Chapter 7.2, Problem 75E

(a)

To determine

If the two sample t-procedures can be used to analyze the provided data for group differences.

Answer to Problem 75E

Solution: A t-test is appropriate to analyze the provided problem.

Explanation of Solution

If the data is assumed to be a simple random sample drawn from a population in which population standard deviations are not known, then a two-sample t-procedure can be used to analyze the data for group differences.

Since, the population standard deviations are not known. It is replaced by sample standard deviations of the samples. Also, the sample sizes are almost equal in the provided problem, which is $n_1=202,n_2=200$, hence it is appropriate to use the two-sample t-procedures.

(b)

To determine

The appropriate null and alternative hypotheses to compare the two groups in terms of consumption of fats.

Answer to Problem 75E

Solution: The null and alternative hypotheses are:

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

Explanation of Solution

The significance test is to compare the two groups in terms of consumption of fats. Hence the null hypothesis is that the consumption of fats in the two groups do not differ significantly as against the alternative that assumes that the consumption of fats in the two groups differs significantly.

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, ${\mu }_{\text{Early}}$ and ${\mu }_{\text{Late}}$ represent the actual average composition of fats in early and late eaters, respectively.

(c)

To determine

To test: A significance test to determine the difference in consumption of fats in two groups.

Answer to Problem 75E

Solution: The t- test statistic is obtained as $\underset{\_}{t=1.62}$ with the p-value as $\underset{\_}{0.1068}$ for 199 degrees of freedom.

Explanation of Solution

Calculation: To test the hypothesis formulated in part (b), the following test statistic is formulated,

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

Where,

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

$\begin{array}{c}{n}_1=\text{Sample size of early eaters}\\ n_2=\text{Sample size of late eaters}\\ {\mu }_1-{\mu }_2=\text{Difference of true means in the two groups}\end{array}$

The difference of 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,

$\begin{array}{c}t=\frac{\left({\overline{x}}_1-{\overline{x}}_2\right)-\left({\mu }_1-{\mu }_2\right)}{\sqrt{\frac{s_1^2}{n_1}+\frac{s_2^2}{n_2}}}\\ =\frac{\left(23.1-21.4\right)-0}{\sqrt{\frac{{12.5}^2}{202}+\frac{{8.2}^2}{200}}}\\ =\frac{1.7}{1.05}\\ =1.62\end{array}$

The p-value for the provided one-sided test is calculated as $P\left(T\ge 1.62\right)$. For the second approximation, the degree 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 degree of freedom is 199. Use Excel function to determine the exact p- value. The Excel function to determine the p- value from t-test statistic is mentioned below and the screenshot is attached,

$\text{tdist}\left(\text{t, df, tails}\right)=\text{tdist}\left(1.62,199,2\right)$

Therefore, the p-value is obtained as 0.1068.

To determine

To explain: A summary of the conclusion of the test performed above.

Answer to Problem 75E

Solution: The two groups of early eaters and late eaters have a similar average fat consumption.

Explanation of Solution

The t-test statistic is obtained as 1.62 in the previous part. For 199 degrees of freedom, the p-value is obtained as 0.1068 using Excel. Since $P>0.05$, do not reject the null hypothesis and hence it is concluded that there is no significant difference between the two groups in terms of fats 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 fats.

Answer to Problem 75E

Solution: The 95% confidence interval is $\underset{\_}{\left(-0.365,3.765\right)}$.

Explanation of Solution

Calculation: The confidence interval for the difference between the means is calculated 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

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

$\begin{array}{c}{n}_1=\text{Sample size of early eaters}\\ n_2=\text{Sample size of late eaters}\\ t^{*}=t\text{ critical value for the provided confidence level}\end{array}$

The t^*–value for 95% confidence level is obtained as 1.97. Substitute the provided values in the above-defined formula to determine the 95% confidence interval as:

$\begin{array}{c}\left(\overline{x_1}-\overline{x_2}\right)\pm t^{*}\sqrt{\frac{s_1^2}{n_1}+\frac{s_2^2}{n_2}}=\left(23.1-21.4\right)\pm \left[\left(1.97\right)\times \sqrt{\frac{{12.5}^2}{202}+\frac{{8.2}^2}{200}}\right]\\ =1.7\pm \left[\left(1.97\right)\times \sqrt{0.774+0.336}\right]\\ =1.7\pm 2.068\\ =\left(1.7-2.068,1.7+2.068\right)\end{array}$

$=\left(-0.368,3.768\right)$

Therefore, the 95% confidence interval is obtained as $\left(-0.368,3.768\right)$.

To determine

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

Answer to Problem 75E

Solution: Both the information obtained from 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.

Explanation of Solution

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

The obtained confidence interval $\left(-0.365,3.765\right)$ shows that the population means difference of zero lies in this interval. So, do not reject the null hypothesis $H_0$. So, it is concluded that the information provided by both confidence interval and the significance test shows that there is no significant difference between the two means.