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 |
Agriculture and Environment
Do problem 7.41, page 377. 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 run the STAT > BASIC STATISTICS > 2 VARIANCES procedure to get the following output. ----------------------------------------- Test for Equal Variances Level1 Loc_1 Level2 Loc_2 ConfLvl 95.0000 Bonferroni confidence intervals for standard deviations Lower Sigma Upper N Factor Levels 0.492454 0.752846 1.51597 10 Loc_1 0.258818 0.395671 0.79674 10 Loc_2 F-Test (normal distribution) Test Statistic: 3.620 P-Value : 0.069
-------------------------------------------------------------------------
From the above printout, we can see that the value of the F-statistic for comparing the variance of Location 1 (.7528) to the variance of Location 2 (0.39567) is 3.62. This value is the 6.9 percentile of an F distribution with 9 and 9 degrees of freedom. The critical value for said F distribution at the a = 0.025 level is Fa/2,9,9=4.03. Since the computed F is smaller than the critical F we DO NOT reject the null hypothesis and conclude that the two groups have similar variances.
We next run the pooled variances t-test to compare the two location means. Using Minitab again we get:
Two-Sample T-Test and CI: Loc_1, Loc_2 Two-sample T for Loc_1 vs Loc_2 N Mean StDev SE Mean Loc_1 10 8.230 0.753 0.24 Loc_2 10 4.090 0.396 0.13 Difference = mu Loc_1 - mu Loc_2 Estimate for difference: 4.140 95% CI for difference: (3.575, 4.705) T-Test of difference = 0 (vs not =): T-Value = 15.39 P-Value = 0.000 DF = 18 Both use Pooled StDev = 0.601
From this we conclude that the two groups differ significantly in their means (p-value of the test is <0.0005). Hence we REJECT the null hypothesis of equal location means and conclude that the two locations have statistically significant means but common variances. Our best estimate of the parameters for location 1 is mean=8.23 and standard deviation 0.601. The best estimate of the parameters for location 2 is mean=4.09 and standard deviation 0.601.
Think of a situation in your studies or research where testing the equality of variances would provide useful information?