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 81 seconds, what sample size should be used? Assume 95% confidence.
To calculate the required sample size, we need to use the formula:
n = ((Z * σ) / E)^2
Where:
n = sample size
Z = Z-score for the desired confidence level (95% confidence level corresponds to a Z-score of 1.96)
σ = standard deviation of the population (given as 4 minutes, or 240 seconds)
E = margin of error (81 seconds)
Plugging in the values, we get:
n = ((1.96 * 240) / 81)^2
n = (470.4 / 81)^2
n = 5.8^2
n ≈ 33.64
Therefore, the sample size should be at least 34 to estimate the population mean time for previews at movie theaters with 95% confidence and a margin of error of 81 seconds.
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 (corresponding to the desired level of confidence)
σ = standard deviation of the population
E = margin of error
In this case, the standard deviation of the amount of time devoted to previews is given as 4 minutes (which we'll convert to seconds), and the margin of error is 81 seconds.
Converting 4 minutes to seconds: 4 minutes * 60 seconds/minute = 240 seconds
Now we have:
σ = 240 seconds
E = 81 seconds
We need to determine the Z-score for a 95% confidence level. In a standard normal distribution, this corresponds to a Z-score of approximately 1.96.
Using these values in the formula:
n = (1.96 * 240 / 81)²
n ≈ (470.4 / 81)²
n ≈ 5.8²
n ≈ 33.64
Therefore, to estimate the population mean time for previews at movie theaters with a margin of error of 81 seconds and 95% confidence, a sample size of approximately 34 should be used.