Statistics Homework Solutions 140

Statistics Homework Solutions 140 - 140 CHAPTER 6 Y z 40 42...

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140 CHAPTER 6 z 40 42 0 12 2 75 15 2 100 0 98. Since the alternate hypothesis is of the form μ X μ Y , the P -value is the area to the left of z 0 98. Thus P 0 1635. We cannot conclude that the mean reduction from drug B is greater than the mean reduction from drug A. 7. (a) X 7 79, s X 1 06, n X 80, Y 7 64, s Y 1 31, n Y 80. Here μ 1 μ X and μ 2 μ Y . The null and alternate hypotheses are H 0 : μ X μ Y 0 versus H 1 : μ X μ Y 0. z 7 79 7 64 0 1 06 2 80 1 31 2 80 0 80. Since the alternate hypothesis is of the form μ X μ Y , the P -value is the area to the right of z 0 80. Thus P 0 2119. We cannot conclude that the mean score on one-tailed questions is greater. (b) The null and alternate hypotheses are H 0 : μ X μ Y 0 versus H 1 : μ X μ Y 0. The z -score is computed as in part (a): z 0 80. Since the alternate hypothesis is of the form μ X μ Y , the P -value is the sum of the areas to the right of z 0 80 and to the left of z 0 80. Thus P 0 2119 0 2119 0 4238.
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