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Difficulty: Very hardStatistical Sampling and Studies

A municipal research group conducted a survey of a random sample of n1n_1 registered voters in a city to estimate the proportion of voters who support a proposed school bond. In this survey, the proportion of voters who supported the bond was 0.520.52, with an associated margin of error of 0.060.06 at a 95%95\% confidence level. The group plans to conduct a follow-up survey of a random sample of n2n_2 registered voters from the same population. Assuming the sample proportion remains approximately the same, and the group wants the margin of error for the follow-up survey to be 0.020.02 at the same 95%95\% confidence level, what is the value of the ratio n2n1\frac{n_2}{n_1}?

Answer: 9

Answer

9
The margin of error for a sample proportion is inversely proportional to the square root of the sample size. Since the target margin of error is one-third of the initial margin of error (from 0.060.06 to 0.020.02), the sample size must be increased by a factor of 32=93^2 = 9. Thus, the ratio of the new sample size to the original sample size is 99.

Step-by-Step Solution

1
State the formula for the margin of error of a sample proportion.
E=z∗p^(1−p^)nE = z^* \sqrt{\frac{\hat{p}(1-\hat{p})}{n}}
This formula defines the mathematical relationship between the margin of error, the confidence level (via the critical value z∗z^*), the sample proportion p^\hat{p}, and the sample size nn.
2
Determine the relationship between margin of error and sample size when other variables are constant.
E∝1nE \propto \frac{1}{\sqrt{n}}
Since the confidence level remains at 95%95\% (constant z∗z^*) and the sample proportion remains approximately the same (constant p^\hat{p}), the margin of error is inversely proportional to the square root of the sample size.
3
Set up the ratio equation for the two surveys.
E1E2=n2n1\frac{E_1}{E_2} = \sqrt{\frac{n_2}{n_1}}
Dividing the expression for the first margin of error by the expression for the second margin of error simplifies to the square root of the ratio of the second sample size to the first sample size.
4
Substitute the given margins of error and solve for the ratio.
n2n1=9\frac{n_2}{n_1} = 9
Substitute E1=0.06E_1 = 0.06 and E2=0.02E_2 = 0.02 to get 0.060.02=3=n2n1\frac{0.06}{0.02} = 3 = \sqrt{\frac{n_2}{n_1}}. Squaring both sides yields the ratio of the sample sizes, which is 99.

Key Concept

The relationship between sample size and margin of error in statistical studies.
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Topics

Statistical Sampling and Studies

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