Although airline schedules and cost are important factors for business travelers when choosing an airline carrier, a USA Today survey found that business travelers list an airline's frequent flyer program as the most important factor. From a sample of \(n=1993\) business travelers who responded to the survey, 618 listed a frequent flyer program as the most important factor. a. What is the point estimate of the proportion of the population of business travelers who believe a frequent flyer program is the most important factor when choosing an airline carrier? b. Develop a \(95 \%\) confidence interval estimate of the population proportion. c. How large a sample would be required to report the margin of error of .01 at \(95 \%\) confidence? Would you recommend that \(U S A\) Today attempt to provide this degree of precision? Why or why not?

To find the point estimate of the proportion of the population, we need to divide the number of business travelers who listed frequent flyer programs as the most important factor (618) by the total number of business travelers sampled (1993). The formula to calculate point estimate is: Point Estimate (p) = \(\frac{\text{Number of success}}{\text{Total number of trials}}\) Point Estimate (p) = \(\frac{618}{1993}\) Now let's calculate the point estimate: Point Estimate (p) = 0.310 Therefore, the point estimate of the proportion of the population of business travelers who believe a frequent flyer program is the most important factor when choosing an airline carrier is 0.310.

To develop a 95% confidence interval estimate of the population proportion, we will need to use the following formula: Confidence Interval (CI) = Point Estimate ± (Z-Value × Standard Error) The standard error for a proportion (SE) can be calculated using the formula: Standard Error (SE) = \(\sqrt{\frac{p(1 - p)}{n}}\) Where: - p is the point estimate of the proportion - n is the total number of trials (sample size) First, let's find the Standard Error (SE): Standard Error (SE) = \(\sqrt{\frac{0.310(1 - 0.310)}{1993}}\) Standard Error (SE) = 0.011 We are asked to develop a 95% confidence interval estimate, and the Z-value for 95% confidence interval is 1.96. Now let's calculate the confidence interval: CI = 0.310 ± (1.96 × 0.011) CI = (0.288, 0.332) Hence, the 95% confidence interval estimate of the population proportion is from 0.288 to 0.332.

To find the required sample size to have a margin of error (ME) of 0.01 at 95% confidence, we can use the following formula: n = \(\frac{(Z-Value)^2 \times p(1 - p)}{(ME)^2}\) Where: - Z-Value is 1.96 (95% confidence interval) - p is the point estimate of the proportion (0.310) - ME is the margin of error (0.01) Now let's calculate the required sample size: n = \(\frac{(1.96)^2 \times 0.310(1 - 0.310)}{(0.01)^2}\) n = 9066.58 Since we cannot have a fractional sample size, we round it up to the nearest whole number which is 9067. Therefore, a sample size of 9067 would be required to report the margin of error of 0.01 at 95% confidence. As for the recommendation, it would depend on whether USA Today has the resources and time to survey 9067 business travelers. If they can afford it, then achieving a margin of error of 0.01 would provide more precise and accurate insights. However, if it is too expensive or time-consuming, the current sample size of 1993 with a margin of error of 0.011 (as calculated in part b) might still provide valuable information for their needs.