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Question: Customers arrive at a movie theater at the advertised movie time only to find that they have to s...

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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 four 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? Assume 95% confidence.

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Solution:

Given ,

= 4 minutes = 240 seconds ..Population SD

E = 78 seconds Margin of error

c = 95% = 0.95 ...confidence level

Find sample size required.

c = 0.95

= 1- c = 1- 0.95 = 0.05

  /2 = 0.025

Using Z table ,

= 1.96

How do you go from .025 to 1.96 the formula does not come to that?

Answer 1