Question

Refer to Exercise 7.8 on page 295.

a. Use your answers from Exercise 7.8(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 7.8(a).

Short answer

Part a. The variable $\overline{x}$has a mean value of ${\mu }_{\overline{x}}=5$for each of the possible sample sizes.

Part b. The population mean is $\mu =5$.

Step-by-step solution

Part (a) Step 1. Given Information   

It is given that the population data is 2,3,5,7,8.

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: 2,3,5,7,8.

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

Sample	$\overline{x}$
2	2
3	3
5	5
7	7
8	8

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

$$\begin{gathered} {\mu }_{\overline{x}}=\frac{2+3+5+7+8}{5} \\ {\mu }_{\overline{x}}=\frac{25}{5} \\ {\mu }_{\overline{x}}=5 \end{gathered}$$

So when the sample size is 1, the variable $\overline{x}$has a mean ${\mu }_{\overline{x}}=5$.

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

For the population data: 2,3,5,7,8.

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

Sample	$\overline{x}$
2,3	$\frac{2+3}{2}=2.5$
2,5	role="math" localid="1652556055867" $\frac{2+5}{2}=3.5$
2,7	role="math" localid="1652556064614" $\frac{2+7}{2}=4.5$
2,8	$\frac{2+8}{2}=5$
3,5	$\frac{3+5}{2}=4$
3,7	$\frac{3+7}{2}=5$
3,8	$\frac{3+8}{2}=5.5$
5,7	$\frac{5+7}{2}=6$
5,8	$\frac{5+8}{2}=6.5$
7,8	$\frac{7+8}{2}=7.5$

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

$$\begin{gathered} {\mu }_{\overline{x}}=\frac{2.5+3.5+4.5+5+4+5+5.5+6+6.5+7.5}{10} \\ {\mu }_{\overline{x}}=\frac{50}{10} \\ {\mu }_{\overline{x}}=5 \end{gathered}$$

So when the sample size is 2, the variable $\overline{x}$has a mean ${\mu }_{\overline{x}}=5$.

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

For the population data: 2,3,5,7,8.

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

Sample	$\overline{x}$
2,3,5	$\frac{2+3+5}{3}=3.33$
2,3,7	$\frac{2+3+7}{3}=4$
2,3,8	$\frac{2+3+8}{3}=4.33$
2,5,7	$\frac{2+5+7}{3}=4.67$
2,5,8	role="math" localid="1652556496326" $\frac{2+5+8}{3}=5$
2,7,8	$\frac{2+7+8}{3}=5.67$
3,5,7	$\frac{3+5+7}{3}=5$
3,5,8	$\frac{3+5+8}{3}=5.33$
3,7,8	$\frac{3+7+8}{3}=6$
5,7,8	$\frac{5+7+8}{3}=6.67$

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

$$\begin{gathered} {\mu }_{\overline{x}}=\frac{3.33+4+4.33+4.67+5+5.67+5+5.33+6+6.67}{10} \\ {\mu }_{\overline{x}}=\frac{50}{10} \\ {\mu }_{\overline{x}}=5 \end{gathered}$$

So when the sample size is 2, the variable $\overline{x}$has a mean ${\mu }_{\overline{x}}=5$.

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

For the population data: 2,3,5,7,8.

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

Sample	$\overline{x}$
2,3,5,7	$\frac{2+3+5+7}{4}=4.25$
2,3,5,8	$\frac{2+3+5+8}{4}=4.5$
2,3,7,8	$\frac{2+3+7+8}{4}=5$
2,5,7,8	$\frac{2+5+7+8}{4}=5.5$
3,5,7,8	$\frac{3+5+7+8}{4}=5.75$

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

$$\begin{gathered} {\mu }_{\overline{x}}=\frac{4.25+4.5+5+5.5+5.75}{5} \\ {\mu }_{\overline{x}}=\frac{25}{5} \\ {\mu }_{\overline{x}}=5 \end{gathered}$$

So when the sample size is 4, the variable $\overline{x}$has a mean ${\mu }_{\overline{x}}=5$.

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

For the population data: 2,3,5,7,8.

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

Sample	$\overline{x}$
2,3,5,7,8	$\frac{2+3+5+7+8}{5}=5$

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

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: 2,3,5,7,8 the population mean can be given as

$$\begin{gathered} \mu =\frac{2+3+5+7+8}{5} \\ \mu =\frac{25}{5} \\ \mu =5 \end{gathered}$$

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