In a recent survey of drinking laws, a random sample of 1000 women showed that 65 % were in favor of increasing the legal drinking age. In a random sample of 1000 men, 60 % favored increasing the legal drinking age. Construct a 95% confidence interval for $\mathrm { p } _ { 1 } - \mathrm { p } _ { 2 }$
A) (0.008,0.092)
B) (0.587,0.912)
C) (-1.423,1.432)
D) (-2.153,1.679)

Question

In a recent survey of drinking laws, a random sample of 1000 women showed that  65 %  were in favor of increasing the legal drinking age. In a random sample of 1000 men,  60 %  favored increasing the legal drinking age. Construct a 95% confidence interval for $\mathrm { p } _ { 1 } - \mathrm { p } _ { 2 }$
A)  (0.008,0.092) 
B)  (0.587,0.912) 
C)  (-1.423,1.432) 
D)  (-2.153,1.679)

Quizplus · Accepted Answer

The confidence interval for the difference between two proportions ($p_1 - p_2$) can be calculated using the formula for the standard error of the difference between two proportions and the Z value for the desired level of confidence (1.96 for 95%). Here, $p_1 = 0.65$ (proportion of women in favor) and $p_2 = 0.60$ (proportion of men in favor). The calculation yields a confidence interval of (0.008, 0.092), indicating that the difference in proportions favoring the increase in legal drinking age between women and men is likely between 0.8% and 9.2%, with 95% confidence.

In a Recent Survey of Drinking Laws, a Random Sample p1−p2\mathrm { p } _ { 1 } - \mathrm { p } _ { 2 }p1​−p2​

In a Recent Survey of Drinking Laws, a Random Sample $\mathrm { p } _ { 1 } - \mathrm { p } _ { 2 }$