Question

A cable TV company wants to estimate the percentage of cable boxes in use during an evening hour. An approximation based on previous surveys is 25 percent. The company wants the new estimate to be at the 95 percent confidence level and within 3 percent of the actual proportion. What sample size is needed

Answer 1

Solution :

Given that,

= 25% = 0.25

1 - = 1 - 0.25 = 0.75

Margin of error = E = 3% = 0.03

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 = 1.96

Sample size = (  Z_/2 / E)² * * (1 - )

= (1.96 / 0.03)² * 0.25 * 0.75

= 800.33 = 801

Sample size = n = 801