Chapter 24, Problem 27E

To determine

To test an appropriate hypothesis and state your conclusions.

Answer to Problem 27E

There is sufficient evidence to support the claim of a difference.

Explanation of Solution

It is given that:

  $\begin{array}{l}{\overline{x}}_1=5.07\\ {\overline{x}}_2=5.68\\ n_1=26\\ n_2=22\\ s_1=1.84\\ s_2=2.91\end{array}$

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 testing:

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

Thus, the value of test statistics is:

  $\begin{array}{l}t=\frac{{\overline{x}}_1-{\overline{x}}_2}{\sqrt{\frac{s_1^2}{n_1}+\frac{s_2^2}{n_2}}}\\ =\frac{5.07-5.68}{\sqrt{\frac{{1.84}^2}{26}+\frac{{2.91}^2}{22}}}\\ =-2.10\end{array}$

And the degree of freedom will be:

  $\begin{array}{l}df=\frac{(\frac{s_1^2}{n_1}+\frac{s_2^2}{n_2})}{\frac{{(\raisebox{1ex}{$s_1^2$}\!\left/ \!\raisebox{-1ex}{$n_1$}\right.)}^2}{n_1-1}+\frac{{(\raisebox{1ex}{$s_2^2$}\!\left/ \!\raisebox{-1ex}{$n_2$}\right.)}^2}{n_2-1}}\\ =34>32\end{array}$

Thus, the P-value will be:

  $0.02<P<0.05$

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

  $P<0.05\Rightarrow \text{Reject }{H}_0$

Thus, we conclude that there is sufficient evidence to support the claim of a difference.