 : 9) 2 35 + (1 : 2) 2 35 = 2 : 074 : At = 0 : 05 level of signicance, z & = 1 : 645 and the rejection region is z > 1 : 645. Since the test statistic z = 2 : 074 is in the rejection region, we reject H . There is su&cient evidence to indicate that the use of vitamin C reduces the mean time required to recover from a common cold and its complications....
View
Full
Document
 Spring '08
 Cheng
 Statistics

Click to edit the document details