HW Solutions for Section 10-1

HW Solutions for Section 10-1 - HW Solutions for Section...

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HW Solutions for Section 10.1 10.9 ) Let 1- Low-fat diet, and 2 – Low-carb diet, where μ refers to the mean number of pounds lost on each diet. Then, H 0 : μ 1 = μ 2 H 1 : μ 1 ≠ μ 2 . This requires a pooled variance t-test with 200 – 2 = 198 degrees of freedom. The appropriate critical values are +/- 1.97 ( or 1.96). The following Minitab output calculates a test statistic of 1.78. Two-Sample T-Test and CI Sample N Mean StDev SE Mean 1 100 7.60 3.20 0.32 2 100 6.70 3.90 0.39 Difference = mu (1) - mu (2) Estimate for difference: 0.900000 95% CI for difference: (-0.094843, 1.894843) T-Test of difference = 0 (vs not =): T-Value = 1.78 P-Value = 0.076 DF = 198 Both use Pooled StDev = 3.5672 Do not reject the null hypothesis. There is not enough evidence to conclude that there is a difference in the mean weight loss between with the two diets. 10.15 ) Same problem as 10.14, except that you use a separate variance test. The test statistic is -4.134 and the p-value is 0.00031. As in problem 14, you would reject the null
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This note was uploaded on 02/24/2010 for the course ACCT 302 taught by Professor Staff during the Spring '08 term at Old Dominion.

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HW Solutions for Section 10-1 - HW Solutions for Section...

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