Why Sample Proportions Vary

Why the Same Sampling Method Can Give Different Proportions

Suppose a population has a fixed proportion $p$ of individuals with a particular characteristic, such as being satisfied with a service. A random sample will not usually reproduce that population proportion exactly. One sample may contain a slightly larger share of satisfied individuals, while another may contain a slightly smaller share. The population proportion is fixed; the sample proportion varies because the particular individuals selected vary.

To understand that variation, imagine repeatedly taking samples of the same size from the same population, using the same random sampling method, and calculating a sample proportion for each sample. The distribution of those proportions is the sampling distribution of the sample proportion. It describes the behavior of the statistic across repetitions, not the spread of individual outcomes within one sample.

Definition: For a binary outcome, let $p$ be the population proportion with the outcome of interest. In a random sample of size $n$, let $X$ be the number of sampled individuals with that outcome. The sample proportion is $\hat{p}=X/n$. Its sampling distribution is the probability distribution of $\hat{p}$ over repeated samples of size $n$ taken by the same sampling procedure.

The sample proportion is a random variable before a particular sample is observed: its value depends on which individuals happen to be selected. After data are collected, the observed value of $\hat{p}$ is one outcome from its sampling distribution. This distinction helps separate the unknown population feature $p$ from the sample statistic $\hat{p}$.

The Binomial Model for a Sample Proportion

First consider independent observations, each with the same probability $p$ of being a success. “Success” is simply the chosen outcome of interest; it need not mean something desirable. For example, a success could be a patient experiencing a side effect or a manufactured item failing inspection. Under this model, the number of successes in $n$ observations follows a binomial distribution:

X\sim\operatorname{Binomial}(n,p), \qquad \hat{p}=\frac{X}{n}.

Because $\hat{p}$ is the count $X$ divided by $n$, it can take only the values $0,1/n,2/n,\ldots,1$. Its probability at a possible value $k/n$ is the probability of observing exactly $k$ successes:

P\left(\hat{p}=\frac{k}{n}\right) =P(X=k) =\binom{n}{k}p^k(1-p)^{n-k}, \qquad k=0,1,\ldots,n.

The binomial model requires that the observations be independent and that each observation have the same success probability $p$. These conditions are appropriate, for example, for independent observations from a large population or for sampling with replacement. A sample drawn without replacement from a finite population is dependent; its sampling variance has a finite-population adjustment, so the independent binomial formula does not apply exactly.

Worked Example: Listing a Small Sampling Distribution

Suppose each of four independent items has probability $p=0.5$ of meeting a specified standard. Let $\hat{p}$ be the proportion of the four items that meet the standard. Find its full sampling distribution.

The number $X$ meeting the standard has a binomial distribution with $n=4$ and $p=0.5$. Since $\hat{p}=X/4$, the possible values of $\hat{p}$ and their probabilities are:

Number of successes $k$	Sample proportion $k/4$	Probability
0	0	$\binom{4}{0}(0.5)^0(0.5)^4=0.0625$
1	0.25	$\binom{4}{1}(0.5)^1(0.5)^3=0.2500$
2	0.50	$\binom{4}{2}(0.5)^2(0.5)^2=0.3750$
3	0.75	$\binom{4}{3}(0.5)^3(0.5)^1=0.2500$
4	1	$\binom{4}{4}(0.5)^4(0.5)^0=0.0625$

The probabilities add to $0.0625+0.2500+0.3750+0.2500+0.0625=1.0000$, as required for a probability distribution. A sample proportion of 0.50 is most likely here, but proportions of 0.25 and 0.75 also occur. The distribution makes the sample-to-sample variation explicit.

The Center and Spread of the Sampling Distribution

Under the independent Bernoulli model, the sampling distribution has a particularly useful center and spread. The sample proportion is centered at the population proportion $p$, and its variance decreases as the sample size increases.

Theorem: If the $n$ binary observations are independent and each has success probability $p$, then $$ E(\hat{p})=p,\qquad \operatorname{Var}(\hat{p})=\frac{p(1-p)}{n},\qquad \operatorname{SD}(\hat{p})=\sqrt{\frac{p(1-p)}{n}}. $$ The standard deviation here is the spread of the sampling distribution of $\hat{p}$, measured in proportion units.

The expectation and variance follow by writing the sample proportion as an average of indicator variables. For each sampled individual, define $Y_i=1$ if the individual has the outcome and $Y_i=0$ otherwise. Then $E(Y_i)=p$, $\operatorname{Var}(Y_i)=p(1-p)$, and $\hat{p}=(Y_1+\cdots+Y_n)/n$.

E(\hat{p}) =E\left(\frac{1}{n}\sum_{i=1}^{n}Y_i\right) =\frac{1}{n}\sum_{i=1}^{n}E(Y_i) =\frac{np}{n} =p.

For independent observations, the variance of their sum is the sum of their variances. Dividing a random variable by $n$ divides its variance by $n^2$. Therefore:

\operatorname{Var}(\hat{p}) =\operatorname{Var}\left(\frac{1}{n}\sum_{i=1}^{n}Y_i\right) =\frac{1}{n^2}\sum_{i=1}^{n}\operatorname{Var}(Y_i) =\frac{n\,p(1-p)}{n^2} =\frac{p(1-p)}{n}.

Taking the square root gives the standard deviation. This derivation explains two important features. First, repeated sample proportions are centered on $p$: on average, they equal the population proportion. Second, their spread shrinks at the rate $1/\sqrt{n}$, not $1/n$. Multiplying the sample size by four halves the standard deviation, all else equal.

Worked Example: Finding the Mean and Spread

Assume independent sampling from a population in which $p=0.30$ of individuals have a particular characteristic. For samples of size $n=100$, find the mean, variance, and standard deviation of $\hat{p}$.

The independent Bernoulli model applies because the setup specifies that observations are independent and share the same success probability. Substituting $p=0.30$ and $n=100$:

E(\hat{p})=p=0.30,

\operatorname{Var}(\hat{p}) =\frac{p(1-p)}{n} =\frac{(0.30)(0.70)}{100} =0.00210,

\operatorname{SD}(\hat{p}) =\sqrt{0.00210} \approx 0.0458.

Thus the distribution is centered at 0.30, or 30%, and its standard deviation is about 0.0458, equivalent to 4.58 percentage points. This standard deviation describes the typical scale of sample-to-sample variation; it does not say that every sample proportion must lie within 4.58 percentage points of 0.30.

How Sample Size Changes Variation

The factor $p(1-p)$ describes how much variability is possible for a binary outcome, while the divisor $n$ makes variance smaller as more observations are collected. The population proportion does not move toward 0.5 or change when $n$ changes. Instead, larger samples tend to produce sample proportions more tightly concentrated around that same $p$.

For a fixed $n$, the quantity $p(1-p)$ is largest when $p=0.5$. To see why, complete the square:

p(1-p)=p-p^2 =\frac{1}{4}-\left(p-\frac{1}{2}\right)^2 \leq \frac{1}{4}.

So, among binary populations, the sampling distribution has its greatest variance at $p=0.5$, and less variance when $p$ is closer to 0 or 1. This is because outcomes are more mixed near $p=0.5$; when nearly everyone has the outcome or nearly no one does, sample counts have less room to fluctuate.

Worked Example: Comparing Two Sample Sizes

Suppose $p=0.20$, and compare independent samples of size $n=25$ and $n=100$. Find the standard deviation of the sample proportion for each size, then describe how the spread changes.

For $n=25$:

\operatorname{SD}(\hat{p}) =\sqrt{\frac{(0.20)(0.80)}{25}} =\sqrt{0.0064} =0.0800.

For $n=100$:

\operatorname{SD}(\hat{p}) =\sqrt{\frac{(0.20)(0.80)}{100}} =\sqrt{0.0016} =0.0400.

The larger sample has four times as many observations, while its sampling-distribution standard deviation is half as large: 0.0400 rather than 0.0800. In percentage-point terms, those spreads are 4 and 8 percentage points. Both distributions remain centered at 0.20; only their spread changes.

Shape of the Sampling Distribution

For small samples, the sampling distribution is discrete and may be noticeably uneven or skewed. The binomial model gives its exact probabilities. As the sample size grows, the distribution of $\hat{p}$ is often approximated by a normal distribution centered at $p$, with standard deviation $\sqrt{p(1-p)/n}$. The normal approximation is appropriate when the expected numbers of successes and failures are both sufficiently large; the success-failure condition is a practical check for this approximation.

The approximation does not change the exact mean or variance under the model. It provides a simpler description of the distribution's shape and can help approximate probabilities. If the condition is not met, a normal curve may describe extreme or skewed probabilities poorly, even when the formula for the mean and variance remains valid.

Why This Variation Matters

A single observed $\hat{p}$ is not generally equal to $p$, even when the sampling method is unbiased. Its difference from $p$ can arise purely from which individuals landed in the sample. Understanding this variability is essential when judging whether a reported sample proportion is informative: the same observed proportion is less precise when based on a smaller sample than when based on a larger one.

The standard deviation formula above uses the population proportion $p$. In practice, $p$ is usually unknown, so the spread must be estimated from sample information when constructing an interval. That is a distinct calculation from describing the theoretical sampling distribution and will require care about which value is substituted for $p$.

Common mistake: “A sample proportion of 0.34 when the population proportion is 0.30 means the population proportion must have changed.” This confuses a sample statistic with the population parameter. If sampling is random, the observed difference can result from sampling variability. The population proportion $p$ is fixed under the model; repeated samples generate different values of $\hat{p}$, centered at $p$.

Another mistake is to say that quadrupling the sample size cuts variability to one quarter. It cuts the variance to one quarter, but the standard deviation—the spread in the original proportion units—becomes one half as large because it is the square root of the variance.

Key takeaway: Under independent Bernoulli sampling, $\hat{p}$ varies from sample to sample even though $p$ is fixed. Its sampling distribution has mean $p$ and standard deviation $\sqrt{p(1-p)/n}$, so larger samples concentrate more closely around the population proportion.

Check Your Understanding

Use the definitions and results above to answer these questions.

A population has $p=0.40$. For an independent sample of size $n=50$, what is the mean of the sampling distribution of $\hat{p}$?
For the setting in question 1, calculate the variance and standard deviation of $\hat{p}$.
If the sample size increases from 50 to 200 while $p$ stays fixed, how does the standard deviation change?
For $n=5$ and $p=0.20$, what values can $\hat{p}$ take? State why the distribution is discrete.
Explain why observing $\hat{p}\ne p$ does not by itself show that the population proportion changed.

Number of successes \(k\)	Sample proportion \(k/4\)	Probability
0	0	\(\binom{4}{0}(0.5)^0(0.5)^4=0.0625\)
1	0.25	\(\binom{4}{1}(0.5)^1(0.5)^3=0.2500\)
2	0.50	\(\binom{4}{2}(0.5)^2(0.5)^2=0.3750\)
3	0.75	\(\binom{4}{3}(0.5)^3(0.5)^1=0.2500\)
4	1	\(\binom{4}{4}(0.5)^4(0.5)^0=0.0625\)

What You'll Learn