Assignment III for Biostats Course VHM 801 at AVC - Winter semester 2018

The assignment is worth 10% of the final course mark. Please be aware that by handing in the home assignment you implicitly acknowledge to have read and accepted the instructions for home assignments as described on the VHM 801 homepage.

The data for the assignment are from a study utilizing a registry of twins born in a certain geographical region in a certain time period. The registry contained 1030 pairs of female twins that were eligible by the inclusion criteria of the study. First, each twin pair had been classified as either monozygotic or dizygotic from background information such as photographs, data on physical similarity and frequency of confusion as children, and in some instances also blood samples. Next, interviews were conducted (separately) with each of the 2060 women, and they were all classified (no/yes) as meeting the criteria for alcoholism according to three different definitions: narrow (alcoholism with dependence), intermediate (alcoholism without dependence) and broad (problem drinking). The alcoholism classifications are only available in summarized form, as the number of twin pairs with 0, 1 and 2 twins classified as positive to alcoholism according to each of the three criteria/definitions. The table below gives these counts, for the monozygotic and dizygotic twin pairs.

Type of twins	Monozygotic			Dizygotic
Alcoholism criteria	Narrow	Intermed	Broad	Narrow	Intermed	Broad
neither positive (0)	537	510	443	377	361	301
one positive (1)	45	65	102	59	68	113
both positive (2)	8	15	45	4	11	26
Total	590			440

For example, there were 537 monozygotic twin pairs with neither of the twins classified as alcoholic according to the narrow criteria, out of the 590 monozygotic twin pairs; similarly, there were 65 monozygotic twin pairs with one of the two twins classified as alcoholic according to the intermediate criteria. The counts in the table are available as a data set in Minitab format and as a comma-separated file, for import into Stata and other statistical software. The home assignment has four questions (a)-(d) which should all be answered. In general, the assumptions of every statistical procedure used should be stated (formally or informally) and checked (where possible), and every statistical analysis should be summarized in a conclusion.

1. Using the narrow criteria of alcoholism, estimate the probability that in a randomly chosen monozygotic twin pair with at least one alcoholic twin, actually both twins in the pair are alcoholics. Supplement the estimate with a 95% confidence interval. Repeat for the intermediate and broad criteria (still for monozygotic twins); what happens to these probabilities when the criteria are changed?

2. Assume that you have a monozygotic twin who has been classified as an alcoholic according to the narrow criteria. Estimate the probability that you would also be classified as an alcoholic using the narrow criteria. In other words, our interest is in the probability that a twin has the condition if her (his) co-twin has the condition. Explain your calculation, and why this probability is not the same as in (a). Repeat also here for the other criteria for alcoholism (for monozygotic twins), and compare the results. As our focus is on the estimates, there is no need to supplement with confidence intervals.

3. Under each criteria of alcoholism, compare the results for monozygotic and dizygotic twin pairs to statistically assess whether monozygotic or dizygotic twin pairs are equally likely to contain 0, 1 and 2 alcoholics. What do your results tell you about a possible genetic factor behind alcoholism? Do the results and conclusions depend on the chosen criteria for alcoholism?

4. In addition to comparing monozygotic and dizygotic twin pairs, it is also of interest to assess directly whether the data show evidence of a dependence between alcoholism in the two twins in a pair. You are required to compute a chi-square test for the null hypothesis of independence by the following steps. Use for this part only the data for dizygotic twins and the broad criteria.
   • Estimate the probability that a randomly chosen person among all the twin pairs is an alcoholic.
   • Using the estimated probability and the assumption of independence between twins, estimate the probabilities that in a randomly chosen twin pair 0, 1, and 2 of the twins are alcoholics, respectively.
   • From these three estimated probabilities, compute the expected number of twin pairs with 0, 1, and 2 alcoholics (under the null hypothesis of independence).
   • Compute the chi-square statistic by adding up terms of the familiar form "(observed-expected) squared divided by expected" for the three categories of twin pairs.
   • Assess the significance of the test statistic computed this way in a chi-square distribution with one degree of freedom.
   • Draw conclusions about the null hypothesis.