Chapter 7, Problem 134E

(a)

Section1:

To determine

To test: The null hypothesis that there is no significant difference in mean between Athletes and Sedentary students for body fat.

Answer to Problem 134E

Solution: The null hypothesis gets rejected. So, there is a significant difference in mean between Athletes and Sedentary students for body fat.

Explanation of Solution

Calculation: The null and alternative hypotheses are:

$\begin{array}{l}{H}_0:{\mu }_1={\mu }_2\\ H_a:{\mu }_1\ne {\mu }_2\end{array}$

The t-statistic is used to test the significance of the difference in mean. The pooled standard deviation is used for two standard deviation as ${\sigma }_{\max }<2\times {\sigma }_{\min }$, which is obtained as follow:

$\begin{array}{c}\text{pooled standard deviation=}\sqrt{\frac{\left(n_1-1\right)\times s_1^2+\left(n_2-1\right)\times s_2^2}{n_1+n_2-2}}\\ =\sqrt{\frac{79\times {5.54}^2+79\times {8.05}^2}{80+80-2}}\\ =6.90992\end{array}$

Degree of freedom for the two samples can be calculated by following formula:

$\begin{array}{c}\text{D}{\text{.F = n}}_1+{\text{n}}_2-2\\ \text{D}\text{.F = 80+80}-2\\ \text{D}\text{.F = }158\end{array}$

Now the t-value can be obtained as follow:

$\begin{array}{l}t=\frac{\left({\overline{x}}_1-{\overline{x}}_2\right)}{s_p\sqrt{\frac{1}{n_1}+\frac{1}{n_2}}}\\ =\frac{\left(25.61-32.51\right)}{6.90992\sqrt{\frac{1}{80}+\frac{1}{80}}}\\ =\frac{-6.9}{1.092554}\\ =-6.3155\end{array}$

The P-value can be calculated by using the function $\text{=T}\text{.DIST}\text{.RT(}x,\text{ deg\_freedom, tails)}$ in Excel. The screenshot is given below:

Conclusion: The P-value is less than $0.05$. Hence the null hypothesis gets rejected, which states that there is a significant difference in mean between Athletes and Sedentary students for body fat.

Section 2:

To determine

To test: The null hypothesis that there is no difference in mean between Athletes and Sedentary students for body mass index.

Answer to Problem 134E

Solution: The null hypothesis gets rejected. So, there is a significant difference in mean between Athletes and Sedentary students for body mass index.

Explanation of Solution

Calculation: The null and alternative hypotheses are:

$\begin{array}{l}{H}_0:{\mu }_1={\mu }_2\\ H_a:{\mu }_1\ne {\mu }_2\end{array}$

The t-statistic is used to test the significance of the difference in mean. The pooled standard deviation is used for two standard deviation as ${\sigma }_{\max }<2\times {\sigma }_{\min }$, which is obtained as follow:

$\begin{array}{c}\text{pooled standard deviation=}\sqrt{\frac{\left(n_1-1\right)\times s_1^2+\left(n_2-1\right)\times s_2^2}{n_1+n_2-2}}\\ =\sqrt{\frac{79\times {2.46}^2+79\times {2.73}^2}{80+80-2}}\\ =2.59851\end{array}$

Degree of freedom for the two samples can be calculated by following formula:

$\begin{array}{c}\text{D}{\text{.F = n}}_1+{\text{n}}_2-2\\ \text{D}\text{.F = 80+80}-2\\ \text{D}\text{.F = }158\end{array}$

The t-value is obtained as follow:

$\begin{array}{l}t=\frac{\left({\overline{x}}_1-{\overline{x}}_2\right)}{s_p\sqrt{\frac{1}{n_1}+\frac{1}{n_2}}}\\ =\frac{\left(21.6-26.41\right)}{2.59851\sqrt{\frac{1}{80}+\frac{1}{80}}}\\ =-11.7071\end{array}$

The P-value can be calculated by using the function $\text{=T}\text{.DIST}\text{.RT(}x,\text{ deg\_freedom, tails)}$ in Excel. The screenshot is given below:

Conclusion: The P-value is less than $0.05$. Hence the null hypothesis gets rejected, which states that there is a significant difference in mean between Athletes and Sedentary students for body mass index.

Section 3:

To determine

To test: The null hypothesis that there is no significant difference in mean between Athletes and Sedentary students for calcium deficit.

Answer to Problem 134E

Solution: The null hypothesis gets rejected. So, there is a difference in mean between Athletes and Sedentary students for calcium deficit.

Explanation of Solution

Calculation: The null and alternative hypotheses are:

$\begin{array}{l}{H}_0:{\mu }_1={\mu }_2\\ H_a:{\mu }_1\ne {\mu }_2\end{array}$

The t-statistic is used to test the significance of the difference in mean. The pooled standard deviation is used for two standard deviation as ${\sigma }_{\max }<2\times {\sigma }_{\min }$, which is obtained as follow:

$\begin{array}{c}\text{pooled standard deviation=}\sqrt{\frac{\left(n_1-1\right)\times s_1^2+\left(n_2-1\right)\times s_2^2}{n_1+n_2-2}}\\ =\sqrt{\frac{79\times {516.63}^2+79\times {372.77}^2}{80+80-2}}\\ =450.480\end{array}$

Degree of freedom for the two samples can be calculated by following formula:

$\begin{array}{c}\text{D}{\text{.F = n}}_1+{\text{n}}_2-2\\ \text{D}\text{.F = 80+80}-2\\ \text{D}\text{.F = }158\end{array}$

The t-value is obtained as follow:

$\begin{array}{l}t=\frac{\left({\overline{x}}_1-{\overline{x}}_2\right)}{s_p\sqrt{\frac{1}{n_1}+\frac{1}{n_2}}}\\ =\frac{\left(297.13-580.54\right)}{450.480\sqrt{\frac{1}{80}+\frac{1}{80}}}\\ =-3.979\end{array}$

The P-value can be calculated by using the function $\text{=T}\text{.DIST}\text{.RT(}x,\text{ deg\_freedom, tails)}$ in Excel. The screenshot is given below:

Conclusion: The P-value is less than $0.05$. Hence the null hypothesis gets rejected, which states that there is a significant difference in mean between Athletes and Sedentary students for calcium deficit.

Section 4:

To determine

To test: The null hypothesis that there is no significant difference in mean between Athletes and Sedentary students for glasses of milk/day.

Answer to Problem 134E

Solution: The null hypothesis does not get rejected. So, there is no difference in mean between Athletes and Sedentary students for glasses of milk/day.

Explanation of Solution

Calculation: The null and alternative hypotheses are:

$\begin{array}{l}{H}_0:{\mu }_1={\mu }_2\\ H_a:{\mu }_1\ne {\mu }_2\end{array}$

The t-statistic is used to test the significance of the difference in mean. The pooled standard deviation is used for two standard deviation as ${\sigma }_{\max }<2\times {\sigma }_{\min }$, which is obtained as follow:

$\begin{array}{c}\text{pooled standard deviation=}\sqrt{\frac{\left(n_1-1\right)\times s_1^2+\left(n_2-1\right)\times s_2^2}{n_1+n_2-2}}\\ =\sqrt{\frac{79\times {1.46}^2+79\times {1.24}^2}{80+80-2}}\\ =1.35447\end{array}$

Degree of freedom for the two samples can be calculated by following formula:

$\begin{array}{c}\text{D}{\text{.F = n}}_1+{\text{n}}_2-2\\ \text{D}\text{.F = 80+80}-2\\ \text{D}\text{.F = }158\end{array}$

The t-value is obtained as follow:

$\begin{array}{l}t=\frac{\left({\overline{x}}_1-{\overline{x}}_2\right)}{s_p\sqrt{\frac{1}{n_1}+\frac{1}{n_2}}}\\ =\frac{\left(2.21-1.82\right)}{1.35447\sqrt{\frac{1}{80}+\frac{1}{80}}}\\ =1.821\end{array}$

The P-value can be calculated by using the function $\text{=T}\text{.DIST}\text{.RT(}x,\text{ deg\_freedom, tails)}$ in Excel. The screenshot is given below:

Conclusion: The P-value is greater than 0.05. So, the null hypothesis does not get rejected, which states that there is no significant difference in mean between Athletes and Sedentary students for glasses of milk/day.

(b)

To determine

The summary of the hypothesis test results.

Answer to Problem 134E

Solution: The null hypotheses get rejected for all the characteristic except the Glasses of milk/day.

Explanation of Solution

With reference to part (a), the P-values are:

$\begin{array}{l}P{\text{-value}}_{\text{Body fat}}=0<0.05\\ P{\text{-value}}_{\text{Calcium deficit}}=0<0.05\\ P{\text{-value}}_{\text{Body mass index}}=0<0.05\\ P{\text{-value}}_{\text{Glasses of milk/day}}=0.0705>0.05\end{array}$

The P-value are less than $0.05$ for all the characteristic except for Glasses of milk/day. The null hypotheses are rejected for all except for Glasses of milk/day.