Some parts of the question are missing. The full question is;
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.
a. 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
Answer:
A) n = 39
B) n = 61
Explanation:
We are given that;
Confidence level (c) = 95% = 0.95
Population standard deviation (σ) = 4 minutes = 4 x 60 seconds = 240 seconds
Margin of error (E) = 75 seconds
A) Formula for sample size is given as;
n = (Z_(α/2)•(σ)/E)²
Now, for confidence level,
1 - α = 0.95,we can find (Z_(α/2)
So, α = 1 - 0.95 =0.05
Thus, α/2 = 0.05/2 = 0.025
Thus,(Z_(α/2) = Z_0.025
Looking at the table i attached, at (Z_(α/2) = Z_0.025, we have a value of 1.96
Hence, we can now find the sample size.
So, n = (Z_(α/2)•(σ)/E)² = ((1.96 x 240)/75)² = 39.33 ≈ 39
B) C remains 0.95
(σ) remains 240 seconds
While E is now = 1 minute = 60 seconds
From above, (Z_(α/2) = 1.96
Thus,
from n = (Z_(α/2)•(σ)/E)²,
n = ((1.96 x 240)/60)² = 61.47 ≈ 61