Inference on p, Comparisons on p, and Chi-Square for Contingency Tables and GOF

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1 Inference on p, Comparisons on p, and Chi-Square for Contingency Tables and GOF 1. In a study of human blood types in nonhuman primates, a sample of 71 orangutans were tested and 14 were found to be blood type B. a. Use R to construct a 95% confidence interval for the proportion (probability) of blood type B in the orangutan population. > prop.test(14,71,correct=false) 1-sample proportions test without continuity correction data: 14 out of 71, null probability 0.5 X-squared = , df = 1, p-value = 3.34e-07 alternative hypothesis: true p is not equal to p b. Interpret the interval you just computed in part (a). We are 95% confident that the proportion of type B blood in the orangutan population is no less than and no more than Experimental studies of cancer often use strains of animals that have a naturally high incidence of tumors. In one such experiment, tumor prone mice were kept in a sterile environment with one group of mice maintained entirely germ free and the other group of mice exposed to the intestinal bacterium Eschericbia coli. The accompanying table shows the incidence of liver tumors. Let p S and p E.coli represent the probabilities of liver tumors under the sterile and the E. coli conditions, respectively. a. Use R to construct a 95% confidence interval for difference in proportions p S p E.coli > prop.test(t(mice),correct=false) 2-sample test for equality of proportions without continuity correction Sterile E. coli Liver Tumor 19 8 No Tumor 30 5 b. Interpret the confidence interval from part (a). With 95% confidence, we are unsure under which condition the probability of liver tumor is larger. If the probability is higher under the germ free condition, it is by as much as If it is higher under the E. coli. condition, it is by as much as c. Conduct a hypothesis test at the 0.05 significance level to test whether there is a difference in the probability of liver tumor under E.coli versus control conditions. (1) α = 0.05 (2) H O : p S p E.coli = 0 H A : p S p E.coli 0 > prop.test(t(mice),correct=false)

2 2-sample test for equality of proportions without continuity correction OR > chisq.test(t(mice),correct=false) (3) χ s 2 = (Z s = since Z s is the square root of the Chi-square statistic with the sign of the difference in p-hats) (4) P = (5) P = > 0.05 = α, fail to reject H O (do not have significant evidence for H A ) (6) There is not significant evidence to conclude that there is a difference in the probability of liver tumors in tumor prone mice under the E. coli or sterile condition. 3. Estrus synchronization products are used to bring cows into heat at a predictable time so that they can be reliably impregnated by artificial insemination. In a study of two estrus synchronization products, 42 mature cows (aged 4 8 years) were randomly allocated to receive either product A or product B, and then all cows were bred by artificial insemination. The table shows how many of the inseminations resulted in pregnancy. a. Construct a 95% confidence interval for the difference in (p A p B ) > prop.test(t(cows),correct=false) 2-sample test for equality of proportions without continuity correction data: t(cows) X-squared = , df = 1, p-value = < p A p B < Product A Product B Number of Cows Pregnant 8 15 Cows Pregnant Not Pregnant 13 7 b. Interpret the interval you just computed in part (a). We are 95% confident that the probability of mature cows (aged 4-8 years) which become pregnant by artificial insemination using Product B is larger than that of Product A by as little as 0.02 or as much as A population-based study, reported online in Stroke: Journal of the American Heart Association, related transient ischemic attack (TIA a mini-stroke ) and a person's risk for a subsequent myocardial infarction. The estimated relative risk for patients whose TIA occurred before age 60 was calculated to be 15.1, with a 95% confidence interval of (4.11,38.6). Interpret this interval. We are 95% confident that the long run risk of myocardial infarction for persons with a previous TIA is 4.11 to 38.6 times greater than that for those who have not had a TIA.

3 5. A portion (and slightly edited version) of a review of Workplace: Occupation and bladder cancer in a population-based case control study in Northern New England Occupational and Environmental Medicine 2011;68: Review by N Costet, C M Villanueva, J J K Jaakkola, M Kogevinas, K P Cantor, W D King, C F Lynch, M J Nieuwenhuijsen, S Cordier. Although the overall prevalence of bladder cancer is small (< 0.02%), it leads to death in approximately 20% of cases, so bladder cancer is an active research area. Several epidemiological studies suggested an association between the risk of bladder cancer and the exposure to trihalomethanes (THMs), the main disinfection byproducts (DBPs) of chlorinated water. The authors analyzed data from pooling three European case-control studies from France, Finland, and Spain (5467 individuals: 2381 cases and 3086 controls). Individual exposure to THMs was calculated combining information on residential history, estimates of the average total THMs (TTHM) level in tap water at the successive residences and personal water consumption. An odds-ratio was calculated for men exposed to an average residential TTHM level > 50 μg/l (OR=1.47 (1.05; 2.05)) when compared to men exposed to levels 50 μg/l. a. Interpret the confidence interval in the context of the setting. We are 95% confident that the long run odds of bladder cancer for men exposed to an average residential TTHM level > 50 g/l is 1.05 to 2.05 times the odds for men exposed to levels 50 g/l. b. Based on this confidence interval, can you determine whether the odds of bladder cancer are significantly larger at the 0.05 significance level? Why or why not? Yes. The interval does not contain the number 1 (which would indicate the long run odds may be the same under either exposure level) so a non-directional test at the 0.05 level would reject the null hypothesis. c. Why did the researchers report a confidence interval on the odds ratio rather than relative risk? A case control study does not give an estimate of the prevalence of bladder cancer (since we choose the number of cases and controls ), so we cannot estimate Pr{bladder cancer exposure level}. So, we cannot compute the risk we are interested in. Because of the numeric equivalence of the disease odds ratio and the exposure odds ratio, we can compute an estimate of the odds ratio we want. d. Can we use this sample odds ratio of 1.47 as an estimate of the relative risk? Why or why not? Yes, we can since the prevalence of bladder cancer is very close to zero and this makes the odds ratio a good approximation to the relative risk. (But, if we calculated a confidence interval on the odds ratio, the interpretation should be about the odds ratio not relative risk.) 6. Each person in a sample of 276 healthy adult volunteers was asked about the variety of social networks that they were in (e.g., relationships with parents, close neighbors, workmates, etc.). They were then given nasal drops containing a rhinovirus and were quarantined for 5 days. Numbers of subjects who developed colds were recorded in the table below. a. Test for an association between the number of types of social relationships and developing a cold at α = > cold<-matrix(c(57,66,52,101),nrow=2) > chisq.test(t(cold),correct=false) data: t(cold) X-squared = , df = 1, p-value = ) α = ) H0: cold associated with number of types of social relationships HA: cold not associated with number of types of social relationships 3) X 2 s = ) P = ) Since P < α, we reject H0. Five or Fewer Six or More Cold No Cold ) There is significant evidence to conclude that the developing a cold is associated with the number of types of social relationships.

4 b. Why did we do a test for association (rather than a test for the probability of a cold being equal under the two categories of number of types of social relationships)? One big sample was recruited into this study (everyone was exposed to the cold virus there were no treatment groups). Since the researchers did not assign the number of relationships or whether the subjects developed a cold or not, this is an observational study we can test for association. 7. One explanation for the widespread incidence of the hereditary condition known as sickle-cell trait is that the trait confers some protection against malarial infection. In one investigation, 543 African children were checked for the trait and for malaria. The results are shown in the table. Do the data provide evidence (α = 0.05) that malaria and having the sickle cell trait are negatively associated? Heavily Infected Light or Non Infected Sickle Cell Trait No Sickle Cell Trait Herpes simplex virus type 2 (HSV-2) is a sexually transmitted disease. As part of the third National Health and Nutrition Examination Survey (NHANES III), prevalence of HSV-2 was determined in four regions of the United States. The data are given in the following table Note, this is a slightly rearranged version of the table given in the text. Northeast Midwest South West HSV No HSV a. Use a chi-square test to compare the prevalence rates at α = (This is a question about the distribution of HSV-2 being different across the four regions, but worded in terms of prevalence rates). > hsv<-matrix(c(323,1165,381,1689,1320,4003,712,1986),nrow=2) > chisq.test(t(hsv),correct=false) data: t(hsv) X-squared = , df = 3, p-value = 8.949e-11 (1) α = 0.01 (2) H0: HSV-2 prevalence is the same in the four regions HA: HSV-2 prevalence differs across the four regions (3) X 2 s = (4) P = 8.95E-11 = which we interpret as P < (5) P < α, reject H0 (6) There is significant evidence to conclude that HSV-2 prevalence differs across the four regions b. Why did we do a test on prevalence rates (rather than a test for association)? Separate samples were taken from each region, so we can do a test for the distribution of HSV-2 across the regions rather than just a test for association. 9. The appearance of leaf pigment glands in the seedling stage of cotton plants is genetically controlled. According to one theory of the control mechanism, the population ratio of glandular to glandless plants resulting from a certain cross should be 11:5; according to another theory it should be 13:3. In one experiment, the cross produced 89 glandular and 36 glandless plants. Use goodness-of-fit tests (at α = 0.10) to determine whether these data are consistent with the 11:5 theory > chisq.test(c(89,36),p=c(11/16,5/16)) Chi-squared test for given probabilities data: c(89, 36)

5 X-squared = , df = 1, p-value = (1) α = 0.10 (2) H0: The 11:5 theory is correct HA: The 11:5 theory is not correct (3) X 2 s = (4) P = (5) P > α, fail to reject H0 (6) There is not significant evidence that the 11:5 theory is incorrect; the data are consistent with the 11:5 theory.