Len KellyMATH302 MATH302 Suggested Practice ProblemsChapter 7 CH7 Review Exercises 3. A U.S. Travel Data Center survey reported that Americans stayed an average of 7.5 nights when they went on vacation. The sample size was 1500. Thus, a point estimate for the mean number of nights that Americans spend on vacation is X 7.5. If the sample standard deviation is 0.8 night, then a 95% confidence interval estimate for the mean number of nights that Americans spend on vacation, , can be found using X z 2 X z 2 n n Since the sample size is 1500, we can substitute s = 0.8 for . Additionally, because of symmetry, we know that the area to the left of z 2 is .975. Therefore, we can find z 2 using normsinv(.975). In this case, z 2 1.96, and our 95% confidence interval estimate for the population mean becomes 7.5 1.96 0.8 7.5 1.96 1500 0.8 7.46 7.54 1500 4. The lengths, in minutes, for a random sample of 11 popular children's animated films are listed below 93 78 83 100 76 78 92 76 77 75 81 A point estimate for the population mean is X 82.64 Because the sample size is small, a 95% confidence interval estimate for the population mean s s is found using X t 2 X t 2 . In this case, we have n n 8.488 8.488 82.64 2.228 (76.9,88.3) . 11 11 82.64 2.228 Please note that to find the t-multiplier, I used =tinv(.05,10). Notice that the tinv function works differently than the norminv function. In the tinv function the first entry is the probability of being in both tails. That is, one minus the confidence level. In this case, 1 - .95 = .05. See CH7_5th_MATH302.xls. 1 Len KellyMATH302 7. A researcher wishes to estimate, within $25, the true average amount of postage a community college spends each year. Assuming that the standard deviation is known to be $80, if she wishes to be 90% confident, how large a sample is necessary? In this case, we want X to be within $25 of with 90% confidence. We know that a 90% confidence interval estimate for is given by 80 80 X 1.65 X 1.65 n n 80 25. Solving, we find n So, we want to find n such that 1.65 80 25 25 n 1.65(80) n 1.65 2 n 1.65(80) 1.65(80) n 27.87 25 25 Thus, always rounding up, we would need to take a sample of 28 expenditures for postage. 11. A federal report stated that 88% of children under age 18 were covered by health insurance in the year 2000. How large a sample is needed to estimate the true proportion of covered children with 90% confidence with a confidence interval 0.05 wide. This problem statement is a little confusing for me. I am going to assume that \"with a confidence interval 0.05 wide\" means that we wish to ensure that our point estimate, p , is within 0.025 of the true population proportion, p. Now, we know that a (1 - )100% confidence interval estimate for p is given by pq pq p z 2 p p z 2 n n pq Thus, if we want p to be within 0.025 of p, then z 2 must be equal to 0.025. Thus, n z 2 pq .025 n n pq .025 z 2 2 pq .025 pq .025 n z 2 n z 2 z n pq 2 .025 2 2 see page 378. 2 1.65 Given the particulars of this problem, we find n .88(.12) 459.99 . Always .025 rounding up we would need a sample size of n = 460. 2 Len KellyMATH302 16. The lengths, in minutes, for a random sample of 11 popular children's animated films are listed below 93 78 83 100 76 78 92 76 77 75 81 To find a 99% confidence interval estimate for the variance of the length for all popular children's animiated films we use n 1 s 2 2 right 2 n 1 s 2 2 left 2 2 Where right CHIINV ( 2, n 1) and left CHIINV (1 2, n 1) Notice that the chiinv function works differently than the tinv or norminv functionsproof that Microsoft should employ more statisticians and fewer programmers. In this case, we have 2 2 CHIINV (.005,10) 25.188 and left CHIINV (.995,10) 2.156 right Thus, the 99% confidence interval for becomes 11 1 8.4882 25.188 2 11 1 8.4882 2.156 28.6 2 334.2 Thus, the 99% confidence interval for becomes 11 1 8.488 2 25.188 11 1 8.488 2 2.156 5.3 18.3 3