Construct a 98% confidence interval for $\mathrm { p } _ { 1 } - \mathrm { p } _2$ Assume the samples are random and independent. Sample statistics: $\mathrm { n } _ { 1 } = 1000 , \mathrm { x } _ { 1 } = 250 , \text { and } \mathrm { n } _ { 2 } = 1200 , \mathrm { x } _ { 2 } = 195$
A)(0.047, 0.128)
B)(-0.621, 0.781)
C)(0.581, 1.819)
D)(1.516, 3.021)

Question

Quizplus · Accepted Answer

To construct a 98% confidence interval for $p_1 - p_2$, first calculate the sample proportions: $p_1 = \frac{x_1}{n_1} = \frac{250}{1000} = 0.25$ and $p_2 = \frac{x_2}{n_2} = \frac{195}{1200} = 0.1625$. Then, use the formula for the confidence interval of two proportions: $\hat{p}_1 - \hat{p}_2 \pm Z_{\alpha/2} \sqrt{\frac{\hat{p}_1(1-\hat{p}_1)}{n_1} + \frac{\hat{p}_2(1-\hat{p}_2)}{n_2}}$. With a 98% confidence level, $Z_{\alpha/2} \approx 2.33$. Plugging in the values gives the interval (0.047, 0.128).

Construct a 98% Confidence Interval For p1−p2\mathrm { p } _ { 1 } - \mathrm { p } _2p1​−p2​

Construct a 98% Confidence Interval For $\mathrm { p } _ { 1 } - \mathrm { p } _2$