Question

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 4 minutes. Use that as a planning value for the standard deviation in answering the following questions. Round your answer to next whole number

Answer 1

This question is incomplete, the complete 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 preview 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 4 minutes. Use that as a planning value for the standard deviation in answering the following questions. Round your answer to next whole number,

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) the required Sample Size is 40

b) the required Sample Size is 62

Step-by-step explanation:

Given the data in the question;

standard deviation σ = 4 minutes

a)

margin of error E = 75 seconds = ( 75 / 60 )minutes = 1.25 minutes

And 95% confidence.

Now, Critical Value of z for 0.95 confidence interval;

∝ = 1 - 0.95 = 0.05

$Z_{\alpha /2$ = 1.96

so, sample size n will be;

n = [ $Z_{\alpha /2$ × σ/E ]²

we substitute;

n = [ 1.96 × (4/1.25) ]²

n =  [ 1.96 × 3.2 ]²

n = [ 6.272 ]²

n = 39.338

Since we referring to a number of sample, its approximately becomes 40

Therefore, the required Sample Size is 40

b)

margin of error E = 1 minutes

And 95% confidence.

Now, Critical Value of z for 0.95 confidence interval;

∝ = 1 - 0.95 = 0.05

$Z_{\alpha /2$ = 1.96

so, sample size n will be;

n = [ $Z_{\alpha /2$ × σ/E ]²

we substitute;

n = [ 1.96 × (4/1) ]²

n =  [ 1.96 × 4 ]²

n = [ 7.84 ]²

n = 61.466

Since we referring to a number of sample, its approximately becomes 62

Therefore, the required Sample Size is 62