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6166 UNIT 2 Section 3 Answers
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Engineering Sciences
Do problem 7.30, page 375. Concentrate on doing the F-test for the difference in population variances. Given the results of this test, perform the associated two independent sample t-test for means? What do you conclude about the distributions of the two groups from these two tests?
Using Minitab we get the following using the STAT > BASIC STATISTICS > 2 VARIANCES procedure. ----------------------------------- Test for Equal Variances Level1 Procedure1 Level2 Procedure2 ConfLvl 95.0000 Bonferroni confidence intervals for standard deviations Lower Sigma Upper N Factor Levels 7.95151 12.4074 26.4507 9 Procedure1 3.99936 6.2405 13.3039 9 Procedure2 F-Test (normal distribution) Test Statistic: 3.953 P-Value : 0.069
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From this conclude that although the variance for Procedure 1 (12.4074) is
almost twice that of Procedure 2 (6.2405), the calculated F-value of 3.953
is less than the critical value for a one-sided variance test at the a
= 0.05 level. That is, F=3.953 is less than Fa/2.8,8
= 4.43 hence we DO NOT reject the null hypothesis of equal variances
and conclude that the two groups have similar variances.
From this we proceed to perform the pooled variance t-test to compare the procedure means. From Minitab we get:
-------------------------------------------------- Two-Sample T-Test and CI: Procedure1, Procedure2 Two-sample T for Procedure1 vs Procedure2 N Mean StDev SE Mean Procedur 9 86.8 12.4 4.1 Procedur 9 70.22 6.24 2.1 Difference = mu Procedure1 - mu Procedure2 Estimate for difference: 16.56 95% CI for difference: (6.74, 26.37) T-Test of difference = 0 (vs not =): T-Value = 3.58 P-Value = 0.003 DF = 16 Both use Pooled StDev = 9.82
From this we conclude that the means of the two groups are statistically different with p-value of 0.003. That is, to avoid not rejecting the null hypothesis our Type I error rate would have to be set at less than 0.003. Since we typically set a at 0.05 or 0.01, it is clear that we reject the null hypothesis of equal means and conclude that the two group means are different. Hence our best estimate is that Procedure 1 has mean 86.8 and standard deviation 9.82 whereas Procedure 2 has mean 70.22 with standard deviation 9.82.
Think of a situation in your studies or research where testing the equality of variances would provide useful information?