Chapter 22, Problem 32E

To determine

To explain: the proof of a real age-based difference on the basis of the information provided.

Explanation of Solution

Given:

  $\begin{array}{c}{n}_1=223\\ {\widehat{p}}_1=0.69\\ n_2=248\\ {\widehat{p}}_2=0.62\end{array}$

Formula used:

  $\begin{array}{c}{\widehat{p}}_{Pooled}=\frac{y_1+y_2}{n_1+n_2}\\ SE\left({\widehat{p}}_1-{\widehat{p}}_2\right)=\sqrt{\frac{{\widehat{p}}_{Pooled}{\widehat{q}}_{Pooled}}{n_1}+\frac{{\widehat{p}}_{Pooled}{\widehat{q}}_{Pooled}}{n_2}}\end{array}$

Calculation:

Here there is required to test who's playing electronic or online games. Suppose $p_1$ represent the proportion of online or electronic games played by boys aged 12-14 and suppose $p_2$ represent the proportion of online or electronic games played by boys aged 15-17. If the proportion difference occurs, then $p_1$ should not be equivalent to $p_2$ . therefore, the hypotheses are going to be

  $\begin{array}{l}{H}_0:p_1-p_2=0\\ H_1:p_1-p_2\ne 0\end{array}$

  $n_1=223,{\widehat{p}}_1=0.69,n_2=248,{\widehat{p}}_2=0.62$. So,

  $\begin{array}{l}{y}_1=223\times 0.69=154\\ y_2=248\times 0.62=154\end{array}$

Pool the sample data.

  $\begin{array}{c}{\widehat{p}}_{Pooled}=\frac{y_1+y_2}{n_1+n_2}\\ =\frac{154+154}{223+248}\\ =\frac{308}{417}\\ =0.6531\end{array}$

Using the pooled SE to estimate $SD\left(p_1-p_2\right)$ . Therefore,

  $\begin{array}{c}SE\left({\widehat{p}}_1-{\widehat{p}}_2\right)=\sqrt{\frac{{\widehat{p}}_{Pooled}{\widehat{q}}_{Pooled}}{n_1}+\frac{{\widehat{p}}_{Pooled}{\widehat{q}}_{Pooled}}{n_2}}\\ =\sqrt{\frac{0.6531\left(1-0.6531\right)}{223}+\frac{0.6531\left(1-0.6531\right)}{248}}\\ =\sqrt{\frac{0.6531\left(0.3469\right)}{223}+\frac{0.6531\left(0.3469\right)}{248}}\\ =0.0439\end{array}$

The observed difference in sample proportions is

  $\begin{array}{c}{\widehat{p}}_1-{\widehat{p}}_2=0.69-0.62\\ =0.07\end{array}$

The z-score is

  $\begin{array}{c}z=\frac{\left({\widehat{p}}_1-{\widehat{p}}_2\right)}{S{E}_{Pooled}\left({\widehat{p}}_1-{\widehat{p}}_2\right)}\\ =\frac{0.07-0}{0.0439}\\ =1.59\end{array}$

The P-value is

  $\begin{array}{c}\text{P-value}=2P\left(z>1.59\right)\\ =2\left(0.0555\right)\\ =0.1110\end{array}$

To consider the null hypothesis, the P-value = 0.1110 is high enough. It may conclude that solid evidence exists that online or electronic games are played similarly by boys aged 12-14 and boys aged 15-17.