Question

At Rachel’s 11th birthday party, eight girls were timed to see how long (in seconds) they could hold their breath in a

relaxed position. After a two-minute rest, they timed themselves while jumping. The girls thought that the mean difference

between their jumping and relaxed times would be zero. Test their hypothesis.

Short answer

Since p- the value is greater than $\alpha =0.05,$we cannot reject the hypothesis $H_0$and we can conclude that it is insufficient

evidence to conclude that the average difference is not zero.

Step-by-step solution

Given Information 

We need to test the hypothesis

Explanation 

We can take $X_{1-}{X}_2$to represent the difference between jumping and relaxed time.

We want to test this hypothesis:

$$\begin{gathered} H_0:{\mu }_1={\mu }_2 \\ {H}_{1:}{\mu }_1\ne {\mu }_2 \end{gathered}$$

In order to test the hypothesis, we are using the Student's test.

Simplification 

We need to find the mean and stand deviation of both samples:

$$\begin{gathered} {\mu }_1=32.375,s_1=9.620477 \\ {\mu }_2=30.625,s_2=8.331309 \end{gathered}$$

The differences between jumping and relaxed times are given by:

5, 7, 2, 1, - 2, 2, 2 -3

The number of differences is $\alpha =1.75$:

$\underset{}{{\displaystyle \sum _{i-1}^8}}{(d_i-a)}^{2=}75.5$

Standard deviation differences are:

$8=\sqrt{\frac{1}{n-1}}\sum _{\mathcal{i}-1}^8{(d\mathcal{i}-a)}^2=\sqrt{\frac{75.5}{7}}=3.284$

The standard error is: $SE=\stackrel{8}{\sqrt{\mathcal{n}}}=1.161$

We can find the value of the test statistic is:

$t=\frac{{\mu }_1-{\mu }_2}{SE}=1.507$

Graphical Presentation 

Using the applet, we get that p-value is 0.17554: