Question

Facebook As part of the Pew Internet and American Life Project, researchers conducted two surveys. The first survey asked a random sample of $1060$U.S. teens about their use of social media. A second survey posed similar questions to a random sample of $2003$U.S. adults. In these two studies, $71.0\%$of teens and $58.0\%$of adults used Facebook. Let plocalid="1654194806576" $T^{\frac{305}{1526}=0.200=20.0\%}$pT= the true proportion of all U.S. teens who use Facebook and p$A^{\frac{305}{1526}=0.200=20\%}$pA = the true proportion of all U.S. adults who use Facebook. Calculate and interpret a $99\%$confidence interval for the difference in the true proportions of U.S. teens and adults who use Facebook.

Short answer

We are $99\%$confident that the true proportion of U.S. teens who use Facebook is between 0.0842 and 0.1758 higher than the true proportion of U.S. adults who use Facebook.

Step-by-step solution

Given information

We have given

$n_1$= Sample size = 1060

${\hat{p}}_1$= Sample proportion = $71.0\%$= $0.710$

$n_2$= Sample size = 2003

${\hat{p}}_2$= Sample proportion = $58.0\%$= $0.580$

c = confidence level = $99\%$= $0.99$

Now we will the conditions for the calculation of hypothesis test for the population proportion p i.e. Random, Independent, Normal.

Randon: As samples are random, hense satisfied.

Independent: Samples are independent as well because $1060$U.S. teens are less than $10\%$of all U.S. teens and the $2003$U.S. adults are less than $10\%$of all U.S. adults.

Normal: $n_1{\hat{p}}_1$= $1060(0.710)=752.6$, ${n_1(1-\hat{p}}_1)=1060(1-0.710)=307.4$, $n_2{\hat{p}}_2=2003(0.580)=1161.74$, $n_2(1-{\hat{p}}_2)=2003(1-0.580)=841.26$

all values are less than $10$. Hence, satisfied.

As all the conditions satisfies, we will determine confience level for $p_1-p_2$

Confidence Level

Given Confidence level $1-\alpha =0.99$,

we determine $z_{\frac{\alpha }{2}}=z_{0.005}$using normal probability table, we get

$z_{\frac{\alpha }{2}}=2.575$

Now,

Margin of error, E = $$\begin{gathered} z_{\frac{\alpha }{2}}.\sqrt{\frac{{{\displaystyle \hat{p}}}_1(1-{{\displaystyle \hat{p}}}_1)}{n_1}+\frac{{{\displaystyle \hat{p}}}_2(1-{{\displaystyle \hat{p}}}_2)}{n_2}} \\ =2.575\sqrt{\frac{0.710(1-0.710)}{1060}+\frac{0.580(1-0.580)}{2003}} \\ \approx 0.0458 \end{gathered}$$

Endpoints for the confidence level are now:

$$\begin{gathered} ({\hat{p}}_1-{\hat{p}}_2)-E=(0.710-0.580)-0.0458=0.130-0.0458\approx 0.0842 \\ ({\hat{p}}_1-{\hat{p}}_2)+E=(0.710-0.580)+0.0458=0.130+0.0458\approx 0.1758 \end{gathered}$$

Therefore, we are$99\%$ confident that the trye proportion of U.S. teens who use Facebook is between $0.0842$and $0.1758$higher than the true proportion of U.S. adults who use Facebook.