# Solved: The Cincinnati Enquirer reported that, in the United States, 66% of adults and 87% of youths ages 12 to 17 use the Inter

#### ByDr. Raju Chaudhari

Oct 14, 2020

The Cincinnati Enquirer reported that, in the United States, 66% of adults and 87% of youths ages 12 to 17 use the Internet (The Cincinnati Enquirer, February 7, 2006). Use the reported numbers as the population proportions and assume that samples of 300 adults and 300 youths will be used to learn about attitudes toward Internet security.

a. Show the sampling distribution of $\overline{p}$ where $\overline{p}$ is the sample proportion of adults using the Internet.
b. What is the probability that the sample proportion of adults using the Internet will be within $\pm .04$ of the population proportion?
c. What is the probability that the sample proportion of youths using the Internet will be within $\pm .04$ of the population proportion?
d. Is the probability different in parts (b) and (c)? If so, why?
e. Answer part (b) for a sample of size 600. Is the probability smaller? Why?

## Solution

(a) The sampling distribution of proportion of adults using the internet $\overline{p}$ is

 $$E(\overline{p}) = 0.66$$

 \begin{aligned} \sigma_{\overline{p}} &=\sqrt{\frac{p*(1-p)}{n}}\\ &=\sqrt{\frac{0.66(1-0.66)}{300}}\\ &= 0.0273496 \end{aligned}

(b) Within $\pm 0.04$ means $0.62\leq \overline{p}\leq 0.70$.

 \begin{aligned} P(0.62\leq \overline{p}\leq 0.70) &= P\bigg(\frac{0.62-0.66}{0.0273}\leq \frac{\overline{p}- p}{\sigma_{\overline{p}}}\leq \frac{0.62-0.66}{0.0273}\bigg)\\ &= P(-1.47 \leq Z \leq 1.47)\\ &= P(Z\leq 1.47) - P(Z\leq -1.47)\\ &= 0.9292 - 0.0708\\ &= 0.8584\\ \end{aligned}

(c) The sampling distribution of proportion of youth using the internet is $\overline{p}$

 $$E(\overline{p}) = 0.87$$

 \begin{aligned} \sigma_{\overline{p}} &=\sqrt{\frac{p*(1-p)}{n}}\\ &=\sqrt{\frac{0.87(1-0.87)}{300}}\\ &= 0.0194165 \end{aligned}

Within $\pm 0.04$ means $0.83\leq \overline{p}\leq 0.91$.

 \begin{aligned} P(0.83\leq \overline{p}\leq 0.91) &= P\bigg(\frac{0.83-0.87}{0.0194}\leq \frac{\overline{p}- p}{\sigma_{\overline{p}}}\leq \frac{0.83-0.87}{0.0194}\bigg)\\ &= P(-2.06 \leq Z \leq 2.06)\\ &= P(Z\leq 2.06) - P(Z\leq -2.06)\\ &= 0.9803 - 0.0197\\ &= 0.9606\\ \end{aligned}

(d) Yes, the probability of being within $\pm 0.04$ is higher for the sample of youth users. This is because the standard error is smaller for the population proportion as it gets closer to 1.

(e) For $n= 600$

 $$E(\overline{p}) = 0.66$$

 \begin{aligned} \sigma_{\overline{p}} &=\sqrt{\frac{p*(1-p)}{n}}\\ &=\sqrt{\frac{0.66(1-0.66)}{600}}\\ &= 0.0193391 \end{aligned}

Within $\pm 0.04$ means $0.62\leq \overline{p}\leq 0.70$.

 \begin{aligned} P(0.62\leq \overline{p}\leq 0.70) &= P\bigg(\frac{0.62-0.66}{0.0193}\leq \frac{\overline{p}- p}{\sigma_{\overline{p}}}\leq \frac{0.62-0.66}{0.0193}\bigg)\\ &= P(-2.07 \leq Z \leq 2.07)\\ &= P(Z\leq 2.07) - P(Z\leq -2.07)\\ &= 0.9808 - 0.0192\\ &= 0.9616\\ \end{aligned}
The probability is larger than in part (b). This is because the larger sample size has reduced the standard error.