Question

An advertising executive wants to estimate the mean weekly amount of time consumers spend watching traditional television daily. Based on previous​ studies, the standard deviation is assumed to be 21 minutes. The executive wants to​ estimate, with​ 99% confidence, the mean weekly amount of time to within ± 6 minutes.

a. What sample size is​ needed?

b. If​ 95% confidence is​ desired, how many consumers need to be​ selected?

Answer 1

Solution :

Given that,

standard deviation = = 21

margin of error = E = 6

a ) 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 = 2.576

Sample size = n = ((Z_/2 * ) / E)²

= ((2.576 * 21) / 6)²

= 81.288

= 81

Sample size = 81

b ) 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 = 1.960

Sample size = n = ((Z_/2 * ) / E)²

= ((1.960 * 21) / 6)²

= 47.0596

= 47

Sample size = 47