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 90 percent confidence level and within 2 percent of the actual proportion. What sample size is needed?

Answer 1

n = 1269

Explanation:

Percentage of cable boxes in use during an evening hour is 25%.

Therefore, P = 0.25

Margin of error = 2%, E = 0.02

Confidence level = 90%

for 90% confidence level , critical Z value = 1.645

We have to find sample size (n) the formula is :

`n=P(1-P)* ((Z)/(E))^2`

`n=0.25(1-0.25)* ((1.645)/(0.02))^2`

`=0.25(0.75)* ((1.645)/(0.02))^2`

= 1268.5492 ≈ $1269

n = 1269