Confidence Interval for a Population Proportion: More on Family Planning

If the random variable has the binomial distribution, then the estimate of the population proportion p is p with hat on top .

The website for UN Gender Statistics contains data on 52 quantitative and 11 qualitative indicators addressing relevant issues related to gender equality and /or women's empowerment.

Example. Suppose that in a random sample of 1000 women ages 15 - 49, 790 have access to family planning resources. The estimate of p is $LaTeX: \hat{p}=\frac{790}{1000}=0.79$ $\hat{p}=\frac{790}{1000}=0.79$ .

To construct the confidence interval we use the normal approximation of the binomial distribution if the criteria for approximation are satisfied. Use p with hat on top instead of p (Since p is unknown).

The confidence interval will be centered around p with hat on top . The margin of error is the radius of the interval, it depends on the standard error of the sampling distribution of and the critical value z subscript alpha divided by 2 end subscript .

If the criteria for approximation of the binomial distribution by the normal distribution are met, then

The standard error is where is the point estimate for p and . n is the sample size.
The margin of error: .
The confidence interval: $\hat{p}-z_{\frac{\alpha}{2}}\sqrt{\frac{\hat{p}\hat{q}}{n}}<p<\hat{p}+z_{\frac{\alpha}{2}}\sqrt{\frac{\hat{p}\hat{q}}{n}}$ $\hat{p}-z_{\frac{\alpha}{2}}\sqrt{\frac{\hat{p}\hat{q}}{n}}<p<\hat{p}+z_{\frac{\alpha}{2}}\sqrt{\frac{\hat{p}\hat{q}}{n}}$ .

Example. Suppose that in a random sample of 1000 adults, 200 are smokers. Construct a 95% confidence interval for the population proportion.

The estimate of p is $LaTeX: \hat{p}=\frac{790}{1000}=0.79$ $\hat{p}=\frac{790}{1000}=0.79$ , and $LaTeX: \hat{q}=1-\hat{p}=1-0.79=0.21$ $\hat{q}=1-\hat{p}=1-0.79=0.21$ , as noted earlier. alpha equals 0.05 space a n d space z subscript alpha divided by 2 end subscript equals 1.96 .

The confidence interval is $LaTeX: 0.79-1.96\times\sqrt{\frac{0.79\times.21}{1000}}<p<0.79+1.96\times\sqrt{\frac{0.79\times.21}{1000}}$ $0.79-1.96\times\sqrt{\frac{0.79\times.21}{1000}}<p<0.79+1.96\times\sqrt{\frac{0.79\times.21}{1000}}$ or 0.7648<p<0.8152.

Interpretation of the result: we are 95% confident that the true population proportion of women ages 15 to 49 years old have access to family planning resources is between 76.5% and 81.5% .