Confidence interval for a proportion

Question

In a survey, 240 of 400 randomly chosen respondents prefer product A. Construct a 95% confidence interval for the true proportion who prefer product A.

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Setup. For a proportion, the interval is \hat{p} \pm z^{}\sqrt{\frac{\hat{p}(1-\hat{p})}{n}}, \qquad z^{} = 1.96 	ext{ at } 95\%. The sample proportion is \hat{p} = 240/400 = 0.6. Compute. The standard error is \sqrt{\frac{0.6 	imes 0.4}{400}} = \sqrt{0.0006} = 0.0245. Margin of error = 1.96 	imes 0.0245 = 0.048, so 0.6 \pm 0.048 = [0.552,\ 0.648]. Interpret. With 95% confidence the true preference rate lies roughly between 55% and 65%. The count of 400 is large and \hat{p} is near the middle, so this normal-based (Wald) interval is trustworthy here. Takeaway. A proportion interval mirrors…

Confidence interval for a proportion

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