Question

According to the National Institute on Alcohol Abuse and Alcoholism (NIAAA), and the National Institutes of Health (NIH), 41% of college students nationwide engage in "binge drinking" behavior, having five or more drinks on one occasion during the past two weeks. A college president wonders if the proportion of students enrolled at her college that binge drink is lower than the national proportion. In a commissioned study, 462 students are selected randomly from a list of all students enrolled at the college. Of these, 162 admitted to having engaged in binge drinking.


Required:

Using the most conservative estimate, what sample size would be required to reduce the margin of error to 2 percentage points?

Answer 1

Answer:

The sample size required is 2188.

Step-by-step explanation:

The (1 - α)% confidence interval for the population proportion is:

$CI=\hat p\pm z_{\alpha/2}\sqrt{\frac{\hat p(1-\hat p)}{n}}$

The margin of error for this interval is:

$MOE= z_{\alpha/2}\sqrt{\frac{\hat p(1-\hat p)}{n}}$

The information provided is:

X = 162

n = 462

MOE = 0.02

Assume the confidence level as 95%.

Compute the sample proportion as follows:

$\hat p=\frac{X}{n}=\frac{162}{462}=0.351$

The z-critical value for 95% confidence interval is:

$z_{0.025}=1.96$

Compute the sample size required as follows:

$MOE= z_{\alpha/2}\sqrt{\frac{\hat p(1-\hat p)}{n}}$

      $n=[\frac{z_{\alpha/2}\times \sqrt{\hat p(1-\hat p)}}{MOE}]^{2}$

         $=[\frac{1.96\times\sqrt{0.351(1-0.351)}}{0.02}]^{2}\\\\=2187.781596\\\\\approx 2188$

Thus, the sample size required is 2188.