Chapter 22, Problem 31E

To determine

To find: that these findings provide proof that there are more parents checking.

Explanation of Solution

Given:

  $\begin{array}{c}{n}_{2004}=868\\ {\widehat{p}}_{2004}=0.33\\ n_{2006}=811\\ {\widehat{p}}_{2006}=0.41\end{array}$

Formula used:

  $\begin{array}{c}{\widehat{p}}_{Pooled}=\frac{y_{2004}+y_{2006}}{n_{2004}+n_{2006}}\\ S{E}_{Pooled}\left({\widehat{p}}_{2006}-{\widehat{p}}_{2004}\right)=\sqrt{\frac{{\widehat{p}}_{Pooled}{\widehat{q}}_{Pooled}}{n_{2006}}+\frac{{\widehat{p}}_{Pooled}{\widehat{q}}_{Pooled}}{n_{2004}}}\end{array}$

Calculation:

There is required to test more parents here to check on the online activities of their teenage children. Let $p_{2004}$ indicate the percentage of parents checking on their teen's online activities in 2004, and let $p_{2006}$ indicate the percentage of parents checking on their teen's online activities in 2006. $p_{2006}$ could be higher than $p_{2004}$ as more parents are checking up on their teenage online activities. So, our hypotheses are going to be

  $\begin{array}{c}{H}_0:p_{2006}-p_{2004}=0\\ H_1:p_{2006}-p_{2004}>0\end{array}$

  $\begin{array}{c}{y}_{2004}=868\times 0.33=286\\ y_{2006}=811\times 0.41=333\end{array}$

Pool the sample data.

  $\begin{array}{c}{\widehat{p}}_{Pooled}=\frac{y_{2004}+y_{2006}}{n_{2004}+n_{2006}}\\ =\frac{286+333}{868+811}\\ =\frac{619}{1679}\\ =0.3686\end{array}$

Estimating $SD\left(p_{2006}-p_{2004}\right)$ . Therefore,

  $\begin{array}{c}S{E}_{Pooled}\left({\widehat{p}}_{2006}-{\widehat{p}}_{2004}\right)=\sqrt{\frac{{\widehat{p}}_{Pooled}{\widehat{q}}_{Pooled}}{n_{2006}}+\frac{{\widehat{p}}_{Pooled}{\widehat{q}}_{Pooled}}{n_{2004}}}\\ =\sqrt{\frac{0.3686\left(1-0.3686\right)}{811}+\frac{0.3686\left(1-0.3686\right)}{868}}\\ =\sqrt{\frac{0.3686\left(0.6314\right)}{811}+\frac{0.3686\left(0.6314\right)}{868}}\\ =0.0236\end{array}$

The observed difference is

  $\begin{array}{c}{\widehat{p}}_{2006}-{\widehat{p}}_{2004}=0.41-0.33\\ =0.08\end{array}$

The z-score is

  $\begin{array}{c}z=\frac{\left({\widehat{p}}_{2006}-{\widehat{p}}_{2004}\right)-0}{S{E}_{Pooled}\left({\widehat{p}}_{2006}-{\widehat{p}}_{2004}\right)}\\ =\frac{0.08-0}{0.0236}\\ =3.40\end{array}$

The P-value is

  $\begin{array}{c}\text{P-value =}P\left(z>3.40\right)\\ =0.0003\end{array}$

P-value = 0.0003 is quite small; the null hypothesis should not be accepted. It might assume that there is ample proof that more parents are verifying the online activities of their teenagers.