Refer to Exercise 7.10 on page 295.

a. Use your answers from Exercise 7.10(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 7.10a).

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

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

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

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

$$\begin{gathered} {\mu }_{\overline{x}}=\frac{2+3+5+5+7+8}{6} \\ {\mu }_{\overline{x}}=\frac{30}{6} \\ {\mu }_{\overline{x}}=5 \end{gathered}$$

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

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

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

$$\begin{gathered} {\mu }_{\overline{x}}=\frac{2.5+3.5+3.5+4.5+5+4+4+5+5.5+5+6+6.5+6+6.5+7.5}{15} \\ {\mu }_{\overline{x}}=\frac{75}{15} \\ {\mu }_{\overline{x}}=5 \end{gathered}$$

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

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

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

$$\begin{gathered} {\mu }_{\overline{x}}=\frac{3.33+3.33+4+4.33+4+4.67+5+4.67+5+5.67+4.33+5+5.33+5+5.33+6+5.67+6+6.67+6.67}{20} \\ {\mu }_{\overline{x}}=\frac{100}{20} \\ {\mu }_{\overline{x}}=5 \end{gathered}$$

So when the sample size is 3, the variable $\overline{x}$has a mean ${\mu }_{\overline{x}}=5$.

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

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

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

$$\begin{gathered} {\mu }_{\overline{x}}=\frac{3.75+4.25+4.5+4.25+4.5+5+4.75+5+5.5+5.5+5+5.25+5.75+5.75+6.25}{15} \\ {\mu }_{\overline{x}}=\frac{75}{15} \\ {\mu }_{\overline{x}}=5 \end{gathered}$$

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

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

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

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

$$\begin{gathered} {\mu }_{\overline{x}}=\frac{4.4+4.6+5+5+5.4+5.6}{6} \\ {\mu }_{\overline{x}}=\frac{30}{6} \\ {\mu }_{\overline{x}}=5 \end{gathered}$$

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

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

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

So when the sample size is 6, 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.

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

$$\begin{gathered} \mu =\frac{2+3+5+5+7+8}{6} \\ \mu =\frac{30}{6} \\ \mu =5 \end{gathered}$$

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