A simple random sample of 60 households in City 1 is taken.In the sample,there are 45 households that decorate their houses with lights for the holidays.A simple random sample of 50 households is also taken from the neighboring City 2.In the sample,there are 40 households that decorate their houses.We wish to estimate the difference in proportions of households that decorate their houses with lights for the holidays,with a 95% confidence interval.What is a 95% confidence interval for the difference in population proportions of households that decorate their houses with lights for the holidays?
A)(-0.181,0.081)
B)(-0.206,0.106)
C)(-0.231,0.138)
D)(-0.255,0.155)

Question

Quizplus · Accepted Answer

First, we must calculate the point estimate for the difference in proportions:
$\hat{p}_1 = \frac{45}{60} = 0.75$
$\hat{p}_2 = \frac{40}{50} = 0.8$
$\hat{p}_1 - \hat{p}_2 = 0.75 - 0.8 = -0.05$
Next, we must calculate the standard error:
$s.e. = \sqrt{\frac{\hat{p}_1(1-\hat{p}_1)}{n_1} + \frac{\hat{p}_2(1-\hat{p}_2)}{n_2}} = \sqrt{\frac{0.75(0.25)}{60} + \frac{0.8(0.2)}{50}} \approx 0.0806$
Using a 95% confidence level, we can find the critical value for a two-tailed test using a t-distribution with $(n_1-1)+(n_2-1) = 108$ degrees of freedom: $t^* = 1.984$
Finally, we can calculate the confidence interval:
$\hat{p}_1 - \hat{p}_2 \pm t^* 	imes s.e.$
$-0.05 \pm 1.984 	imes 0.0806$
$-0.05 \pm 0.1601$
$(-0.21, 0.11)$
Therefore, the best choice is B, (-0.206,0.106).

A Simple Random Sample of 60 Households in City 1