Question

a random sample of $n_1 = 217$ people who live in a city were selected and 72 identified as blue collar workers. a random sample of $n_2 = 101$ people who live in a rural area were selected and 58 identified as blue collar workers. find the 98% confidence interval for the difference in the proportion of people that live in a city who identify as blue collar workers and the proportion of people that live in a rural area who identify as blue collar workers. round answers to 2 decimal places, use confidence interval notation.

Explanation:

Step1: Calculate sample proportions

For city sample, $n_1 = 217$, $x_1=72$, so $\hat{p}_1=\frac{x_1}{n_1}=\frac{72}{217}\approx 0.33$. For rural - area sample, $n_2 = 101$, $x_2 = 58$, so $\hat{p}_2=\frac{x_2}{n_2}=\frac{58}{101}\approx 0.57$.

Step2: Determine z - value

For a 98% confidence interval, the significance level $\alpha=1 - 0.98 = 0.02$, and $\alpha/2=0.01$. The $z -$value $z_{\alpha/2}=z_{0.01}\approx 2.33$.

Step3: Calculate the standard error

The standard error $SE=\sqrt{\frac{\hat{p}_1(1 - \hat{p}_1)}{n_1}+\frac{\hat{p}_2(1 - \hat{p}_2)}{n_2}}=\sqrt{\frac{0.33\times(1 - 0.33)}{217}+\frac{0.57\times(1 - 0.57)}{101}}\approx\sqrt{\frac{0.33\times0.67}{217}+\frac{0.57\times0.43}{101}}\approx\sqrt{\frac{0.2211}{217}+\frac{0.2451}{101}}\approx\sqrt{0.00102 - 0.00243}\approx\sqrt{0.00345}\approx 0.06$.

Step4: Calculate the confidence - interval limits

The lower limit $LL=(\hat{p}_1-\hat{p}_2)-z_{\alpha/2}\times SE=(0.33 - 0.57)-2.33\times0.06=- 0.24-0.14=-0.38$. The upper limit $UL=(\hat{p}_1-\hat{p}_2)+z_{\alpha/2}\times SE=(0.33 - 0.57)+2.33\times0.06=-0.24 + 0.14=-0.10$.

Answer:

$(-0.38,-0.10)$