STA
6166 UNIT 2 Section 3 Answers

Welcome  <  Begin  <  <  Unit 2 Section 3 Exercises 
To Ag and Env. Answers 
To Tox and Health Answers 
To Social and Education Answers 
To Engineering Answers 
Engineering Sciences
Do problem 7.30, page 375. Concentrate on doing the Ftest for the difference in population variances. Given the results of this test, perform the associated two independent sample ttest 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 FTest (normal distribution) Test Statistic: 3.953 PValue : 0.069

From this conclude that although the variance for Procedure 1 (12.4074) is
almost twice that of Procedure 2 (6.2405), the calculated Fvalue of 3.953
is less than the critical value for a onesided variance test at the a
= 0.05 level. That is, F=3.953 is less than F_{a/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 ttest to compare the procedure means. From Minitab we get:
 TwoSample TTest and CI: Procedure1, Procedure2 Twosample 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) TTest of difference = 0 (vs not =): TValue = 3.58 PValue = 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 pvalue 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?