Exercises questions answers

EXERCISES

10.1 Give definitions of the following terms in your own words:

a.     Sample proportion

b.     Population proportion

c.      Binomial variable

d.     Bernoulli trial

e.      Continuity correction

f.       Confidence interval for a proportion

10.2 Peripheral neuropathy is a complication of uncontrolled diabetes. The num-ber of cases of peripheral neuropathy among a control group of 35 diabetic patients was 12. Among a group of 11 patients who were taking an oral agent to prevent hyperglycemia, there were three cases of peripheral neu-ropathy. Is the proportion of patients with peripheral neuropathy compara-ble in both groups? Perform the test at a = 0.05.

10.3 Construct exact 95% confidence intervals for the proportion of patients with peripheral neuropathy in the medication group and the proportion of pa-tients in the control group in the previous exercise. Construct two confi-dence intervals for each proportion, one with correction for continuity and the other without correction for continuity.

10.4 Referring to Exercise 10.2, construct an approximate 95% confidence inter-val for the difference between the proportions of patients affected by pe-ripheral neuropathy in the control group and in the medication group.

10.5 A dental researcher investigated the occurrence of edentulism (defined in the research study as loss of two or more permanent teeth, not including loss of prophylactically extracted wisdom teeth) in a rural Latin American village. A total of 34 out of 100 sampled adults had lost at least two teeth. A study of a U.S. city found that the rate of loss of at least two teeth was 14%. Was the proportion of persons who had edentulism higher in the Latin American village than in the U.S. city? Conduct the test at the a = 0.05 level.

10.6 Calculate an exact 95% confidence interval for the proportion of edentulous persons in the Latin American village (refer to Exercise 10.5.)

10.7 For the data in Exercise 10.5, compute a 99% confidence interval using the normal approximation with continuity correction. Is the result close to the exact interval found in Exercise 10.6? Explain why or why not.

10.8 In a British study of social class and health, a total of 171 out of 402 lower social class persons were classified as overweight. The percent of over-weight persons in the general population was 39%. Based on these findings, would you assert that low social class is related to being overweight? Test this hypothesis at the α = 0.01 level.

10.9 A longitudinal study of occupational status and smoking behavior among women reported at baseline that 170 per 1000 professional/managerial women were nicotine dependent. The corresponding rate among blue collar women was 310 per 1000. At the a = 0.05 level, determine whether there is a significant difference in nicotine dependence between the proportion of women who are classified as professional/managerial workers in compari-son to those who are classified as blue collar workers. Then compute the ap-proximate 99% confidence interval for the difference between these two proportions.

10.10 An epidemiologic study examined risk factors associated with pediatric AIDS. In a small study of 30 cases and 30 controls, a positive history of substance abuse occurred among 11 of the cases and 6 of the controls. Based on these data, can the investigator assert that substance abuse is sig-nificantly associated with pediatric AIDS at the a = 0.05 level? Compute the approximate 95% confidence interval for the difference between the proportions of substance abuse found in the case and control groups.

Answers:

10.2 Z’ = (W₁ – W₂ )/ √{[W_c (1 – W_c )/n₁ + W_c (1 – W_c )/n₂ ]}, where W_c = (X₁ + X₂ )/(n₁ + n₂) and X₁ = 12, the number with peripheral neuropathy out of n₁ = 35 in the con-trol group of diabetic patients and X₂ = 3, out of the 11 patients taking an oral agent to prevent hyperglycemia, so n₂ = 11. Z’ is approximately standard normal under the null hypothesis. W_c = 15/46 = 0.3261. W₁ = 12/35 = 0.3429 and W₂ = 3/11 = 0.2727. So Z’ = 0.07/ √{0.3261(0.6739)/35 + 0.2727(0.7273)/11} = 0.07/0.1559 = 0.4490. In this case, the p-value (two-sided) is approximately 2(0.5 – 0.1736) = 0.6528. So we cannot detect a significant difference between these two proportions.

10.6 The number of Latin American patients with edentulism is x = 34. The sample size is n = 100. The confidence level 1 – a = 0.90. The formula for the confidence in-terval is by Clopper–Pearson [{1 + (100 – 34 + 1)F(0.95:200 – 68 + 2, 68)/34}^–1, {1 + (100 – 34)/{35F(0.95:68 + 2, 2(100 – 34)}^–1] = [1/{1 + 67F(0.95:134, 68)/34}, 1/{1 + 66/(35F(0.95:70, 132))}]. F(0.95:134, 68) 1.45 and F(0.95:70, 132) 1.42. So the interval is [0.259, 0.430]

10.8 The sample proportion p = 171/402 = 0.425. We are testing the hypothesis that p = 0.39 against the alternative p > 0.39 that the proportion overweight in the lower social class in Britain is higher than for the general British population. Take Z = (0.425 – 0.39)/{ √ (0.39) √ (0.61)/ √402} = 0.035/0.02433 = 1.439. This is non-significant. For a one-sided test at the 0.01 significance level, the critical Z = 2.33.