Chapter 9.3, Problem 14E

To determine

To construct:

And interpret a 99% confidence interval for the true mean difference between the cholesterol levels for people who take the new drug.

Answer to Problem 14E

Solution:

The required confidence interval for the mean of paired differences is. $\left(-12.31,1.3139\right)$.

Explanation of Solution

Given Information:

A pharmaceutical company is running tests to see how well its new drug lowers cholesterol. Ten adults volunteer to participate in the study. The total cholesterol level of each participant (in mg/dL) is recorded once at the start of the study and then again after three months of taking the drug. The results are given in the following table.

Total Cholesterol level(in mg/dL)
Initial level	Level after 3 months
210	201
200	195
215	208
194	197
206	200
221	203
203	190
189	188
208	210
211	210

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,

And 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:

Initial Level	Level after 3 months	$d_i$	$\left(d_i-\overline{d}\right)$	${\left(d_i-\overline{d}\right)}^2$
210	201	$-9$	$-3.5$	12.25
200	195	$-5$	0.5	0.25
215	208	$-7$	$-1.5$	2.25
194	197	3	8.5	72.25
206	200	$-6$	$-0.5$	0.25
221	203	$-18$	$-12.5$	156.25
203	190	$-13$	$-7.5$	56.25
189	188	$-1$	4.5	20.25
208	210	2	7.5	56.25
211	210	$-1$	4.5	20.25
	Sum	$-55$		396.5

The paired difference is calculated as,

$d_i=x_2-x_1$

Substitute 210 for $x_1$ and 201 for $x_2$ in the above equation.

$\begin{array}{rll}{d}_1& =& 201-210\\ & =& -9\end{array}$

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 10 that is $n=10$. Then the mean of the paired differences is calculated as,

$\begin{array}{rll}\overline{d}& =& \frac{{\displaystyle \sum d_i}}{n}\\ & =& \frac{\left(-9\right)+\left(-5\right)+\mathrm{...}+2+\left(-1\right)}{10}\\ & =& \frac{-55}{10}\\ & =& -5.5\end{array}$

Substitute $-9$ for $d_1$ and $-5.5$ for $\overline{d}$ in $\left(d_1-\overline{d}\right)$

$\begin{array}{rll}\left(d_1-\overline{d}\right)& =& \left(-9-\left(-5.5\right)\right)\\ \left(d_1-\overline{d}\right)& =& -3.5\end{array}$

Square the both sides of the equation.

$\begin{array}{rll}{\left(d_1-\overline{d}\right)}^2& =& {\left(-3.5\right)}^2\\ {\left(d_1-\overline{d}\right)}^2& =& 12.25\end{array}$

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,

$\begin{array}{rll}{{\displaystyle \sum \left(d_i-\overline{d}\right)}}^2& =& 12.25+\mathrm{0.25.......56.25}+20.25\\ & =& 396.5\end{array}$

The standard deviation of the paired difference is,

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

Substitute 396.5 for ${\displaystyle \sum {\left(d_i-\overline{d}\right)}^2}$ and 10 for n in the above equation

$\begin{array}{rll}{s}_d& =& \sqrt{\frac{396.5}{9}}\\ & =& \sqrt{44.055}\\ & =& 6.63\end{array}$

It is given that $c=0.99$.

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

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 10 so the degree of freedom is

$\begin{array}{rll}n-1& =& 10-1\\ & =& 9\end{array}$

Then,

$\begin{array}{rll}{t}_{\frac{\alpha }{2}}& =& t_{\frac{0.01}{2}}\\ & =& 3.250\end{array}$

The margin of error is calculated as,

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

Substitute 3.250 for $t_{\frac{\alpha }{2}}$, 6.63 for $s_d$ and 10 for n,

$\begin{array}{rll}E& =& 3.250\times \frac{6.63}{\sqrt{10}}\\ & =& 6.8139\end{array}$

The confidence interval is,

$\begin{array}{rll}\left(\overline{d}\pm E\right)& =& \left(-5.5-6.8139,-5.5+6.8139\right)\\ & =& \left(-12.31,1.3139\right)\end{array}$

Interpretation:

Since one of the end point is negative so this implies that researchers are 99% confident that the mean difference between the scores was negative and lies between -12.31 and 1.3139.and thus the new drug has reduced the cholesterol level in participants.

Conclusion:

Thus the confidence interval for the mean of paired differences is. $\left(-12.31,1.3139\right)$.