Question

The Harvard College Alcohol Study finds that $67\%$of college students support efforts to "crack down on underage drinking." The study took a random sample of almost $15,000$students, so the population proportion who support a crackdown is close to $p=0.67$The administration of a local college surveys an SRS of $100$students and finds that $62$support a crackdown on underage drinking.

(a) Suppose that the proportion of all students attending this college who support a crackdown is $67\%$, the same as the national proportion. What is the probability that the proportion in an SRS of $100$students is as small as or smaller than the result of the administration’s sample? Follow the four-step process.

(b) A writer in the college’s student paper says that “support for a crackdown is lower at our school than nationally.” Write a short letter to the editor explaining why the survey does not support this conclusion

Short answer

(a) $P(\hat{p}\le 0.62)=0.1446$

(b) Since the probability is greater than $0.05$, it is likely to obtain a sample with sample proportion of $0.62$if the true population proportion is $0.67$or $67\%$. Thus there is not sufficient evidence of a lower proportion.

Step-by-step solution

Part(a) Step 1: Given Information

Given

$$\begin{gathered} p=67\% \\ =0.67 \end{gathered}$$

$x=62$

$n=100$

Part(a) Step 2: Explanation

The sample proportion is the number of successes divided by the sample size:

$$\begin{gathered} \hat{p}=\frac{x}{n} \\ =\frac{62}{100} \\ =0.62 \end{gathered}$$

The mean of the sampling distribution of $\hat{p}$is equal to the population proportion $p$:

localid="1650095153168" $$\begin{gathered} {\mu }_{\hat{p}}=p \\ =0.67 \end{gathered}$$

The standard deviation of the sampling distribution of $\hat{p}$is:

localid="1650095171714" $$\begin{gathered} {\sigma }_{\hat{p}}=\sqrt{\frac{p(1-p)}{n}} \\ =\sqrt{\frac{0.67(1-0.67)}{100}} \\ \approx 0.0470213 \end{gathered}$$

The z-score is the value decreased by the mean, divided by the standard deviation:

localid="1650095190764" $$\begin{gathered} z=\frac{x-\mu }{\sigma } \\ =\frac{0.62-0.67}{0.0470213} \\ \approx -1.06 \end{gathered}$$

Determine the corresponding probability using table A:

localid="1650095208985" $$\begin{gathered} P(\hat{p}\le 0.62)=P(z<-1.06) \\ =0.1446 \end{gathered}$$

Part(b) Step 1: Given Information

Given

$$\begin{gathered} p=67\% \\ =0.67 \end{gathered}$$

$x=62$

$n=100$

Part(b) Step 2: Explanation

The sample proportion is the number of successes divided by the sample size:

$$\begin{gathered} \hat{p}=\frac{x}{n} \\ =\frac{62}{100} \\ =0.62 \end{gathered}$$

The mean of the sampling distribution of $\hat{p}$is equal to the population proportion $p$:

localid="1650095264695" $$\begin{gathered} {\mu }_{\hat{p}}=p \\ =0.67 \end{gathered}$$

The standard deviation of the sampling distribution of $\hat{p}$is:

localid="1650095291522" $$\begin{gathered} {\sigma }_{\hat{p}}=\sqrt{\frac{p(1-p)}{n}} \\ =\sqrt{\frac{0.67(1-0.67)}{100}} \\ \approx 0.0470213 \end{gathered}$$

The z-score is the value decreased by the mean, divided by the standard deviation:

localid="1650095308205" $$\begin{gathered} z=\frac{x-\mu }{\sigma } \\ =\frac{0.62-0.67}{0.0470213} \\ \approx -1.06 \end{gathered}$$

Determine the corresponding probability using table A:

localid="1650095325932" $$\begin{gathered} P(\hat{p}\le 0.62)=P(z<-1.06) \\ =0.1446 \end{gathered}$$