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 Proportion

a. For the population, find the proportion of odd numbers.

b. Table 6-2 describes the sampling distribution of the sample mean. Construct a similar table representing the sampling distribution of the sample proportion of odd numbers. Then combine values of the sample proportion 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 proportion of odd numbers.

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

Short answer

a. Population Proportion: 0.67

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

Sample	Sample proportion	Probability
(4,4)	0	$\frac{1}{9}$
(4,5)	0.5	$\frac{1}{9}$
(4,9)	0.5	$\frac{1}{9}$
(5,4)	0.5	$\frac{1}{9}$
(5,5)	1	$\frac{1}{9}$
(5,9)	1	$\frac{1}{9}$
(9,4)	0.5	$\frac{1}{9}$
(9,5)	1	$\frac{1}{9}$
(9,9)	1	$\frac{1}{9}$

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

Sample proportion	Probability
0.0	$\frac{1}{9}$
0.5	$\frac{4}{9}$
1.0	$\frac{4}{9}$

c.The mean of the sampling distribution of the sample proportions is equal to 0.67.

d. Since the mean value of the sampling distribution of the sample proportion is equal to the population proportion, the sample proportion of odd numbers can be considered an unbiased estimator of the population proportion of odd numbers.

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 proportion

a.

The observations are {4,5,9}.

The total number of values (n) is equal to3.

The number of odd values (x) is equal to 2.

The population proportion of odd numbers is equal to

$$\begin{gathered} p=\frac{x}{n} \\ =\frac{2}{3} \\ =0.67 \end{gathered}$$

Thus, the population proportion of odd numbers is equal to 0.67.

Sampling distribution of sample proportions

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 number of odd values in each of the nine samples is tabulated below.

Sample	Number of odd values
(4,4)	0
(4,5)	1
(4,9)	1
(5,4)	1
(5,5)	2
(5,9)	2
(9,4)	1
(9,5)	2
(9,9)	2

The following formula is used to compute the sample proportions:

$\hat{p}=\frac{\text{Number}\text{of}\text{odd}\text{numbers}}{\text{Sample}\text{Size}}$

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

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

Sample	Sample proportion	Probability
(4,4)	$$\begin{gathered} {\hat{p}}_1=\frac{0}{2} \\ =0 \end{gathered}$$	$\frac{1}{9}$
(4,5)	$$\begin{gathered} {\hat{p}}_2=\frac{1}{2} \\ =0.5 \end{gathered}$$	$\frac{1}{9}$
(4,9)	$$\begin{gathered} {\hat{p}}_3=\frac{1}{2} \\ =0.5 \end{gathered}$$	$\frac{1}{9}$
(5,4)	$$\begin{gathered} {\hat{p}}_4=\frac{1}{2} \\ =0.5 \end{gathered}$$	$\frac{1}{9}$
(5,5)	$$\begin{gathered} {\hat{p}}_5=\frac{2}{2} \\ =1 \end{gathered}$$	$\frac{1}{9}$
(5,9)	$$\begin{gathered} {\hat{p}}_6=\frac{2}{2} \\ =1 \end{gathered}$$	$\frac{1}{9}$
(9,4)	$$\begin{gathered} {\hat{p}}_7=\frac{1}{2} \\ =0.5 \end{gathered}$$	$\frac{1}{9}$
(9,5)	$$\begin{gathered} {\hat{p}}_8=\frac{2}{2} \\ =1 \end{gathered}$$	$\frac{1}{9}$
(9,9)	$$\begin{gathered} {\hat{p}}_9=\frac{2}{2} \\ =1 \end{gathered}$$	$\frac{1}{9}$

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

Sample Median	Probability
0	$\frac{1}{9}$
0.5	$\frac{4}{9}$
1	$\frac{4}{9}$

Mean of the sample proportions

c.

The mean of the sample proportions is computed below:

$$\begin{gathered} \text{Mean}\text{of}\hat{p}=\frac{{\hat{p}}_1+{\hat{p}}_2+.....+{\hat{p}}_9}{9} \\ =\frac{0+0.5+......+1}{9} \\ =0.67 \end{gathered}$$

Thus, the mean of the sampling distribution of the sample proportion is equal to 0.67.

Unbiased estimator

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 proportion (0.67) is not equal to the population proportion (0.67).

Thus, the sample proportion of odd numbers can be considered an unbiased estimator of the population proportion of odd numbers.