Chapter 7.1, Problem 36E

(a)

To determine

To find: The weight change for each subject.

Answer to Problem 36E

Solution: The table for weight change is as follows,

Weight before	Weight after	Weight change
55.7	61.7	6
54.9	58.8	3.9
59.6	66	6.4
62.3	66.2	3.9
74.2	79	4.8
75.6	82.3	6.7
70.7	74.3	3.6
53.3	59.3	6
73.3	79.1	5.8
63.4	66	2.6
68.1	73.4	5.3
73.7	76.9	3.2
91.7	93.1	1.4
55.9	63	7.1
61.7	68.2	6.5
57.8	60.3	2.5

Explanation of Solution

Calculation:

To calculate weight change follow the below mentioned steps in Minitab;

Step 1: Enter the variable name as ‘Weight before’ in column C1 and enter the data of ‘weight before’ and enter the variable name as ‘Weight after’ in column C2 and enter the data of ‘Weight after’.

Step 2: Enter variable name as ‘Weight change’ in column C3.

Step 3: Go to $"\text{Calc}"\to "\text{Calculater}"$

Step 4: In the dialog box that appears, select ‘Weight Change’ in ‘Store result in variable’.

Step 5: Enter 'Weight after'-'Weight before' in ‘Expression’. Click ‘OK’ on dialog box.

From Minitab result, the table for weight change is as follows,

Weight before	Weight after	Weight change
55.7	61.7	6
54.9	58.8	3.9
59.6	66	6.4
62.3	66.2	3.9
74.2	79	4.8
75.6	82.3	6.7
70.7	74.3	3.6
53.3	59.3	6
73.3	79.1	5.8
63.4	66	2.6
68.1	73.4	5.3
73.7	76.9	3.2
91.7	93.1	1.4
55.9	63	7.1
61.7	68.2	6.5
57.8	60.3	2.5

(b)

Section 1:

To determine

To find: The mean of ‘weight change’.

Answer to Problem 36E

Solution: The required mean is $4.731\text{ kg}$.

Explanation of Solution

Calculation:

To calculate the mean of ‘weight change’ follow the below mentioned steps in Minitab;

Step 1: Follow the step 1 to 5 performed in part (a).

Step 2: Go to $\text{"Stat"}\to "\text{Basic}\text{statistics"}\to \text{"Display Discriptive statistics"}$ and click on ‘OK’

Step 3: In dialog box that appears select ‘Weight change’ under the field marked as ‘Variables’ and click on ‘Statistics’.

Step 4: In dialog box that appears select ‘None’ and then ‘Mean’ and click OK twice.

From the Minitab output, the mean is $4.731\text{ kg}$.

Section 2:

To determine

To find: The standard deviation of weight change.

Answer to Problem 36E

Solution: The required standard deviation is 1.746 kg.

Explanation of Solution

Calculation:

To calculate standard deviation of weight change follow the below mentioned steps in Minitab;

Step 1: Follow the step 1 to 5 performed in part (a).

Step 2: Go to $\text{"Stat"}\to "\text{Basic}\text{statistics"}\to \text{"Display Discriptive statistics"}$ and click on ‘OK’

Step 3: In dialog box that appears select ‘Weight change’ under the field marked as ‘Variables’ and then click on ‘Statistics’.

Step 4: In dialog box that appears select ‘None’ and then ‘standard deviation’ and click ‘OK’ twice.

From the Minitab results, the standard deviation is 1.746.

(c)

Section 1:

To determine

To find: The standard error of the weight change.

Answer to Problem 36E

Solution: The standard error is 0.436 kg.

Explanation of Solution

Calculation:

To obtain the standard error of weight change follow the below mentioned steps in Minitab;

Step 1: Follow the step 1 to 5 performed in part (a).

Step 2: Go to $\text{"Stat"}\to \text{'}\text{Basic}\text{statistics"}\to "\text{Display Discriptive statistics"}$. Then click on ‘OK’

Step 3: In dialog box that appears select ‘Weight change’ under the field marked as ‘Variables’. Then click on ‘Statistics’.

Step 4: In dialog box that appears select ‘None’ and then ‘SE of mean’. Then click ‘Ok’ on both dialog box.

From Minitab result, the standard error is 0.436 kg.

Section 2:

To determine

To find: The margin of error for 95% confidence interval for mean weight change.

Answer to Problem 36E

Solution: The margin of error is 0.929 kg.

Explanation of Solution

Calculation:

The formula to calculate the margin of error is as follows,

$\text{Margin}\text{of}\text{error}\text{= }{t}_{\frac{\alpha }{2}}\times \frac{s}{\sqrt{n}}$

Where $\alpha$ is significance level, s is standard deviation of sample, $t_{\frac{\alpha }{2}}$ is the critical value of the t test statistic at the $\alpha$ level of significance and n is sample size. So, the margin of error is computed as,

$\begin{array}{c}\text{Margin}\text{of}\text{error}\text{= }{t}_{0.025}\times \frac{s}{\sqrt{n}}\\ =2.13\times \frac{1.746}{\sqrt{16}}\\ =0.929\end{array}$

Hence the margin of error for mean weight change is 0.929 kg.

Section 3:

To determine

To find: The $95\%$ confidence interval for mean weight change.

Answer to Problem 36E

Solution: The required confidence interval is $\left(3.801,5.661\right)$.

Explanation of Solution

Calculation:

The confidence interval is an interval for which there are 95% chances that it contains the population parameter (population mean).

To calculate confidence interval, follow the below mentioned steps in Minitab;

Step 1: Enter the provided data into Minitab and enter variable name as ‘Weight change’

Step 2: Go to ‘Stat’ then point on ‘Basic Statistics’ and select ‘1-sample-t’.

Step 3: In the dialog box that appears select ‘Weight change’ under the field marked as ‘sample in columns’ and click on ‘option’.

Step 4: In the dialog box that appears, enter ‘95.0’ under the field marked as ‘confidence level’ and select ‘not equal’ under the field marked as ‘Alternative hypothesis’.

From Minitab results, 95% confidence interval in is $\left(3.801,5.661\right)$.

(d)

Section 1:

To determine

To find: The mean weight gain in pounds.

Answer to Problem 36E

Solution: The mean weight decrement in pounds is 10.4082 pounds.

Explanation of Solution

Calculation:

Consider ${\mu }_{\text{pounds}}$ is the mean weight gain in pounds and ${\mu }_{\text{kilogram}}$ mean weight gain in kilograms. So use the change of scale to find the mean weight gain in pounds. Change of scale means all values of data is multiplied or divided by some constant number. The effect of change of scale for mean is as follows,

${\mu }_{\text{pounds}}={\mu }_{\text{kilogram}}\times c$

Where c is the constant.

Also,

$1\text{K}g=2.2\text{ pounds}$

So here,

$\begin{array}{c}c=2.2\\ {\mu }_{\text{pounds}}={\mu }_{\text{kilograms}}\times 2.2\\ {\mu }_{\text{pounds}}=4.731\times 2.2\\ =10.4082\end{array}$

Hence, the mean weight decrement in pounds will be 10.4082 pounds.

Section 2:

To determine

To find: The standard deviation of weight gains in pounds.

Answer to Problem 36E

Solution: The standard deviation of weight gain in pounds is 3.8412 pounds.

Explanation of Solution

Calculation:

Consider ${\sigma }_{\text{pounds}}$ is the standard deviation weight gain in pounds and ${\sigma }_{\text{kilogram}}$ standard deviation of weight gain in kilograms. So use the change of scale to find the mean weight gain in pounds. Change of scale means all values of data is multiplied or divided by some constant number. The effect of change of scale for mean is as follows,

${\sigma }_{\text{pounds}}={\sigma }_{\text{kilogram}}\times c$

Where c is the constant.

Also,

$1\text{K}g=2.2\text{ pounds}$

So here,

$\begin{array}{c}c=2.2\\ {\sigma }_{\text{pounds}}={\sigma }_{\text{kilograms}}\times 2.2\\ {\sigma }_{\text{pounds}}=1.746\times 2.2\\ =3.8412\end{array}$

Hence, the standard deviation of weight gain in pounds will be 3.8412 pounds.

Section 3:

To determine

To find: The confidence interval of weight gain in pounds.

Answer to Problem 36E

Solution: The confidence interval of weight gain in pounds is $\left(8.362,12.454\right)$

Explanation of Solution

Calculation:

The lower limit and the upper limit of the confidence interval are changed in pounds by multiplying them with 2.2. So the new confidence interval in pounds will be,

The lower limit,

$3.801\times 2.2=8.3622$

The upper limit,

$5.661\times 2.2=12.454$

So the confidence interval will be $\left(8.362,12.454\right)$

(e)

Section 1:

To determine

To explain: The hypothesis to test whether the weight change is equal to 16 lb.

Answer to Problem 36E

Solution: The null hypothesis is,

$H_0\text{:}\text{}\mu =16$

And the alternative hypothesis is,

$H_a\text{: }\mu \ne 16$

Explanation of Solution

The null hypothesis is the hypothesis of no difference. So the null hypothesis assumes that the mean weight $\left(\mu \right)$ is 16 lb and is stated as,

$H_0\text{:}\text{}\mu =16$

Against the alternative hypothesis which assumes that the mean weight $\left(\mu \right)$ is not equal to 16 lb which is stated as,

$H_a\text{: }\mu \ne 16$

Section 2:

To determine

To find: The value of test statistic.

Answer to Problem 36E

Solution: The value of test statistic is $-5.83$.

Explanation of Solution

Calculation:

To calculate the test statistic, follow the below mentioned steps in Minitab;

Step 1: Go to $\text{"Stat"}\to "\text{Basic}\text{statistics"}\to "\text{1-sample t"}$

Step 2: In the dialog box that appears click on ‘Summarized data’ Enter the values of sample size, mean, standard deviation. Then click on ‘Option’.

Step 3: Click on ‘Perform hypothesis test’ and enter ‘16’ under the field as ‘Hypothesized mean’.

Step 4: In the dialog box that appears, enter ‘95.0’ in confidence level and select ‘not equal’ under the field as ‘Alternative hypothesis’. Then click ‘Ok’.

From Minitab results, the value of test statistic is $-5.83$.

Section 3:

To determine

To find: The P- value of the test statistic.

Answer to Problem 36E

Solution: The P- value is 0.000.

Explanation of Solution

Calculation:

To calculate the test statistic, follow the below mentioned steps in Minitab;

Step 1: Go to $\text{"Stat"}\to "\text{Basic}\text{statistics"}\to "\text{1-sample }t"$

Step 2: In the dialog box that appears click on ‘Summarized data’ Enter the values of sample size, mean, standard deviation. Then click on ‘Option’.

Step 3: Click on ‘Perform hypothesis test’ and enter ‘16’ under the field as ‘Hypothesized mean’.

Step 4: In the dialog box that appears, enter ‘95.0’ in confidence level and select ‘not equal’ under the field as ‘Alternative hypothesis’. Then click ‘Ok’.

From Minitab results, the P-value is 0.000

Section 4:

To determine

To explain: The test results.

Answer to Problem 36E

Solution: The null hypothesis is rejected and it is concluded that the mean weight gain is not 16 lb.

Explanation of Solution

The P-value is 0.000 is less than significance level (0.05) So, the null hypothesis is rejected and it is concluded that the mean weight is different from 16 lb.

(f)

To determine

To explain: The obtained results.

Answer to Problem 36E

Solution: The mean of ‘weight change’ is $-4.731$ kg and 10.40 pounds that means on an average each person’s weight is decreased to 4.371 kg or 10.40 ponds. The standard deviation in kilograms is 1.746 kg and in pounds is 3.81 pounds. That means on an average, the weight change varies 1.746 kg (3.81 pounds) from the mean value. The standard error is 0.436 kg. The margin of error is 0.929 kg. The $95\%$ confidence interval is $\left(3.801,5.661\right)$ in Kg and $\left(8.362,12.454\right)$ in pounds. The P- value is 0.000 and the null hypothesis is rejected.

Explanation of Solution

The mean of ‘weight change’ in kilograms is $-4.731$ kg and in pounds is -10.40 pounds that means on an average each person’s weight has been decreased to 4.371 kg or 10.40 pounds by feeding them extra calories. The standard deviation in kilograms is 1.746 kg and in pounds is 3.81 pounds. That means on an average the weight change varies 1.746 kg (3.81 pounds) from the mean value. The standard error is 0.436 kg. The margin of error is 0.929 kg which expresses the maximum expected difference between the true population parameter. The $95\%$ confidence interval is $\left(3.801,5.661\right)$ in kg and is $\left(8.362,12.454\right)$ in pounds. It means there is 95% chances that the interval $\left(3.801,5.661\right)$ contains the true population parameter (population mean). The $95\%$ confidence interval is $\left(8.362,12.454\right)$ (in pounds). It means there is 95% chances that the inetrval $\left(8.362,12.454\right)$ contains the population parameter (population mean). The P- value is 0.000. The test statistic value is said to be statistically significant when the P-value is more than the provided significance level. Here, the P- value is less than significance level which is 5% so the value of the test statistic is insignificant and the null hypothesis is rejected and it is concluded that the mean weight change is not equal to 16 lb.