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

Answer:

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%).

Step-by-step 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:

$p=\dfrac{X_1+X_2}{n_1+n_2}=\dfrac{494+552}{1000+1000}=\dfrac{1046}{2000}=0.523$

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

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

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).