Question

7.34 Refer to Exercise 7.4 on page 295.

a. Use your answers from Exercise 7.4(b) to determine the mean, ${\mu }_5$, of the variable $\tilde{x}$ for each of the possible sample sizes.

b. For each of the possible sample sizes, determine the mean, ${\mu }_5$, of the variable $\overrightarrow{x}$, using only your answer from Exercise 7.4(a).

Short answer

a). The variable $\overline{x}$has a mean value of $5$ (${\mu }_{\overline{x}}$).

b). $5$is the population average.

Step-by-step solution

Part (a) Step 1: Given Information

Data on the population:$2,5,8$.

Part (a) Step 2: Explanation

For the variable $x$, we have population data, which is $2,5,8$.

The mean ${\mu }_{\overline{x}}$of the variable $\overline{x}$for each of the samples is calculated as follows:

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

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

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

${\mu }_{\overline{x}}=\frac{2+5+8}{3}$

$=\frac{15}{3}$

$=5$

The variable $\overline{x}$has a mean ${\mu }_{\overline{x}}$ of $5$.

Part (a) Step 3: Explanation

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

Sample	$\overline{x}$
$2,5$	$\frac{2+5}{2}=3.5$
$2,8$	$\frac{2+8}{2}=5$
$5,8$	$\frac{5+8}{2}=6.5$

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

role="math" localid="1650970365589" ${\mu }_{\overline{x}}=\frac{3.5+2+6.5}{3}$

role="math" localid="1650970431300" $=\frac{15}{3}$

$=5$

The variable $\overline{x}$ has a mean value of $5$ (${\mu }_{\overline{x}}$).

Part (a) Step 4: Explanation

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

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

Interpretation: We can see from the foregoing that the mean of all potential sample means is the same.

Part (b) Step 1: Given Information

Data on the population:$2,5,8$.

Part (b) Step 2: Explanation

The following is a definition of the population mean:

$\mu =\frac{\sum x_i}{N}$

$=\frac{2+5+8}{3}$

$=5$

$5$is the population average.

We can see that $\mu =5$ based on the results of parts (a) and (b).

The population mean is equal to the mean of all feasible sample means.