Question

Which sample size will produce a margin of error of ±4.1%?

856

595

782

362

Answer 1

Answer on Question #62838 – Math – Statistics and Probability

Question

Which sample size will produce a margin of error of $\pm 4.1\%$?

856

595

782

362

Solution

For a $95\%$ confidence level, standard deviation $\sigma$ and sample size $n$, the margin of error is given by

$$\pm 1.96 \frac{\sigma}{\sqrt{n}} = \pm E = \pm 0.041.$$

Hence

$$n = \left(\frac{1.96\sigma}{E}\right)^2 = \frac{1.96^2 \sigma^2}{E^2} = 2285.306 \sigma^2.$$

For a $95\%$ confidence level $E \approx \frac{0.98}{\sqrt{n}}$, hence

$$n \approx \left(\frac{0.98}{E}\right)^2 = \left(\frac{0.98}{0.041}\right)^2 \approx 571$$

For a $99\%$ confidence level $E \approx \frac{1.29}{\sqrt{n}}$, hence

$$n \approx \left(\frac{1.29}{E}\right)^2 = \left(\frac{1.29}{0.041}\right)^2 \approx 990$$

For a $90\%$ confidence level $E \approx \frac{0.82}{\sqrt{n}}$, hence

$$n \approx \left(\frac{0.82}{E}\right)^2 = \left(\frac{0.82}{0.041}\right)^2 \approx 400$$

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