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? 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.
To estimate the sample size needed to estimate the population mean time for previews at movie theaters, we can use the formula:
n = (Z * σ / E)^2
Where:
n = Sample size
Z = Z-score (determined by the desired confidence level)
σ = Standard deviation
E = Margin of error
First, let's calculate the Z-score for a 95% confidence level. In a standard normal distribution, the Z-score for a 95% confidence level is approximately 1.96.
a. To estimate the sample size with a margin of error of 78 seconds:
Using σ = 5 minutes = 300 seconds,
E = 78 seconds,
Z = 1.96,
n = (1.96 * 300 / 78)^2
n ≈ 119.4
Therefore, a sample size of approximately 120 should be used.
b. To estimate the sample size with a margin of error of 1 minute:
Using σ = 5 minutes = 300 seconds,
E = 1 minute = 60 seconds,
Z = 1.96,
n = (1.96 * 300 / 60)^2
n ≈ 245.16
Therefore, a sample size of approximately 246 should be used.