Question

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 Standard Deviation For the following, round results to three decimal places.

a. Find the value of the population standard deviation

b. Table 6-2 describes the sampling distribution of the sample mean. Construct a similar table representing the sampling distribution of the sample standard deviation s. Then combine values of s 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 standard deviation. d. Based on the preceding results, is the sample standard deviation an unbiased estimator of the population standard deviation? Why or why not?

Short answer

a. Population standard deviation: 2.160

b. The following table represents the sampling distribution of the sample standard deviation.

Sample	Sample standard deviation$\left(s^2\right)$	Probability
(4,4)	0	$\frac{1}{9}$
(4,5)	0.707	$\frac{1}{9}$
(4,9)	3.536	$\frac{1}{9}$
(5,4)	0.707	$\frac{1}{9}$
(5,5)	0	$\frac{1}{9}$
(5,9)	2.828	$\frac{1}{9}$
(9,4)	3.536	$\frac{1}{9}$
(9,5)	2.828	$\frac{1}{9}$
(9,9)	0	$\frac{1}{9}$

Combining all the same values of $s$, the following table is obtained.

Sample standard deviation	Probability
0.0	$\frac{3}{9}$
0.707	$\frac{2}{9}$
3.536	$\frac{2}{9}$
2.828	$\frac{2}{9}$

c.The mean of the sampling distribution of the sample standard deviation is equal to 1.571.

d. Since the mean value of the sampling distribution of the sample standard deviation is not equal to the population standard deviation, the sample standard deviation cannot be considered an unbiased estimator of the population standard deviation.

Step-by-step solution

Given information

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

Population standard deviation

a.

The population mean is computed as shown below:

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

The value of the population standard deviation is computed as follows:

$$\begin{gathered} \sigma =\sqrt{\frac{{\sum }_{i=1}^n{(x_i-\mu )}^2}{n}} \\ =\sqrt{\frac{{\left(4-6\right)}^2+{\left(5-6\right)}^2+{\left(9-6\right)}^2}{3}} \\ =2.160 \end{gathered}$$

Thus, the population standard deviation $\left(\sigma \right)$is equal to 2.160.

Sampling distribution of sample standard deviations

b.

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

(4,4)	(4,5)	(4,9)
(5,4)	(5,5)	(5,9)
(9,4)	(9,5)	(9,9)

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

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

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

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

The following formula of the sample standard deviation is utilized to compute the value of s for each of the nine samples:

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

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

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

Sample	Sample standard deviation(s)	Probability
(4,4)	$$\begin{gathered} s_1=\sqrt{\frac{{\left(4-4\right)}^2+{\left(4-4\right)}^2}{2-1}} \\ =0 \end{gathered}$$	$\frac{1}{9}$
(4,5)	$$\begin{gathered} s_2=\sqrt{\frac{{\left(4-4.5\right)}^2+{\left(5-4.5\right)}^2}{2-1}} \\ =0.707 \end{gathered}$$	$\frac{1}{9}$
(4,9)	$$\begin{gathered} s_3=\sqrt{\frac{{\left(4-6.5\right)}^2+{\left(9-6.5\right)}^2}{2-1}} \\ =3.536 \end{gathered}$$	$\frac{1}{9}$
(5,4)	$$\begin{gathered} s_4=\sqrt{\frac{{\left(5-4.5\right)}^2+{\left(4-4.5\right)}^2}{2-1}} \\ =0.707 \end{gathered}$$	$\frac{1}{9}$
(5,5)	$$\begin{gathered} s_5=\sqrt{\frac{{\left(5-5\right)}^2+{\left(5-5\right)}^2}{2-1}} \\ =0 \end{gathered}$$	$\frac{1}{9}$
(5,9)	$$\begin{gathered} s_6=\sqrt{\frac{{\left(5-7\right)}^2+{\left(9-7\right)}^2}{2-1}} \\ =2.828 \end{gathered}$$	$\frac{1}{9}$
(9,4)	$$\begin{gathered} s_7=\sqrt{\frac{{\left(9-6.5\right)}^2+{\left(4-6.5\right)}^2}{2-1}} \\ =3.536 \end{gathered}$$	$\frac{1}{9}$
(9,5)	$$\begin{gathered} s_8=\sqrt{\frac{{\left(9-7\right)}^2+{\left(5-7\right)}^2}{2-1}} \\ =2.828 \end{gathered}$$	$\frac{1}{9}$
(9,9)	$$\begin{gathered} s_9=\sqrt{\frac{{\left(9-9\right)}^2+{\left(9-9\right)}^2}{2-1}} \\ =0 \end{gathered}$$	$\frac{1}{9}$

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

Sample standard deviation$\left(s^2\right)$	Probability
0	$\frac{3}{9}$
0.707	$\frac{2}{9}$
3.536	$\frac{2}{9}$
2.828	$\frac{2}{9}$

Mean of the sample standard deviations

c.

The mean of the sample standard deviations is computed below:

$$\begin{gathered} \overline{s}=\frac{s_1+s_2+.....+s_9}{9} \\ =\frac{0+0.707+......+0}{9} \\ =1.571 \end{gathered}$$

Thus, the mean of the sampling distribution of the sample standard deviation is equal to 1.571.

Unbiased estimator vs biased estimator

d.

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

Similarly, a biased estimator is a sample statistic whose sampling distribution has a mean value not equal to the population parameter.

The mean value of the sampling distribution of the sample standard deviation is not equal to the population standard deviation.

Thus, the sample standard deviation cannot be considered an unbiased estimator of the population standard deviation.