Question

DirectTV is interested in conducting a study to determine the percentage of all their subscribers would be willing to pay $90 per month for a premium cable package. What is the minimum size sample needed to estimate the population proportion with a margin of error of 0.03 or less at 96% confidence?

Answer 1

Solution:

Given:

Margin of Error = E = 0.03

Confidence level = c = 96% = 0.96

We have to find minimum sample size to estimate the population proportion.

Formula for sample size is:

Since previous estimate of proportion is not given , we use p = 0.5

We need to find z_c value for c=96% confidence level.

Find Area = ( 1 + c ) / 2 = ( 1 + 0.96) /2 = 1.96 / 2 = 0.9800

Look in z table for Area = 0.9800 or its closest area and find z value.

Area = 0.9798 is closest to 0.9800 and it corresponds to 2.0 and 0.05 , thus z critical value = 2.05

That is : Z_c = 2.05

thus

( Sample size n is always rounded up)