Why Enumerate a Sampling Distribution?
A statistic changes from sample to sample, even when the population stays fixed. For a small population, we can see exactly how this happens: list every possible sample of a chosen size, calculate the statistic for each one, and record how often each value occurs. The resulting probability distribution is the exact sampling distribution for that statistic under the stated sampling plan.
Enumeration is more than a way to get an answer for a tiny example. It makes the probability model visible. In particular, it distinguishes the possible samples from the possible values of a statistic: different samples can give the same mean or proportion, and their probabilities must be combined.
Samples and Their Probabilities
Suppose a finite population has \(N\) distinct units, with a fixed value \(x_i\) recorded for unit \(i\). A simple random sample without replacement (SRSWOR) of size \(n\) selects \(n\) different units so that every subset of \(n\) population units is equally likely. The sample is treated as an unordered set: selecting units A and B is the same sample as selecting B and A.
The statistic values do not generally have equal probabilities. Equal probability applies to the sample subsets. If several subsets yield the same statistic value, that value receives the sum of their probabilities.
This equal-subset model follows from the definition of simple random sampling. If the \(n\) units are drawn one at a time without replacement, each ordered sequence of distinct units has probability \(1/[N(N-1)\cdots(N-n+1)]\). Any particular subset has exactly \(n!\) possible orders. Therefore every subset has the same probability, \(n!/[N(N-1)\cdots(N-n+1)]=1/\binom{N}{n}\).
Exact Distribution of a Sample Mean
For each possible sample subset \(S\), calculate the sample mean by adding its \(n\) recorded values and dividing by \(n\). The population values remain fixed throughout; the randomness is which units enter the sample. When different subsets have equal means, count each subset toward that mean’s probability.
Worked Example: All Sample Means from Four Units
Consider a hypothetical finite population with four distinct units whose measured values, in units, are 2, 4, 6, and 10. Draw an SRSWOR of size \(n=2\). The population size is \(N=4\), so there are \(\binom{4}{2}=4!/[2!2!]=6\) equally likely samples. Enumerate them and calculate their means.
| Sample values | Sample mean calculation | Sample mean |
|---|---|---|
| 2, 4 | (2 + 4) / 2 = 3 | 3 |
| 2, 6 | (2 + 6) / 2 = 4 | 4 |
| 2, 10 | (2 + 10) / 2 = 6 | 6 |
| 4, 6 | (4 + 6) / 2 = 5 | 5 |
| 4, 10 | (4 + 10) / 2 = 7 | 7 |
| 6, 10 | (6 + 10) / 2 = 8 | 8 |
All six sample means are different, and each is produced by one of the six equally likely subsets. Thus each mean has probability \(1/6\), approximately 0.1667. The exact distribution is
The values in the distribution are means, not individual population measurements. For example, 3 occurs as a sample mean even though neither population unit has value 3.
Exact Distribution of a Sample Proportion
A sample proportion is a sample mean of zero-one indicators: assign 1 to a unit with the attribute of interest and 0 to a unit without it. If \(K\) sampled units have the attribute, then \(\hat{p}=K/n\). In a finite population with \(M\) units having the attribute, the population proportion is \(p=M/N\).
Worked Example: A Proportion from Five Units
Suppose a hypothetical population contains five distinct units: two have a specified attribute and three do not. Draw an SRSWOR of size \(n=2\), and find the exact distribution of the sample proportion \(\hat{p}\) with the attribute. There are \(\binom{5}{2}=5!/[2!3!]=10\) equally likely sample subsets.
Count samples according to the number \(K\) of attribute-having units. For \(K=0\), both units must come from the three without the attribute, giving \(\binom{2}{0}\binom{3}{2}=1(3)=3\) samples. For \(K=1\), choose one of the two with the attribute and one of the three without it, giving \(\binom{2}{1}\binom{3}{1}=2(3)=6\) samples. For \(K=2\), choose both attribute-having units, giving \(\binom{2}{2}\binom{3}{0}=1(1)=1\) sample. The counts total \(3+6+1=10\), as required.
The corresponding proportions are \(0/2=0\), \(1/2=0.5\), and \(2/2=1\). Dividing each count by the ten equally likely samples gives:
For instance, the probability that the sample proportion is 0.5 is 0.6, because six of the ten possible samples contain exactly one unit with the attribute. The probabilities sum to \(0.3+0.6+0.1=1\). This calculation relies on sampling two distinct units from the five; it would describe a different sampling plan if units could be selected more than once.
Combining Samples with the Same Statistic
A distribution table should have one row for each distinct statistic value, not necessarily one row for each sample. The next example includes repeated population values, which makes it especially important to keep track of distinct units even when their measurements match.
Worked Example: Repeated Means in a Larger Enumeration
Consider five distinct hypothetical units with values \(1,3,3,7,9\), measured in units. Label them A, B, C, D, and E in that order. Take an SRSWOR of size \(n=3\). There are \(\binom{5}{3}=5!/[3!2!]=10\) equally likely subsets. The table lists every subset, its sum, and its mean.
| Sample | Sum | Sample mean |
|---|---|---|
| ABC | 1 + 3 + 3 = 7 | 7/3 |
| ABD | 1 + 3 + 7 = 11 | 11/3 |
| ABE | 1 + 3 + 9 = 13 | 13/3 |
| ACD | 1 + 3 + 7 = 11 | 11/3 |
| ACE | 1 + 3 + 9 = 13 | 13/3 |
| ADE | 1 + 7 + 9 = 17 | 17/3 |
| BCD | 3 + 3 + 7 = 13 | 13/3 |
| BCE | 3 + 3 + 9 = 15 | 15/3 = 5 |
| BDE | 3 + 7 + 9 = 19 | 19/3 |
| CDE | 3 + 7 + 9 = 19 | 19/3 |
The mean \(11/3\) occurs for two subsets, so its probability is \(2/10=0.2\). The mean \(13/3\) occurs for three subsets, so its probability is \(3/10=0.3\). Grouping all ten rows by mean gives the exact distribution:
The probabilities total \((1+2+3+1+1+2)/10=10/10=1\). Units B and C both have value 3, but they are different population units: subsets ABC and ABE, for example, are distinct possible samples. Ignoring unit identity could lead to an incorrect count of equally likely samples.
A Reliable Enumeration Procedure
For a small population, a systematic sequence of steps prevents missed samples and probability errors. Keep the population units identifiable, even when some recorded values are equal.
State the finite population, the sample size \(n\), and that sampling is simple random sampling without replacement. Check that the question gives fixed values for all \(N\) units and that every size-\(n\) subset is equally likely.
Calculate \(\binom{N}{n}\). This is the number of rows your complete enumeration must contain before any repeated statistic values are combined.
For a mean, add the sampled values and divide by \(n\). For a proportion, count attribute-having units and divide by \(n\).
Count the subsets giving each distinct statistic value, then divide each count by \(\binom{N}{n}\).
Confirm that the sample count is \(\binom{N}{n}\) and that the probabilities over all distinct statistic values sum to 1.
Why the Counting Rule Works
The rule \(P(T=t)=m_t/\binom{N}{n}\) follows directly from the sample space. There are \(\binom{N}{n}\) equally likely subsets, each with probability \(1/\binom{N}{n}\). The event \(T=t\) consists of exactly the \(m_t\) subsets that produce \(t\). Since these subsets are mutually exclusive, their probabilities add:
Every possible subset produces one statistic value, so grouping the subsets by their values partitions the full sample space. In symbols, the counts across all distinct \(t\) values add to \(\binom{N}{n}\); dividing by \(\binom{N}{n}\) shows that all probabilities add to 1. This is both a justification of the formula and a useful check on hand calculations.
Common Mistakes and Why Exact Distributions Matter
- Giving equal probability to distinct statistic values. A mean that occurs for one subset and a mean that occurs for three subsets do not have the same probability. Divide occurrence counts by the total number of subsets.
- Counting values instead of units. Two units with the same measurement are still distinct possible selections. Identify units while enumerating, then combine samples only when their statistic values match.
- Using ordered draws as separate samples. Under the SRSWOR definition used here, a subset is one sample regardless of the order in which its units might have been drawn. Counting orders is possible only if the probabilities are handled consistently.
- Treating sample proportions as population proportions. A sample proportion varies by sample; the population proportion \(p=M/N\) is fixed. The exact distribution gives probabilities for possible \(\hat{p}\) values, not uncertainty about which population value was recorded.
- Assuming the distribution has a familiar smooth shape. Enumeration gives the actual finite set of possible statistic values and their probabilities. It may be discrete, uneven, or concentrated at only a few values.
Exact enumeration is useful because it lets us inspect the sampling behavior without approximating it. It also provides a direct check on claims about possible statistic values and their probabilities. For larger populations, listing every subset may be impractical, but the same probability model remains the reference point for understanding what a sampling distribution describes.
Check Your Understanding
Use equally likely sample subsets and distinguish the samples from the statistic values.
- A population has four distinct units with values 1, 2, 5, and 8. How many size-two samples are possible? What is the probability of each subset under an SRSWOR?
- If three of ten possible subsets produce the same sample mean, what is the probability of that mean?
- A population has six units, two of which have a specified attribute. An SRSWOR of size two is taken. How many samples contain exactly one unit with the attribute?
- Why must two units with equal recorded values still be treated as distinct units when counting possible samples?
- What two checks can help identify an incomplete or incorrectly weighted exact distribution?