Chapter 9.3, Problem 11E

To determine

To construct:

And interpret a 99% confidence interval for the true mean difference between the weights to determine the mean amount of weight lost by doing on the diet.

Answer to Problem 11E

Solution:

The required confidence interval for the mean of paired differences is. $\left(-12.19,-7.3476\right)$.

Explanation of Solution

Given Information:

A dietician wants to see how much weight a certain diet can help patients lose. Nine people agree to participate in her study. She weighs each patient at the beginning of the study and after 30 days of being on diet. The results are shown in table.

Patients’ Weights (in Pounds)
Starting Weight	180	176	152	230	183	214	250	197	201
Ending Weight	172	167	144	219	172	201	239	186	195

Formula Used:

The paired difference for any pair data values of the two dependent samples are given by

$d=x_2-x_1$

Where $x_2$ is a value of the second sample,

And $x_1$ is a value of the first sample which is paired with $x_2$.

The mean of the paired differences is calculated as,

$\overline{d}=\frac{{\displaystyle \sum d_i}}{n}$

Where $d_i$ is the paired difference of the $i^{th}$ pair of values.

Also n is the total number of data values.

The sample standard deviation of the paired differences is calculated as,

$s_d=\sqrt{\frac{{\displaystyle \sum \left(d_i-\overline{d}\right)}}{n-1}}$

Where $d_i$ is the paired difference of the $i^{th}$ pair of values,

$\overline{d}$ is the mean of the pair differences,

And n is the total number of values given in the data.

For Margin of error we need critical value $t_{\frac{\alpha }{2}}$,

$E=\left(t_{\frac{\alpha }{2}}\right)\left(\frac{s_d}{\sqrt{n}}\right)$

The confidence interval is calculated as,

$\left(\overline{d}\pm E\right)$

Where the lower endpoint is $\overline{d}-E$ and the upper endpoint is $\overline{d}+E$.

Calculation:

Starting Weight $x_1$	Ending Weight $x_2$	$d_i$	$\left(d_i-\overline{d}\right)$	${\left(d_i-\overline{d}\right)}^2$
180	172	$-8$	1.77	3.1329
176	167	$-9$	0.77	0.5929
152	144	$-8$	1.77	3.1329
230	219	$-11$	$-1.23$	1.5129
183	172	$-11$	$-1.23$	1.5129
214	201	$-13$	$-3.23$	10.4329
250	239	$-11$	$-1.23$	1.5129
197	186	$-11$	$-1.23$	1.5129
201	195	$-6$	3.77	14.2129
	Sum	$-88$		37.5561

The paired difference is calculated as,

$d_i=x_2-x_1$

Substitute 180 for $x_1$ and 172 for $x_2$ in the above equation.

$\begin{array}{rll}{d}_1& =& 172-180\\ & =& -8\end{array}$

Proceed in the same manner to calculate $d_i$ for the rest of the data and refer table for the rest of the $d_i$ values calculated. The size of each sample is 9 that is $n=9$. Then the mean of the paired differences is calculated as,

$\begin{array}{rll}\overline{d}& =& \frac{{\displaystyle \sum d_i}}{n}\\ & =& \frac{\left(-8\right)+\left(-9\right)+\mathrm{......}+\left(-11\right)+\left(-6\right)}{9}\\ & =& \frac{-88}{9}\\ & =& -9.77\end{array}$

Substitute $-8$ for $d_1$ and $-9.77$ for $\overline{d}$ in $\left(d_1-\overline{d}\right)$.

$\begin{array}{rll}\left(d_1-\overline{d}\right)& =& \left(-8-\left(-9.77\right)\right)\\ \left(d_1-\overline{d}\right)& =& 1.77\end{array}$

Square the both sides of above equation.

$\begin{array}{rll}{\left(d_1-\overline{d}\right)}^2& =& {\left(1.77\right)}^2\\ {\left(d_1-\overline{d}\right)}^2& =& 3.1329\end{array}$

Proceed in the same manner to calculate ${\left(d_i-\overline{d}\right)}^2$ for all the $1\le i\le n$ for the rest data and refer table for the rest of the ${\left(d_i-\overline{d}\right)}^2$ values calculated. Then the value of ${{\displaystyle \sum \left(d_i-\overline{d}\right)}}^2$ is calculated as,

$\begin{array}{rll}{{\displaystyle \sum \left(d_i-\overline{d}\right)}}^2& =& 3.1329+0.5929+\mathrm{3.1329.....}+14.2129\\ & =& 37.5561\end{array}$

The standard deviation of the paired difference is calculated as,

$s_d=\sqrt{\frac{{\displaystyle \sum \left(d_i-\overline{d}\right)}}{n-1}}$

Substitute 37.5561 for ${\displaystyle \sum {\left(d_i-\overline{d}\right)}^2}$ and 9 for n in the above equation.

$\begin{array}{rll}{s}_d& =& \sqrt{\frac{37.5561}{8}}\\ & =& \sqrt{4.6945}\\ & =& 2.166\end{array}$

It is given that $c=0.99$ and,

$\begin{array}{rll}\alpha & =& 1-0.99\\ & =& 0.01\end{array}$

To calculate the margin of error we need $t_{\frac{\alpha }{2}}$, the critical value. The value of $\alpha$ is given to be 0.01 and n is given to be 9 so the degree of freedom is,

$\begin{array}{rll}n-1& =& 9-1\\ & =& 8\end{array}$

Then,

$\begin{array}{rll}{t}_{\frac{\alpha }{2}}& =& t_{\frac{0.01}{2}}\\ & =& 3.355\end{array}$

The margin of error is calculated as,

$E=\left(t_{\frac{\alpha }{2}}\right)\left(\frac{s_d}{\sqrt{n}}\right)$

Substitute 3.355 for $t_{\frac{\alpha }{2}}$, 2.166 for $s_d$ and 9 for $n$.

$\begin{array}{rll}E& =& 3.355\times \frac{2.166}{\sqrt{9}}\\ & =& 2.42231\end{array}$

The confidence interval is,

$\begin{array}{rll}\left(\overline{d}\pm E\right)& =& \left(-9.77-2.42231,-9.77+2.42231\right)\\ & =& \left(-12.19,-7.3476\right)\end{array}$

Interpretation:

Since both the end points are negative so this implies that researchers are 99% confident that the mean difference between the weights was negative and lies between $-12.19$ and $-7.3476$ and hence this implies that the weight of the patients has reduced by going on the diet.

Conclusion:

Thus the confidence interval for the mean of paired differences is. $\left(-12.19,-7.3476\right)$.