In Exercises 7–10, use the same population of {4, 5, 9} that was used in Examples 2 and 5. As in Examples 2 and 5, assume that samples of size n = 2 are randomly selected with replacement.

Sampling Distribution of the Sample Variance

a. Find the value of the population variance ${\sigma }^2$.

b. Table 6-2 describes the sampling distribution of the sample mean. Construct a similar table representing the sampling distribution of the sample variance $s^2$. Then combine values of $s^2$that are the same, as in Table 6-3 (Hint: See Example 2 on page 258 for Tables 6-2 and 6-3, which describe the sampling distribution of the sample mean.)

c. Find the mean of the sampling distribution of the sample variance.

d. Based on the preceding results, is the sample variance an unbiased estimator of the population variance? Why or why not?

A population of ages of three children is considered. Samples of size equal to 2 are extracted from this population with replacement.

a.

The population mean is computed as shown below:

$$\begin{gathered} \mu =\frac{\sum x}{n} \\ =\frac{4+5+9}{3} \\ =6 \end{gathered}$$

The value of the population variance is computed as follows:

$$\begin{gathered} {\sigma }^2=\frac{{\sum }_{i=1}^n{(x_i-\mu )}^2}{n} \\ =\frac{{\left(4-6\right)}^2+{\left(5-6\right)}^2+{\left(9-6\right)}^2}{3} \\ =4.7 \end{gathered}$$

Thus, the population variance $\left({\sigma }^2\right)$ is equal to 4.7.

b.

All possible samples of size 2 selected with replacement are tabulated below:

The sample means of all the nine samples are computed below:

$$\begin{gathered} {\overline{x}}_1=\frac{4+4}{2} \\ =4 \\ {\overline{x}}_2=\frac{4+5}{2} \\ =4.5 \\ {\overline{x}}_3=\frac{4+9}{2} \\ =6.5 \end{gathered}$$

$$\begin{gathered} {\overline{x}}_4=\frac{5+4}{2} \\ =4.5 \\ {\overline{x}}_5=\frac{5+5}{2} \\ =5 \\ {\overline{x}}_6=\frac{5+9}{2} \\ =7 \end{gathered}$$

$$\begin{gathered} {\overline{x}}_7=\frac{9+4}{2} \\ =6.5 \\ {\overline{x}}_8=\frac{9+5}{2} \\ =7 \\ {\overline{x}}_9=\frac{9+9}{2} \\ =9 \end{gathered}$$

$s^2=\frac{{\sum }_{i=1}^n{(x_i-\overline{x})}^2}{n-1}$

Since there are nine samples, the probability of the nine sample variances is written as $\frac{1}{9}$.

The following table shows all possible samples of size equal to 2, the corresponding sample variances, and the probability values.

Combining the values of $s^2$ that are the same, the following probability values are obtained.

c.

The mean of the sample variances is computed below:

$$\begin{gathered} {\overline{s}}^2=\frac{s_1^2+s_2^2+.....+s_9^2}{9} \\ =\frac{0+0.5+......+0}{9} \\ =4.7 \end{gathered}$$

Thus, the mean of the sampling distribution of the sample variance is equal to 4.7.

d.

An unbiased estimator is a sample statistic whose sampling distribution has a mean value equal to the population parameter.

The mean value of the sampling distribution of the sample variance is equal to the population variance.

Thus, the sample variance can be considered as an unbiased estimator of the population variance.