Question

9. + -/4.16 points ASWSBE14 8.E.029.MI. My Notes + Ask Your Teacher You may need to use the appropriate appendix table or technology to answer this 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 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. Need Help? Read It Master It Talk to a Tutor

Answer 1

a. Margin of Error is E=105 seconds = 1.75 minutes

Further standard deviation=6 minutes

So we will find n using formula of E

$E=z*\frac{\sigma }{\sqrt{n}}$

So $n=(\frac{z*\sigma}{E})^2=(\frac{1.96*6}{1.75})^2=45.16=46$

b. Now E=1, so $n=(\frac{z*\sigma}{E})^2=(\frac{1.96*6}{1})^2=138.30=139$