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 72 seconds, what sample size should be used? Assume 95% confidence.
b. If we want to estimate the population mean time for previews at movie theaters with a margin of error of 1 minute, what sample size should be used? Assume 95% confidence.

Given that customers 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.

[tex]\sigma =5[/tex]

Since population std deviation is given we can use Z critical values for finding the margin of errors.

a) Margin of error = 72 seconds = 1.2 minutes

For 95% , [tex]1.96*\frac{5}{\sqrt{n} } =1.2\\n = 66.61\\n =67[/tex]

Sample size =67

b) Margin of error = 1

[tex]1.96*\frac{5}{\sqrt{n} } =1\\n = 96.04\\n =96[/tex]

Sample size = 96