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 Median

a. Find the value of the population median.

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

Short answer

a. Population Median: 5.0

b. The following table represents the sampling distribution of the sample medians.

Sample	Sample Median	Probability
(4,4)	4.0	$\frac{1}{9}$
(4,5)	4.5	$\frac{1}{9}$
(4,9)	6.5	$\frac{1}{9}$
(5,4)	4.5	$\frac{1}{9}$
(5,5)	5.0	$\frac{1}{9}$
(5,9)	7.0	$\frac{1}{9}$
(9,4)	6.5	$\frac{1}{9}$
(9,5)	7.0	$\frac{1}{9}$
(9,9)	9.0	$\frac{1}{9}$

By combining all the same values of medians, the following table is obtained.

Sample Median	Probability
4.0	$\frac{1}{9}$
4.5	$\frac{2}{9}$
5.0	$\frac{1}{9}$
6.5	$\frac{2}{9}$
7.0	$\frac{2}{9}$
9.0	$\frac{1}{9}$

c.The mean of the sampling distribution of the sample median is equal to 6.0.

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

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 median

a.

The observations are {4,5,9}.

Since the population size is odd (n = 3), the following formula is used to compute the population median:

$$\begin{gathered} \text{Population}\text{Median}={\left(\frac{n+1}{2}\right)}^{th}\text{observation} \\ ={\left(\frac{3+1}{2}\right)}^{th}\text{observation} \\ =2^{nd}\text{observation} \\ =5 \end{gathered}$$

Thus, the population median is equal to 5.

Sampling distribution of sample medians

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)

Note that for a sample of size 2, the median has the following formula:

$$\begin{gathered} Median=\frac{{\left(\frac{n}{2}\right)}^{th}obs+{\left(\frac{n}{2}+1\right)}^{th}obs}{2} \\ =\frac{{\left(\frac{2}{2}\right)}^{th}obs+{\left(\frac{2}{2}+1\right)}^{th}obs}{2} \\ =\frac{1^{st}obs+2^{nd}obs}{2} \end{gathered}$$

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

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

Sample	Sample median	Probability
(4,4)	$$\begin{gathered} \text{Sample}{\text{Median}}_1=\frac{4+4}{2} \\ =4.0 \end{gathered}$$	$\frac{1}{9}$
(4,5)	$$\begin{gathered} \text{Sample}{\text{Median}}_2=\frac{4+5}{2} \\ =4.5 \end{gathered}$$	$\frac{1}{9}$
(4,9)	$$\begin{gathered} \text{Sample}{\text{Median}}_3=\frac{4+9}{2} \\ =6.5 \end{gathered}$$	$\frac{1}{9}$
(5,4)	$$\begin{gathered} \text{Sample}{\text{Median}}_4=\frac{5+4}{2} \\ =4.5 \end{gathered}$$	$\frac{1}{9}$
(5,5)	$$\begin{gathered} \text{Sample}{\text{Median}}_5=\frac{5+5}{2} \\ =5.0 \end{gathered}$$	$\frac{1}{9}$
(5,9)	$$\begin{gathered} \text{Sample}{\text{Median}}_6=\frac{5+9}{2} \\ =7.0 \end{gathered}$$	$\frac{1}{9}$
(9,4)	$$\begin{gathered} \text{Sample}{\text{Median}}_7=\frac{9+4}{2} \\ =6.5 \end{gathered}$$	$\frac{1}{9}$
(9,5)	$$\begin{gathered} \text{Sample}{\text{Median}}_8=\frac{9+5}{2} \\ =7.0 \end{gathered}$$	$\frac{1}{9}$
(9,9)	$$\begin{gathered} \text{Sample}{\text{Median}}_9=\frac{9+9}{2} \\ =9.0 \end{gathered}$$	$\frac{1}{9}$

By combining the values of medians that are the same, the following probability values are obtained.

Sample median	Probability
4.0	$\frac{1}{9}$
4.5	$\frac{2}{9}$
5.0	$\frac{1}{9}$
6.5	$\frac{2}{9}$
7.0	$\frac{2}{9}$
9.0	$\frac{1}{9}$

Mean of the sample medians

c.

The mean of the sample medians is computed below:

$$\begin{gathered} \text{Mean}\text{of}\text{Sample}\text{Medians}=\frac{\text{Sample}{\text{Med}}_{\text{1}}+\text{Sample}{\text{Med}}_{\text{2}}+.....+\text{Sample}{\text{Med}}_{\text{9}}}{9} \\ =\frac{4+4.5+......+9}{9} \\ =6.0 \end{gathered}$$

Thus, the mean of the sampling distribution of the sample median is equal to 6.0.

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 median (6.0) is not equal to the population median (5.0).