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 26 minutes. The executive wants to​ estimate, with​ 99% confidence, the mean weekly amount of time to within ±33 minutes.

a. What sample size is​ needed?

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

a. The sample size required for​ 99% confidence is ___

Answer 1

Solution :

Given that,

Population standard deviation = = 26 minutes

Margin of error = E = 3 minutes

a) At 99% confidence level the z is,

= 1 - 99%

= 1 - 0.99 = 0.01

/2 = 0.005

Z/2 = Z0.005 = 2.576

sample size = n = [Z/2* / E] 2

n = [2.576 * 26 / 3 ]2

n = 498.42

Sample size = n = 499

b) At 95% confidence level the z is,

= 1 - 95%

= 1 - 0.95 = 0.05

/2 = 0.025

Z/2 = Z0.025 = 1.96  

sample size = n = [Z/2* / E] 2

n = [1.96 * 26 / 3 ]2

n = 288.54

Sample size = n = 289