Construct a 95 % confidence interval for $\mathrm { p } _ { 1 } - \mathrm { p } _ { 2 }$ for a survey that finds 30 % of 240 males and 41 % of 200 females are opposed to the death penalty.
A) (-0.200,-0.021)
B) (-1.532,1.342)
C) (-0.561,0.651)
D) (-1.324,1.512)

Question

Construct a  95 %  confidence interval for $\mathrm { p } _ { 1 } - \mathrm { p } _ { 2 }$ for a survey that finds  30 %  of 240 males and  41 %  of 200 females are opposed to the death penalty.
A)  (-0.200,-0.021) 
B)  (-1.532,1.342) 
C)  (-0.561,0.651) 
D)  (-1.324,1.512)

Quizplus · Accepted Answer

To construct a 95% confidence interval for the difference between two proportions, $\mathrm { p } _ { 1 } - \mathrm { p } _ { 2 }$, we use the formula:$$\left( \hat{p}_1 - \hat{p}_2 \right) \pm Z_{\alpha/2} \sqrt{ \frac{\hat{p}_1(1-\hat{p}_1)}{n_1} + \frac{\hat{p}_2(1-\hat{p}_2)}{n_2} }$$Given:- $\hat{p}_1 = 0.30$ (proportion of males opposed to the death penalty)- $n_1 = 240$ (number of males surveyed)- $\hat{p}_2 = 0.41$ (proportion of females opposed to the death penalty)- $n_2 = 200$ (number of females surveyed)- $Z_{\alpha/2} = 1.96$ for a 95% confidence levelPlugging in the values:$$\begin{align*}\text{Margin of Error} &= 1.96 \sqrt{ \frac{0.30(1-0.30)}{240} + \frac{0.41(1-0.41)}{200} } \&= 1.96 \sqrt{ \frac{0.21}{240} + \frac{0.2419}{200} } \&= 1.96 \sqrt{ 0.000875 + 0.0012095 } \&= 1.96 \sqrt{ 0.0020845 } \&\approx 1.96 \times 0.045655 \&\approx 0.0894842\end{align*}$$The difference in proportions is $0.30 - 0.41 = -0.11$.The 95% confidence interval is:$$(-0.11 \pm 0.0894842) = (-0.1994842, -0.0205158)$$Rounded to three decimal places, the interval is approximately $(-0.200, -0.021)$, which matches choice A.

Construct a 95 % Confidence Interval For p1−p2\mathrm { p } _ { 1 } - \mathrm { p } _ { 2 }p1​−p2​

Construct a 95 % Confidence Interval For $\mathrm { p } _ { 1 } - \mathrm { p } _ { 2 }$