Answers to the Review questions from Homework 4

Review Exercise Answers (Chapter 19)

6. No. Different students are more or less likely to walk through different parts of campus at different times and calculating the chances of getting a representative sample is difficult or impossible.`

Review Exercise Answers (Chapter 20)

3. (a) there should be 50,000 tickets. The box is the population
(b) each ticket has a 0 (gross income under 50K) or a 1 (gross income over 50K)
(c) False. The SD of the boxs Square Root (.2 x .8) = 0.4
(d) True. The draws are the sample
(e) The number of sample forms with gross incomes over 50K is likethe sum of 900 draws from the box. The expected value is 180. The SE of this is SQRT(900) x 0.4 = 12. The number of sample forms with incomes over 50K is around 180 give or 12.
12 out of 900 is about 1.33%. The percentage of forms in the sample with gross incomes over 50K will be 20% give or take 1.33% or so. The chance of getting between 19% and 21% is Z= -.75 to z= .75 or about 55%.

11. (a) 357, 340
(b) 71.4%, 68%
expected values are calculated from the "box" and observed values come from the sample.

Review Exercise Answers (Chapter 21)

1. (a) The sample is like 500 draws from a box with 25,000 tickets. Each ticket is marked 1 (has computer) or 0 (does not have). The number of sample households is like the sum of the draws. The fraction of 1's in the box is unknown but can be estimated from the sample as 79/500 or 15.8%. Based on sample information, the SD of the box is SQRT(.158 x .842) = .36. The SE for the number of households is SQRT(500) x .36 or 8. The SE for the percentage is 8/500 x 100 = 1.6%. The percentage of households in the town with computers is estimated as 15.8% give or take about 1.6%.
You can't use the national figure of 14.8% as the estimate for the 25,000 in this particular town because you can't make the argument that the 25,000 isexactly representative of the nation. In other words, 14.8% may be the US figure, but if that town of 25,000 is Beverly Hills, 14.8% doesn't apply.

(b) The 95% confidence interval is 15.8% +/- (2 * 1.6%)

3. In this sample (172+207)/500 = 379/500 = 75.8% own cars. The percentage of households in the town with cars is estimated as 75.8% give or take 1.9% or so.