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).

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.

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.

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 ${\mu }_{\overline{x}}=5.5$.

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.

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 ${\mu }_{\overline{x}}=5.5$.

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.

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 ${\mu }_{\overline{x}}=5.5$.

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.

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.

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 ${\mu }_{\overline{x}}=\mu$.