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 24 minutes. The executive wants to​ estimate, with​ 99% confidence, the mean weekly amount of time to within plus or minus ±66 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,

Population standard deviation = = 24

Margin of error = E = 6

a)

Z/2 = 2.576

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

n = [2.576 * 24/ 6]2

n = 106.17

Sample size is required = 107

b)

Z/2 = 1.96

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

n = [1.96 * 24/ 6]2

n = 61.46

Consumers need to be​ selected = 62