Chapter 24, Problem 36E

(a)

To determine

To explain does it appear that it is better to study while listening to Mozart than to rap music and test an appropriate hypothesis and state your conclusion.

Answer to Problem 36E

It is not better to study while listening to Mozart than to rap music.

Explanation of Solution

It is given that:

  $\begin{array}{l}{\overline{x}}_1=10\\ {\overline{x}}_2=10.72\\ n_1=20\\ n_2=29\\ s_1=3.19\\ s_2=3.99\end{array}$

Now, let us define the hypotheses for testing:

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

Thus, the value of test statistics is:

  $\begin{array}{l}t=\frac{{\overline{x}}_1-{\overline{x}}_2}{\sqrt{\frac{s_1^2}{n_1}+\frac{s_2^2}{n_2}}}\\ =\frac{10-10.72}{\sqrt{\frac{{3.19}^2}{20}+\frac{{3.99}^2}{29}}}\\ =-0.70\end{array}$

And the value of degree of freedom is:

  $\begin{array}{l}df=\frac{(\frac{s_1^2}{n_1}+\frac{s_2^2}{n_2})}{\frac{{(\raisebox{1ex}{$s_1^2$}\!\left/ \!\raisebox{-1ex}{$n_1$}\right.)}^2}{n_1-1}+\frac{{(\raisebox{1ex}{$s_2^2$}\!\left/ \!\raisebox{-1ex}{$n_2$}\right.)}^2}{n_2-1}}\\ =45\end{array}$

Thus, the P-value will be:

  $P>0.10$

And 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, there is not sufficient evidence to support the claim of an improvement. So, it is not better to study while listening to Mozart than to rap music.

(b)

To determine

To create a $90\%$ confidence interval for the mean difference in memory score between students who study to Mozart and those who listen to no music at all and interpret the interval.

Answer to Problem 36E

We are $90\%$ confidence that the mean memory score of students who study to Mozart is between $-5.3518\text{ and }-0.1882$ lower than the mean memory score of students who listen to no music at all.

Explanation of Solution

It is given that:

  $\begin{array}{l}{\overline{x}}_1=10\\ {\overline{x}}_2=12.77\\ n_1=20\\ n_2=13\\ s_1=3.19\\ s_2=4.73\end{array}$

Thus, the degree of freedom will be 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$}\right.)}^2}{n_2-1}}\\ =19\end{array}$

And the t -value will be:

  $t=1.729$

Thus, the confidence interval will be:

  $\begin{array}{l}({\overline{x}}_1-{\overline{x}}_2)-t_{\alpha /2}\times \sqrt{\frac{s_1^2}{n_1}+\frac{s_2^2}{n_2}}\\ =(10-12.77)-1.729\times \sqrt{\frac{{3.19}^2}{20}+\frac{{4.73}^2}{13}}\\ =-5.3518\\ ({\overline{x}}_1-{\overline{x}}_2)+t_{\alpha /2}\times \sqrt{\frac{s_1^2}{n_1}+\frac{s_2^2}{n_2}}\\ =(10-12.77)+1.729\times \sqrt{\frac{{3.19}^2}{20}+\frac{{4.73}^2}{13}}\\ =-0.1882\end{array}$

Thus, we conclude that we are $90\%$ confidence that the mean memory score of students who study to Mozart is between $-5.3518\text{ and }-0.1882$ lower than the mean memory score of students who listen to no music at all.