Chapter 8, Problem 21CE

a.

To determine

Give all the possible samples of size 2.

Find the mean of every sample.

Answer to Problem 21CE

All possible samples of size 2 are $\left(1,1\right)$, $\left(1,2\right)$, $\left(1,3\right)$, $\left(2,1\right)$, $\left(2,2\right)$, $\left(2,3\right)$, $\left(3,1\right)$, $\left(3,2\right)$ and $\left(3,3\right)$.

The mean of every sample is 1, 1.5, 2, 1.5, 2, 2.5, 2, 2.5 and 3.

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
$\left(1,1\right)$	$\frac{1+1}{2}=1$
$\left(1,2\right)$	$\frac{1+2}{2}=1.5$
$\left(1,3\right)$	$\frac{1+3}{2}=2$
$\left(2,1\right)$	$\frac{2+1}{2}=1.5$
$\left(2,2\right)$	$\frac{2+2}{2}=2$
$\left(2,3\right)$	$\frac{2+3}{2}=2.5$
$\left(3,1\right)$	$\frac{3+1}{2}=2$
$\left(3,2\right)$	$\frac{3+2}{2}=2.5$
$\left(3,3\right)$	$\frac{3+3}{2}=3$

Thus, all possible samples of size 2 are $\left(1,1\right)$, $\left(1,2\right)$, $\left(1,3\right)$, $\left(2,1\right)$, $\left(2,2\right)$, $\left(2,3\right)$, $\left(3,1\right)$, $\left(3,2\right)$ and $\left(3,3\right)$.

Thus, the mean of every sample is 1, 1.5, 2, 1.5, 2, 2.5, 2, 2.5 and 3.

b.

To determine

Compute the means of the distribution of the sample mean and the population mean.

Give the comparison of the mean of the distribution of the sample mean with the population mean.

Answer to Problem 21CE

The means of the distribution of the sample mean and the population mean are 2 and 2.

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

Explanation of Solution

Population mean is calculated as follows:

$\begin{array}{c}\text{Mean}=\frac{\text{Sum of population values}}{\text{Number of values}}\\ =\frac{1+2+3}{3}\\ =\frac{6}{3}\\ =2\end{array}$

Then, the population mean is 2.

The mean of the sample means is calculated as follows:

$\begin{array}{c}\text{Mean}=\frac{\text{Sum}\text{of}\text{all}\text{the}\text{sample}\text{means}}{\text{Number}\text{of}\text{samples}}\\ =\frac{1+1.5+2+1.5+2+2.5+2+2.5+3}{9}\\ =\frac{18}{9}\\ =2\end{array}$

Then, mean of the distribution of the sample mean is 2.

Thus, the means of the distribution of the sample mean and the population mean are 2 and 2.

Comparison:

The mean of the distribution of the sample mean is 2 and the population mean is 2. The two means are exactly same.

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

c.

To determine

Give the dispersion of the population with that of the sample mean.

Answer to Problem 21CE

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

Explanation of Solution

The variance of the population is

$\begin{array}{c}\text{Variance}=\frac{{\left(1-2\right)}^2+{\left(2-2\right)}^2+{\left(3-2\right)}^2}{3}\\ =\frac{1+0+1}{3}\\ =\frac{2}{3}\end{array}$

The variance of the sample means is

$\begin{array}{c}\text{Variance}=\frac{\begin{array}{l}{\left(1-2\right)}^2+{\left(1.5-2\right)}^2+{\left(2-2\right)}^2+\\ {\left(1.5-2\right)}^2+{\left(2-2\right)}^2+{\left(2.5-2\right)}^2+\\ {\left(2-2\right)}^2+{\left(2.5-2\right)}^2+{\left(3-2\right)}^2\end{array}}{9}\\ =\frac{1+0.25+0+0.25+0+0.25+0+0.25+1}{9}\\ =\frac{3}{9}\\ =\frac{1}{3}\end{array}$

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

d.

To determine

Give the shapes of the two distributions.

Answer to Problem 21CE

The shape of the population distribution is Uniform.

The shape of the distribution of the sample means is triangular.

Explanation of Solution

The frequency distribution of the population is:

Values	Frequency	Probability
1	1	$\frac{1}{3}=0.33$
2	1	$\frac{1}{3}=0.33$
3	1	$\frac{1}{3}=0.33$
	$N=3$

Software procedure:

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

• Choose Stat > 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 Values.
• Click OK.

Output using MINITAB software is given below:

From the MINITAB output, it can be observed that the shape of the population distribution is uniform.

Thus, the shape of the population distribution is Uniform.

The frequency distribution of the sample means is:

Sample mean	Frequency	Probability
1	1	$\frac{1}{9}=0.11$
1.5	2	$\frac{2}{9}=0.22$
2	3	$\frac{3}{9}=0.33$
2.5	2	$\frac{2}{9}=0.22$
3	1	$\frac{1}{9}=0.11$
	$N=9$

Software procedure:

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

• Choose Stat > 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 MINITAB output, it can be observed that the shape of the distribution of the sample means is triangular.

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