z = +/- 2.68

68.6 – 2.68 * 8.2 / √50 = 65.49

68.6 + 2.68 * 8.2 / √50 = 71.71 b. Does the 99% confidence interval include the value suggested by the NCAA? Interpret this result.

Yes, the 99% confidence interval includes the value suggested by the NCAA, which is 70. Therefore the result makes sense.

c. Suppose you decided to switch from a 99% to a 95% confidence interval. Without performing any calculations, will the interval increase, decrease, or stay the same? Which of the values in the formula will change?

Switching the confidence interval to 95% would decrease width of the interval and therefore the z level will be smaller than the +/-2.68 determined with the 99% confidence interval.

Question 42. As a condition of employment, Fashion Industries applicants must pass a drug test. Of the last 220 applicants 14 failed the test. Develop a 99% confidence interval for the proportion of applicants that fail the test. Would it be reasonable to conclude that more than 10% of the applicants are now failing the test?

14/220 = 0.64 InvNorm = 2.5758

Standard Error = 2.5758 * √(0.64 * 0.36 / 220) = 0.08336

CI 99% = 0.64 – 0.08336 < u < 0.64 + 0.08336

CI 99% = 0.5566 < u < 0.7234 It is not be reasonable to conclude that more than 10% of the applicants are now failing the test.

Question 50. Families USA, a monthly magazine that discusses issues related to health and health costs, surveyed 20 of its subscribers. It found that the annual health insurance premiums for a family with coverage through an