Question

Direct Satellite Television It is believed that

25%

of U.S. homes have a direct satellite television receiver. Round up your final answers to the next whole number.

How large a sample is necessary to estimate the true population of homes that have a direct satellite television receiver with

90%

confidence and within

3

percentage points?

Answer 1

Solution,

Given that,

= 0.25

1 - = 1 - 0.25 = 0.75

margin of error = E = 0.03

At 90% confidence level

= 1 - 90%

= 1 - 0.90 =0.10

/2 = 0.05

Z_/2 = Z_0.05 = 1.645

sample size = n = (Z _{/ 2} / E )² * * (1 - )

= (1.645 / 0.03)² * 0.25 * 0.75

= 563.75

sample size = n = 564