Chapter 8.2, Problem 19E

To determine

To test: Whether there is any sufficient evidence to infer that the mean number of calories burned by tennis players is less than 546 calories based on the p-value and test statistic value.

Answer to Problem 19E

There is sufficient evidence to infer that the mean number of calories burned by tennis players is less than 546 calories.

Explanation of Solution

Given info:

To examine the average number of calories burned per hour by the male tennis players, a random sample of 36 male tennis players were tested, the results of the number of calories based on the survey is $\overline{x}=544.8$. The additional information is about the population parameters $\mu =546$ $\sigma =3$ and level of significance $\alpha =0.01$.

Calculation:

The testing hypotheses are given below:

Null hypothesis $H_0$:

$H_0:\mu =546$

Alternative hypothesis $H_1$:

$H_1:\mu <546$

Requirements for z-test about mean when population standard deviation $\left(\sigma \right)$known:

• The sample must be drawn randomly from the desired population.
• The sample must be greater than or equal to 30, if not the population must be approximately normal.

Here, the sample size is greater than 30 and the sample is drawn randomly.

Thus, z-test about mean is valid in this case.

Decision rule based on classical approach:

If $z<-z_{\alpha }$, then reject the null hypothesis $H_0$.

Critical value:

For the level of significance,

$\alpha =0.01$

Hence, the cumulative area to the left is,

$\begin{array}{c}\text{Area}\text{to}\text{the}\text{left}=1-\text{Area}\text{to}\text{the}\text{right}\\ =1-0.01\\ =0.99\end{array}$

From Table E of the standard normal distribution from Appendix A, the critical value is 2.33.

Thus, the left tailed critical value of z for the level of significance $\alpha =0.01$ is $\left(-z_{\alpha }\right)=-2.33$.

Test statistic:

The test statistic for the large sample single mean test is,

$z=\frac{\left(\overline{x}-\mu \right)}{\frac{\sigma }{\sqrt{n}}}$

The test statistic is obtained as follows:

Here, $\overline{x}=544.8$ $n=36$ and $\mu =546$ $\sigma =3$

$\begin{array}{c}z=\frac{\left(\overline{x}-\mu \right)}{\frac{\sigma }{\sqrt{n}}}\\ =\frac{544.8-546}{\frac{3}{\sqrt{36}}}\\ =\frac{-1.2}{0.5}\\ =-2.4\end{array}$

Thus, the test statistic is $z=-2.4$.

Conclusion based on classical approach:

The test statistic value is $z=-2.4$ and the critical value is $\left(-z_{\alpha }\right)=-2.33$.

Here, test statistic value is less than the critical value.

That is, $-2.4\left(=z\right)<-2.33\left(=-z_{\alpha }\right)$

Hence, reject the null hypothesis $H_0$ by the rejection rule.

Thus, it can be concluded that there is sufficient evidence to infer that the mean number of calories burned by tennis players is less than 546 calories.

Decision rule based on classical approach:

If $p-value\le \alpha$, then reject the null hypothesis $H_0$.

If $p-value>\alpha$, then accept the null hypothesis $H_0$.

P-value:

The value of p can be obtained using the MINITAB software.

Statistical procedure:

The step by step procedure to find the p-value using the MINITAB software is given below:

• Select Graphs, then Click on View probability>click ok.
• Under Distribution, choose Normal.
• Under shaded area, select Define shaded area by X-value.
• Enter the X-value as -2.4.
• Select Left tail and click OK.

The output obtained by MINITAB is given below:

From the MINITAB output, it can be seen that the value of p is 0.0082.

Thus the value of p is $p=0.0082$.

Conclusion based on classical approach:

The p value is $p=0.0082$ and the level of significance is $\alpha =0.01$.

Here, the p value is less than the level of significance value.

That is, $0.0082\left(=p\right)<0.01\left(=\alpha \right)$

Hence, reject the null hypothesis $H_0$ by the rejection rule.

Thus, it can be concluded that there is sufficient evidence to infer that the mean number of calories burned by tennis players is less than 546 calories.