Question

A sociologist wishes to conduct a poll to estimate the percentage of Americans who favor affirmative action program for women and minorities for admission to colleges and universities. What sample size should be obtained if she wishes the estimate to be within a margin of error of 0.04, with 95% confidence, if she uses a 2003 estimate of 0.55 obtained from a Gallup Youth Survey

Answer 1

A sample of 595 should be obtained.

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 zscore that has a pvalue of
`1 - (\alpha)/(2)`.

The margin of error is:

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

Estimate of 0.55

This means that
`\pi = 0.55`

95% confidence level

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

What sample size should be obtained if she wishes the estimate to be within a margin of error of 0.04?

This is n for which M = 0.04. So

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

`0.04 = 1.96\sqrt{(0.55*0.45)/(n)}`

`0.04√(n) = 1.96√(0.55*0.45)`

`√(n) = (1.96√(0.55*0.45))/(0.04)`

`(√(n))^2 = ((1.96√(0.55*0.45))/(0.04))^2`

`n = 594.2`

Rounding up,

A sample of 595 should be obtained.