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

Answer:

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

Step-by-step 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 $\hat{p}_1=\frac{87}{249}\approx0.3494$ and $\hat{p}_2=\frac{58}{113}\approx0.5133$. 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:

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

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.