IE 241 Solutions 2

1. IE 241 Solutions 2

2. 1. Ho: pf = pn Ha: pf ≠ pn 46/200 = 0.23 fraternity proportion 51/300 = 0.17 nonfraternity proportion 97/500 = 0.19 overall proportion Because of the large sample size, we can use t.05 = z.05 = 1.96for α = .05for this 2-tailed test. Because the observed t < 1.96, we cannot reject the null hypothesis and we conclude that no difference of opinion on this issue has been demonstrated.

3. 2. Ho: py– pa = .10 Ha: py– pa > .10 50/200 = .25 youth proportion 60/500 = .12 adult proportion 110/700 = .157 overall proportion Because of the large sample size, we can use t.05 = z.05 = 1.65 at α =.05 for this 1-tailed test. Since the observed t < 1.65, we can not reject the null hypothesis.

4. 3. Ho: μA – μB = 0 Ha: μA – μB ≠ 0 sA = 8 sB = 10 nA = 100 nB = 100 For this 2-tailed test at α = .05, t = z = 1.96. Since the observed t > 1.96, we reject Ho and conclude that there is a difference between the teaching methods with 95% confidence.

5. Expected ratio: q2 : p2+2pq : r2+2qr : 2pr 0.16 : 0.48 : 0.20 : 0.16 when p = .4, q = .4, r = .2. Χ2 test of goodness of fit For 3 df, χ2 = 7.815 at α= .05, so we can reject Ho with 95% confidence and declare that the observed frequencies are not compatible with the pattern.

6. Ho: Drug and Effect are not related Ha: Drug and Effect are related χ2 test of independence in contingency table The expected frequencies are obtained by (ni.)(n.j) / n and are shown in parentheses in the table.

7. 5 (continued) The df here = (2-1)(3-1) = 2 and at α =.05, the critical value of χ2 = 5.99. Since the observed χ2 < 5.99, we cannot reject Ho.

8. 6(a). Ho: σ = 8 Ha: σ < 8 = 42 n = 20, s = 5 χ2 test that variance = specified value The critical value of χ2 for 19 df where α =.05 is 30.1. Since the observed χ2 is much less than 30.1, we cannot reject the null hypothesis. So the data do not support the proposition that the variance is < 8.

9. 6(b) Ho: μ = 45 Ha: μ ≠ 45 = 42 n = 20, s = 5 The critical value of t for 19 df in this 2-tailed test at α =.05 is 2.093. Since the observed value of t < -2.093, we can reject the null hypothesis that μ = 45.

10. 7. Ho: μ1 = μ2 Ha: μ1 ≠ μ2 The t test is The critical t for (10-1) + (12 -1) = 20 df is 2.086 for this two-tailed test at α = .05. Since the observed t < 2.086, we cannot reject the null hypothesis.

11. 7 (continued) Testing the validity of the t test by ensuring that the two variances are not significantly different. The critical value of F for (12-1) and (10-1) df at α= .05 is 3.10. Since the observed F < 3.10, the two variances are not significantly different and the t test is valid.

12. 8. For a sample of 20 rats, 10 in each group: Ho: μraw = μroasted Ha: μraw ≠ μroasted The critical t value for (10-1) + (10-1) = 18 df at α = .05 for this two-tailed test is 2.101. Since the observed t < 2.101, we cannot reject the null hypothesis.

13. 8 (continued) Testing the validity of the t test by ensuring that the two variances are not significantly different. The critical value of F for (10-1) and (10-1) df at α= .05 is 3.18. Since the observed F < 3.18, the two variances are not significantly different and the t test is valid.

14. 9. Is the variability of wood affected by weathering? The critical value of F = 1.598 for α = .05 for (100-1) and (40-1) df. Since the observed F < 1.598, the data do not justify rejecting the null hypothesis.

15. 10. Ho: μC – μU = 0 Ha: μC – μU ≠ 0 The test statistic is The critical t value for (20-1)+(20-1) = 38 df at α = .05 for this two-tailed test is -2.024. Since the observed t <-2.024, we can reject the null hypothesis.

16. 10 (continued) The 95% confidence interval for the difference between the two means is So the 95% confidence interval is .66 ≤ μdiff ≤ 13.24 Since this confidence interval does not include 0, it confirms the result of the hypothesis test.

17. 11. Are gender and color-blindness related? χ2 test of independence for contingency table The expected frequencies are obtained by (ni.)(n.j)/n, so the expected frequency for cell11 is (956)(480)/1000 = 458.88 = 459. The expected frequencies for the other cells are shown in parentheses.

18. 11 (continued) The critical value of χ2 for =(2-1)(2-1)=1 df is 3.84. Since the observed χ2 >> 3.84, we can reject the null hypothesis and declare that gender and color-blindness are related.

19. 12. 12 pairs of runners matched for ability Ho: μdiff = 0 Ha: μdiff > 0 The critical t value for (12-1) df for this 1-tailed test at α = .05 is 1.796. Since the observed t > 1.796, we can reject the null hypothesis and declare the stimulant effective.

20. 12 (continued) The 95% confidence interval for the mean difference is So the 95% confidence interval is 0.572 < μ < 1.762 Since the interval does not cover 0, it confirms the result for the t test.

21. 13. Ho: μ1 = μ2 Ha: μ1 < μ2 The critical value of t for 68 df = z for this 1-tailed test at α = .05 is 1.645. Since the observed t < -1.645, the null hypothesis is rejected and method 1 is declared better (takes less time) than method 2.

22. 13 (continued) The 95% confidence interval for the time difference is and the interval is -0.2354 < μ < 3.5854 This interval clearly covers 0 even though the difference is significant because …