Which statement about confidence intervals for a proportion is true?

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Multiple Choice

Which statement about confidence intervals for a proportion is true?

Explanation:
The key idea is that the precision of a confidence interval for a proportion improves as the sample size grows. The interval’s width depends on the standard error of the proportion, which is sqrt(p̂(1 − p̂)/n). As n increases, this standard error shrinks roughly like 1/√n, so the margin of error (the part added and subtracted from p̂) gets smaller and the interval becomes narrower. In typical large-sample practice, you use p̂ ± z* sqrt(p̂(1 − p̂)/n); doubling the sample size reduces the margin of error, making the interval tighter. This is why larger samples generally produce narrower intervals. Smaller or fixed samples don’t automatically produce the same precision, and the interval width does depend on how much data you’ve collected.

The key idea is that the precision of a confidence interval for a proportion improves as the sample size grows. The interval’s width depends on the standard error of the proportion, which is sqrt(p̂(1 − p̂)/n). As n increases, this standard error shrinks roughly like 1/√n, so the margin of error (the part added and subtracted from p̂) gets smaller and the interval becomes narrower. In typical large-sample practice, you use p̂ ± z* sqrt(p̂(1 − p̂)/n); doubling the sample size reduces the margin of error, making the interval tighter. This is why larger samples generally produce narrower intervals. Smaller or fixed samples don’t automatically produce the same precision, and the interval width does depend on how much data you’ve collected.

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