Chapter 7.2, Problem 58E

Section 1:

To determine

To test: The significance test to test whether there is a difference in sprint speed of Elite players from the Canadian National team and a university squad.

Answer to Problem 58E

Solution: There is a significant difference between the sprint speed of both types of players. The t -statistic is 2.89 which lies between $P=0.01\text{ and }P=0.02$ for 12 degrees of freedom.

Explanation of Solution

Calculation: The hypothesis is considered as that there is no difference in sprint speed of elite players and a university squad against the alternative that there is difference in the sprint speeds of the elite players and the university squad. Hence, the hypotheses are formulated as:

$\begin{array}{l}{H}_0:{\mu }_E={\mu }_U\\ H_a:{\mu }_E\ne {\mu }_U\end{array}$

The two-sample t- test statistic is defined as:

$t=\frac{\left(\overline{x_1}-\overline{x_2}\right)-\left({\mu }_1-{\mu }_2\right)}{\sqrt{\frac{s_1^2}{n_1}+\frac{s_2^2}{n_2}}}$

Where,

$\begin{array}{c}{\overline{x}}_1=\text{Mean of the treatment group}\\ {\overline{x}}_2=\text{Mean of the control group}\\ s_1=\text{Standard deviation of treatment group}\\ s_2=\text{Standard deviation of control group}\end{array}$

$\begin{array}{c}{n}_1=\text{Size of treatment group}\\ n_2=\text{Size of control group}\\ {\mu }_1-{\mu }_2=\text{Difference of means}\end{array}$

The difference of means is considered as 0 points as the null hypothesis states that there is no difference between the two sets of players. Substitute the provided values in the above-defined formula to compute the two sample t statistic. So,

$\begin{array}{c}t=\frac{\left(\overline{x_1}-\overline{x_2}\right)-\left({\mu }_1-{\mu }_2\right)}{\sqrt{\frac{s_1^2}{n_1}+\frac{s_2^2}{n_2}}}\\ =\frac{\left(27.3-26\right)-0}{\sqrt{\frac{{0.7}^2}{16}+\frac{{1.5}^2}{13}}}\\ =\frac{1.3}{0.45}\\ =2.89\end{array}$

The P-value for the provided one-sided test is $P\left(T\ge 2.89\right)$. For the second approximation, the degrees of freedom k is the smaller of $n_1-1\text{ and }{n}_2-1$.

$\begin{array}{c}{n}_1-1=16-1\\ =15\end{array}$

$\begin{array}{c}{n}_2-1=13-1\\ =12\end{array}$

So, the degree of freedom is 12. Compare the obtained value of t-statistic with the values in the table D for 12 degrees of freedom. The table D shows that the value of $t=2.89$ lies between $P=0.005\text{ and }P=0.01$. As the alternative is two-sided, double the tail area p. Hence, the value $t=2.89$ lies between $P=0.01\text{ and }P=0.02$.

To determine

To explain: The conclusion of the performed significance test.

Answer to Problem 58E

Solution: The data strongly suggests that there is a significant difference between the sprint speed of the elite players of Canadian National team and a university squad.

Explanation of Solution

The t-test statistic is obtained as 2.89 in the previous part. For 12 degrees of freedom, the Table-D shows that the P lies between 0.01 and 0.02. This shows that $P<0.05$. So, the null hypothesis is rejected. Hence, it can be concluded that the data strongly suggest that there is a significant difference between the sprint speeds of both the types of players.

Section 2:

To determine

To test: The Significance test to test whether there is difference in peak heart rate of Elite players from the Canadian National team and the university squad.

Answer to Problem 58E

Solution: The data strongly suggests that there is no significant difference between the peak heart rate of the elite players of Canadian National team and a university squad. The t-statistic is obtained as $t=-0.45$ with $P-\text{value}=0.661$ for 12 degrees of freedom.

Explanation of Solution

Calculation: The hypothesis is considered as that there is no difference in peak heart rate of elite players and a university squad against the alternative that there is difference in peak heart rate of the elite players and university squad. Hence, the hypotheses are formulated as:

$\begin{array}{l}{H}_0:{\mu }_E={\mu }_U\\ H_a:{\mu }_E\ne {\mu }_U\end{array}$

The two-sample t- test statistic is defined as:

$t=\frac{\left(\overline{x_1}-\overline{x_2}\right)-\left({\mu }_1-{\mu }_2\right)}{\sqrt{\frac{s_1^2}{n_1}+\frac{s_2^2}{n_2}}}$

Where,

$\begin{array}{c}{\overline{x}}_1=\text{Mean of the treatment group}\\ {\overline{x}}_2=\text{Mean of the control group}\\ s_1=\text{Standard deviation of treatment group}\\ s_2=\text{Standard deviation of control group}\end{array}$

$\begin{array}{c}{n}_1=\text{Size of treatment group}\\ n_2=\text{Size of control group}\\ {\mu }_1-{\mu }_2=\text{Difference of means}\end{array}$

The difference of means is considered as 0 points as the null hypothesis states that there is no difference between the two sets of players. Substitute the provided values in the above-defined formula to compute the two sample t statistic. So,

$\begin{array}{c}t=\frac{\left(\overline{x_1}-\overline{x_2}\right)-\left({\mu }_1-{\mu }_2\right)}{\sqrt{\frac{s_1^2}{n_1}+\frac{s_2^2}{n_2}}}\\ =\frac{\left(192.0-193.0\right)-0}{\sqrt{\frac{6^2}{16}+\frac{6^2}{13}}}\\ =\frac{-1}{2.24}\\ =-0.45\end{array}$

The P-value for the provided one-sided test is $P\left(T\ge -0.45\right)$. For the second approximation, the degrees of freedom k is the smaller of $n_1-1\text{ and }{n}_2-1$.

$\begin{array}{c}{n}_1-1=16-1\\ =15\end{array}$

$\begin{array}{c}{n}_2-1=13-1\\ =12\end{array}$

So, the degree of freedom is 12. Compare the obtained value of t- statistic with the values in the table D for 12 degrees of freedom. The table D shows that the value of $t=-0.45$ lies beyond 0.25. Use the formula =TDIST (t, d.f, tails) to determine the exact P- value in Excel software. The screenshot is provided below:

Hence, the P-value is obtained as 0.661.

To determine

To explain: The conclusion of the performed significance test.

Answer to Problem 58E

Solution: The data strongly suggests that there is no significant difference between the peak heart rate of the elite players of Canadian National team and a university squad.

Explanation of Solution

The t-test statistic is obtained as -0.45 in the previous part. For 12 degrees of freedom, the P-value is obtained as 0.661 by using Excel in the previous part. This shows that $P>0.05$. So, the null hypothesis is not rejected. Hence, it can be concluded that the data strongly suggest that there is no significant difference between the peak heart rate of both types of players.

Section 3:

To determine

To test: Significance test to test whether there is difference in intermittent recovery test of Elite players from the Canadian National team and university squad.

Answer to Problem 58E

Solution: The data strongly suggests that there is a significant difference between the intermittent recovery test of the elite players of Canadian National team and university squad. The t- test statistic is obtained as $t=6.35$ with $P-\text{value}=0.000$ for 12 degrees of freedom.

Explanation of Solution

Calculation: The hypothesis is considered as that there is no difference in intermittent recovery test of elite players and a university squad against the alternative that there is difference in intermittent recovery test of the elite players and university squad. Hence, the hypotheses are formulated as:

$\begin{array}{l}{H}_0:{\mu }_E={\mu }_U\\ H_a:{\mu }_E\ne {\mu }_U\end{array}$

The two-sample t- test statistic is defined as:

$t=\frac{\left(\overline{x_1}-\overline{x_2}\right)-\left({\mu }_1-{\mu }_2\right)}{\sqrt{\frac{s_1^2}{n_1}+\frac{s_2^2}{n_2}}}$

Where

$\begin{array}{c}{\overline{x}}_1=\text{Mean of the treatment group}\\ {\overline{x}}_2=\text{Mean of the control group}\\ s_1=\text{Standard deviation of treatment group}\\ s_2=\text{Standard deviation of control group}\end{array}$

$\begin{array}{c}{n}_1=\text{Size of treatment group}\\ n_2=\text{Size of control group}\\ {\mu }_1-{\mu }_2=\text{Difference of means}\end{array}$

The difference of means is considered as 0 points as the null hypothesis states that there is no difference between the two sets of players. Substitute the provided values in the above-defined formula to compute the two sample t-statistic. So,

$\begin{array}{c}t=\frac{\left(\overline{x_1}-\overline{x_2}\right)-\left({\mu }_1-{\mu }_2\right)}{\sqrt{\frac{s_1^2}{n_1}+\frac{s_2^2}{n_2}}}\\ =\frac{\left(1160-781\right)-0}{\sqrt{\frac{{191}^2}{16}+\frac{{129}^2}{13}}}\\ =\frac{379}{59.67}\\ =6.35\end{array}$

For the second approximation, the degrees of freedom k is the smaller of $n_1-1\text{ and }{n}_2-1$.

$\begin{array}{c}{n}_1-1=16-1\\ =15\end{array}$

$\begin{array}{c}{n}_2-1=13-1\\ =12\end{array}$

So, the degree of freedom is 12. Compare the obtained value of t -statistic with the values in the table D for 12 degrees of freedom. The table D shows that the value of $t=6.35$ lies beyond the table. Use the formula =TDIST (t, d.f, tails) to determine the exact P- value in Excel software. The screenshot is provided below:

Therefore, the P-value is obtained as 0.000.

To determine

To explain: The conclusion of the performed significance test.

Answer to Problem 58E

Solution: The data strongly suggests that there is a significant difference between the intermittent recovery test of the elite players of Canadian National team and university squad.

Explanation of Solution

The t-test statistic is obtained as 6.35 in the previous part. For 12 degrees of freedom, the P-value is obtained as 0.000 by using Excel in the previous part. This shows that $P<0.05$. So, the null hypothesis is rejected. Hence, it can be concluded that the data strongly suggests that there is a significant difference between the intermittent recovery tests of both the types of players.