Chapter 7, Problem 131E

(a)

Section 1:

To determine

To find: The difference in mean and standard error for body weight.

Answer to Problem 131E

Solution: The difference in mean for body weight is $\underset{\_}{-0.7}$ and standard error is $\underset{\_}{2.29}$.

Explanation of Solution

Calculation: The difference of mean can be calculated as follows:

$\begin{array}{c}{\overline{x}}_1-{\overline{x}}_2=0.4-1.1\\ =-0.7\end{array}$

The standard error (SE) of difference of mean can be calculated as follows:

$\begin{array}{c}\text{SE}=\frac{s}{\sqrt{n}}\\ =\frac{8.6}{\sqrt{14}}\\ =2.298447\\ \approx 2.29\end{array}$

Section 2:

To determine

To find: The difference in mean and standard error for calorie intake.

Answer to Problem 131E

Solution: The difference in mean for body weight is $\underset{\_}{14}$ and standard error is $\underset{\_}{56.125}$.

Explanation of Solution

Calculation:

The difference of mean can be calculated as follows:

$\begin{array}{c}{\overline{x}}_1-{\overline{x}}_2=2589-2575\\ =14\end{array}$

The standard error (SE) of difference of mean can be calculated as follows:

$\begin{array}{c}\text{SE}=\frac{s}{\sqrt{n}}\\ =\frac{210}{\sqrt{14}}\\ =56.1248\\ \approx 56.125\end{array}$

(b)

Section 1:

To determine

To test: The significant differences in body weight.

Answer to Problem 131E

Solution: There is no significant difference in body weight.

The t-statistic for the test hypothesis is obtained as t-value $=-0.305$.

The P-statistic for the test hypothesis is obtained as P-value $=0.76418$.

Explanation of Solution

Calculation: The hypothesis for study is defined as

$H_0$: The difference of body weight is $=$ $0$.

$H_a$: The difference of body weight is $\ne 0$.

To test the hypothesis that there is no significant difference in body weight, t-statistic is used to determine the significance of the difference. The t-value is obtained as follows:

$\begin{array}{c}t=\frac{\left(\overline{x}-\mu \right)}{\frac{s}{\sqrt{n}}}\\ =\frac{\left(-0.7-0\right)}{\frac{8.6}{\sqrt{14}}}\\ =\frac{-0.7}{2.29845}\\ =-0.30455\end{array}$

$\approx 0.305$

Now, the P-value can be obtained by using the standard normal table for $t=-0.305$ and the P-value $=0.76418$ for two-tailed t-statistic.

Conclusion: The P-value for t-test is greater than $0.05$. So, the null hypothesis cannot be rejected significantly, which states that there is no significant difference in body weight.

Section 2:

To determine

To test: The significant differences in calorie intake.

Answer to Problem 131E

Solution: There is no significant difference in calorie intake.

The t-statistic for the test hypothesis is obtained as t-value $=0.249$.

The P-statistic for the test hypothesis is obtained as P-value $=0.80258$.

Explanation of Solution

Calculation: The hypothesis for study is defined as

$H_0$: The difference of calorie intake is $=0$.

$H_a$: The difference of calorie intake is $\ne 0$.

To test the hypothesis, t-statistic is used to determine the significance of the difference. The t- value is obtained as follow:

$\begin{array}{c}t=\frac{\left(\overline{x}-\mu \right)}{\frac{s}{\sqrt{n}}}\\ =\frac{\left(14-0\right)}{\frac{210}{\sqrt{14}}}\\ =\frac{14}{56.12486}\\ =0.24944\end{array}$

$\approx 0.249$

Now, the P-value can be obtained by using the standard normal table for $t=0.24944$. The P-value is $0.80258$ for two-tailed t-statistic.

Conclusion: The P-value for t-test is greater than $0.05$. So, the null hypothesis cannot be rejected significantly, which states that there is no significant difference in calorie intake.

(c)

Section 1:

To determine

To find: The $95\%$ confidence interval for the difference of body weight.

Answer to Problem 131E

Solution: The required $95\%$ confidence interval is $\underset{\_}{\text{confidence interval=}\left(-5.66,4.26\right)}$.

Explanation of Solution

Calculation: The confidence interval for the difference of body weight can be obtained by first calculating the margin of error. The margin of error is obtained as follow:

$\begin{array}{c}\text{Margin of error}=t*\times \frac{s}{\sqrt{n}}\\ =2.16\times \frac{8.6}{\sqrt{14}}\\ =2.16\times 2.298\\ =4.965\end{array}$

Now, the confidence interval can be obtained as follows:

$\begin{array}{c}\text{Confidence interval}=\overline{x}\pm \text{Margin of error}\\ \text{= }-\text{0}\text{.7}\pm 4.96\\ =\left(-5.66,4.26\right)\end{array}$

Interpretation: As the hypothesized mean of $0$ lies inside the $95\%$ confidence interval, the null hypothesis cannot be rejected, which states that difference of body weight is within the $95\%$ confidence interval significantly.

Section 2:

To determine

To find: The $95\%$ confidence interval for the difference of calorie intake.

Answer to Problem 131E

Solution: The required $95\%$ confidence interval is $\underset{\_}{\left(-107.23,135.23\right)}$.

Explanation of Solution

Calculation: The confidence interval for the difference of calorie intake can be obtained by first calculating the margin of error. The margin of error is obtained as follow:

$\begin{array}{c}\text{Margin of error}=t*\times \frac{s}{\sqrt{n}}\\ =2.16\times \frac{210}{\sqrt{14}}\\ =2.16\times 56.125\\ =121.23\end{array}$

Now, the confidence interval can be obtained as follows:

$\begin{array}{c}\text{Confidence interval=}\overline{x}\pm \text{Margin of error}\\ \text{=14}\pm 121.23\\ \left(-107.23,135.23\right)\end{array}$

Interpretation: As the hypothesized mean of $0$ lies inside the $95\%$ confidence interval, the null hypothesis cannot be rejected, which states that difference of calorie intake is within the $95\%$ confidence interval significantly.

(d)

To determine

The summary of the results.

Answer to Problem 131E

Solution: The result cannot be generalized and the outcome of study will also have affected due to the violation of instruction by three subjects as the sample size is too small.

Explanation of Solution

Here, the study is biased for a specific city, the model cannot be generalized for a whole population, and there are three violations out of fourteen subjects. The effect could be a serious concern in the study due to small sample size.