Question

The NWBC found that 42.1% of women-owned businesses provided retirement plans contributions.

What sample size could be 95% confident that the estimated (sample) proportion is within 4 percentage points of the true population proportion?

Answer 1

Sample size of 586 or higher.

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)}`

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 could be 95% confident that the estimated (sample) proportion is within 4 percentage points of the true population proportion?

Sample size of at least n when
`M = 0.04`

42.1% of women-owned businesses provided retirement plans contributions, which means that
`p = 0.421`. So

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

`0.04 = 1.96\sqrt{(0.421*0.579)/(n)}`

`0.04√(n) = 0.9677`

`√(n) = (0.9677)/(0.04)`

`√(n) = 24.19`

`√(n)^(2) = (24.19)^(2)`

`n = 585.2`

We need a sample size of 586 or higher.