STA 6166 UNIT 2 Section 3 Answers
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# Unit 2 Section 3 Answers

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?