Question

A College Alcohol Study has interviewed random samples of students at four-year colleges. In the most recent study, 494 of 1000 women reported drinking alcohol and 552 of 1000 men reported drinking alcohol. What is the 95% confidence interval of the drinking alcohol percentage difference between women and men

Answer 1

The 95% confidence interval for the difference between the proportion of women who drink alcohol and the proportion of men who drink alcohol is (-0.102, -0.014) or (-10.2%, -1.4%).

Explanation:

We want to calculate the bounds of a 95% confidence interval of the difference between proportions.

For a 95% CI, the critical value for z is z=1.96.

The sample 1 (women), of size n1=1000 has a proportion of p1=0.494.

`p_1=X_1/n_1=494/1000=0.494`

The sample 2 (men), of size n2=1000 has a proportion of p2=0.552.

`p_2=X_2/n_2=552/1000=0.552`

The difference between proportions is (p1-p2)=-0.058.

`p_d=p_1-p_2=0.494-0.552=-0.058`

The pooled proportion, needed to calculate the standard error, is:

`p=(X_1+X_2)/(n_1+n_2)=(494+552)/(1000+1000)=(1046)/(2000)=0.523`

The estimated standard error of the difference between means is computed using the formula:

`s_(p1-p2)=\sqrt{(p(1-p))/(n_1)+(p(1-p))/(n_2)}=\sqrt{(0.523*0.477)/(1000)+(0.523*0.477)/(1000)}\\\\\\s_(p1-p2)=√(0.000249+0.000249)=√(0.000499)=0.022`

Then, the margin of error is:

`MOE=z \cdot s_(p1-p2)=1.96\cdot 0.022=0.0438`

Then, the lower and upper bounds of the confidence interval are:

`LL=(p_1-p_2)-z\cdot s_(p1-p2) = -0.058-0.0438=-0.102\\\\UL=(p_1-p_2)+z\cdot s_(p1-p2)= -0.058+0.0438=-0.014`

The 95% confidence interval for the difference between proportions is (-0.102, -0.014).