Chapter 9.4, Problem 20E

To determine

To Construct:

Construct and interpret a 95% confidence interval for the true differences between the proportions of students who pass the test and play video games for two hours on the night before the test and those who pass the test but do not play games.

Answer to Problem 20E

Solution:

The true differences between the proportions of students who pass the test and play video games for two hours on the night before the test and those who pass the test but do not play games is between -0.30268 to 0.05878. There is not sufficient evidence to indicate that playing video games negatively impacts students’ performance.

Explanation of Solution

Given:

Ann, a teacher, is concerned with the obsession many of her students have with video games. She is afraid that the video games have a negative impact on her students’ performance in the classroom. To test the theory, Ann sends home a letter explaining the study, and 82 parents agree to let their children participate. Ann then randomly divides the students into two equal groups. Group A is required to play video games for two hours one evening, while Group B is not allowed any time to play the games. The following day, the students are given a review test over previously learned material. From Group A, 29 students pass the test. From Group B, 34 students pass the test.

Formula Used:

When the samples taken are independent, simple random samples, the conditions for a binomial distribution are met for both samples, and the sample sizes are large enough to ensure that $n_1{\widehat{p}}_1\ge 5,$$n_1(1-{\widehat{p}}_1)\ge 5,$$n_2{\widehat{p}}_2\ge 5,$$n_2(1-{\widehat{p}}_2)\ge 5$, the margin of error of a confidence interval for the differences between two population proportions is given by

$E=z_{\raisebox{1ex}{$\alpha $}\!\left/ \!\raisebox{-1ex}{$2$}\right.}\sqrt{\frac{{\widehat{p}}_1(1-{\widehat{p}}_1)}{n_1}+\frac{{\widehat{p}}_2(1-{\widehat{p}}_2)}{n_2}}$

Where $z_{\raisebox{1ex}{$\alpha $}\!\left/ \!\raisebox{-1ex}{$2$}\right.}$ is the critical value for the level of significance, $c=1-\alpha$, such that the area under the standard normal distribution to the right of $z_{\raisebox{1ex}{$\alpha $}\!\left/ \!\raisebox{-1ex}{$2$}\right.}$ is equal to $\frac{\alpha }{2}$, ${\widehat{p}}_{1}and{\widehat{p}}_2$ are two sample proportions, and $n_{1}and{n}_2$ are two sample sizes.

The confidence interval for the difference between two population proportions is given by

$\left({\widehat{p}}_1-{\widehat{p}}_2\right)-E<p_1-p_2<\left({\widehat{p}}_1-{\widehat{p}}_2\right)+E$ or $\left(\left({\widehat{p}}_1-{\widehat{p}}_2\right)-E,\left({\widehat{p}}_1-{\widehat{p}}_2\right)+E\right)$

Where ${\widehat{p}}_{1}and{\widehat{p}}_2$ are two sample proportions, and $n_{1}and{n}_2$ are two sample sizes, $\left({\widehat{p}}_1-{\widehat{p}}_2\right)$ is the point estimate for the difference between population proportions, E is the margin of error.

Calculation:

The given values are,

$\begin{array}{rll}{x}_1& =& 29\\ n_1& =& 41\\ x_2& =& 34\\ n_2& =& 41\end{array}$

At 95% level of confidence so $z_{\alpha /2}=z_{0.05/2}=1.96$

Using these sample sizes to calculate the sample proportions

$\begin{array}{rll}{\widehat{p}}_1& =& \frac{x_1}{n_1}\\ & =& \frac{29}{41}\\ & =& 0.70731\end{array}$$$

$\begin{array}{rll}{\widehat{p}}_2& =& \frac{x_2}{n_2}\\ & =& \frac{34}{41}\\ & =& 0.82926\end{array}$

Subtracting these two sample proportions produces the point estimate.

$\begin{array}{rll}\left({\widehat{p}}_1-{\widehat{p}}_2\right)& =& 0.70731-0.82926\\ & =& -0.12195\end{array}$

The margin of error is calculated as follows-:

$\begin{array}{rll}E& =& z_{\raisebox{1ex}{$\alpha $}\!\left/ \!\raisebox{-1ex}{$2$}\right.}\sqrt{\frac{{\widehat{p}}_1(1-{\widehat{p}}_1)}{n_1}+\frac{{\widehat{p}}_2(1-{\widehat{p}}_2)}{n_2}}\\ & =& 1.96\sqrt{\frac{0.70731(1-0.70731)}{41}+\frac{0.82926(1-0.82926)}{41}}\\ & =& 0.18073\end{array}$

Subtracting the margin of error from the point estimate and then adding the margin of error to the point estimate gives the following endpoints of the confidence interval

Lower end point:

$\begin{array}{rll}\left({\widehat{p}}_1-{\widehat{p}}_2\right)-E& =& -0.12195-0.18073\\ & =& -0.30268\end{array}$

Upper end point:

$\begin{array}{rll}\left({\widehat{p}}_1-{\widehat{p}}_2\right)+E& =& -0.12195+0.18073\\ & =& 0.05878\end{array}$

Thus, the 95% confidence interval for the difference between the two population ranges from -0.30268 to 0.05878. The confidence interval can be written mathematically using either inequality symbols or interval notation, as follows.

$-0.30268<p_1-p_2<0.05878$ or $(-0.30268,0.05878)$

Interpretation:

The true differences between the proportions of students who pass the test and play video games for two hours on the night before the test and those who pass the test but do not play games is between -0.30268 and 0.05878. Since the confidence interval contains 0, there is not sufficient evidence to indicate that playing video games negatively impacts students’ performance.