Chapter 10.2, Problem 56E

(a)

To determine

To explain why the sample results give evidence for the alternative hypothesis.

Explanation of Solution

It is given that:

  $\begin{array}{l}{\overline{x}}_1=15.833\\ {\overline{x}}_2=13.714\\ n_1=42\\ n_2=42\\ s_1=3.944\\ s_2=3.550\end{array}$

The given claim that: difference in the means.

Now, we have to find out the appropriate hypotheses for performing a significance test.

Thus, the claim is either the null hypothesis or the alternative hypothesis. The null hypothesis states that the population proportions are equal. If the null hypothesis is the claim then the alternative hypothesis states the opposite of the null hypothesis.

Therefore, the appropriate hypotheses for this is:

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

Where we have,

  ${\mu }_1=$ the true mean number of turns required to finish the memory game for students who listen to music.

  ${\mu }_2=$ the true mean number of turns required to finish the memory game for students who do not listen to music.

We then know that the sample means $15.833\text{ and 13}\text{.714}$ differ which agrees with the alternative hypothesis $H_a:{\mu }_1\ne {\mu }_2$ that the means are different and thus the sample results give evidence for the alternative hypothesis.

(b)

To determine

To calculate the standardized test statistics and P -value.

Answer to Problem 56E

The P -value is $0.01<P<0.02\text{ or }P=0.0134$ and the $t=2.585$ .

Explanation of Solution

It is given that:

  $\begin{array}{l}{\overline{x}}_1=15.833\\ {\overline{x}}_2=13.714\\ n_1=42\\ n_2=42\\ s_1=3.944\\ s_2=3.550\end{array}$

The given claim that: difference in the means.

Now, we have to find out the appropriate hypotheses for performing a significance test.

Thus, the claim is either the null hypothesis or the alternative hypothesis. The null hypothesis states that the population proportions are equal. If the null hypothesis is the claim then the alternative hypothesis states the opposite of the null hypothesis.

Therefore, the appropriate hypotheses for this is:

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

Where we have,

  ${\mu }_1=$ the true mean number of turns required to finish the memory game for students who listen to music.

  ${\mu }_2=$ the true mean number of turns required to finish the memory game for students who do not listen to music.

Now, find the test statistics:

  $\begin{array}{l}t=\frac{({\overline{x}}_1-{\overline{x}}_2)-({\mu }_1-{\mu }_2)}{\sqrt{\frac{s_1^2}{n_1}+\frac{s_2^2}{n_2}}}\\ =\frac{15.833-13.714-0}{\sqrt{\frac{{3.944}^2}{42}+\frac{{3.550}^2}{42}}}\\ =2.585\end{array}$

Now, the degree of freedom will be:

  $\begin{array}{l}df=\min (n_1-1,n_2-1)\\ =\min (42-1,42-1)\\ =41\end{array}$

Since student’s T distribution table in the appendix does not contains the value of $df=41$ so we will take the nearest value $df=40$ . So the P -value will be:

  $0.01=2(0.005)<P<2(0.01)=0.02$

On the other hand by using the calculator command: $2\times \text{tcdf(2}\text{.585,1E99,41)}$ which results in the P -values as $0.0134$ .

Thus, the P -value is $0.01<P<0.02\text{ or }P=0.0134$ and the $t=2.585$ .

(c)

To determine

To explain what conclusion would you make.

Explanation of Solution

From part (a) and (b), we have,

  $\begin{array}{l}{\overline{x}}_1=15.833\\ {\overline{x}}_2=13.714\\ n_1=42\\ n_2=42\\ s_1=3.944\\ s_2=3.550\end{array}$

The appropriate hypotheses for this is:

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

And the P -value is $0.01<P<0.02\text{ or }P=0.0134$ and the $t=2.585$ .

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

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

Thus, we conclude that there is convincing evidence that listening to music affects the average number of turns required to finish the memory game for students.