A random sample of 100 students at a high school was asked whether they would ask their father or mother for help with a financial problem. A second sample of 100 different students was asked the same question regarding a dating problem. Forty-three students in the first sample and 47 students in the second sample replied that they turned to their mother rather than their father for help. Construct a 98 % confidence interval for $\mathrm { p } _ { 1 } - \mathrm { p } _ { 2 }$
A) (-0.204,0.124)
B) (-1.324,1.521)
C) (-0.591,0.762)
D) (-1.113,1.311)

Question

A random sample of 100 students at a high school was asked whether they would ask their father or mother for help with a financial problem. A second sample of 100 different students was asked the same question regarding a dating problem. Forty-three students in the first sample and 47 students in the second sample replied that they turned to their mother rather than their father for help. Construct a  98 %  confidence interval for  $\mathrm { p } _ { 1 } - \mathrm { p } _ { 2 }$
A)  (-0.204,0.124) 
B)  (-1.324,1.521) 
C)  (-0.591,0.762) 
D)  (-1.113,1.311)

Quizplus · Accepted Answer

The confidence interval for the difference between two proportions $\mathrm { p } _ { 1 } - \mathrm { p } _ { 2 }$ can be calculated using the formula: $$CI = \left( \hat{p}_1 - \hat{p}_2 \right) \pm Z_{\alpha/2} \sqrt{ \frac{\hat{p}_1(1-\hat{p}_1)}{n_1} + \frac{\hat{p}_2(1-\hat{p}_2)}{n_2} }$$ where $\hat{p}_1 = \frac{43}{100}$, $\hat{p}_2 = \frac{47}{100}$, $n_1 = n_2 = 100$, and $Z_{\alpha/2}$ for a 98% confidence level is approximately 2.33 (since the Z-score for a 99% confidence level in one tail is 2.33, and 98% confidence level means 1% in each tail in a two-tailed test). Plugging in the values: $$CI = (0.43 - 0.47) \pm 2.33 \sqrt{ \frac{0.43(0.57)}{100} + \frac{0.47(0.53)}{100} }$$ $$CI = -0.04 \pm 2.33 \sqrt{ \frac{0.2451}{100} + \frac{0.2491}{100} }$$ $$CI = -0.04 \pm 2.33 \sqrt{0.004942}$$ $$CI = -0.04 \pm 0.164$$ $$CI = (-0.204, 0.124)$$ This interval suggests that the difference in proportions of students turning to their mother for financial versus dating problems could range from -0.204 to 0.124, indicating that there is no significant preference for either parent for these types of problems within the given confidence interval.

A Random Sample of 100 Students at a High School p1−p2\mathrm { p } _ { 1 } - \mathrm { p } _ { 2 }p1​−p2​

A Random Sample of 100 Students at a High School $\mathrm { p } _ { 1 } - \mathrm { p } _ { 2 }$