What is the sample proportion p-hat?

For a random sample of size n from a population with proportion of successes p, the number of successes is X, which is binomial with parameters n and p. The sample proportion is p-hat equals X divided by n. Because X varies from sample to sample, p-hat is a random variable.

What values can the sample proportion take?

Since X can be 0, 1, 2, up to n, the sample proportion p-hat equals X over n takes the values k over n for k equals 0, 1, up to n. For example, when n equals 5 the possible values are 0, 0.2, 0.4, 0.6, 0.8 and 1 - proportions, not counts.

How do you find the exact probability that the sample proportion equals k over n?

Because p-hat equals X over n and X is binomial, P(p-hat equals k over n) equals P(X equals k) equals n choose k times p to the power k times one minus p to the power n minus k. For example, for a fair coin tossed 4 times, P(p-hat equals one half) equals P(X equals 2) equals 6 times (0.5) to the fourth equals 0.375.

What are the mean and standard deviation of the sample proportion?

The mean of the sample proportion is E(p-hat) equals p, and its standard deviation, the standard error, is the square root of p times one minus p divided by n. For example, when p equals 0.2 and n equals 100, the mean is 0.2 and the standard deviation is the square root of 0.0016, which is 0.04.

When is the exact distribution of the sample proportion symmetric?

The column graph of p-hat is symmetric only when p equals 0.5, because then P(p-hat equals k over n) equals P(p-hat equals n minus k over n). When p is less than 0.5 the columns lean towards the small proportions, and when p is greater than 0.5 they lean towards the large proportions. In every case the columns are centred on the mean p.

Is the exact distribution the same as the normal approximation?

No. The exact distribution comes directly from the binomial probabilities P(X equals k), so it is a set of columns at the values k over n. The normal approximation is a separate idea used only for large samples, where the distribution of p-hat is approximately a bell curve with mean p and standard deviation sqrt(p(1-p)/n).

What is the sample proportion p-hat?

For a random sample of size n from a population with proportion of successes p, the number of successes X is binomial with parameters n and p, and the sample proportion is p-hat equals X divided by n. Because X varies between samples, p-hat is a random variable.

What values can the sample proportion take?

The value p-hat equals X over n takes the values k over n for k equals 0, 1, up to n. For n equals 5 these are 0, 0.2, 0.4, 0.6, 0.8 and 1 - proportions, not counts.

How do you find the exact probability that p-hat equals k over n?

Use P(p-hat equals k over n) equals P(X equals k) equals n choose k times p to the power k times one minus p to the power n minus k. For a fair coin tossed 4 times, P(p-hat equals one half) equals 6 times (0.5) to the fourth equals 0.375.

What are the mean and standard deviation of p-hat?

The mean is E(p-hat) equals p and the standard deviation (standard error) is the square root of p times one minus p over n. For p equals 0.2 and n equals 100, the mean is 0.2 and the standard deviation is 0.04.

When is the exact distribution of p-hat symmetric?

Only when p equals 0.5, since then P(p-hat equals k over n) equals P(p-hat equals n minus k over n). Smaller p leans the columns towards the small proportions, larger p towards the large proportions, but the columns always centre on p.

Is the exact distribution the same as the normal approximation?

No. The exact distribution is the set of binomial columns at the values k over n. The normal approximation is a separate large-sample idea in which p-hat is treated as approximately normal with mean p and standard deviation sqrt(p(1-p)/n).

Year 12 Methods (Unit 3 & 4) Sampling and estimation

The exact distribution of the sample proportion

20 practice questions 0 video lessons Theory + worked examples

In Year 12 Mathematical Methods (Queensland, QCAA), the sample proportion \(\hat p=\dfrac{X}{n}\) records the fraction of successes in a random sample of size \(n\), where \(X\sim B(n,p)\). Its exact distribution lists the values \(\hat p=\dfrac{k}{n}\) with the binomial probabilities \(P\!\left(\hat p=\dfrac{k}{n}\right)=P(X=k)=\dbinom{n}{k}p^{k}(1-p)^{n-k}\). The mean is \(E(\hat p)=p\) and the standard deviation (the standard error) is \(\sqrt{\dfrac{p(1-p)}{n}}\) — a distribution you can tabulate and graph exactly for small \(n\).

Take a random sample of size \(n\) from a large population in which a proportion \(p\) are successes. The number of successes \(X\) is a binomial random variable, \(X\sim B(n,p)\). The sample proportion \(\hat p=\dfrac{X}{n}\) is the fraction of the sample that are successes. Because \(X\) changes from one sample to the next, \(\hat p\) is itself a random variable.

Since \(X\) can equal \(0,1,2,\ldots,n\), the sample proportion takes the values \(\hat p=\dfrac{k}{n}\) — the proportions \(0,\dfrac{1}{n},\dfrac{2}{n},\ldots,1\), not the counts. Each has an exact probability read straight from the binomial: \(P\!\left(\hat p=\dfrac{k}{n}\right)=P(X=k)=\dbinom{n}{k}p^{k}(1-p)^{n-k}\). For small \(n\) this can be set out as a table or drawn as a column graph.

The mean of the sample proportion is \(E(\hat p)=p\) (it does not depend on \(n\)), and the standard deviation — called the standard error — is \(\text{SD}(\hat p)=\sqrt{\dfrac{p(1-p)}{n}}\). A larger sample gives a smaller standard error, so \(\hat p\) clusters more tightly around \(p\). The shape is symmetric when \(p=0.5\) and leans towards the small or large proportions otherwise.

Key idea. \(\hat p=\dfrac{X}{n}\) with \(X\sim B(n,p)\) takes the values \(\dfrac{k}{n}\) with probability \(\dbinom{n}{k}p^{k}(1-p)^{n-k}\); \(E(\hat p)=p\) and \(\text{SD}(\hat p)=\sqrt{\dfrac{p(1-p)}{n}}\). This is the exact distribution — the normal approximation for large samples is a separate idea.

\(p=0.5\): the exact distribution of \(\hat p\) is symmetric about \(\hat p=0.5\)

\(p=0.2\): leans towards the small proportions, centred on \(\hat p=0.2\)

For a sample of size \(n\) with \(X\sim B(n,p)\), the exact probability of each value of \(\hat p\):

\[P\!\left(\hat p=\dfrac{k}{n}\right)=P(X=k)=\dbinom{n}{k}p^{k}(1-p)^{n-k}\]

P (\hat{p} = k / n) = (n k) p^{k} {(1 - p)}^{n - k}

The mean and standard deviation (standard error) of the sample proportion:

\[E(\hat p)=p,\qquad \text{SD}(\hat p)=\sqrt{\dfrac{p(1-p)}{n}}\]

E (\hat{p}) = p

To reverse these for \(p\) or \(n\) from a given mean and standard deviation:

\[p=E(\hat p),\qquad n=\dfrac{p(1-p)}{\big(\text{SD}(\hat p)\big)^{2}}\]

n = \frac{p (1 - p)}{{SD}^{2}}

Shape rule. The columns sit at \(\hat p=\dfrac{k}{n}\) and are centred on \(p\). The distribution is symmetric when \(p=0.5\), leans towards the small proportions when \(p<0.5\), and towards the large proportions when \(p>0.5\).

Working with the exact distribution of \(\hat p\)

Identify \(n\) and \(p\). The sample size is \(n\) and the population proportion is \(p\); the number of successes is \(X\sim B(n,p)\).
List the values. The sample proportion takes \(\hat p=\dfrac{k}{n}\) for \(k=0,1,\ldots,n\) — proportions, not counts.
Exact probabilities. \(P\!\left(\hat p=\dfrac{k}{n}\right)=\dbinom{n}{k}p^{k}(1-p)^{n-k}\). Add several of these for an interval such as \(P(\hat p\le a)\).
Mean and standard deviation. \(E(\hat p)=p\) and \(\text{SD}(\hat p)=\sqrt{\dfrac{p(1-p)}{n}}\) (take the square root; do not stop at \(\dfrac{p(1-p)}{n}\)).
Shape and technology. Compare \(p\) with \(0.5\) for the shape; a calculator's binomial menu gives each \(P(X=k)\) and cumulative probabilities directly.

Technology. With QCAA-approved technology, compute \(P(X=k)\) with a binomial pdf and \(P(X\le k)\) with a binomial cdf; convert the count \(k\) to the proportion \(\dfrac{k}{n}\) at the end.

Example 1 — An exact probability

A fair coin is tossed \(n=4\) times. Find \(P\!\left(\hat p=\dfrac{1}{2}\right)\) for the sample proportion of heads.

Solution

\(\hat p=\tfrac{1}{2}\)	\(\Rightarrow\)	\(X=2\)
\(P(X=2)\)	\(=\)	\(\dbinom{4}{2}(0.5)^{4}\)
	\(=\)	\(\dfrac{6}{16}=0.375\)

P = 0.375

Example 2 — Mean and standard error

A sample of size \(n=100\) is taken from a population with \(p=0.2\). Find the mean and standard deviation of \(\hat p\).

Solution

\(E(\hat p)\)	\(=\)	\(p=0.2\)
\(\text{SD}(\hat p)\)	\(=\)	\(\sqrt{\dfrac{0.2\times 0.8}{100}}\)
	\(=\)	\(\sqrt{0.0016}=0.04\)

SD = 0.04

Example 3 — An interval probability

A fair coin is tossed \(n=5\) times. Find \(P(\hat p\le 0.4)\) for the sample proportion of heads.

Solution

The counts \(\dbinom{5}{k}\) are \(1,5,10,10,5,1\), and each height is \(\dfrac{1}{32}\dbinom{5}{k}\).

\(P(\hat p\le 0.4)\)	\(=\)	\(P(X\le 2)\)
	\(=\)	\(\dfrac{1+5+10}{32}=\dfrac{16}{32}\)
	\(=\)	\(0.5\)

P = 0.5

Example 4 — Find the sample size

A population has \(p=0.5\). What sample size \(n\) gives \(\text{SD}(\hat p)=0.05\)?

Solution

Square both sides of the standard-error formula, then solve for \(n\).

\(\dfrac{0.5\times 0.5}{n}\)	\(=\)	\(0.05^{2}=0.0025\)
\(n\)	\(=\)	\(\dfrac{0.25}{0.0025}=100\)

n = 100

Common pitfalls

\(\hat p\) takes proportions, not counts. The values are \(\dfrac{k}{n}\) (e.g. \(0,0.2,0.4,\ldots,1\) for \(n=5\)), not \(0,1,2,\ldots,n\). Convert the count \(k\) to \(\dfrac{k}{n}\).

The standard deviation needs the square root. \(\text{SD}(\hat p)=\sqrt{\dfrac{p(1-p)}{n}}\); \(\dfrac{p(1-p)}{n}\) alone is the variance. Do not forget to take the root.

The mean is \(p\), not \(\hat p\) or \(np\). \(E(\hat p)=p\), the population proportion, for every sample size. The count \(X\) has mean \(np\), but the proportion \(\hat p\) has mean \(p\).

Keep the binomial coefficient. \(P\!\left(\hat p=\dfrac{k}{n}\right)=\dbinom{n}{k}p^{k}(1-p)^{n-k}\); dropping \(\dbinom{n}{k}\) leaves only \(p^{k}(1-p)^{n-k}\), the probability of one particular arrangement.

Frequently asked questions

What is the sample proportion \(\hat p\)?

For a sample of size \(n\) with \(X\sim B(n,p)\) successes, \(\hat p=\dfrac{X}{n}\). Since \(X\) varies between samples, \(\hat p\) is a random variable.

What values can \(\hat p\) take?

The proportions \(\dfrac{k}{n}\) for \(k=0,1,\ldots,n\). For \(n=5\): \(0,0.2,0.4,0.6,0.8,1\).

How do you find \(P\!\left(\hat p=\dfrac{k}{n}\right)\)?

Use \(P(X=k)=\dbinom{n}{k}p^{k}(1-p)^{n-k}\). For a fair coin tossed \(4\) times, \(P(\hat p=\tfrac{1}{2})=\dbinom{4}{2}(0.5)^{4}=0.375\).

What are the mean and standard deviation of \(\hat p\)?

\(E(\hat p)=p\) and \(\text{SD}(\hat p)=\sqrt{\dfrac{p(1-p)}{n}}\). For \(p=0.2\), \(n=100\): mean \(0.2\), standard deviation \(0.04\).

When is the distribution symmetric?

Only when \(p=0.5\), since then \(P(\hat p=\tfrac{k}{n})=P(\hat p=\tfrac{n-k}{n})\). Otherwise it leans towards the small (\(p<0.5\)) or large (\(p>0.5\)) proportions.

Is this the same as the normal approximation?

No. This is the exact distribution built from the binomial columns. The normal approximation is a separate large-sample result, treated in the next subtopic.

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