Chapter 24, Problem 22E

To determine

To explain do these results provide evidence that the program is worthwhile if we consider these students representative of all students who might attend this summer school in other years.

Answer to Problem 22E

No, they do not provide evidence.

Explanation of Solution

In this question, we have to find out do these results provide evidence that the program is worthwhile if we consider these students representative of all students who might attend this summer school in other years. So, we have,

  $\begin{array}{l}{\overline{x}}_1=60.1667\\ {\overline{x}}_2=65.5\\ n_1=6\\ n_2=6\\ s_1=7.2778\\ s_2=8.5264\end{array}$

The mean and the standard deviation is calculated by using excel by going to the data tab and then to the analysis tab and then selecting the data analysis option. Then we go to the t -test option in the dialogue box that appeared on screen. Then we select the data and the result will come out.

Now, let us define the hypotheses for test as:

  $\begin{array}{l}{H}_0:{\mu }_1={\mu }_2\\ H_1:{\mu }_1<{\mu }_2\end{array}$

Thus, the value of test statistics is as:

  $\begin{array}{l}t=\frac{{\overline{x}}_1-{\overline{x}}_2}{\sqrt{\frac{s_1^2}{n_1}+\frac{s_2^2}{n_2}}}\\ =\frac{60.1667-65.5}{\sqrt{\frac{{7.2778}^2}{6}+\frac{{8.5264}^2}{6}}}\\ =-1.17\end{array}$

And the degree of freedom is as:

  $\begin{array}{l}df=\frac{(\frac{s_1^2}{n_1}+\frac{s_2^2}{n_2})}{\frac{{(\raisebox{1ex}{$s_1^2$}\!\left/ \!\raisebox{-1ex}{$n_1$}\right.)}^2}{n_1-1}+\frac{{(\raisebox{1ex}{$s_2^2$}\!\left/ \!\raisebox{-1ex}{$n_2$}\right.)}^2}{n_2-1}}\\ =9\end{array}$

And the P-value will be as:

  $P>0.10$

As we know that if the P-value is less than or equal to the significance level then the null hypothesis is rejected. So,

  $P>0.05\Rightarrow \text{Fail to Reject }{H}_0$

Thus, we conclude that there is not a sufficient evidence to support the claim of an improvement.