Chapter 10, Problem 7CRE

(a)

To determine

To write: the few sentences differentiating the performance of these two groups.

Explanation of Solution

Given:

  

The centre the treatment group is greater than the treatment group for the control group, the reason is that the median lies more to the right of the box of the box plot.

The spread for both groups looks to be about the same, because the width between the whiskers of the box plots are the approximately same for both box pots.

The distribution of the control group seeps to be right − skewed, where the distribution of the treatment group seems symmetric.

(b)

To determine

To construct: and explain a 90% confidence interval for the mean difference in gains between treatment and control.

Answer to Problem 7CRE

(0.0306, 6.2694)

Explanation of Solution

Given:

The mean is

  $\begin{array}{l}{\overline{x}}_1=11.4\\ {\overline{x}}_2=8.25\end{array}$

Formula used:

Lower and upper limit is

  $\begin{array}{l}\left({\overline{x}}_1-{\overline{x}}_2\right)-t_{\alpha /2}\times \sqrt{\frac{s_1^2}{n_1}+\frac{s_2^2}{n_2}}\\ \left({\overline{x}}_1-{\overline{x}}_2\right)+t_{\alpha /2}\times \sqrt{\frac{s_1^2}{n_1}+\frac{s_2^2}{n_2}}\end{array}$

Calculation:

The mean is the sum of all values divided by the number of values:

  $\begin{array}{l}{\overline{x}}_1=11.4\\ {\overline{x}}_2=8.25\end{array}$

The variance is

  $\begin{array}{l}{s}_1=3.1693\\ s_2=3.6936\end{array}$

  $c=90\%$

The degrees of freedom:

  $df=\min (n_1-1,n_2-1)=\min \left(10-1,8-1\right)=7$

Determine $t_c$ with $df=7$ using table B:

  $t_c=1.895$

The endpoints of the confidence interval for ${\mu }_1-{\mu }_2$ are:

  $\begin{array}{l}\left({\overline{x}}_1-{\overline{x}}_2\right)-t_{\alpha /2}\times \sqrt{\frac{s_1^2}{n_1}+\frac{s_2^2}{n_2}}\\ =\left(11.4-8.25\right)-1.895\times \sqrt{\frac{{3.1693}^2}{10}+\frac{{3.6936}^2}{8}}\\ =0.0306\end{array}$

  $\begin{array}{l}\left({\overline{x}}_1-{\overline{x}}_2\right)+t_{\alpha /2}\times \sqrt{\frac{s_1^2}{n_1}+\frac{s_2^2}{n_2}}\\ =\left(11.4-8.25\right)+1.895\times \sqrt{\frac{{3.1693}^2}{10}+\frac{{3.6936}^2}{8}}\\ =6.2694\end{array}$

There are 90% confident that the mean difference is between 0.0306 and 6.2694.

(c)

To determine

To Explain: reason for generalize these results to the population of all students who failed the mathematics.

Answer to Problem 7CRE

No

Explanation of Solution

No, because the students admit to participate in the study and therefore the sample is not a arbitrary sample. Although the arbitrary requirement is not satisfied, it is not possible to generalize to the all the population.