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

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.

b.

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

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.

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

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.

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).