9-4 Two Variances or Standard Deviations 483 SOLUTION REQUIREMENT CHECK (1) The two populations are independent of each other. The two samples are not matched in any way. (2) Given the design for the study, we assume that the two samples can be treated as simple random samples. (3) A normal quantile plot of each set of sample weights shows that both samples appear to be from populations with a normal distribution. The requirements are satisfied. Here are the statistics for the above data sets: ■ ANSUR I 1988: n = 12, s = 12.4194 kg ■ ANSUR II 2012: n = 10, s = 15.5306 kg Because we stipulate that in this section the larger variance is denoted by s2 1, we switch the above two data sets and we let s2 1 = 15.5306 2 and n 1 = 10. By default, s2 2 = 12.4194 2 and n 2 = 12. Step 1: The claim that the variation did not change from ANSUR I 1988 to ANSUR II 2012 can be expressed symbolically as s 2 1 = s 2 2 or as s1 = s2. We will use s1 = s2. Step 2: If the original claim is false, then s1 ≠ s2. Step 3: Because the null hypothesis is the statement of equality and because the alternative hypothesis cannot contain equality, we have H0: s1 = s2 (original claim) H1: s1 ≠ s2 Step 4: The significance level is a = 0.05. Step 5: Using Technology Steps 5 and 6 can be skipped using technology. See the accompanying StatCrunch display showing that the test statistic is F = 1.5637842 and the P-value is 0.4779. StatCrunch Manual Calculations If technology is not available, first note that because this test involves two population variances, we use the F distribution. Step 6: The test statistic is F = s2 1 s2 2 = 15.53062 12.41942 = 1.5638 P-Value Method Due to the format of Table A-5, the P-value method is a bit tricky without technology, but here we go. For a two-tailed test with significance level 0.05, there is an area of 0.025 in the right tail, so we use the two pages for the F distribution (Table A-5) with “0.025 in the right tail.” The degrees of freedom are found as follows: ■ Numerator degrees of freedom = n1 - 1 = 10 - 1 = 9 ■ Denominator degrees of freedom = n2 - 1 = 12 - 1 = 11 With numerator degrees of freedom = 9 and denominator degrees of freedom = 11, we find that the critical value of F is 3.5879. The test statistic of F = 1.5638 is not greater than the critical value, so we know that the area to the right of the test statistic is more than 0.025, and it follows that for this two-tailed test, P@value 7 0.05. continued
RkJQdWJsaXNoZXIy NjM5ODQ=