Construct a 95% confidence interval for $\mathrm { p } _ { 1 } - \mathrm { p } _ { 2 }$ The sample statistics listed below are from independent samples.
Sample statistics: $\mathrm { n } _ { 1 } = 50 , \mathrm { x } _ { 1 } = 35 , \text { and } \mathrm { n } _ { 2 } = 60 , \mathrm { x } _ { 2 } = 40$
A) (-0.141,0.208)
B) (-0.871,0.872)
C) (-1.341,1.781)
D) (-2.391,3.112)

Question

Construct a 95% confidence interval for $\mathrm { p } _ { 1 } - \mathrm { p } _ { 2 }$ The sample statistics listed below are from independent samples.
Sample statistics: $\mathrm { n } _ { 1 } = 50 , \mathrm { x } _ { 1 } = 35 , \text { and } \mathrm { n } _ { 2 } = 60 , \mathrm { x } _ { 2 } = 40$
A)  (-0.141,0.208) 
B)  (-0.871,0.872) 
C)  (-1.341,1.781) 
D)  (-2.391,3.112)

Quizplus · Accepted Answer

To construct a 95% confidence interval for $p_1 - p_2$, where $p_1$ and $p_2$ are the proportions of successes in two independent samples, we first calculate the sample proportions: $p_1 = \frac{x_1}{n_1} = \frac{35}{50} = 0.7$ and $p_2 = \frac{x_2}{n_2} = \frac{40}{60} = 0.6667$. The standard error (SE) of the difference in proportions is calculated using the formula: $\sqrt{\frac{p_1(1-p_1)}{n_1} + \frac{p_2(1-p_2)}{n_2}}$. Plugging in the values, we get $\sqrt{\frac{0.7(0.3)}{50} + \frac{0.6667(0.3333)}{60}} \approx 0.0887$. Using a Z-score of 1.96 for a 95% confidence interval, the margin of error (ME) is $1.96 \times SE \approx 0.174$. The confidence interval is then $p_1 - p_2 \pm ME = (0.7 - 0.6667) \pm 0.174 = 0.0333 \pm 0.174$, which gives us (-0.141, 0.208).

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

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