Sample Exam #2 BioStats Instructions. No books, notes, or calculators are allowed. 1. (15 points) Find the minimum sample size required to estimate a population proportion, p, given that the margin of error is four percentage points for a 95% confidence interval where p is estimated as pˆ = 12.5%. Solution: 2 z0.05/2 pˆ(1 − pˆ) E2 = (1.96)2 (0.125)0.875 = 262.6094 so n = 263. 0.042 2. (20 points) The Common Murre population of birds became endangered in California due to an oil spill. Biologists want to know if their efforts to restore the colony have been successful. Since the breeding colony of birds is site dependent, the mean number of breeding pairs of Common Murres were observed at a specific rock known to be a breeding site. Find the p–value of a test that the mean number of breeding pairs of Common Murre is greater than 400. Use the following data collected from seven days of observations: 404 398 402 406 399 405 403 Assume the data comes from a normal distribution. Solution: Let H0 : µ = 400 and H1 : µ > 400. One calculates x¯ = 402.4286 √ = 2.147487. The p–value and s2 = 8.952381. The t(7 − 1) statistic is x¯s/−400 7 = 1 − 0.9623164 = 0.03768361 (between 0.025 and 0.05 using the Table A3). 3. (15 points) Medical researchers reviewing the risk of death for elderly patients taking dementia drugs found the following data. Number of Patients Total Deaths Dementia Drugs 500 23 Placebo 450 12 BioStats Sample Exam #2 Find a 95% confidence interval for the difference in death rate proportions. Solution: s 23 477 23 12 lower bound = − − 1.96 500 500 + 500 450 500 s 23 477 23 12 upper bound = − + 1.96 500 500 + 500 450 500 12 450 438 450 = −0.004304544 450 12 450 438 450 450 = 0.04297121 Thus the 95% confidence interval is (−0.0043, 0.043) 4. (15 points) Using the following data, use a significance level of 0.05 to determine whether there is a significant linear correlation between the two variables. x y Solution: 0 8 5 X n = 5, 3 2 4 6 5 9 xj = 24, j=1 5 X j=1 x2j = 194, 5 X j=1 12 12 5 X yj = 37, j=1 yj2 = 329, 5 X xj yj = 219. j=1 Thus 5(219) − 24(37) p = 0.6277225. r=p 5(194) − 242 5(329) − 372 r The t(5 − 2) statistic is q = 1.3967. For H0 : ρ = 0 versus H1 : ρ 6= 0, 1−r2 5−2 one has a p–value of 0.2569 (or greater than 0.20 using Table A3). There is not sufficient evidence to support the conclusion of a significant linear correlation. BioStats Sample Exam #2 5. (20 points) It is known that 41% of Americans have type A blood, 9% have type B, 4% has type AB and 46% have type O. A random sample of 200 stomach cancer patients yielded 92 with blood type A, 20 with type B, 4 with type AB and 84 having type O. Is the data significant at the 5% level to enable us to reject the null hypothesis that the blood type distribution of stomach cancer suffers is the same as the general population? Solution: One has o1 = 92, o2 = 20, o3 = 4, o4 = 84 and e1 e2 e3 e4 = = = = np1 np2 np4 np4 = 200(0.41) = 82, = 200(0.09) = 18, = 200(0.04) = 8 and = 200(0.46) = 92. Thus the test statistic is 4 X (92 − 82)2 (20 − 18)2 (4 − 8)2 (84 − 92)2 (oj − ej )2 = + + + = 4.1374. ej 82 18 8 92 j=1 Using Table A-4, since χ20.05 (4 − 1) = 7.815 > 4.1374 we cannot reject at 5% significance H0 , that the distribution of blood types for stomach cancer patients is the same as the general population. 6. (15 points) Consider the following Two-Way ANOVA table (missing some values). Source df Iron-form 1 Dosage 2 Iron-form:Dosage 2 Error — Total 107 SS MS F P 2.074 2.074 5.99 0.0161 15.588 —– —– 0.0007 —– 0.405 1.17 0.3145 35.296 —– 53.768 BioStats Sample Exam #2 Fill in the missing values. Is there an effect from interaction between Iron–form and Dosage? Solution: df for Error = 107 − 1 − 2 − 2 = 102 SS for Iron-form:Dosage = 2(0.405) = 0.810 15.588 MS for Dosage = = 7.794 2 25.296 MS for Error = = 0.346 102 7.794 F for Dosage = = 22.53 0.346 Source df SS MS F Iron-form 1 2.074 2.074 5.99 Dosage 2 15.588 7.794 22.53 Iron-form:Dosage 2 0.810 0.405 1.17 Error 102 35.296 0.346 Total 107 53.768 P 0.0161 0.0007 0.3145 There is not interactive effect (p–value of 0.3145 for H0 : no effect).