Question

Music and memory Does listening to music while studying help or hinder students’ learning? Two statistics students designed an experiment to find out. They selected a random sample of $30$students from their medium-sized high school to participate. Each subject was given $10$minutes to memorize two different lists of $20$words, once while listening to music and once in silence. The order of the two word lists was determined at random; so was the order of the treatments. The difference (Silence − Music) in the number of words recalled was recorded for each subject. The mean difference was $1.57$and the standard deviation of the differences was $2.70$.

a. If the result of this study is statistically significant, can you conclude that the difference in the ability to memorize words was caused by whether students were performing the task in silence or with music playing? Why or why not?

b. Do the data provide convincing evidence at the$\alpha =0.01$ significance level that the number of words recalled in silence or when listening to music differs, on average, for students at this school?

c. Based on your conclusion in part (a), which type of error—a Type I error or a Type II error—could you have made? Explain your answer.

Short answer

a. Yes we can conclude that the difference in the ability to memorize words was caused by whether students were performing the task in silence or with music playing.

b. Yes, there is convincing evidence at the $\alpha =0.01$significance level that the number of words recalled in silence or when listening to music differs, on average, for students at this school.

c. We have made a Type I error.

Step-by-step solution

Part (a): Step 1: Given information

We have been given that we have a random sample of $30$students and the mean difference was $1.57$and the standard deviation of the differences was $2.70$.

We need to find out whether the difference in the ability to memorize words was caused by whether students were performing the task in silence or with music playing.

Part (a): Step 2: Explanation

We note that the selected students are a random sample and the order of treatments was randomly assigned to the students and this implies that the result needs to be due to the treatments themselves, as the randomization ensures that the result cannot be due to other variables except the treatment.

Part (b): Step 1: Given information

We have been given that we have a random sample of $30$students and the mean difference was $1.57$and the standard deviation of the differences was $2.70$.

We need to find out do the data provide convincing evidence at the $\alpha =0.01$significance level that the number of words recalled in silence or when listening to music differs, on average, for students at this school.

Part (b): step 2: Explanation

Given:

$$\begin{gathered} n=Samplesize=30 \\ {\overline{x}}_d=Meandifferences=1.57 \\ {s}_d=S\tan darddeviationofdifferences=2.70 \\ \alpha =Significancelevel=0.01 \end{gathered}$$

Now we will carry out a hypothesis test for the population mean difference:

$$\begin{gathered} H_0=Nullhypothesis \\ (H{)}_a=Alternativehypothesis \\ {H}_0:{\mu }_d=0 \\ {H}_a:{\mu }_d\ne 0 \\ Nowwewilldeterminethevalueofteststatistics: \\ t=\frac{{\overline{x}}_d-{\mu }_d}{{\displaystyle \raisebox{1ex}{$s_d$}\!\left/ \!\raisebox{-1ex}{$\sqrt{n}$}\right.}} \\ t=\frac{1.57-0}{{\displaystyle \raisebox{1ex}{$2.70$}\!\left/ \!\raisebox{-1ex}{$\sqrt{30}$}\right.}}\approx 3.185 \end{gathered}$$

The P-value is the probability of obtaining the value of test statistics.

Degree of freedom $n-1=30-1=29$

The test is a two-tailed test so we double the boundaries of the P-value.

$$\begin{gathered} 0.002=2\left(0.001\right)<P<2\left(0.0025\right)=0.005 \\ nowatt=3.185anddegreeoffreedom=30-1=29,pvalue=0.00344 \end{gathered}$$

If P-value is less than the significance level then reject the null hypothesis

$P<0.01\Rightarrow Reject{H}_0$

Yes, there is convincing evidence at the $\alpha =0.01$significance level that the number of words recalled in silence or when listening to music differs, on average, for students at this school

Part (c): Step 1: Given information

We have been given that we have a random sample of $30$students and the mean difference was $1.57$and the standard deviation of the differences was $2.70$.

We need to find out that based on part a which type of error—a Type I error or a Type II error—could we have made.

Part (c): Step 2: Explanation

We could have only made a Type I error (if we made an error) because we have rejected the null hypothesis.