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 (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 72 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 1

(a) The sample size required is 43.

(b) The sample size required is 62.

Explanation:

The (1 - α) % confidence interval for population mean is:

`CI=\bar x\pm z_(\alpha/2)\ (\sigma)/(√(n))`

The margin of error for this interval is:

`MOE=z_(\alpha/2)\ (\sigma)/(√(n))`

(a)

The information provided is:

σ = 4 minutes

MOE = 72 seconds = 1.2 minutes

Confidence level = 95%

α = 5%

Compute the critical value of z for α = 5% as follows:

`z_(\alpha/2)=z_(0.05/2)=z_(0.025)=1.96`

*Use a z-table.

Compute the sample size required as follows:

`MOE=z_(\alpha/2)\ (\sigma)/(√(n))`

`n=[(z_(\alpha/2)* \sigma)/(MOE) ]^(2)`

`=[(1.96* 4)/(1.2)]^(2)\\\\=42.684\\\\\approx 43`

Thus, the sample size required is 43.

(b)

The information provided is:

σ = 4 minutes

MOE = 1 minute

Confidence level = 95%

α = 5%

Compute the critical value of z for α = 5% as follows:

`z_(\alpha/2)=z_(0.05/2)=z_(0.025)=1.96`

*Use a z-table.

Compute the sample size required as follows:

`MOE=z_(\alpha/2)\ (\sigma)/(√(n))`

`n=[(z_(\alpha/2)* \sigma)/(MOE) ]^(2)`

`=[(1.96* 4)/(1)]^(2)\\\\=61.4656\\\\\approx 62`

Thus, the sample size required is 62.