stats 363 HW-Section-9-4

# stats 363 HW-Section-9-4 - Section 9.4 Exercise 9.19 . The...

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Unformatted text preview: Section 9.4 Exercise 9.19 . The hypotheses are H : &amp; 1 &amp; &amp; 2 = 0 versus H a : &amp; 1 &amp; &amp; 2 &lt; : 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 &amp; 1 : 31 q : 0560 36 + : 0540 45 = &amp; 1 : 334 : Hence, the p-value is p-value = P f z &lt; &amp; 1 : 334 g : 0918 : Since the p-value &gt; : 05, H cannot be rejected and there is no su&amp;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 : &amp; 1 &amp; &amp; 2 = 0 versus H a : &amp; 1 &amp; &amp; 2 &gt; : This is a right-tailed 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 &amp; 5 : 8 r (2 : 9) 2 35 + (1 : 2) 2 35 = 2 : 074 : At = 0 : 05 level of signicance, z &amp; = 1 : 645 and the rejection region is z &gt; 1 : 645. Since the test statistic z = 2 : 074 is in the rejection region, we reject H . There is su&amp;cient evidence to indicate that the use of vitamin C reduces the mean time required to recover from a common cold and its complications....
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## stats 363 HW-Section-9-4 - Section 9.4 Exercise 9.19 . The...

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