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Statistics Homework Solutions 139

Statistics Homework Solutions 139 - 139 SECTION 6.5 We can...

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SECTION 6.5 139 We can conclude that the mean warpage is less than 2 mm. 11. X 1 25, s 0 624500, n 4. There are 4 1 3 degrees of freedom. The null and alternate hypotheses are H 0 : μ 2 5 versus H 1 : μ 2 5. t 1 25 2 5 0 624500 4 4 003. Since the alternate hypothesis is of the form μ μ 0 , the P -value is the area to the left of t 4 003. From the t table, 0 01 P 0 025. A computer package gives P 0 014. We can conclude that the mean amount absorbed is less than 2.5%. 13. (a) StDev = (SE Mean) N 1 8389 11 6 0989. (b) t 10 025 2 228. The lower 95% confidence bound is 13 2874 2 228 1 8389 9 190. (c) t 10 025 2 228. The upper 95% confidence bound is 13 2874 2 228 1 8389 17 384. (d) t 13 2874 16 1 8389 1 475. Section 6.5 1. X 8 5, s X 1 9, n X 58, Y 11 9, s Y 3 6, n Y 58. The null and alternate hypotheses are H 0 : μ X μ Y 0 versus H 1 : μ X μ Y 0. z 8 5 11 9 0 1 9 2 58 3 6 2 58 6 36. Since the alternate hypothesis is of the form μ X μ Y , the P -value is the area to the left of
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