Teens' online profiles Over half of all American teens (ages 12 to 17 years) have an online profile, mainly on Facebook. A random sample of 487 teens with profiles found that 385 included photos of themselves. ${}^{13}$

(a) Construct and interpret a $95\%$confidence interval for $p$. Follow the four-step process.

(b) Is it plausible that the true proportion of American teens with profiles who have posted photos of themselves is $0.75?$ Use your result from part (a) to support your answer.

By using a four-step process we need to construct and interpret a$95\%$confidence level for p.

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

$$\begin{gathered} \hat{p}=\frac{x}{n} \\ =\frac{385}{487} \\ \approx 0.7906 \end{gathered}$$

Determine $z_{\alpha /2}=z_{0.025}$using table A (search up $0.025$in the table, the z-score is then the found z-score with opposite sign) with confidence level $1-\alpha =95\%=0.95$

$z_{\alpha /2}=z_{0.025}=1.96$

As a result, the margin of error is:

localid="1650247541501" $$\begin{gathered} E=z_{\alpha /2}·\sqrt{\frac{\hat{p}(1-\hat{p})}{n}} \\ =1.96·\sqrt{\frac{0.7906(1-0.7906)}{487}} \\ \approx 0.0361 \end{gathered}$$

As a result, the confidence interval is:

localid="1650247564168" $$\begin{gathered} 0.7545=0.7906-0.0361 \\ =\hat{p}-E<p<\hat{p}+E \\ =0.7906+0.0361 \\ =0.8267 \end{gathered}$$

We are 95 % confident that the true population proportion is between $0.7545$and $0.8267.$

By using part(a) we need to find whether it is sufficient evidence or not.

Since the confidence interval does not contain $0.75$, there is sufficient evidence to reject the claim that the true proportion of American teens with profiles who have posted photos of themselves is $0.75$.