Question

Abstain from drinking In a Harvard School of Public Health survey, $2105$ of $10,904$ randomly selected U.S. college students were classified as abstainers (nondrinkers).

(a) Construct and interpret a $99\%$ confidence interval for$p$. Follow the four-step process.

(b) A newspaper article claims that $25\%$ of U.S. college students are nondrinkers. Use your result from (a) to comment on this claim.

Short answer

a)$0.1833<p<0.2027$

b) There is sufficient evidence to reject the claim.

Step-by-step solution

Part (a) Step 1: Given Information

Calculating a confidence interval for p by using four-step process.

Part (a) Step 2: Explanation

The sample proportion is calculated by dividing the number of successes by the sample size:

$$\begin{gathered} \hat{p}=\frac{x}{n} \\ =\frac{2105}{10904} \\ \approx 0.1930 \end{gathered}$$

Determine $z_{\alpha /2}=z_{0.005}$using table A (search up $0.005$in the table, the z-score is then the found z-score with opposite sign) with confidence level $1-\alpha =99\%=0.99$:

$z_{\alpha /2}=z_{0.005}=2.575$

As a result, the margin of error is:

localid="1650247703364" $$\begin{gathered} E=z_{\alpha /2}·\sqrt{\frac{\hat{p}(1-\hat{p})}{n}} \\ =2.575·\sqrt{\frac{0.1930(1-0.1930)}{10904}} \\ \approx 0.0097 \end{gathered}$$

As a result, the confidence interval is:

localid="1650247736847" $$\begin{gathered} 0.1833=0.1930-0.0097 \\ =\hat{p}-E<p<\hat{p}+E \\ =0.1930+0.0097 \\ =0.2027 \end{gathered}$$

We're $99$percent sure the true population proportion is between$0.1833and0.2027$

Part (b) Step 1: Given Information

By using part (a) we need to find the result for $25\%$of U.S. college students.

Part (b) Step 2: Explanation 

There is adequate evidence to refute the assertion that $25\%$ of U.S. college students are nondrinkers because the confidence interval does not contain$25\%$ text (or 0.25 text).