1. Section 7.4 Table 7.4-1 summarizes hypothesis tests about one variance (standard deviation). (These tests assume that the sample is taken from a normal distribution, but the tests are not robust when the sample is taken from a non-normal distribution.) Table 7.4-2 summarizes hypothesis tests to compare two variances (standard deviations). These tests assume that each sample is taken from a normal distribution and are not robust when each sample is taken from a non-normal distribution. 1.In Class Exercise 7.2-4, the hypothesis test about the mean in Text Exercise 7.2-8 was performed under the assumption that the variance in weight of home-born babies is  X 2 = 525 2. The relevant SPSS output produced from the data was as follows:

2. (a) (b) State the test statistic and describe the critical region for testing H 0 :  X 2 = 525 2 versus  X 2 < 525 2 with  = 0.05. The test statistic is  2 = The one-sided critical region with  = 0.05 is (n – 1)s 2 ——— = 525 2  2  3.940. 10s 2 —— 525 2 Calculate the value of the test statistic, and write a summary of the results. From the SPSS output, we find s =336.31606. 10 (336.31606) 2  2 = ——————— = 525 2 4.104 Since  2 = 4.104 >  2 0.95 (10) = 3.940, we fail to reject H 0. We conclude that the variance in weight (grams) of home-born babies is not less than 525 2.

3. 1. - continued (c) What is the approximate p-value of this test? The p-value of the test is P(Q  4.104) is between 0.05 and 0.10, from Table IV in Appendix B. 10s 2 ——  4.104;  2 = 525 2 = 525 2 P a  2 (10) random variable 2. (a) (b) Do parts (a) and (b) of Text Exercise 7.4-7; then do part (c) stated below. The test statistic is  2 = The critical region of size  = 0.05 is (n – 1)s 2 ——— = 30  2  28.87. 18s 2 —— 30  = P{  2  28.87 ;  2 = 80} = 18s 2 P ——  28.87 ;  2 = 80 = 30

4. 18s 2 30 P ——  28.87 — ;  2 = 80 = 80 80 18s 2 P ——  10.826 ;  2 = 80 = 30 slightly less than 0.10, from Table IV in Appendix B. (c)Suppose s 2 = 43.8 for a random sample of size n = 19. State the results of the hypothesis test, and find the p-value of the test. With s 2 = 43.8, we have 18 (43.8)  2 = ———— = 30 26.28 Since  2 = 26.28 <  2 0.05 (18) = 28.87, we fail to reject H 0. We conclude that  2 = 30. The p-value of the test is P(Q  26.28) = between 0.05 and 0.10, from Table IV in Appendix B. 18s 2 ——  26.28 ;  2 = 30 = 30 P a  2 (18) random variable

5. 3.In Class Exercise 7.3-1 (Text Exercise 7.3-2), we assumed that  X 2 =  Y 2. Now, test H 0 :  X 2 =  Y 2 versus H 1 :  X 2   Y 2 by finding a test statistic, defining a critical region, and writing the results of the test. Use  = 0.05. The test statistic is The two-sided critical region with  = 0.05 is s 2 x — and its inverse. s 2 y We fail to reject H 0. We conclude that the variance in weight is not different for male and female gallinules. Note: Levene’s f Test for equality of variances, available from SPSS, is a much more robust test. s 2 x —  s 2 y —  s 2 x f 0.025 (15,12) = 3.18 orf 0.025 (12,15) = 2.96 s 2 x 1356.75 — = ———— = 1.96 and s 2 y 692.21 — = ———— = 0.51 s 2 x 1356.75

6. 4. Since f = 0.675 < f 0.05 (1,26) = 4.23, we fail to reject H 0. We conclude that the variance in birth weight is not different for mothers who have five or fewer prenatal visits and mothers who have had six or more prenatal visits. In Class Exercise 7.3-2 (Text Exercise 7.3-5), we assumed that the variances were equal. Now, with  = 0.05, write the results for Levene’s f Test for equality of variances from the SPSS output (reproduced below). obtained with linear interpolation