91Ó°ÊÓ

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

We need to determine the mean, ${\mathrm{μ}}_s$, of the variable $\stackrel{¯}{x}$for each of the possible sample sizes.

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

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

The variable $\stackrel{¯}{x}$has the following mean

$$\begin{gathered} {\mathrm{μ}}_{\stackrel{¯}{x}}=\frac{1+2+3+4}{4} \\ {\mathrm{μ}}_{\stackrel{¯}{x}}=\frac{10}{4} \\ {\mathrm{μ}}_{\stackrel{¯}{x}}=2.5 \end{gathered}$$

So when the sample size is 1, the variable $\stackrel{¯}{x}$has a mean ${\mathrm{μ}}_{\stackrel{¯}{x}}=2.5$.

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

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

The variable $\stackrel{¯}{x}$has the following mean

$$\begin{gathered} {\mathrm{μ}}_{\stackrel{¯}{x}}=\frac{1.5+2+2.5+2.5+3+3.5}{6} \\ {\mathrm{μ}}_{\stackrel{¯}{x}}=\frac{15}{6} \\ {\mathrm{μ}}_{\stackrel{¯}{x}}=2.5 \end{gathered}$$

So when the sample size is 2, the variable $\stackrel{¯}{x}$has a mean ${\mathrm{μ}}_{\stackrel{¯}{x}}=2.5$.

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

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

The variable $\stackrel{¯}{x}$has the following mean

$$\begin{gathered} {\mathrm{μ}}_{\stackrel{¯}{x}}=\frac{2+2.33+2.67+3}{4} \\ {\mathrm{μ}}_{\stackrel{¯}{x}}=\frac{10}{4} \\ {\mathrm{μ}}_{\stackrel{¯}{x}}=2.5 \end{gathered}$$

So when the sample size is 3, the variable $\stackrel{¯}{x}$has a mean ${\mathrm{μ}}_{\stackrel{¯}{x}}=2.5$.

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

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 $\stackrel{¯}{x}$has a mean ${\mathrm{μ}}_{\stackrel{¯}{x}}=2.5$.

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

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

$$\begin{gathered} \mathrm{μ}=\frac{1+2+3+4}{4} \\ \mathrm{μ}=\frac{10}{4} \\ \mathrm{μ}=2.5 \end{gathered}$$

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