Question

Refer to Exercise 7.6 on page 295.

a. Use your answers from Exercise 7.6(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 7.6(a).

Short answer

Part a. The variable $\overline{x}$has a mean value of ${\mu }_{\overline{x}}=5.5$for each of the possible sample sizes.

Part b. The population mean is $\mu =5.5$.

Step-by-step solution

Part (a) Step 1. Given Information  

It is given that the population data is 3,4,7,8.

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: 3,4,7,8.

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

Sample	$\overline{x}$
3	3
4	4
7	7
8	8

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

$$\begin{gathered} {\mu }_{\overline{x}}=\frac{3+4+7+8}{4} \\ {\mu }_{\overline{x}}=\frac{22}{4} \\ {\mu }_{\overline{x}}=5.5 \end{gathered}$$

So when the sample size is 1, the variable $\overline{x}$has a mean ${\mu }_{\overline{x}}=5.5$.

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

For the population data: 3,4,7,8.

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

Sample	$\overline{x}$
3,4	$\frac{3+4}{2}=3.5$
3,7	$\frac{3+7}{2}=5$
3,8	$\frac{3+8}{2}=5.5$
4,7	$\frac{4+7}{2}=5.5$
4,8	$\frac{4+8}{2}=6$
7,8	$\frac{7+8}{2}=7.5$

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

$$\begin{gathered} {\mu }_{\overline{x}}=\frac{3.5+5+5.5+5.5+6+7.5}{6} \\ {\mu }_{\overline{x}}=\frac{33}{6} \\ {\mu }_{\overline{x}}=5.5 \end{gathered}$$

So when the sample size is 2, the variable $\overline{x}$has a mean ${\mu }_{\overline{x}}=5.5$.

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

For the population data: 3,4,7,8.

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

Sample	$\overline{x}$
3,4,7	$\frac{3+4+7}{3}=4.67$
3,4,8	$\frac{3+4+8}{3}=5$
3,7,8	$\frac{3+7+8}{3}=6$
4,7,8	$\frac{4+7+8}{3}=6.33$

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

$$\begin{gathered} {\mu }_{\overline{x}}=\frac{4.67+5+6+6.33}{4} \\ {\mu }_{\overline{x}}=\frac{22}{4} \\ {\mu }_{\overline{x}}=5.5 \end{gathered}$$

So when the sample size is 3, the variable $\overline{x}$has a mean ${\mu }_{\overline{x}}=5.5$.

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

For the population data: 3,4,7,8.

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

Sample	$\overline{x}$
3,4,7,8	$\frac{3+4+7+8}{4}=5.5$

So when the sample size is 4, the variable$\overline{x}$has a mean ${\mu }_{\overline{x}}=5.5$.

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: 3,4,7,8 the population mean can be given as

$$\begin{gathered} \mu =\frac{3+4+7+8}{4} \\ \mu =\frac{22}{4} \\ \mu =5.5 \end{gathered}$$

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