91Ó°ÊÓ

$$\begin{gathered} {\stackrel{¯}{x}}_1=15.833 \\ {\stackrel{¯}{x}}_2=13.714 \\ {n}_1=42 \\ {n}_2=42 \\ {s}_1=3.944 \\ {s}_2=3.550 \end{gathered}$$

The given claim is that a difference in the means.

Now we must determine the most appropriate hypotheses for a significance test.

As a result, either the null hypothesis or the alternative hypothesis is the claim. According to the null hypothesis, the population proportions are equal. If the claim is the null hypothesis, the alternative hypothesis is the polar opposite of the null hypothesis.

The appropriate hypotheses for this are:

$$\begin{gathered} H_0:{\mathrm{μ}}_1={\mathrm{μ}}_2 \\ {H}_0:{\mathrm{μ}}_{1}notequalto{\mathrm{μ}}_2 \end{gathered}$$

For students who listen to music, ${\mathrm{μ}}_1=$the true mean number of turns is required to complete the memory game.

For students who do not listen to music, ${\mathrm{μ}}_2$is the true mean number of turns required to complete the memory game.

We then know that the sample means $15.833$and $13.714$difference which agrees with the alternative hypothesis $H_a:{\mathrm{μ}}_1≠{\mathrm{μ}}_2$that the means are different and thus the sample results give evidence for the alternative hypothesis.

From part (a)

We have,

$$\begin{gathered} H_0:{\mathrm{μ}}_1={\mathrm{μ}}_2 \\ {H}_0:{\mathrm{μ}}_{1}notequalto{\mathrm{μ}}_2 \end{gathered}$$

Now, find the test statistics:

$$\begin{gathered} t=\frac{\left({\stackrel{¯}{x}}_1-{\stackrel{¯}{x}}_2\right)-\left({\mathrm{μ}}_1-{\mathrm{μ}}_2\right)}{\sqrt{\frac{{\text{1}}_1^2}{n_1}+\frac{{\frac{2}{2}}_2^n}{n_2}}} \\ =\frac{15.833-13.714-0}{\sqrt{\frac{3.{944}^2}{42}+\frac{{3550}^2}{42}}} \\ =2.585 \end{gathered}$$

The degree of liberty will now be:

$$\begin{gathered} df=min(n_1-1,n_2-1) \\ =min(42-1.42-1) \\ =41 \end{gathered}$$

Hence, the 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: $2×tcdf(2.585,1\mathrm{E}99,41)$which results in the $P$-values as: $0.0134$

Therefore the $P$-value is $0.01<P<0.02$or $P=0.0134$and the$t=2.585$

From part (a) and part (b)

We have,

The $P$-value is $0.01<P<0.02orP=0.0134$and the $t=2.585$

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

$P<0.05⇒\text{Reject}{H}_0$

Therefore, We conclude that listening to music has a significant impact on the average number of turns required to complete the memory game for students.