Question

It is believed that 25% of U.S. homes have a direct satellite television receiver. How large a sample is necessary to estimate the true population of homes which do with 95% confidence and within 3 percentage points?

Answer 1

A sample of 801 is needed.

Explanation:

In a sample with a number n of people surveyed with a probability of a success of
`\pi`, and a confidence level of
`1-\alpha`, we have the following confidence interval of proportions.

`\pi \pm z\sqrt{(\pi(1-\pi))/(n)}`

In which

z is the z-score that has a p-value of
`1 - (\alpha)/(2)`.

The margin of error is of:

`M = z\sqrt{(\pi(1-\pi))/(n)}`

95% confidence level

So
`\alpha = 0.05`, z is the value of Z that has a p-value of
`1 - (0.05)/(2) = 0.975`, so
`Z = 1.96`.

25% of U.S. homes have a direct satellite television receiver.

This means that
`\pi = 0.25`

How large a sample is necessary to estimate the true population of homes which do with 95% confidence and within 3 percentage points?

This is n for which M = 0.03. So

`M = z\sqrt{(\pi(1-\pi))/(n)}`

`0.03 = 1.96\sqrt{(0.25*0.75)/(n)}`

`0.03√(n) = 1.96√(0.25*0.75)`

`√(n) = (1.96√(0.25*0.75))/(0.03)`

`(√(n))^2 = ((1.96√(0.25*0.75))/(0.03))^2`

`n = 800.3`

Rounding up:

A sample of 801 is needed.