Question

An advertising executive wants to estimate the mean amount of time that consumers spend with digital media daily. From past​ studies, the standard deviation is estimated as 42 minutes.

A. What sample size is needed if the executive wants to be 90% confident of being correct to within plus or minus 4 minutes?

B. If 99% confidence is desired, how many consumers need to be selected?

Answer 1

Solution :

Given that,

Population standard deviation = = 42

Margin of error = E = 4

A) At 90% confidence level the z is,

= 1 - 90%

= 1 - 0.90 = 0.10

/2 = 0.05

Z/2 = Z0.05 = 1.645

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

n = [1.645 * 42 / 4]2

n = 298.33

Sample size = n = 299

B) 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 * 42 / 4]2

n = 731.59

Sample size = n = 732