Question

Suppose xequals the number of heads observed when asingle coin is tossed; that is, x= 0 or x= 1. The population corresponding to xis the set of 0s and 1s generated when thecoin is tossed repeatedly a large number of times. Supposewe select n= 2 observations from this population. (That is,we toss the coin twice and observe two values of x.)

1. List the three different samples (combinations of 0s and1s) that could be obtained.
2. Calculate the value of $\overline{\mathbf{X}}$ffor each of the samples.
3. Show that the sample proportion of 1s, $\hat{\mathbf{p}}$, is equal to $\overline{\mathbf{X}}$.
4. List the values that$\hat{\mathbf{p}}$can assume, and find the probabilitiesof observing these values.
5. Construct a graph of the sampling distribution of$\hat{\mathbf{p}}$.

Short answer

a.

b.

Sample	Sample mean
(0,1)	½
(1,0)	½
(1,1)	1
(0,0)	0

c. Proved.

d.

Sample	Sample mean	Probability
(0,1)	½	¼
(1,0)	½	¼
(1,1)	1	¼

e. We construct a plot of the sampling distribution and see that the line is centered on the sample mean.

Step-by-step solution

Given information

We tossed an unbiased coin twice and observed the values, so, sample size n = 2. X is the value of heads, that is X =0or X =1.

List the different samples 

a.

As we tossed the coin twice, so, each of the two data values in the sample is equally likely to be a 0 or a 1.

Therefore, there are possible samples of size n = 2.

Thus, the samples can be written as,$\left[\left(0,1\right),\left(1,0\right),\left(1,1\right),\left(0,0\right)\right]$

Calculate the sample means

b.

The sample means are the sum of all values in the sample divided by the total number of observations.

Therefore, the sample means are,

Sample	Sample mean
(0,1)	½
(1,0)	½
(1,1)	1
(0,0)	0

Proof of the statement

c.

Here the total sample mean is

$\begin{array}{rcl}\overline{X}& =& \frac{1}{4}\sum _1^{4}samplemean\\ & =& \frac{\left(\frac{1}{2}+\frac{1}{2}+1+0\right)}{4}\\ & =& \frac{1}{2}\\ & & \end{array}$

Now, the sample proportion of 1s is,

$\begin{array}{rcl}\hat{p}& =& \frac{1\times Totalnumberof1s}{totalobsevations}\\ & =& \frac{4}{8}\\ & =& \frac{1}{2}\\ & & \end{array}$

Therefore, from the above results, we can conclude that the sample proportion of 1s, $\hat{p}$, is equal to$\overline{X}$ .

List the assumed values and calculate their probabilities 

d.

As we proved earlier that the sample proportion is equal to the mean, so, we can list the values that $\overline{X}$can assume.

Each of these samples is equally likely to occur. Thus, there is a 1 in 4 chance for each sample.

Sample	Sample mean	Probability
(0,1)	½	¼
(1,0)	½	¼
(1,1)	1	¼
(0,0)	0	1/4

Now, we can obtain the sampling distribution of the sample mean$\overline{X}$ which is the sample proportion for 1s, $\hat{p}$, by adding the probabilities that correspond to the same sample mean

Sample mean	Probability
1/2	¼ + ¼ = ½
1	¼
0	¼

 Step 6: Construction of a graph of the sampling distribution 

e.

Here, we construct a plot of the sampling distribution and see that the line is centered on the sample mean.

Thus, the graph clearly represents the sampling distribution of $\hat{\mathbf{p}}$.