Question

Hybridization A hybridization experiment begins with four peas having yellow pods and one pea having a green pod. Two of the peas are randomly selected with replacement from this population.

a. After identifying the 25 different possible samples, find the proportion of peas with yellow pods in each of them, then construct a table to describe the sampling distribution of the proportions of peas with yellow pods.

b. Find the mean of the sampling distribution.

c. Is the mean of the sampling distribution [from part (b)] equal to the population proportion of peas with yellow pods? Does the mean of the sampling distribution of proportions always equal the population proportion?

Short answer

a.

The following table describes the sampling distribution of the sample proportions of yellow peas:

Sample proportion	Probability
0	$\frac{1}{25}$
0.5	$\frac{8}{25}$
1	$\frac{16}{25}$

b. The mean of the sampling distribution of sample proportion of peas with yellow pods is equal to 0.2.

c. The mean of the sampling distribution of sample proportion of yellow peas (0.2) is equal to the population, proportion of yellow peas (0.2).

Yes, the mean of the sampling distribution of proportions is always equal to the population proportion.

Step-by-step solution

Given information

The color of 5 pea pods is noted taking two peas at a time.

Sampling distribution of sample proportions

a.

A total of 5 peas are available.

Sample of 2 peas each is selected.

Let the color of the pea pods be denoted as follows:

• Y for yellow
• G for green

A subscript is attached with the symbol “Y” denoting the serial number of peas selected.

There are 4 yellow peas and one green pea.

All possible samples for the color of the pea pods are given as follows:

$Y_1{Y}_1$	$Y_1{Y}_2$	$Y_1{Y}_3$	$Y_1{Y}_4$	$Y_{1}G$	$GG$	$G{Y}_1$	$Y_2{Y}_1$	$Y_2{Y}_2$
$Y_2{Y}_3$	$Y_2{Y}_4$	$Y_{2}G$	$G{Y}_2$	$Y_3{Y}_1$	$Y_3{Y}_2$	$Y_3{Y}_3$	$Y_3{Y}_4$	$Y_{3}G$
$Y_4{Y}_1$	$Y_4{Y}_2$	$Y_4{Y}_3$	$Y_4{Y}_4$	$Y_{4}G$	$G{Y}_4$	$G{Y}_3$

The sample proportion of yellow peas for each sample has the following formula:

$\hat{p}=\frac{\text{Number}\text{of}\text{Yellow}\text{Peas}}{2}$

The following table shows all possible samples of size equal to 2 and the corresponding sample proportions:

Sample	Sample proportion of girls
$Y_1{Y}_1$	$$\begin{gathered} {\hat{p}}_1=\frac{2}{2} \\ =1 \end{gathered}$$
$Y_1{Y}_2$	$$\begin{gathered} {\hat{p}}_2=\frac{2}{2} \\ =1 \end{gathered}$$
$Y_1{Y}_3$	$$\begin{gathered} {\hat{p}}_3=\frac{2}{2} \\ =1 \end{gathered}$$
$Y_1{Y}_4$	$$\begin{gathered} {\hat{p}}_4=\frac{2}{2} \\ =1 \end{gathered}$$
$Y_{1}G$	$$\begin{gathered} {\hat{p}}_5=\frac{1}{2} \\ =0.5 \end{gathered}$$
$Y_2{Y}_1$	$$\begin{gathered} {\hat{p}}_6=\frac{2}{2} \\ =1 \end{gathered}$$
$Y_2{Y}_2$	$$\begin{gathered} {\hat{p}}_7=\frac{2}{2} \\ =1 \end{gathered}$$
$Y_2{Y}_3$	$$\begin{gathered} {\hat{p}}_8=\frac{2}{2} \\ =1 \end{gathered}$$
$Y_2{Y}_4$	$$\begin{gathered} {\hat{p}}_9=\frac{2}{2} \\ =1 \end{gathered}$$
$Y_{2}G$	$$\begin{gathered} {\hat{p}}_{10}=\frac{1}{2} \\ =0.5 \end{gathered}$$
$Y_3{Y}_1$	$$\begin{gathered} {\hat{p}}_{11}=\frac{2}{2} \\ =1 \end{gathered}$$
$Y_3{Y}_2$	$$\begin{gathered} {\hat{p}}_{12}=\frac{2}{2} \\ =1 \end{gathered}$$
$Y_3{Y}_3$	$$\begin{gathered} {\hat{p}}_{13}=\frac{2}{2} \\ =1 \end{gathered}$$
$Y_3{Y}_4$	$$\begin{gathered} {\hat{p}}_{14}=\frac{2}{2} \\ =1 \end{gathered}$$
$Y_3{Y}_4$	$$\begin{gathered} {\hat{p}}_{14}=\frac{2}{2} \\ =1 \end{gathered}$$
$Y_{3}G$	$$\begin{gathered} {\hat{p}}_{15}=\frac{1}{2} \\ =0.5 \end{gathered}$$
$Y_4{Y}_1$	$$\begin{gathered} {\hat{p}}_{16}=\frac{2}{2} \\ =1 \end{gathered}$$
$Y_4{Y}_2$	$$\begin{gathered} {\hat{p}}_{17}=\frac{2}{2} \\ =1 \end{gathered}$$
$Y_4{Y}_3$	$$\begin{gathered} {\hat{p}}_{18}=\frac{2}{2} \\ =1 \end{gathered}$$
$Y_4{Y}_4$	$$\begin{gathered} {\hat{p}}_{19}=\frac{2}{2} \\ =1 \end{gathered}$$
$Y_{4}G$	$$\begin{gathered} {\hat{p}}_{20}=\frac{1}{2} \\ =0.5 \end{gathered}$$
$G{Y}_1$	$$\begin{gathered} {\hat{p}}_{21}=\frac{1}{2} \\ =0.5 \end{gathered}$$
$G{Y}_2$	$$\begin{gathered} {\hat{p}}_{22}=\frac{1}{2} \\ =0.5 \end{gathered}$$
$G{Y}_3$	$$\begin{gathered} {\hat{p}}_{23}=\frac{1}{2} \\ =0.5 \end{gathered}$$
$G{Y}_4$	$$\begin{gathered} {\hat{p}}_{24}=\frac{1}{2} \\ =0.5 \end{gathered}$$
$GG$	$$\begin{gathered} {\hat{p}}_{25}=\frac{0}{2} \\ =0 \end{gathered}$$

Combining the values of proportions that are the same, the following probability values are obtained:

Sample proportion	Probability
0	$\frac{1}{25}$
0.5	$\frac{8}{25}$
1	$\frac{16}{25}$

Mean of the sample proportions

b.

The mean of the sample proportions of correct answers is computed below:

$$\begin{gathered} Meanof\hat{p}=\frac{{\hat{p}}_1+{\hat{p}}_2+.....+{\hat{p}}_{25}}{25} \\ =\frac{1+1+........+0}{25} \\ =0.8 \end{gathered}$$

Thus, the mean of the sampling distribution of the sample proportions is equal to 0.8.

Population proportion

The population can be described as$\left\{Y_1,Y_2,Y_3,Y_4,G\right\}$.

The population proportion of yellow peas is computed below:

$$\begin{gathered} p=\frac{4}{5} \\ =0.8 \end{gathered}$$

Thus, the population proportion of yellow peas is equal to 0.8.

Comparison

c.

Here, the population proportion of yellow peas (0.8) is equal to the mean of the sampling distribution of sample proportions of yellow peas (0.8).

Yes, the mean of the sampling distribution of sample proportions is always equal to the population proportion.