Worked Comparisons of Sampling Distributions

What a Comparison Should Establish

A sampling distribution describes the values a statistic would take over repeated samples under a specified sampling plan. To compare two sampling distributions, identify what is held fixed and what changes: the population, the sample size, or the sampling design. Then compare their centers, spreads, and, when relevant, shapes. A conclusion about one feature does not automatically establish a conclusion about the others.

For sample means, the population mean determines the center when samples are selected appropriately. The population standard deviation and sample size help determine the spread, and sampling with or without replacement can matter as well. For sample proportions, both the population proportion and sample size affect the distribution. These facts let us make careful comparisons without assuming that a sampling distribution is normal.

Definition: A sampling distribution of a statistic is the probability distribution of the statistic over all samples selected according to a specified sampling plan. A valid comparison names the statistic and sampling plan, then describes how the distributions differ in center, spread, or shape.

For independent observations sampled with replacement from a population with mean \(\mu\) and standard deviation \(\sigma\), the sample mean \(\bar{x}\) has mean \(\mu\) and standard error \(SE=\sigma/\sqrt{n}\). For a simple random sample without replacement from a finite population of size \(N\), when \(\sigma^2\) is defined using divisor \(N\), its standard error is \(\sqrt{\frac{\sigma^2}{n}\frac{N-n}{N-1}}\). For independent binary observations with population proportion \(p\), the sample proportion \(\hat{p}\) has mean \(p\) and standard error \(\sqrt{\frac{p(1-p)}{n}}\). Each formula depends on its stated sampling conditions.

Compare Exact Distributions, Not Just Their Centers

If a population is small enough, we can list all possible samples and calculate the exact distribution of a statistic. This gives a direct comparison and makes clear that two distributions can have the same center but different spreads. In the example below, each draw is made with replacement, so the two draws are independent and each ordered pair of population values is equally likely.

Worked Example: Two Populations with the Same Mean

Consider two hypothetical populations, A with values \(2,4,6\) and B with values \(0,4,8\). From each population, take a sample of size \(n=2\) with replacement and calculate \(\bar{x}\). Compare the resulting sampling distributions.

Population centers and spreads. Both population means are 4. In A, the population variance is \(\sigma_A^2=((2-4)^2+(4-4)^2+(6-4)^2)/3=8/3\). In B, it is \(\sigma_B^2=((0-4)^2+(4-4)^2+(8-4)^2)/3=32/3\). Thus B’s population variance is four times A’s.

Distribution for A. There are \(3^2=9\) equally likely ordered samples. The sample mean can be \(2,3,4,5,\) or \(6\). Counting ordered pairs gives one pair producing 2, two producing 3, three producing 4, two producing 5, and one producing 6.

P(\bar{x}=2,3,4,5,6)=\left(\frac{1}{9},\frac{2}{9},\frac{3}{9},\frac{2}{9},\frac{1}{9}\right)

For example, \(\bar{x}=2\) occurs only for the ordered pair \((2,2)\), so its probability is \(1/9\), not \(3/9\). The mean of this sampling distribution is 4. Its variance is \(4(1/9)+1(2/9)+0(3/9)+1(2/9)+4(1/9)=12/9=4/3\).

Distribution for B. The possible sample means are \(0,2,4,6,\) and \(8\). The same count of ordered pairs gives their respective probabilities \(1/9,2/9,3/9,2/9,\) and \(1/9\). Its mean is 4, and its variance is \(16(1/9)+4(2/9)+0(3/9)+4(2/9)+16(1/9)=48/9=16/3\).

Comparison. Both sampling distributions are centered at 4, but B’s variance is four times A’s, so its standard deviation is twice as large: \(\sqrt{16/3}/\sqrt{4/3}=2\). The formula also confirms the results: \(SE_A=\sqrt{(8/3)/2}=\sqrt{4/3}\), while \(SE_B=\sqrt{(32/3)/2}=\sqrt{16/3}\). The populations’ different spreads change the variability of the sample mean, even though their means match.

Separate Sample-Size Effects from Sampling-Design Effects

Increasing the sample size generally reduces the standard error of a sample mean, but the exact comparison depends on the sampling plan. With replacement, the standard error is \(\sigma/\sqrt{n}\). Without replacement, sampling a substantial fraction of a finite population reduces variability further; the finite population correction accounts for that reduction. The mean remains centered at the population mean in either design when the sample is selected as specified.

Worked Example: A Larger Sample from a Finite Population

Suppose a hypothetical finite population consists of the five values \(1,3,5,7,9\). A simple random sample is taken without replacement. Compare the sampling distributions of \(\bar{x}\) for \(n=2\) and \(n=4\).

The population mean is \(\mu=5\), and its variance, using divisor \(N=5\), is \(\sigma^2=(16+4+0+4+16)/5=8\). For \(n=2\), the standard error is

SE_2=\sqrt{\frac{8}{2}\frac{5-2}{5-1}} =\sqrt{3}\approx 1.732

There are \(\binom{5}{2}=10\) equally likely unordered samples. Their means are \(2,3,4,4,5,5,6,6,7,8\). The distribution is centered at 5, and its variance is \([9+4+1+1+0+0+1+1+4+9]/10=30/10=3\), agreeing with \(SE_2^2=3\).

For \(n=4\), every sample omits exactly one population value. The five equally likely sample means are \(4,4.5,5,5.5,6\): omitting 9 gives \((25-9)/4=4\), omitting 7 gives \(4.5\), omitting 5 gives 5, omitting 3 gives \(5.5\), and omitting 1 gives 6. The mean is again 5, and the variance is \([1+0.25+0+0.25+1]/5=0.5\). The standard error is therefore \(\sqrt{0.5}\approx0.7071\), also given by

SE_4=\sqrt{\frac{8}{4}\frac{5-4}{5-1}} =\sqrt{0.5}\approx0.7071

The larger sample gives a distribution with the same center and substantially less spread. Here the sample of four includes a large fraction of the population, so the finite population correction is important. The comparison does not imply that each possible sample mean for \(n=4\) is closer to 5 than every possible mean for \(n=2\); it establishes that the distribution’s standard error is smaller.

Compare Proportions Using Their Own Parameters

A sample proportion is the mean of binary indicators that equal 1 for a success and 0 otherwise. Under independent sampling, each indicator has mean \(p\) and variance \(p(1-p)\). Averaging \(n\) independent indicators gives a sampling distribution with mean \(p\) and standard error \(\sqrt{p(1-p)/n}\). Thus the population proportion sets the center, while both \(p\) and \(n\) affect the spread.

Worked Example: Different Population Proportions, Same Sample Size

Suppose two hypothetical populations are sampled independently with replacement, and each sample has size \(n=10\). In population A, the proportion with a specified characteristic is \(p_A=0.20\); in population B, it is \(p_B=0.50\). Compare the centers, standard errors, and one exact probability in each sampling distribution.

The expected sample proportions are \(E(\hat{p}_A)=0.20\) and \(E(\hat{p}_B)=0.50\). Their standard errors are

SE_A=\sqrt{\frac{0.20(0.80)}{10}} =\sqrt{0.016}\approx0.1265,\qquad SE_B=\sqrt{\frac{0.50(0.50)}{10}} =\sqrt{0.025}\approx0.1581

The second distribution is centered higher and has greater standard error. The greater spread follows from \(p_B(1-p_B)=0.25\), which exceeds \(p_A(1-p_A)=0.16\) for this comparison; it is not a rule that a larger value of \(p\) always gives a larger standard error.

To compare an exact probability, \(\hat{p}=0\) means that all 10 observations are failures. Under independent sampling, \(P(\hat{p}=0)=(1-p)^{10}\). Therefore, \(P(\hat{p}_A=0)=0.80^{10}\approx0.1074\), whereas \(P(\hat{p}_B=0)=0.50^{10}\approx0.0009766\). These are probabilities of observing no successes in a sample of 10 under the respective population proportions. They do not imply that either population proportion is uncertain or randomly changing.

A Short Justification for the Sample-Size Comparison

The standard-error formula for independent sampling explains why a larger sample narrows a sample-mean distribution. Write the observations as \(X_1,\ldots,X_n\), each with mean \(\mu\) and variance \(\sigma^2\). Since \(\bar{x}=(X_1+\cdots+X_n)/n\), independence means that the variance of the sum is the sum of the variances. Scaling a random variable by \(1/n\) scales its variance by \(1/n^2\). Thus

\operatorname{Var}(\bar{x}) =\frac{1}{n^2}\operatorname{Var}(X_1+\cdots+X_n) =\frac{1}{n^2}(n\sigma^2) =\frac{\sigma^2}{n}, \qquad SE=\sqrt{\operatorname{Var}(\bar{x})} =\frac{\sigma}{\sqrt{n}}.

This derivation requires independent observations with a common finite variance. It does not say that the sampling distribution must have any particular shape. If observations are dependent because sampling is without replacement from a small finite population, use the finite-population standard-error formula instead.

How to Make a Defensible Comparison

A useful comparison begins by stating the sampling plan and the statistic. Then calculate or identify the center and spread under each condition, keeping units attached to means and proportions. If an exact distribution is available, listing its possible values and probabilities can show differences in shape or particular event probabilities that a standard error alone cannot capture.

Specify what is being sampled.
Name the population, sample size, statistic, and whether observations are independent or selected without replacement.

Compare centers.
For the sample mean, use the population mean; for a sample proportion under the stated model, use \(p\). State whether the centers match or differ.

Compare spreads and, if needed, shapes.
Use the standard-error formula appropriate to the design. For exact comparisons, enumerate sample outcomes or use a correctly specified simulation.

A simulation should reproduce the sampling design: each simulated sample must be drawn from the appropriate population, with replacement or without replacement as required, and the statistic must be recalculated for every repetition. The resulting simulated distribution approximates the sampling distribution; it is not a substitute for specifying the model or checking whether the simulation matches the question.

Common Mistakes

Using the population spread as the spread of the sample mean. The standard error is not generally \(\sigma\); with independent sampling it is \(\sigma/\sqrt{n}\).
Assuming equal centers mean equal distributions. Example 1 has matching means but different standard errors and variances.
Ignoring the sampling design. The independent-sampling formula does not account for the reduction in variability from sampling a large fraction of a finite population without replacement.
Calling a standard error a statement about every sample. It summarizes the distribution’s spread; individual sample outcomes may be farther from the center than the standard error.
Assuming standard error always increases as \(p\) increases. For proportions, the relevant factor is \(p(1-p)\), which rises to its maximum at \(p=0.5\) and then falls.

Key takeaway: A sound comparison separates location, spread, and shape. Use the sampling design and the appropriate formula, and do not infer that two sampling distributions are alike just because one feature, such as the center, is the same.

Check Your Understanding

For each question, state the comparison and give a brief justification.

Two independent samples with replacement come from populations with the same \(\mu\) but different \(\sigma\). What can you conclude about the centers of the sample-mean distributions, and what additional quantity determines their spreads?
A sample mean is calculated from independent observations with \(\sigma=12\). Compare its standard error for \(n=9\) and \(n=36\).
Why does sampling four of the five values without replacement in the finite-population example have less variability than sampling two?
For independent samples of equal size, compare the standard errors of sample proportions when \(p=0.10\) and \(p=0.50\). Which is larger, and why?
What must a simulation do differently if the intended sampling plan is without replacement rather than with replacement?