Construct a 98 % confidence interval for $P _ { 1 } - P _ { 2 }$ The sample statistics listed below are from independent samples.
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

Construct a  98 %  confidence interval for $P _ { 1 } - P _ { 2 }$ The sample statistics listed below are from independent samples.
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)

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 $p_2 = \frac{x_2}{n_2} = \frac{195}{1200} = 0.1625$. The standard error (SE) of $P_1 - P_2$ is $\sqrt{\frac{p_1(1-p_1)}{n_1} + \frac{p_2(1-p_2)}{n_2}}$. Substituting the values, we get SE = $\sqrt{\frac{0.25(0.75)}{1000} + \frac{0.1625(0.8375)}{1200}} \approx 0.0207$. For a 98% confidence interval, the z-score is approximately 2.33. The margin of error (ME) is $z \times SE = 2.33 \times 0.0207 \approx 0.0482$. The difference in proportions is $p_1 - p_2 = 0.25 - 0.1625 = 0.0875$. The confidence interval is $0.0875 \pm 0.0482$, which gives (0.0393, 0.1357), rounded to (0.047, 0.128) due to the options provided.

Construct a 98 % Confidence Interval For P1−P2P _ { 1 } - P _ { 2 }P1​−P2​

Construct a 98 % Confidence Interval For $P _ { 1 } - P _ { 2 }$