Stat 1040
Chapter 21 Solutions

  1. The sample is like 500 draws from a box with 25,000 tickets. The tickets are marked 0 (does not have a computer) or 1 (has a computer). We do not know how many tickets are marked 0 and how many are marked 1. We can approximate the proportion of 1's using the sample percentage: (79/500)*100% = 15.8%. We can approximate the SD of the box by sqrt{.158*.842} = .36, so the SE of the sum is sqrt{500}*.36 = 8, and the SE of the percent is (8/500)*100% = 1.6%.
    1. 15.8%, 1.6%.
    2. The confidence interval is 15.8% plus or minus 3.2%. (12.6%, 19.0%).

  2. The sample is like 500 draws from a box with 25,000 tickets. The tickets are marked 0 (does not have a refrigerator) or 1 (has a refrigerator). We do not know how many tickets are marked 0 and how many are marked 1. We can approximate the proportion of 1's using the sample percentage: (498/500)*100% = 99.6%. We can approximate the SD of the box by sqrt{.996*.004} = .063, so the SE of the sum is sqrt{500}*.063 = 1.41, and the SE of the percent is (1.41/500)*100% = 0.3%.
    1. 99.6%, 0.3%.
    2. The confidence interval cannot be found because the box is so lopsided that the normal curve is not valid.

    1. The data are like 6000 draws from a box with 0's (for those students who didn't know that Chaucer wrote it), and 1's (for those students who did know that Chaucer wrote it.) The fraction of 1's in the box can be estimated by the fraction of 1's in the sample, which is 0.361. So, the SD of the box can be estimated as sqrt{0.361*0.639} = 0.48. The SE for the sum is sqrt{6000}*0.48 = 37, and the SE for the percentage is (37/6000)* 100% = 0.6%. So the 95% confidence interval is 36.1% plus or minus 1.2%. (34.9%,37.3%).
    2. 95.2% plus or minus 0.6%. (94.6%,95.8%).

  4. Option (ii) is the best - for example, if the two percentage points is 2 SE's, then 95% of the estimates will be right to within two percentage points.