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Social Case-based Statistics II CSSS 322 Professor: Mark S. Handcock Solutions to Homework 5 Due Thursday, May 16, 2002 Problems to be handed in: 1) In an article on crime in the United States, Newsweek magazine (January 10, 1994) quoted FBI statistics stating that of all blacks slain in 1992, 94% were slain by blacks, and of all whites slain in 1992, 83% were slain by whites. Let Y denote race of victim and X denote race of murderer. a) What conditional distributions do these statistics refer to, those of Y at given levels of X , or those of X at given levels of Y ? Set up a table with race of murderer as rows and race of victim as columns, showing these conditional distributions. Solution(8) : These statistics refer to the conditional distribution of X (race of murderer) at given levels of Y (race of victim). We can represent the data with a contingency table: Murderer (X) Black White Total Black 94% 6% 100% White 17% 83% 100% b) Are X and Y independent random variable or dependent? Explain conceptually what you mean (You need not do a test). Solution(5) : X and Y are dependent, since there is a strong association between race of victim and race of murder. 2) The table below is from the 1991 General Social Survey. White subjects were asked: If your party nominated a (Negro/Black) fro President, would you vote for him or her if he/she were quali ed for the job? and During the last few years, has anyone in your family brought a friend who was a (Negro/Black) home for dinner? Black for President Yes No Don t Know Total Yes 113 6 0 119 Home For Dinner No Don t Know 239 49 13 301 1 0 2 3 Total 353 55 15 423 CSSS 322 Spring 2002 May 21, 2002 Page 1 a) Conduct a 2 test of the hypothesis that these two responses are independent. What is the p -value of the test and your decision. Solution(10) : Data of the question with expected frequencies in parentheses: Black for President Yes No Don t Know Total Yes 113 (99.31) 6 (15.47) 0 (4.22) 119 Home For Dinner No Don t Know 239 (251.19) 49 (39.14) 13 (10.67) 301 1 (2.5) 0 (0.39) 2 (0.11) 3 Total 353 55 15 423 Let fo represent the frequency of observed frequency and fe represent the frequency of expected frequency. Use the information from the table; we then calculate the Chi-Square statistic: (fo fe )2 2 = = 50.49, fe with the degrees of freedom df = (3 1)(3 1) = 4 The p -value of 2 = 50.49 is 4 less than 0.001. There is very strong evidence of an association. b) Some of the expected frequencies are small so the Normal approximation to the binomial may not hold. Hence the 2 test that is based on this approximation might not be valid. It is possible to conduct a small-sample exact test for testing the null hypothesis of independence between the two responses given the marginal totals. This can be done by many statistical software packages and gives a p -value of 0.0000035. Solution(3) : Given that the small-sample exact test for testing the null hypothesis of independence between the two responses given the marginal totals has a p -value of 0.0000035, the test also provides very strong evidence of an association. c) Given the results of the test, interpret the nature of the deviations from the independence assumption. What categories are higher and lower than would be expected if the factors were independent? Solution(4) : The observed frequencies are larger than expected when both responses are Yes or both are No, and less than expected when one is Yes and the other is that No; is, those who have had a black home for dinner are more likely to be willing to vote for a black for president, relative to what is predicted by the hypothesis of independence. d) Using Yes and No categories of each response, estimate the odds ratio. What does it tell us in this case? Solution(5) : The odds ratio = CSSS 322 Spring 2002 113 49 = 3.9 239 6 May 21, 2002 Page 2 The interpretation is that for those who would vote for a black for President, the odds of having invited a Black home for dinner are 3.9 times the odds of an invite for those who would not vote for a Black for President. Extra Credit Problem: 3) According to the U.S. Department of Justice, in 1994 he incarceration rate in the nation s prisons was 646 per 100,000 male residents, 45 per 100,000 female residents, 1471 per 100,000 black residents, and 207 per 100,000 white residents (Bureau of Justice Statistics Bulletin: Prisoners in 1994). a) Find the odds ratio between gender and being incarcerated. Give an interpretation of it. Solution(5) : The probability of incarceration for males is 646/100000 and that of women is 0.00646 45/100000. Hence the odds of incarceration rate for males = (1 0.00646) = 0.0065 and the odds of incarceration rate for females = 0.00045 (1 0.00045) = 0.00045. 0.0065 The odds ratio between gender and whether incarcerated = 0.00045 = 14.4, indicating that males have a much greater (i.e. 14.4 times) odds of being incarcerated than females. This indicates a strong relationship between gender and race in terms of incarceration. b) Find the odds ratio between race and being incarcerated. Give an interpretation of it. Solution(5) : The odds of incarceration rate for blacks = 0.01471 (1 0.01471) = 0.01493 The odds of incar- 0.00207 ceration rate for whites = (1 0.00207) = 0.00207. and whether incarcerated = 0.01493 = 7.2. 0.00207 Hence the odds ratio between race The odds of incarceration for blacks were 7.2 times the odds of whites. c) According to the odds ratio, which has the stronger association with whether incarcerated, gender or race? Solution(2) : Both associations are strong, but association with gender is stronger since the odds ratio is farther from 1 (i.e. 14.4 verses 7.2). d) Conduct a 2 test of independence between the incarceration and race, and incarceration and gender. For these assume that we have samples of 100,000 from each race and gender category. Solution(7) : For this problem we assume that we have populations of 100,000 for each of the CSSS 322 Spring 2002 May 21, 2002 Page 3 groups. Consider rst a comparison of incarceration and gender: Gender Male Female Total Yes Incarceration No 99354 (99654.5) 99955 (99654.5) 199309 646 (345.5) 45 (345.5) 691 The expected frequencies are in parentheses. The test statistic is = 2 (fo fe )2 = 524.53, fe with the degrees of freedom df = (2 1)(2 1) = 1 We have very strong evidence of an association. Next consider a comparison of incarceration and race: Incarceration Race Yes No Black White Total 1471 (839) 207 (99161) 1678 99354 (99161) 99955 (839) 198322 The expected frequencies are in parentheses. The test statistic is 2 = (fo fe )2 = 960.2, fe with the degrees of freedom df = (2 1)(2 1) = 1 We have very strong evidence of an association. Note that the strength of this evidence depends on the size of the populations. Smaller populations leads to smaller test statistics. CSSS 322 Spring 2002 May 21, 2002 Page 4

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