91Ó°ÊÓ

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.

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,$\left[\left(0,1\right),\left(1,0\right),\left(1,1\right),\left(0,0\right)\right]$

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

Therefore, the sample means are,

Here the total sample mean is

$\begin{array}{rcl}\stackrel{¯}{X}& =& \frac{1}{4}\underset{1}{\overset{4}{∑}}samplemean\\ & =& \frac{\left(\frac{1}{2}+\frac{1}{2}+1+0\right)}{4}\\ & =& \frac{1}{2}\\ & & \end{array}$

Now, the sample proportion of 1s is,

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

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

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

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

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

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 $\hat{\mathbf{p}}$.