A December 2013 Gallup poll of 13,186 nationally representative adults found that 8.9% reported being sick with a cold. Based on these results, what is a 95% confidence interval for the proportion p of American adults who were sick with a cold that month?

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Quizplus · Accepted Answer

The 95% confidence interval for a proportion can be calculated using the formula: $CI = \hat{p} \pm Z \sqrt{\frac{\hat{p}(1-\hat{p})}{n}}$, where $\hat{p}$ is the sample proportion, $n$ is the sample size, and $Z$ is the Z-score for the desired confidence level (approximately 1.96 for 95%). Plugging in the given values ($\hat{p} = 0.089$, $n = 13186$), the calculation yields a margin of error approximately equal to 0.005, resulting in a confidence interval of 0.089 ± 0.005.

A December 2013 Gallup Poll of 13,186 Nationally Representative Adults