The 10% Condition for Sample Means

Why the Population Size Matters

In “Standard Deviation of the Sample Mean,” we used \(\sigma_{\bar{x}}=\sigma/\sqrt{n}\) for independent observations. When a sample is drawn without replacement from a finite population, however, selecting one individual affects which individuals remain available for later selections. The observations are not truly independent.

If the sample is a small part of the population, that dependence has little practical effect, and \(\sigma/\sqrt{n}\) is a useful approximation for the standard deviation of \(\bar{x}\). The 10% condition is a check for deciding whether the sample is small enough. It applies to sampling without replacement from a finite population; it is not a requirement when observations are independent, such as independent draws with replacement.

Definition: For a simple random sample of size \(n\) drawn without replacement from a finite population of size \(N\), the 10% condition is met when the sample size is less than 10% of the population size: \(n<0.10N\). If it is met, the observations can be treated as approximately independent when using \(\sigma_{\bar{x}}=\sigma/\sqrt{n}\).

The condition is about the fraction of the population sampled, not whether \(n\) seems large or small by itself. A sample of 100 could be a small fraction of a population of 10,000, but a substantial fraction of a population of 500.

Check the Condition Before Using the Formula

To check the 10% condition, identify both the sample size \(n\) and the population size \(N\). Then compare \(n\) with \(0.10N\). Equivalently, calculate the sampling fraction \(n/N\) and check whether it is less than 0.10. Make sure \(N\) is the number of individuals in the population from which the sample was selected, not the number in some larger or unrelated group.

Identify the sampling design.
Determine whether the data come from a simple random sample without replacement from a finite population. The 10% check addresses the dependence that can arise in this design.

Find \(n\) and \(N\).
Use the number selected for the sample as \(n\) and the total number of individuals in the population as \(N\).

Compare with 10%.
Calculate \(0.10N\), or calculate \(n/N\). The condition is met if \(n<0.10N\), equivalently if \(n/N<0.10\).

Choose the standard deviation formula.
If the condition is met, use \(\sigma/\sqrt{n}\) as the approximate standard deviation of \(\bar{x}\). If it is not met, use the finite-population adjustment when the exact standard deviation is needed.

The exact standard deviation for a simple random sample without replacement from a finite population is the formula introduced in “Standard Deviation of the Sample Mean.” The square-root factor is called the finite-population adjustment. It accounts for the reduced variability that results when a sizable share of the population is sampled.

\sigma_{\bar{x}} = \frac{\sigma}{\sqrt{n}} \sqrt{\frac{N-n}{N-1}}

When the sample is less than 10% of the population, the adjustment factor is close to 1, so \(\sigma/\sqrt{n}\) is a reasonable approximation. When the sample is a larger fraction, the factor can be meaningfully below 1. In that case, \(\sigma/\sqrt{n}\) alone overstates the exact standard deviation for sampling without replacement.

Worked Example: A Random Sample of Neighborhood Homes

A city has 2,500 homes in a particular neighborhood. A researcher selects a simple random sample of 100 homes without replacement and records each home’s monthly water use. The population standard deviation is \(\sigma=12\) gallons. Check the 10% condition and find the approximate standard deviation of the sample mean.

State. The goal is to determine whether \(\sigma/\sqrt{n}\) is appropriate as an approximation for the standard deviation of the sample mean monthly water use.

Plan. The sample is a simple random sample without replacement, so observations are dependent by the sampling design. Check the 10% condition using \(n=100\) and \(N=2500\). If the condition is met, use \(\sigma_{\bar{x}}=\sigma/\sqrt{n}\). We assume the stated population standard deviation is known.

Do. Ten percent of the population is:

0.10N=0.10(2500)=250\text{ homes}

Compare the sample size with that threshold:

$$ n=100<250=0.10N $$

The condition is met. The approximate standard deviation of the sample mean is:

\sigma_{\bar{x}} \approx \frac{\sigma}{\sqrt{n}} = \frac{12}{\sqrt{100}} = \frac{12}{10} = 1.2\text{ gallons}

Conclude. Because the sample is less than 10% of the neighborhood’s homes, treating the observations as approximately independent is reasonable for this calculation. Across repeated samples of 100 homes, sample mean monthly water use typically varies by about 1.2 gallons from the population mean, under this approximation.

When the Sample Is Too Large a Fraction

If a simple random sample without replacement is at least 10% of its population, the 10% condition is not met under the strict check used here. This does not mean the sample was collected incorrectly or that a sample mean cannot be calculated. It means that the usual independence approximation needs more care. Use the finite-population adjustment if the standard deviation for sampling without replacement is required.

Notice that the adjustment is less than 1 when \(n>1\), because \(N-n<N-1\). Thus, the exact standard deviation is smaller than \(\sigma/\sqrt{n}\). Sampling a sizable portion of the population leaves fewer possible values for the remaining draws, reducing the variation among sample means.

Worked Example: Sampling a Large Share of a Small Population

A wildlife team takes a simple random sample without replacement of 20 animals from a population of 100 animals. A measured characteristic has population standard deviation \(\sigma=10\) units. Check the 10% condition and calculate the exact standard deviation of the sample mean.

State. We need to decide whether the usual \(\sigma/\sqrt{n}\) formula is an appropriate approximation under the 10% condition, then find the exact standard deviation for this sampling design.

Plan. The sample is random and drawn without replacement, so check \(n<0.10N\). If the condition fails, calculate the exact standard deviation using the finite-population adjustment. The random-sampling design is stated; the 10% check concerns the remaining dependence from sampling without replacement.

Do. Calculate 10% of the population:

0.10N=0.10(100)=10\text{ animals}

Since \(20\) is not less than \(10\), the 10% condition is not met. Calculate the exact standard deviation:

\sigma_{\bar{x}} = \frac{10}{\sqrt{20}} \sqrt{\frac{100-20}{100-1}} = 10\sqrt{\frac{80}{20(99)}} = \frac{20}{\sqrt{99}} \approx 2.0101\text{ units}

For comparison, using only the usual formula would give:

\frac{10}{\sqrt{20}} \approx 2.2361\text{ units}

Conclude. Because the sample is 20% of the population, the 10% condition is not met, so the unadjusted formula is not the appropriate exact calculation. The exact standard deviation of the sample mean is about 2.0101 units; omitting the finite-population adjustment would give the larger value 2.2361 units.

Use the Condition When Planning a Sample

The same comparison can help when designing a study. Rearranging \(n<0.10N\) gives \(N>10n\). So, for a planned sample size \(n\), the population must be larger than ten times the sample size for the strict 10% condition to be met. This is a quick planning check, not a way to change the actual population size: use the population that the study is intended to describe.

Worked Example: Check a Planned Sample Against Two Possible Populations

A community survey plans to select a simple random sample of 75 registered anglers without replacement. The survey team is considering two possible target populations: a list of 720 anglers or a list of 900 anglers. The population standard deviation of a quantitative measure is \(\sigma=24\) minutes. Determine which population size meets the 10% condition and find the approximate standard deviation of the sample mean when it does.

State. We will check the 10% condition for each proposed population size, then use the standard formula for the population size that meets the condition.

Plan. The planned sample size is \(n=75\), and both designs use simple random sampling without replacement. Compare 75 with 10% of each target population. For a population that passes, calculate \(\sigma/\sqrt{n}\) as an approximation.

Do. For a population of 720 anglers:

0.10(720)=72,\qquad 75\not<72

The condition is not met for the 720-person population. For a population of 900 anglers:

0.10(900)=90,\qquad 75<90

The condition is met for the 900-person population. For that population, the approximate standard deviation is:

\sigma_{\bar{x}} \approx \frac{24}{\sqrt{75}} \approx 2.7713\text{ minutes}

Conclude. A sample of 75 is more than 10% of a population of 720, so the usual formula is not justified by the 10% condition for that population. It is less than 10% of a population of 900, so \(\sigma/\sqrt{n}\) is a reasonable approximation there. For the 900-person population, the standard deviation of the sample mean is approximately 2.7713 minutes.

Common Mistakes and AP Exam Tip

Checking the wrong population size: Use the size of the population from which the sample was actually selected. A large region’s population is not the right \(N\) if the sampling list covers only one town.
Using the sample size alone: The condition compares \(n\) with \(N\). State the comparison, such as \(100<0.10(2500)=250\), rather than simply claiming that the sample is small.
Thinking the 10% condition makes observations truly independent: Sampling without replacement still creates dependence. The condition says that treating the observations as approximately independent is reasonable for the standard-deviation approximation.
Using \(\sigma/\sqrt{n}\) without checking the design: First identify whether observations are independent or whether the sample is without replacement from a finite population. The 10% condition is relevant to the latter situation.
Assuming failure means the sample is invalid: Failure means the approximation needs adjustment, not that the sample mean is unusable. For sampling without replacement, the finite-population adjustment accounts for the difference.
Confusing spread with shape: The 10% condition concerns independence and the standard deviation calculation. It does not establish that the sampling distribution of \(\bar{x}\) is normal.

For full-credit communication, name the sampling design, identify \(n\) and \(N\), show the comparison with 10%, and state what the result permits. For example: “Because 100 is less than 10% of 2,500, the 10% condition is met, so \(\sigma/\sqrt{n}\) is a reasonable approximation to the standard deviation of the sample mean.” If the check fails, say so and explain that the finite-population adjustment is needed for the exact standard deviation.

Key takeaway: For a simple random sample without replacement from a finite population, check whether \(n<0.10N\). If the condition is met, use \(\sigma/\sqrt{n}\) as an approximation for the standard deviation of \(\bar{x}\). If it is not met, account for the finite population using the adjustment factor.

Check Your Understanding

For each question, assume that the sample is a simple random sample drawn without replacement unless stated otherwise.

A sample of 80 students is selected from a school with 1,000 students. Does the 10% condition hold? Show the comparison.
A sample of 45 residents is selected from a population of 400. Is \(\sigma/\sqrt{n}\) justified by the 10% condition? Explain.
A population has standard deviation 15 units. A random sample of 100 is drawn from a population of 2,000. Find the approximate standard deviation of the sample mean and give its units.
In your own words, explain why sampling 20% of a finite population can make \(\sigma/\sqrt{n}\) alone inaccurate for sampling without replacement.
Does meeting the 10% condition show that the sampling distribution of \(\bar{x}\) is normal? Explain what the condition does establish.