Lecture 18

ConfidenceConfidenceIntervalsIntervalsforforProportionProportion

Formula

Another population parameter which we can estimate with a confidence interval is the population proportion. Let $\hat{p}=$ be a proportion of distingusihed cases in a sample. Let $p=$ be the corresponding population proportion. A $100(1-\alpha)\%$ confidence interval for $p$ is given by:$$\hat{p}-z_{\alpha / 2} \sqrt{\frac{\hat{p}(1-\hat{p})}{n}} \leq \,\, p \,\, \leq \hat{p}+z_{\alpha / 2} \sqrt{\frac{\hat{p}(1-\hat{p})}{n}} $$This confidence interval is valid when the sample size is large.

Example 1

In a sample of $85$ wild turkeys, $10$ have histomoniasis (parasitic infection).
Estimate the population proportion of wild turkeys with histonommiasis with a $95 \%$ confidence interval.
Let us compute the sample proportion first: $\hat{p}=\frac{10}{85}=0.118.$ The $95\%$ confidence interval is given by:$$0.118 \pm 1.96 \sqrt{\frac{0.118(1-0.118)}{85}} $$$ \Rightarrow \quad 0.0509 \leq p \leq 0.2243 \quad $ with $95\%$ confidence.

Example 2

In a random sample of $2800$ Canadian voters $98$ indicated that they will vote for the Green`s in the next Federal election. Estimate the population proportion with a $95\%$ confidence interval.
$\hat{p}=\frac{98}{2800}=0.035$
$$0.035 \pm 1.96 \sqrt{\frac{0.035(1-0.035)}{2800}} $$$ \quad \Rightarrow 0.0282 \leq p \leq 0.0418 $
$\quad \Rightarrow 2.82\% \leq p\leq 4.18\% \quad $with $95\%$ confidence.

Half the confidence interval's width is the margin of error of the estimate: $\frac{4.18-2.82}{2}=0.68\%$

Lecture 17 Lecture 19