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 four minutes. Use that as a planning value for the standard deviation in answering the following questions.
If we want to estimate the population mean time for previews at movie theaters with a margin of error of 75 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.
A.) 40
B.) 62
Equation is ((Za/2)^2*STD^2)/MarginError^2
40
To estimate the sample size needed to estimate the population mean time for previews at movie theaters with a given margin of error, we can use the formula:
n = (Z * σ / E)^2
Where:
n = Sample size needed
Z = Z-value corresponding to the desired confidence level (in this case, 95% confidence corresponds to a Z-value of approximately 1.96)
σ = Standard deviation of the population (given as 4 minutes)
E = Margin of error
Let's calculate the sample size for each question:
a. Margin of error = 75 seconds = 1.25 minutes
n = (1.96 * 4 / 1.25)^2
n ≈ 15.57
Since we can't have a fraction of a sample, we need to round up to the nearest whole number. Therefore, a sample size of 16 should be used.
b. Margin of error = 1 minute
n = (1.96 * 4 / 1)^2
n ≈ 61.02
Again, rounding up to the nearest whole number, a sample size of 62 should be used.
Therefore, to estimate the population mean time for previews at movie theaters with a margin of error of 75 seconds, a sample size of 16 should be used. And to estimate the population mean time with a margin of error of 1 minute, a sample size of 62 should be used.