Chapter 7.2, Problem 76E

(b)

To determine

The null and alternative hypotheses for the consumption of proteins.

Answer to Problem 76E

Solution: The null and alternative hypotheses are formulated as

$\begin{array}{c}{H}_0:{\mu }_{\text{Early}}={\mu }_{\text{Late}}\\ H_a:{\mu }_{\text{Early}}\ne {\mu }_{\text{Late}}\end{array}$

Explanation of Solution

The significance test is to compare the two groups in terms of consumption of proteins. Hence, the hypothesis is that the consumption of proteins in the two groups is the same against the alternative that the consumption of proteins in the two groups is not the same.

Therefore, the hypotheses are formulated as:

$\begin{array}{l}{H}_0:{\mu }_{\text{Early}}={\mu }_{\text{Late}}\\ H_a:{\mu }_{\text{Early}}\ne {\mu }_{\text{Late}}\end{array}$

In the above hypotheses ${\mu }_{\text{Early}}$ represents the average protein consumption of early eaters and ${\mu }_{\text{Late}}$ represents the average protein consumption of late eaters.

To determine

To find: The null and alternative hypotheses for the consumption of carbohydrates.

Answer to Problem 76E

Solution: The null and alternative hypotheses are formulated as

$\begin{array}{l}{H}_0:{\mu }_{\text{Early}}={\mu }_{\text{Late}}\\ H_a:{\mu }_{\text{Early}}\ne {\mu }_{\text{Late}}\end{array}$

Explanation of Solution

Calculation: The significance test is to compare the two groups in terms of consumption of carbohydrates. Hence, the hypothesis is that the consumption of carbohydrates in the two groups is the same against the alternative that the consumption of carbohydrates in the two groups is not the same.

Therefore, the hypotheses are formulated as

$\begin{array}{l}{H}_0:{\mu }_{\text{Early}}={\mu }_{\text{Late}}\\ H_a:{\mu }_{\text{Early}}\ne {\mu }_{\text{Late}}\end{array}$

In the above hypothesis ${\mu }_{\text{Early}}$ represents the average carbohydrate consumption of early eaters and ${\mu }_{\text{Late}}$ represents the average carbohydrate consumption of late eaters.

(c)

To determine

To find: The test statistic with the degrees of freedom and the p-value for consumption of proteins.

Answer to Problem 76E

Solution: The t – test statistic is obtained as 2.47 and p – value as 0.0144 for 199 degrees of freedom for consumption of proteins.

Explanation of Solution

Calculation: The two-sample t – test statistic for the hypothesis formulated in part (a) is defined as:

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

Where,

$\begin{array}{c}{\overline{x}}_1=\text{Mean of the protein consumption by early eaters}\\ {\overline{x}}_2=\text{Mean of the protein consumption by late eaters}\\ s_1=\text{Standard deviation of protein consumption by early eaters}\\ s_2=\text{Standard deviation of protein consumption by late eaters}\end{array}$

$\begin{array}{c}{n}_1=\text{Size of protein consumption by early eaters}\\ n_2=\text{Size of protein consumption by late eaters}\\ {\mu }_1-{\mu }_2=\text{Difference of means}\end{array}$

The difference of means is considered as 0 in the null hypothesis. Substitute the provided values in the above defined formula to compute the two sample t statistic. So,

$\begin{array}{c}t=\frac{\left(27.6-25.7\right)-0}{\sqrt{\frac{{8.6}^2}{202}+\frac{{6.8}^2}{200}}}\\ =\frac{1.9}{0.77}\\ =2.47\end{array}$

The p-value for the provided one–sided test is calculated as $P\left(T\ge 2.47\right)$. For the second approximation, the degrees of freedom k is the smaller of $n_1-1\text{ and }{n}_2-1$.

$\begin{array}{c}{n}_1-1=202-1\\ =201\end{array}$

$\begin{array}{c}{n}_2-1=200-1\\ =199\end{array}$

So, the degrees of freedom are 199. The Excel function to determine the p- value from t-test statistic is displayed in the attached screenshot,

Therefore, the p-value is obtained as 0.0144.

To determine

To explain: The conclusion of the performed significance test.

Answer to Problem 76E

Solution: There is significant difference between the early eaters and late eaters in terms of consumption of proteins.

Explanation of Solution

The t-test statistic is obtained as 2.47 in the previous part. For 199 degrees of freedom, p-value is obtained as 0.0144 using Excel. Since, the p-value is less than 0.05 so the null hypothesis will be rejected at 5% significance level and it is concluded that there is a significant difference between the two groups in terms of proteins consumption.

To determine

To find: The test statistic with the degrees of freedom and the p-value for the consumption of carbohydrates.

Answer to Problem 76E

Solution: The t – test statistic is obtained as 0.29 and p – value as 0.772 for 199 degrees of freedom for carbohydrates consumption.

Explanation of Solution

Calculation: The two-sample t – test statistic for the hypothesis formulated in part (a) is defined as:

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

Where,

$\begin{array}{c}{\overline{x}}_1=\text{Mean of the carbohydrate consumption by early eaters}\\ {\overline{x}}_2=\text{Mean of the carbohydrateconsumption by late eaters}\\ s_1=\text{Standard deviation of carbohydrate consumption by early eaters}\\ s_2=\text{Standard deviation of carbohydrate consumption by late eaters}\end{array}$

$\begin{array}{c}{n}_1=\text{Size of carbohydrate consumption by early eaters}\\ n_2=\text{Size of carbohydrate consumption by late eaters}\\ {\mu }_1-{\mu }_2=\text{Difference of means}\end{array}$

The difference of true means is considered as 0 according to the null hypothesis. Substitute the provided values in the above defined formula to compute the two sample t statistic. So,

$\begin{array}{c}t=\frac{\left(64.1-63.5\right)-0}{\sqrt{\frac{{21.0}^2}{202}+\frac{{20.8}^2}{200}}}\\ =\frac{0.6}{2.08}\\ =0.29\end{array}$

The p-value for the provided one-sided test is calculated as $P\left(T\ge 0.29\right)$. For the second approximation, the degrees of freedom k is the smaller of $n_1-1\text{ and }{n}_2-1$.

$\begin{array}{c}{n}_1-1=202-1\\ =201\end{array}$

$\begin{array}{c}{n}_2-1=200-1\\ =199\end{array}$

So, the degrees of freedom are 199. The Excel function to determine the p- value from t-test statistic is displayed in the attached screenshot,

Therefore, the p-value is obtained as 0.772.

To determine

The conclusion of the performed significance test.

Answer to Problem 76E

Solution: There is no significant difference between the early eaters and late eaters in terms of consumption of carbohydrates.

Explanation of Solution

The t-test statistic is obtained as 0.29 in the previous part. For 199 degrees of freedom, p-value is obtained as 0.772 using Excel. Since the p-value is more than 0.05, the null hypothesis will not be rejected at 5% significance level and it is concluded that there is no significant difference between the two groups in terms of carbohydrates consumption.

(d)

To determine

To find: A 95% confidence interval for the difference of means between early eaters and late eaters in terms of consumption of proteins.

Answer to Problem 76E

Solution: A required 95% confidence interval is $\underset{\_}{\left(0.38,3.42\right)}$.

Explanation of Solution

Calculation: The formula for confidence interval for the difference between the means is defined as:

$\text{Confidence interval}=\left({\overline{x}}_1-{\overline{x}}_2\right)\pm t^{*}\sqrt{\frac{s_1^2}{n_1}+\frac{s_2^2}{n_2}}$

where

$\begin{array}{c}{\overline{x}}_1=\text{Mean of the protein consumption by early eaters}\\ {\overline{x}}_2=\text{Mean of the protein consumption by late eaters}\\ s_1=\text{Standard deviation of protein consumption by early eaters}\\ s_2=\text{Standard deviation of protein consumption by late eaters}\end{array}$

$\begin{array}{c}{n}_1=\text{Size of protein consumption by early eaters}\\ n_2=\text{Size of protein consumption by late eaters}\\ t^{*}=t\text{ value for the provided confidence level}\end{array}$

According to the Table D provided in the Appendix, the critical value for a two-tailed test at 95% confidence level is 1.962 for 199 degrees of freedom. Substitute the provided values in the above-defined formula to determine the 95% confidence interval for the difference between the early eaters and late eaters in terms of consumption of proteins. So,

$\begin{array}{c}\left({\overline{x}}_1-{\overline{x}}_2\right)\pm t^{*}\sqrt{\frac{s_1^2}{n_1}+\frac{s_2^2}{n_2}}=\left(27.6-25.7\right)\pm \left[\left(1.962\right)\times \sqrt{\frac{{8.6}^2}{202}+\frac{{6.8}^2}{200}}\right]\\ =1.9\pm \left[\left(1.962\right)\times \sqrt{0.366+0.231}\right]\\ =\left(1.9\pm 1.52\right)\\ =\left(1.9-1.52,1.9+1.52\right)\end{array}$

$=\left(0.38,3.42\right)$

Therefore, the 95% confidence interval for the difference between the means is obtained as $\left(0.38,3.42\right)$.

To determine

To explain: The comparison of information from the obtained confidence interval with the information given by the significance test.

Answer to Problem 76E

Solution: The information provided by both confidence interval and the significance test shows that there is a significant difference between the two means of proteins.

Explanation of Solution

The significance test conducted in part (c) shows that the null hypothesis is rejected. The obtained confidence interval $\left(0.38,3.42\right)$ shows that population mean difference of zero does not lie in this interval. So, reject the null hypothesis $H_0$. The information obtained from confidence interval and the significance test shows that the null hypothesis is rejected and concludes that there is a significant difference between the two means of proteins.

To determine

To find: A 95% confidence interval for the difference of means between early eaters and late eaters in terms of consumption of carbohydrates.

Answer to Problem 76E

Solution: A required confidence interval is $\underset{\_}{\left(-3.49,4.69\right)}$.

Explanation of Solution

Calculation: The formula for confidence interval for the difference between the means is defined as:

$\text{Confidence interval}=\left({\overline{x}}_1-{\overline{x}}_2\right)\pm t^{*}\sqrt{\frac{s_1^2}{n_1}+\frac{s_2^2}{n_2}}$

where

$\begin{array}{c}{\overline{x}}_1=\text{Mean of the carbohydrate consumption by early eaters}\\ {\overline{x}}_2=\text{Mean of the carbohydrate consumption by late eaters}\\ s_1=\text{Standard deviation of carbohydrate consumption by early eaters}\\ s_2=\text{Standard deviation of carbohydrate consumption by late eaters}\end{array}$

$\begin{array}{c}{n}_1=\text{Size of Carbohydrate consumption by early eaters}\\ n_2=\text{Size of Carbohydrate consumption by late eaters}\\ t^{*}=t\text{ value for the provided confidence level}\end{array}$

According to the Table D provided in the Appendix, the critical value for a two-tailed test at 95% confidence level is 1.962 for 199 degrees of freedom. Substitute the provided values in the above defined formula to determine the 95% confidence interval for the difference between the early eaters and late eaters in terms of consumption of carbohydrates. So,

$\begin{array}{c}\left({\overline{x}}_1-{\overline{x}}_2\right)\pm t^{*}\sqrt{\frac{s_1^2}{n_1}+\frac{s_2^2}{n_2}}=\left(64.1-63.5\right)\pm \left[\left(1.962\right)\times \sqrt{\frac{{21.0}^2}{202}+\frac{{20.8}^2}{200}}\right]\\ =0.6\pm \left[\left(1.962\right)\times \sqrt{2.183+2.163}\right]\\ =\left(0.6\pm 4.09\right)\\ =\left(0.6-4.09,0.6+4.09\right)\end{array}$

$=\left(-3.49,4.69\right)$

Interpretation: Therefore, the 95% confidence interval for the difference between the means is obtained as $\left(-3.49,4.69\right)$.

To determine

To explain: The comparison of information from obtained confidence interval with the information given by the significance test.

Answer to Problem 76E

Solution: The information provided by both confidence interval and the significance test shows that there is no significant difference between the two means of carbohydrates.

Explanation of Solution

The significance test conducted in part (c) shows that the null hypothesis is not rejected.

The obtained confidence interval $\left(-3.49,4.69\right)$ shows that population mean difference of zero lies in this interval. So, do not reject the null hypothesis $H_0$. Hence, the information obtained from both confidence interval and the significance test shows that the null hypothesis is not rejected and concludes that there is no significant difference between the two means of carbohydrates.

To determine

To explain: A short summary on the results of the performed significance test and obtained confidence intervals for three dietary compositions.

Answer to Problem 76E

Solution: From the calculated value of the test statistic from the previous exercise for the consumption of fats, the null hypothesis is not rejected and it is concluded that there is no significant difference between the consumption of fats. The obtained confidence interval $\left(-0.37,3.77\right)$ shows that zero is contained in the confidence interval, and hence, the null hypothesis should not be rejected. Thus, both the performed t-test and the obtained confidence interval provide the same results of the analyses.

Explanation of Solution

From the calculated value of the test statistic for the consumption of proteins, the null hypothesis is rejected and it is concluded that there is a significant difference between the consumption of proteins. The obtained confidence interval $\left(0.38,3.42\right)$ shows that zero is not contained in the confidence interval, and hence, the null hypothesis is rejected. Thus, both the performed t-test and the obtained confidence interval provide the same results of the analyses.

From the calculated value of the test statistic for the consumption of carbohydrates, the null hypothesis is not rejected and it is concluded that there is no significant difference between the consumption of carbohydrates. The obtained confidence interval $\left(-3.49,4.69\right)$ shows that zero is contained in the confidence interval, and hence, the null hypothesis is not rejected. Thus, both the performed t-test and the obtained confidence interval provide the same results of the analyses.