Chapter 8, Problem 9E

a.

To determine

Find the possible number of different samples of size 3.

Answer to Problem 9E

The possible number of different samples of size 3 is 20.

Explanation of Solution

From the given information, the number of cases each partner actually tried in court is 3, 6, 3, 3, 0 and 1.

The possible number of different samples of size 3 is obtained by using the following formula:

${}_N{C}_n=\frac{N!}{\left(N-n\right)!n!}$

Substitute 6 for the N and 3 for the n

Then,

$\begin{array}{c}{}_6{C}_3=\frac{6!}{\left(6-3\right)!3!}\\ =\frac{6\times 5\times 4\times 3!}{3!3!}\\ =\frac{6\times 5\times 4}{3\times 2\times 1}\\ =20\end{array}$

Thus, the possible number of different samples of size 3 is 20.

b.

To determine

Give the all possible samples of size 3.

Find the mean of each sample.

Answer to Problem 9E

All possible samples of size 3 are $\left(3,6,3\right)$, $\left(3,6,3\right)$, $\left(3,3,3\right)$, $\left(3,3,0\right)$, $\left(3,3,0\right)$, $\left(3,3,0\right)$, $\left(3,0,1\right)$, $\left(3,0,1\right)$, $\left(3,0,1\right)$, $\left(6,3,3\right)$, $\left(6,3,0\right)$, $\left(6,3,0\right)$, $\left(6,0,1\right)$, $\left(3,6,0\right)$, $\left(3,6,1\right)$, $\left(3,3,1\right)$, $\left(3,3,1\right)$, $\left(3,3,1\right)$, $\left(6,3,1\right)$ and $\left(6,3,1\right)$.

The mean of each sample is 4, 4, 3, 2, 2, 2, 2, 1.33, 1.33, 1.33, 4, 3, 3, 2.33, 3, 3.33, 2.33, 2.33, 2.33, 3.33 and 3.33.

Explanation of Solution

The mean is calculated by using the following formula:

$\text{Mean}=\frac{\text{Sum}\text{of}\text{all}\text{the}\text{observations}}{\text{Number}\text{of}\text{observations}}$

Sample	Mean
$\left(3,6,3\right)$	$\frac{3+6+3}{3}=4$
$\left(3,6,3\right)$	$\frac{3+6+3}{3}=4$
$\left(3,3,3\right)$	$\frac{3+3+3}{3}=3$
$\left(3,3,0\right)$	$\frac{3+3+0}{3}=2$
$\left(3,3,0\right)$	$\frac{3+3+0}{3}=2$
$\left(3,3,0\right)$	$\frac{3+3+0}{3}=2$
$\left(3,0,1\right)$	$\frac{3+0+1}{3}=1.33$
$\left(3,0,1\right)$	$\frac{3+0+1}{3}=1.33$
$\left(3,0,1\right)$	$\frac{3+0+1}{3}=1.33$
$\left(6,3,3\right)$	$\frac{6+3+3}{3}=4$
$\left(6,3,0\right)$	$\frac{6+3+0}{3}=3$
$\left(6,3,0\right)$	$\frac{6+3+0}{3}=3$
$\left(6,0,1\right)$	$\frac{6+0+1}{3}=2.33$
$\left(3,6,0\right)$	$\frac{3+6+0}{3}=3$
$\left(3,6,1\right)$	$\frac{3+6+1}{3}=3.33$
$\left(3,3,1\right)$	$\frac{3+3+1}{3}=2.33$
$\left(3,3,1\right)$	$\frac{3+3+1}{3}=2.33$
$\left(3,3,1\right)$	$\frac{3+3+1}{3}=2.33$
$\left(6,3,1\right)$	$\frac{6+3+1}{3}=3.33$
$\left(6,3,1\right)$	$\frac{6+3+1}{3}=3.33$

Thus, all possible samples of size 3 are $\left(3,6,3\right)$, $\left(3,6,3\right)$, $\left(3,3,3\right)$, $\left(3,3,0\right)$, $\left(3,3,0\right)$, $\left(3,3,0\right)$, $\left(3,0,1\right)$, $\left(3,0,1\right)$, $\left(3,0,1\right)$, $\left(6,3,3\right)$, $\left(6,3,0\right)$, $\left(6,3,0\right)$, $\left(6,0,1\right)$, $\left(3,6,0\right)$, $\left(3,6,1\right)$, $\left(3,3,1\right)$, $\left(3,3,1\right)$, $\left(3,3,1\right)$, $\left(6,3,1\right)$ and $\left(6,3,1\right)$.

Thus, the mean of each sample is 4, 4, 3, 2, 2, 2, 1.33, 1.33, 1.33, 4, 3, 3, 2.33, 3, 3.33, 2.33, 2.33, 2.33, 3.33 and 3.33.

c.

To determine

Compare the mean of the sample means to the population mean.

Answer to Problem 9E

The mean of the distribution of the sample means is equal to the population mean.

Explanation of Solution

The mean of the sample means is calculated as follows:

$\begin{array}{c}\text{Mean}=\frac{\text{Sum}\text{of}\text{all}\text{the}\text{sample}\text{means}}{\text{Number}\text{of}\text{samples}}\\ =\frac{\begin{array}{l}4+4+3+2+2+2+1.33+1.33+1.33+4+\\ 3+3+2.33+3+3.33+2.33+2.33+2.33+3.33+3.33\end{array}}{20}\\ =\frac{53.3}{20}\\ =2.67\end{array}$

Population mean is calculated as follows:

$\begin{array}{c}\text{Mean}=\frac{3+6+3+3+0+1}{6}\\ =\frac{16}{6}\\ =2.67\end{array}$

The population mean is 2.67.

Comparison:

The mean of the distribution of the sample mean is 6.8 and the population mean is 6.8. The two means are exactly same.

Thus, the mean of the distribution of the sample means is equal to the population mean.

d.

To determine

Give the comparison of the dispersion in sample means with that of the population.

Answer to Problem 9E

The dispersion in the population is greater than with that of the sample mean.

Explanation of Solution

A frequency distribution for the sample means is obtained as follows:

Let $\overline{x}$ be the sample mean and f be the frequency.

Sample mean	f	Probability
1.33	3	$\frac{3}{20}=0.15$
2	3	$\frac{3}{20}=0.15$
2.33	4	$\frac{4}{20}=0.2$
3	4	$\frac{4}{20}=0.2$
3.33	3	$\frac{3}{20}=0.15$
4	3	$\frac{3}{20}=0.15$
	$N=20$

Software procedure:

Step-by-step procedure to obtain the bar chart using MINITAB:

• Choose Graph > Bar chart.
• Under Bars represent, enter select Values from a table.
• Under One column of values select Simple.
• Click on OK.
• Under Graph variables enter probability and under categorical variable enter sample mean.
• Click OK.

Output using MINITAB software is given below:

From the bar chart it can be observed that the shape of the distribution of the sample means is normal.

Thus, the shape of the distribution of the sample means is normal.

From the given information, the number of cases by the six partners each is 3, 6, 3, 3, 0 and 1.

Number of cases	f	Probability
0	1	$\frac{1}{6}=0.17$
1	1	$\frac{1}{6}=0.17$
3	3	$\frac{3}{6}=0.5$
6	1	$\frac{1}{6}=0.17$
	$N=6$

Software procedure:

Step-by-step procedure to obtain the bar chart using MINITAB:

• Choose Graph > Bar chart.
• Under Bars represent, enter select Values from a table.
• Under One column of values select Simple.
• Click on OK.
• Under Graph variables enter probability and under categorical variable enter Number of cases.
• Click OK.

Output using MINITAB software is given below:

From the bar chart it can be observed that the shape of the population distribution is uniform.

Thus, the shape of the population distribution is uniform.

The population values are 3, 6, 3, 3, 0 and 1. From the part b, the mean of each sample is 4, 4, 3, 2, 2, 2, 1.33, 1.33, 1.33, 4, 3, 3, 2.33, 3, 3.33, 2.33, 2.33, 2.33, 3.33 and 3.33.

The population values are between 0 and 6. The sample mean values are between 1.33 and 4.

Thus, the dispersion in the population is greater than with that of the sample mean.