Q. 7.39 - Elementary Statistics

Question

Refer to Exercise 7.9 on page 295.

a. Use your answers from Exercise 7.9(b) to determine the mean, ${\mu }_s$, of the variable $\overline{x}$ for each of the possible sample sizes.

b. For each of the possible sample sizes, determine the mean, ${\mu }_s$, of the variable $\overline{x}$, using only your answer from Exercise 7.9(a).

Step-by-Step Solution

Answer

Part a. The variable $\overline{x}$ has a mean value of ${\mu }_{\overline{x}}=3.5$ for each of the possible sample sizes.

Part b. The population mean is $\mu =3.5$.

Part (a) Step 1. Given Information

It is given that the population data is 1,2,3,4,5,6.

We need to determine the mean, ${\mu }_s$, of the variable $\overline{x}$ for each of the possible sample sizes.

Part (a) Step 2. When the sample size is 1

For the population data: 1,2,3,4,5,6.

The sample and sample mean for a sample of size $n=1$ are shown in the table below.

Sample	$\overline{x}$
1	1
2	2
3	3
4	4
5	5
6	6

The variable $\overline{x}$ has the following mean

$$\begin{gathered} {\mu }_{\overline{x}}=\frac{1+2+3+4+5+6}{6} \\ {\mu }_{\overline{x}}=\frac{21}{6} \\ {\mu }_{\overline{x}}=3.5 \end{gathered}$$

So when the sample size is 1, the variable $\overline{x}$ has a mean ${\mu }_{\overline{x}}=3.5$.

Part (a) Step 3. When the sample size is 2

For the population data: 1,2,3,4,5,6.

The sample and sample mean for a sample of size $n=2$ are shown in the table below.

Sample	$\overline{x}$
1,2	$\frac{1+2}{2}=1.5$
1,3	$\frac{1+3}{2}=2$
1,4	$\frac{1+4}{2}=2.5$
1,5	$\frac{1+5}{2}=3$
1,6	$\frac{1+6}{2}=3.5$
2,3	$\frac{2+3}{2}=2.5$
2,4	$\frac{2+4}{2}=3$
2,5	$\frac{2+5}{2}=3.5$
2,6	$\frac{2+6}{2}=4$
3,4	$\frac{3+4}{2}=3.5$
3,5	$\frac{3+5}{2}=4$
3,6	$\frac{3+6}{2}=4.5$
4,5	$\frac{4+5}{2}=4.5$
4,6	$\frac{4+6}{2}=5$
5,6	$\frac{5+6}{2}=5.5$

The variable $\overline{x}$ has the following mean

$$\begin{gathered} {\mu }_{\overline{x}}=\frac{1.5+2+2.5+3+3.5+2.5+3+3.5+4+3.5+4+4.5+4.5+5+5.5}{15} \\ {\mu }_{\overline{x}}=\frac{52.5}{15} \\ {\mu }_{\overline{x}}=3.5 \end{gathered}$$

So when the sample size is 2, the variable $\overline{x}$ has a mean ${\mu }_{\overline{x}}=3.5$.

Part (a) Step 4. When the sample size is 3

For the population data: 1,2,3,4,5,6.

The sample and sample mean for a sample of size $n=3$ are shown in the table below.

Sample	$\overline{x}$
1,2,3	$\frac{1+2+3}{3}=2$
1,2,4	$\frac{1+2+4}{3}=2.33$
1,2,5	$\frac{1+2+5}{3}=2.67$
1,2,6	$\frac{1+2+6}{3}=3$
1,3,4	$\frac{1+3+4}{3}=2.67$
1,3,5	$\frac{1+3+5}{3}=3$
1,3,6	$\frac{1+3+6}{3}=3.33$
1,4,5	$\frac{1+4+5}{3}=3.33$
1,4,6	$\frac{1+4+6}{3}=3.67$
1,5,6	$\frac{1+5+6}{3}=4$
2,3,4	$\frac{2+3+4}{3}=3$
2,3,5	$\frac{2+3+5}{3}=3.33$
2,3,6	$\frac{2+3+6}{3}=3.67$
2,4,5	$\frac{2+4+5}{3}=3.67$
2,4,6	$\frac{2+4+6}{3}=4$
2,5,6	$\frac{2+5+6}{3}=4.33$
3,4,5	$\frac{3+4+5}{3}=4$
3,4,6	$\frac{3+4+6}{3}=4.33$
3,5,6	$\frac{3+5+6}{3}=4.67$
4,5,6	$\frac{4+5+6}{3}=5$

The variable $\overline{x}$ has the following mean

$$\begin{gathered} {\mu }_{\overline{x}}=\frac{2+2.33+2.67+3+2.67+3+3.33+3.33+3.67+4+3+3.33+3.67+3.67+4+4.33+4+4.33+4.67+5}{20} \\ {\mu }_{\overline{x}}=\frac{70}{20} \\ {\mu }_{\overline{x}}=3.5 \end{gathered}$$

So when the sample size is 3, the variable $\overline{x}$ has a mean ${\mu }_{\overline{x}}=3.5$.

Part (a) Step 5. When the sample size is 4

For the population data: 1,2,3,4,5,6.

The sample and sample mean for a sample of size $n=4$ are shown in the table below.

Sample	$\overline{x}$
1,2,3,4	$\frac{1+2+3+4}{4}=2.5$
1,2,3,5	$\frac{1+2+3+5}{4}=2.75$
1,2,3,6	$\frac{1+2+3+6}{4}=3$
1,2,4,5	$\frac{1+2+4+5}{4}=3$
1,2,4,6	$\frac{1+2+4+6}{4}=3.25$
1,2,5,6	$\frac{1+2+5+6}{4}=3.5$
1,3,4,5	$\frac{1+3+4+5}{4}=3.25$
1,3,4,6	$\frac{1+3+4+6}{4}=3.5$
1,3,5,6	$\frac{1+3+5+6}{4}=3.75$
1,4,5,6	$\frac{1+4+5+6}{4}=4$
2,3,4,5	$\frac{2+3+4+5}{4}=3.5$
2,3,4,6	$\frac{2+3+4+6}{4}=3.75$
2,3,5,6	$\frac{2+3+5+6}{4}=4$
2,4,5,6	$\frac{2+4+5+6}{4}=4.25$
3,4,5,6	$\frac{3+4+5+6}{4}=4.5$

The variable $\overline{x}$ has the following mean

$$\begin{gathered} {\mu }_{\overline{x}}=\frac{2.5+2.75+3+3+3.25+3.5+3.25+3.5+3.75+4+3.5+3.75+4+4.25+4.5}{15} \\ {\mu }_{\overline{x}}=\frac{52.5}{15} \\ {\mu }_{\overline{x}}=3.5 \end{gathered}$$

So when the sample size is 4, the variable $\overline{x}$ has a mean ${\mu }_{\overline{x}}=3.5$.

Part (a) Step 6. When the sample size is 5

For the population data: 1,2,3,4,5,6.

The sample and sample mean for a sample of size $n=5$ are shown in the table below.

Sample	$\overline{x}$
1,2,3,4,5	$\frac{1+2+3+4+5}{5}=3$
1,2,3,4,6	$\frac{1+2+3+4+6}{5}=3.2$
1,2,3,5,6	$\frac{1+2+3+5+6}{5}=3.4$
1,2,4,5,6	$\frac{1+2+4+5+6}{5}=3.6$
1,3,4,5,6	$\frac{1+3+4+5+6}{5}=3.8$
2,3,4,5,6	$\frac{2+3+4+5+6}{5}=4$

The variable $\overline{x}$ has the following mean

$$\begin{gathered} {\mu }_{\overline{x}}=\frac{3+3.2+3.4+3.6+3.8+4}{6} \\ {\mu }_{\overline{x}}=\frac{21}{6} \\ {\mu }_{\overline{x}}=3.5 \end{gathered}$$

So when the sample size is 5, the variable $\overline{x}$ has a mean ${\mu }_{\overline{x}}=3.5$.

Part (a) Step 7. When the sample size is 6

For the population data: 1,2,3,4,5,6.

The sample and sample mean for a sample of size $n=6$ are shown in the table below.

Sample	$\overline{x}$
1,2,3,4,5,6	$\frac{1+2+3+4+5+6}{6}=3.5$

So when the sample size is 6, the variable $\overline{x}$ has a mean ${\mu }_{\overline{x}}=3.5$.

Thus it can be seen that the mean of all potential sample means is the same. 

Part (b) Step 1. Find the population mean

For the given population data: 1,2,3,4,5,6 the population mean can be given as

$$\begin{gathered} \mu =\frac{1+2+3+4+5+6}{6} \\ \mu =\frac{21}{6} \\ \mu =3.5 \end{gathered}$$

So from the results, it can be observed that the population mean is equal to the mean of all potential sample means that is ${\mu }_{\overline{x}}=\mu$.