Hw5a - df = 16, t = -2.33 p = 0.016 ( or 0.018 if they are...

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STAT 350 Homework #5 Solution Addendum 7.46 – Test for equal variances: p – value = 0.027. Therefore it is NOT appropriate to assume equal variances 7.48 - Test for equal variances: p – value = 0.015. Therefore it is NOT appropriate to assume equal variances 7.52 – this problem should be analyzed using a paired t-test, so the issue of equal variances is irrelevant. 8.26 – Test for equal variances: p -value = 0.205, therefore it IS appropriate to use the test with equal variances. Therefore the following would also be acceptable. Then the following would also be acceptable:
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Unformatted text preview: df = 16, t = -2.33 p = 0.016 ( or 0.018 if they are using the t table from the text) Fail to reject null hypothesis at = 0.01. Two-Sample T-Test and CI Sample N Mean StDev SE Mean 1 8 1.710 0.530 0.19 2 10 2.530 0.870 0.28 Difference = mu (1) - mu (2) Estimate for difference: -0.820 95% upper bound for difference: -0.207 T-Test of difference = 0 (vs <): T-Value = -2.33 P-Value = 0.016 DF = 16 Both use Pooled StDev = 0.7407 8.40- this problem should be analyzed using a paired t-test, so the issue of equal variances is irrelevant....
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