Question

1. if you wanted to give a survey that had a 3% margin of error, how many people would you have to survey?

○ 10
○ 30
○ 500
○ 1112

Explanation:

Step1: Recall Margin of Error Formula

The formula for the margin of error \( E \) in a proportion - based survey is \( E = z_{\alpha/2}\sqrt{\frac{p(1 - p)}{n}} \), where \( z_{\alpha/2} \) is the z - score corresponding to the confidence level, \( p \) is the estimated proportion of the population, and \( n \) is the sample size. For a 95% confidence level (a common confidence level), \( z_{\alpha/2}\approx1.96 \). If we assume \( p = 0.5 \) (this is the worst - case scenario that maximizes the product \( p(1 - p) \) and thus gives the largest required sample size for a given margin of error), the formula simplifies to \( E=1.96\sqrt{\frac{0.5\times(1 - 0.5)}{n}}=1.96\sqrt{\frac{0.25}{n}}=\frac{1.96\times0.5}{\sqrt{n}}=\frac{0.98}{\sqrt{n}} \)

Step2: Solve for n when \( E = 0.03 \)

We set \( E = 0.03 \) and solve for \( n \):

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If we use a 99% confidence level, \( z_{\alpha/2}\approx2.576 \), and with \( p = 0.5 \), the formula is \( E = 2.576\sqrt{\frac{0.25}{n}}=\frac{2.576\times0.5}{\sqrt{n}}=\frac{1.288}{\sqrt{n}} \)
Setting \( E = 0.03 \):

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But among the given options, 1112 is the closest to the sample size required for a reasonable confidence level (for example, if we use a slightly different approximation or a different method of calculating sample size for surveys, the formula for sample size in surveys can also be approximated as \( n=\frac{1}{E^{2}} \) when \( p = 0.5 \) and using a simplified z - score approximation. If \( E = 0.03 \), \( n=\frac{1}{(0.03)^{2}}=\frac{1}{0.0009}\approx1111 \), which is very close to 1112.

Answer:

1112 (the option with 1112)