HW5-Solutions

# HW5-Solutions - STAT 101 - Agresti Homework 5 Solutions...

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Unformatted text preview: STAT 101 - Agresti Homework 5 Solutions 10/18/10 Chapter 7 7.1. These are independent samples, because the subjects in the two samples are different, with no matching between one sample with the other sample. 7.5. (a) 2 1 164 140 24 y y- =- = ; ( 29 ( 29 2 2 2 2 2.83 se = + = . (b) 164/140 = 1.17, which is a 17% increase. (c) 2 1 191 166 25 y y- =- = ; There was an estimated 25 pound increase in the mean weight of adult American men between 1962 and 2002. 191/166 = 1.15; There was an estimated 15% increase in the mean weight of adult American men between 1962 and 2002. 7.6. (a) The response variable is median net worth in 2002. The explanatory variable is race (white, black). (b) (i) \$89,000 - \$6,000 = \$83,000; There is an estimated difference of \$83,000 in median net worth between white and black households in 2002. (ii) \$89,000/\$6,000 = 14.8; The median net worth in white households in 2002 was estimated to be 14.8 times the median net worth in black households in 2002. 7.8. (a) 0.00164 0.00015 = 0.00149; This estimated difference of 0.001 seems very small. (b) 0.00164/0.00015 = 10.9; The probability that a black male is a homicide victim is 10.9 times the probability that a white male is a homicide victim, which is a very large effect. (c) The relative risk better summarizes results, especially when both proportions are very small. 7.9. (a) We can be 95% confident that the interval 0.18 to 0.26 contains the difference between the population proportion of teens who listen to lots of music with degrading sexual messages and have intercourse and the population proportion of teens who listen to little or no music with degrading sexual messages and have intercourse. (b) If the two population proportions were equal, it would be very unlikely to observe a difference as large as we did. It appears that the proportion of teens who listen to lots of music with degrading sexual messages and have intercourse is greater than the proportion of teens who listen to little or no degrading music and have intercourse. 7.11. (a) ( 29 ( 29 ( 29 ( 29 1 1 1 2 2 2 1 1 0.399 0.601 12,708 0.482 0.518 8783 0.0069 se n n =- +- = + = (b) ( 29 ( 29 ( 29 ( 29 2 1 0.482 0.399 1.96 0.0069 0.083 0.014 0.069 to 0.097. z se - =- = = We can be 95% confident that interval 0.069 to 0.097 contains the difference in the population proportion of college students who drank to get drunk in 2001 and the population proportion of college students who drank to get drunk in 1993. 7.12. (a) We have two independent random samples with at least 10 observations in each category for each group. Let 1 = the proportion of students in 1993 who said they engaged in unplanned sexual activities because of drinking alcohol and 2 = the proportion of students in 2001 who said they engaged in unplanned sexual activities because of drinking alcohol. The null hypothesis is 2 1 : H - = , and the alternative hypothesis is 2 1 : a H - . (b) If the true difference in proportions were really 0, the ....
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## This note was uploaded on 07/14/2011 for the course STA 101 taught by Professor Alan during the Fall '10 term at University of Florida.

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HW5-Solutions - STAT 101 - Agresti Homework 5 Solutions...

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