Q.91 - The Practice of Statistics for AP

Question

Does listening to music while studying help or hinder students’ learning? Two AP 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. A boxplot of the differences in the number of words recalled (music- silent) is shown below, along with some Minitab output from a one-sample t test. Perform a complete analysis of the students’ data. Include a confidence interval.

Step-by-Step Solution

Answer

The $95\%$ confidence interval is $(-1.561,-0.298)$.

Step 1: Given information

Given in the question that, Two AP 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. A boxplot of the differences in the number of words recalled (music -silent) is shown below, along with some Minitab output from a one-sample t test. Perform a complete analysis of the students’ data. Include a confidence interval

Step 2: Explanation

Given

The boxplot and output are

The formula to compute the t-confidence interval for population mean is:

$Cl=\overline{x}\pm t_a/2df\times \frac{s}{\sqrt{n}}$

Step 3:Alternative hypothesis

The null and the alternative hypothesis are: 

$$\begin{gathered} H_0:\mu =0 \\ {H}_a:\mu \ne 0 \end{gathered}$$

From the provided output, the $\mathrm{P}$-value is less than the significance level $(0.05)$. Thus, the null hypothesis is rejected and it implies that there is enough evidence to agree with the provided claim at $5\%$significance level.

Here, the degree of freedom is:

$$\begin{gathered} D f=n-1 \\ =30-1 \\ =29 \end{gathered}$$

The confidence interval using $\mathrm{Ti}-83$ output is:

Thus, the $95\%$ confidence interval is $(-1.561,-0.298)$