Question

In: Statistics and Probability

Customers arrive at a movie theater at the advertised movie time only to find that they have to sit through several previews and pre-preview ads before the movie starts.

 

Customers arrive at a movie theater at the advertised movie time only to find that they have to sit through several previews and pre-preview 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 6 minutes. Use that as a planning value for the standard deviation in answering the following questions. (Round your answers up to the nearest whole number.)

(a)

If we want to estimate the population mean time for previews at movie theaters with a margin of error of 105 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.

-You may need to use the appropriate appendix table or technology to answer this question.

How large a sample should be selected to provide a 95% confidence interval with a margin of error of 10? Assume that the population standard deviation is 50. (Round your answer up to the nearest whole number.)

Solutions

Expert Solution

a)

for95% CI crtiical Z          = 1.960
standard deviation σ= 6.000
margin of error E =105/60= 1.75 minutes
required sample size n=(zσ/E)2                  = 46

b)

for95% CI crtiical Z          = 1.960
standard deviation σ= 6.000
margin of error E = 1
required sample size n=(zσ/E)2                  = 139

c)

for95% CI crtiical Z          = 1.960
standard deviation σ= 50.000
margin of error E = 10
required sample size n=(zσ/E)2                  = 97

Related Solutions

Customers arrive at a movie theater at the advertised movie time only to find that they...
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....
Customers arrive at a movie theater at the advertised movie time only to find that they...
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....
Customers arrive at a movie theater at the advertised movie time only to find that they...
Customers arrive at a movie theater at the advertised movie time only to find that they have to sit through several previews and pre-preview 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 6 minutes. Use that as a planning value for the standard deviation in answering the following questions....
ADVERTISEMENT
ADVERTISEMENT
ADVERTISEMENT