Question

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.

Answer 1

Solution :

(a)Given that,

= 6

E = 105 sec = 1.45 min

At 95% confidence level the z is ,

= 1 - 95% = 1 - 0.95 = 0.05

/ 2 = 0.05 / 2 = 0.025

Z_/2 = Z_0.025 = 1.96

sample size n = ( Z_/2* / E)²

= (1.96 *6 / 1.45)²

^{n = 66}

(b)Solution :

Given that,

= 6

E = 1 min

At 95% confidence level the z is ,

= 1 - 95% = 1 - 0.95 = 0.05

/ 2 = 0.05 / 2 = 0.025

Z_/2 = Z_0.025 = 1.96

sample size n = ( Z_/2* / E)²

= (1.96 *6 / 1)²

^{n =138.29}

^{n = 138}