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. a. If we want to estimate the population mean time for previews at movie theaters with a margin of error of seconds, what sample size should be used

Answer 1

Answer:

$n=35$

Step-by-step explanation:

From the question we are told that:

Standard Deviation $\sigma=4min$

Let

$CI=95\%$

Since

Significance level $\alpha$

$\alpha =1-CI$

$\alpha =1-0.95$

Therefore

$Z_{\alpha/2}=Z_{0.025$

$Z_{\alpha/2}}=1.96$

Generally the equation for Sample size is mathematically given by

$n = (Z_{\alpha/2}* \frac{\sigma}{E})^2$

$n= \frac{1.96 * 3}{1}^2$

$n=35$