Question

An advertising executive wants to estimate the mean weekly amount of time 18- to 24-year-olds spend watching traditional television in a large city. Based on studies in other cities, the standard deviation is assumed to be 10 minutes. The executive wants to estimate, within 99% confidence, the mean weekly amount of time within ±3 minutes.

a. What sample size is needed?

b. If 95% confidence is desired, how many 18- to 24-year-olds need to be selected?

c. If she wants to estimate the mean within ±2 minutes, what size is needed for a 99% and 95% confidence level, respectively? Please show work

Answer 1

Solution :

Given that,

Population standard deviation = = 10

Margin of error = E = 3

a) At 99% confidence level the z is,

= 1 - 99%

= 1 - 0.99 = 0.01

/2 = 0.005

Z_/2 = Z_0.005 = 2.576

sample size = n = [Z_/2* / E] ²

n = [2.576 * 10 / 3 ]²

n = 73.73

Sample size = n = 74

b) At 95% confidence level the z is,

= 1 - 95%

= 1 - 0.95 = 0.05

/2 = 0.025

Z_/2 = Z_0.025 = 1.96  

sample size = n = [Z_/2* / E] ²

n = [1.96 *10 / 3]²

n = 42.68

Sample size = n = 43

c) Margin of error = E = 2

At 99% confidence level the z is,

= 1 - 99%

= 1 - 0.99 = 0.01

/2 = 0.005

Z_/2 = Z_0.005 = 2.576

sample size = n = [Z_/2* / E] ²

n = [2.576 * 10 / 2 ]²

n = 165.89

Sample size = n = 166

At 95% confidence level the z is,

= 1 - 95%

= 1 - 0.95 = 0.05

/2 = 0.025

Z_/2 = Z_0.025 = 1.96  

sample size = n = [Z_/2* / E] ²

n = [1.96 *10 / 2]²

n = 96.04

Sample size = n = 97