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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?

User HenriqueMS
by
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1 Answer

7 votes

Answer:

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.

User Mgigirey
by
7.2k points
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