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. 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. Use that as a planning value for the standard deviation in answering the following questions. a. If we want to estimate the population mean time for previews at movie theaters with a margin of error of 78 seconds, what sample size should be used

57

Step-by-step explanation:

Given that :

Standard deviation (σ) = 5 minutes = (5 *60) = 300 seconds

Margin of Error, E = 78

Assume a confidence interval of 95%;

Using the relation :

((Zα/2 * σ) / E)²

(1 - α)/2 = (1 - 0.95)/2 = 0.05 /2 = 0.025

Z0.025 = 1.96

Sample size = ((1.96 * 300) / 78)²

Sample size = (588 / 78)^2

Sample size = 7.5384615^2

Sample size = 56.828

= 57 samples

ghjmutmum

step-by-step explanation:

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step-by-step explanation: