STA 6166 UNIT 2 Section 3 Answers
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Toxicology and Health Sciences
Carbon monoxide (CO) concentrations (in ppm) were measured at two locations in San Jose, CA over a two-month period. Summary data are presented below. Perform an 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?
Class n Sample_Mean Sample_Std_Dev I 24 19.6 8.1 II 18 29.8 6.6
To answer this question does not require the computer but to simply implement the equations for the F-test for 2 variances followed by the appropriate t-test.
Since the Class 1 group has the largest sample standard deviation we will consider it population 1 for the F-statistic (i.e. Class I variance will be on the top in the ratio.) Note that we were given the sample standard deviations, hence these values will have to be squared for the test (very common mistake here).
The calculated statistic value is F=65.61/43.56 =1.506. Assume the test will be performed at the a=0.05 type I error rate (this is typical for testing variances prior to performing a two sample t-test). Thus the F statistic value will be compared to F0.025,23,17 critical value. Note that the table in our book does not have the value for df1=23, but it does have df1=20 and df1=24. The value of F0.025,23,17 should be somewhere between 2.62 and 2.56. Since our observed F-statistic value is much less than the smallest of these, we would conclude to DO NOT reject the null hypothesis of equal variances and conclude that the groups have similar variances.
The previous test suggests that the two sample t-test for pooled variances would be appropriate for testing the differences in the Class means. We first need to compute the pooled variance and degrees of freedom.
The t statistics is then calculated as:
The problem description does not specify, hence we will test the two sided hypothesis at the type I error rate of a=0.05. The critical value for this test is t0.025,40 = 2.021. We reject the null hypothesis since the absolute value of the calculated t-statistic, 4.362, is much greater than the critical value of 2.021. Hence we conclude that the two Classes have significantly different means but common variance. The best estimate for the mean and standard deviation for Class I is 19.6 and 7.5 respectively. The best estimate for the mean and standard deviation for Class II is 29.8 and 7.5 respectively.
Think of a situation in your studies or research where testing the equality of variances would provide useful information?