Sampling Distribution of p

Assume that in an election race between Candidate $A$ and Candidate $B, 0.60$ of the voters prefer Candidate $A$. If a random sample of $10$ voters were polled, it is unlikely that exactly $60 \%$ of them $(6)$ would prefer Candidate $A$. By chance the proportion in the sample preferring Candidate $A$ could easily be a little lower than $0.60$ or a little higher than $0.60$. The sampling distribution of $p$ is the distribution that would result if you repeatedly sampled $10$ voters and determined the proportion $(p)$ that favored Candidate $A$.

The sampling distribution of $p$ is a special case of the sampling distribution of the mean. Table 1 shows a hypothetical random sample of $10$ voters. Those who prefer Candidate $A$ are given scores of $1$ and those who prefer Candidate $B$ are given scores of $0$. Note that seven of the voters prefer candidate $A$ so the sample proportion $(\mathrm{p})$ is

$p=7 / 10=0.70$

As you can see, $p$ is the mean of the $10$ preference scores.

Table 1. Sample of voters.

The distribution of $\mathrm{p}$ is closely related to the binomial distribution. The binomial distribution is the distribution of the total number of successes (favoring) Candidate $A$, for example) whereas the distribution of $p$ is the distribution of the mean number of successes. The mean, of course, is the total divided by the sample size, $N$. Therefore, the sampling distribution of $p$ and the binomial distribution differ in that $p$ is the mean of the scores $(0.70)$ and the binomial distribution is dealing with the total number of successes (7).

The binomial distribution has a mean of:

$\mu=N \Pi$

Dividing by $N$ to adjust for the fact that the sampling distribution of $p$ is dealing with means instead of totals, we find that the mean of the sampling distribution of $p$ is:

$\mu_{\mathrm{p}}=\Pi$

The standard deviation of the binomial distribution is:

$\sqrt{N \pi(1-\pi)}$

Dividing by $N$ because $p$ is a mean not a total, we find the standard error of $p$:

$\sigma_{p}=\frac{\sqrt{N \pi(1-\pi)}}{N}=\sqrt{\frac{\pi(1-\pi)}{N}}$

Returning to the voter example, $\Pi=0.60$ and $N=10$. (Don't confuse $\Pi=0.60$, the population proportion and $p=0.70$, the sample proportion.) Therefore, the mean of the sampling distribution of $\mathrm{p}$ is $0.60$. The standard error is

$\sigma_{p}=\sqrt{\frac{0.60(1-.60)}{10}}=0.155$

The sampling distribution of $\mathrm{p}$ is a discrete rather than a continuous distribution. For example, with an $N$ of $10$, it is possible to have a $p$ of $0.50$ or a $p$ of $0.60$ but not a $p$ of $0.55$.

The sampling distribution of $p$ is approximately normally distributed if $N$ is fairly large and $\pi$ is not close to $0$ or $1$. A rule of thumb is that the approximation is good if both $\mathrm{N} \pi$ and $\mathrm{N}(1-\pi)$ are greater than $10$. The sampling distribution for the voter example is shown in Figure 1. Note that even though $N(1-\pi)$ is only $4$, the approximation is quite good.

Figure 1. The sampling distribution of $p$. Vertical bars are the probabilities; the smooth curve is the normal approximation.

Source: David M. Lane, https://onlinestatbook.com/2/sampling_distributions/samp_dist_p.html
This work is in the Public Domain.