Question

Customers arrive at a movie theater at the advertised movie time only to find that they have to sit through several previews and prepreview ads before the movie starts. Many complain that the time devoted to previews is too long (The Wall Street Journal, October 12, 2012). A preliminary sample conducted by The Wall Street Journal showed that the standard deviation of the amount of time devoted to previews was five minutes.How large a sample should be taken to estimate the meantime for previews at movie theaters to within 90 seconds (1.5 minutes) with 90% confidence?

round z or t value to 3 decimal places

Answer 1

At 90% confidence level, the critical value is z_0.05 = 1.645

Margin of error = 1.5

or, z_0.05 * $\sigma/\sqrt n$ = 1.5

or, 1.645 * 5/$\sqrt n$ = 1.5

or, n = (1.645 * 5/1.5)^2

or. n = 31