Question

A random sample of n 1 = 249 people who live in a city were selected and 87 identified as a democrat. a random sample of n 2 = 113 people who live in a rural area were selected and 58 identified as a democrat. find the 98% confidence interval for the difference in the proportion of people that live in a city who identify as a democrat and the proportion of people that live in a rural area who identify as a democrat.

Answer 1

`CI=\{-0.2941,-0.0337\}`

Explanation:

Assuming conditions are met, the formula for a confidence interval (CI) for the difference between two population proportions is
`\displaystyle CI=(\hat{p}_1-\hat{p}_2)\pm z^*\sqrt{\frac{\hat{p}_1(1-\hat{p}_1)}{n_1}+\frac{\hat{p}_2(1-\hat{p}_2)}{n_2}` where
`\hat{p}_1` and
`n_1` are the sample proportion and sample size of the first sample, and
`\hat{p}_2` and
`n_2` are the sample proportion and sample size of the second sample.

We see that
`\hat{p}_1=(87)/(249)\approx0.3494` and
`\hat{p}_2=(58)/(113)\approx0.5133`. We also know that a 98% confidence level corresponds to a critical value of
`z^*=2.33`, so we can plug these values into the formula to get our desired confidence interval:

`\displaystyle CI=(\hat{p}_1-\hat{p}_2)\pm z^*\sqrt{\frac{\hat{p}_1(1-\hat{p}_1)}{n_1}+\frac{\hat{p}_2(1-\hat{p}_2)}{n_2}}\\\\CI=\biggr((87)/(249)-(58)/(113)\biggr)\pm 2.33\sqrt{((87)/(249)(1-(87)/(249)))/(249)+((58)/(113)(1-(58)/(113)))/(113)}\\\\CI=\{-0.2941,-0.0337\}`

Hence, we are 98% confident that the true difference in the proportion of people that live in a city who identify as a democrat and the proportion of people that live in a rural area who identify as a democrat is contained within the interval {-0.2941,-0.0337}

The 98% confidence interval also suggests that it may be more likely that identified democrats in a rural area have a greater proportion than identified democrats in a city since the differences in the interval are less than 0.