# 9.8: Sampling Distribution of p

Skills to Develop

• Compute the mean and standard deviation of the sampling distribution of $$p$$
• State the relationship between the sampling distribution of p and the normal distribution

Assume that in an election race between $$\text{Candidate A}$$ and $$\text{Candidate B}$$, $$0.60$$ of the voters prefer $$\text{Candidate A}$$. If a random sample of $$10$$ voters were polled, it is unlikely that exactly $$60\%$$ of them ($$6$$) would prefer $$\text{Candidate A}$$. By chance the proportion in the sample preferring $$\text{Candidate A}$$ 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 $$\text{Candidate A}$$.

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

$p = \frac{7}{10} = 0.70$

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

Table $$\PageIndex{1}$$: Sample of voters

Voter Preference
1 1
2 0
3 1
4 1
5 1
6 0
7 1
8 0
9 1
10 1

The distribution of $$p$$ is closely related to the binomial distribution. The binomial distribution is the distribution of the total number of successes (favoring $$\text{Candidate A}$$, 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:

$\mu _p=\pi$

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 $$p$$ is $$0.60$$. The standard error is

$\sigma _p=\sqrt{\frac{0.60(1-0.60)}{10}}=0.155$

The sampling distribution of $$p$$ 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 $$N\pi$$ and $$N(1-\pi )$$ are greater than $$10$$. The sampling distribution for the voter example is shown in Figure $$\PageIndex{1}$$. Note that even though $$N(1-\pi )$$ is only $$4$$, the approximation is quite good.

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

### Contributor

• Online Statistics Education: A Multimedia Course of Study (http://onlinestatbook.com/). Project Leader: David M. Lane, Rice University.