Question

An advertising agency that serves a major radio station wants to estimate the mean amount of time that the station's audience spends listening to the radio daily. From past studies, the standard deviation is estimated as 48 minutes. A. What sample size is needed if the agency wants to be 95% confident of being correct to within +/-5 minutes? A sample size of ____ listeners is needed. ( Round up to the nearest integer.) B. If 99% confidence is desired, how many listeners need to be selected? If 99% confidence is desired, ___listeners need to be selected. (Round up to the nearest integer.)

Answer 1

Solution :

Given that,

standard deviation = = 48

margin of error = E = 5

A)

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 * 48) / 5)²

= 354

A sample size of 354 listeners is needed .

B)

At 99% confidence level the z is ,

= 1 - 99% = 1 - 0.99 = 0.01

/ 2 = 0.01 / 2 = 0.005

Z_/2 = Z_0.005 = 2.575

Sample size = n = ((Z_/2 * ) / E)²

= ((2.575 * 48) / 5)²

= 611

A sample size of 611 listeners is needed .