I found this one a tricky one: An airline was concerned about passengers arriving too late at the airport to allow for the additional security measures. Based on a survey of 1,000 passengers, the mean time from arrival at the airport to reaching the boarding gate was 1 hour, 20 minutes, with a standard deviation of 30 minutes. If the airline wants to make sure at the 95 percent confidence level that passengers have sufficient time to catch their flight, how much time ahead of their flight should passengers be advised to arrive at the airport? A) One hour, fifty minutes. B) Two hours, ten minutes. C) Two hours, thirty minutes. D) Two hours, forty-five minutes.

This is an interesting question, moreso than I thought at first glance. I choose A. Population variance is unknown, implying t-distribution, but sample size of 1000 is so large z-distribution can be used. A single-tail z-distribution at 95% confidence has critical value of 1.645. The part that stumped me at first was what sample standard deviation to use. At first I wanted to use sigma_m = 30/sqrt(1000) which is about one minute. But then I didn’t see any answers that followed through this course of reasoning. So therefore I decided that the standard error of the sample is pretty low, so we can assume the mean of 80 minutes pretty well (+/- one minute). And then use the variance of the sample to fairly decently gauge the variance of the population. Thus, 1.645 * 30 = 50 minutes, which added to the mean gives one hour, fifty minutes.

> Thus, 1.645 * 30 = 50 minutes, which added to the > mean gives one hour, fifty minutes. I’m an idiot, I added incorrectly. 50 minutes + 80 minutes is 130 minutes, or two hours, 10 minutes. Thus choice B.

Probably too easy. But 2*30/2 + (1 hour 20min) = 1hour 50min Wrong? I guess so…