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Sampling distributions for proportions · Tutorial 407 of 1000

When the 10 Percent Condition Applies

See when the 10% condition lets you use the usual standard deviation formula for a sample proportion from a finite population.

Intermediate 9 min read

What You'll Learn

  • Explain why sampling without replacement makes successive selections dependent.
  • Check whether a sample is no more than 10% of a school or city population.
  • Use the 10% condition to decide when the usual standard deviation formula is appropriate.
  • Distinguish the 10% condition from the Large Counts condition for a normal model.
  • Describe what the condition does not establish about bias or representativeness.

Why the Sampling Method Matters

In Standard Deviation of p-hat Formula, you learned that \(\sigma_{\hat{p}}=\sqrt{p(1-p)/n}\) when the observations can be treated as independent. For a sample drawn without replacement from a finite population, however, the selections are not exactly independent. Once an individual is selected, that individual cannot be selected again, and the composition of the remaining population changes slightly.

The 10% condition is a practical rule for deciding when that dependence is small enough to use the usual standard deviation formula. It asks whether the sample size is no more than 10% of the population size. If it is, AP Statistics treats the observations as approximately independent for this calculation.

Condition: When sampling without replacement from a finite population of size \(N\), check that \(n\leq0.10N\). Equivalently, check that the sample is no more than 10% of the population.

Here, \(n\) is the number of individuals in the sample, and \(N\) is the number of individuals in the population from which the sample is drawn. Use the population relevant to the question—not a larger region or a different group that happens to contain the target population.

What Sampling Without Replacement Does

Imagine drawing student names from a school roster without putting each name back. After one student is selected, there is one fewer student in the remaining roster. The chance of selecting a student with a particular characteristic can therefore change from draw to draw. The selections are dependent: knowing which student was selected first gives some information about the remaining pool.

If the sample is a small fraction of the population, each selection changes the remaining pool only a little. The probability of selecting a person with the characteristic stays close to the population proportion \(p\), and the dependence between selections is small. This is the idea behind the 10% condition. It does not make sampling without replacement literally independent; it supports treating the observations as approximately independent for the standard deviation calculation.

The usual formula is:

$$ \sigma_{\hat{p}}=\sqrt{\frac{p(1-p)}{n}} $$

When the 10% condition is met, this formula gives the AP Statistics standard deviation for the sampling distribution of \(\hat{p}\), provided the sample is random and the sample size is fixed. As in Mean of the Sampling Distribution of p-hat, the center is \(p\); this tutorial focuses on when the standard deviation formula is justified.

Key idea: Sampling without replacement creates some dependence. If \(n\leq0.10N\), the sample is small enough relative to the population to treat that dependence as negligible for the usual standard deviation formula.

How to Check the Condition

There are two equivalent ways to check. Calculate 10% of the population and compare it with the sample size, or calculate the sample’s fraction of the population and check that it is at most 0.10. The first method is often quickest when the population size is given.

$$ n\leq 0.10N \qquad\text{or, equivalently,}\qquad \frac{n}{N}\leq0.10 $$

For example, if a school has 1,800 students, then 10% of the population is \(0.10(1{,}800)=180\). A random sample of 120 students satisfies the condition because \(120\leq180\). A random sample of 200 does not, because \(200>180\).

When comparing several samples from the same population, check the largest sample size. If the largest sample meets the condition, all smaller sample sizes meet it too. Make sure the population count matches the sampling frame: a survey of students enrolled at one school should use that school’s enrollment as \(N\), not the population of the city where it is located.

Worked Example: A Random Sample from a School

Worked Example: A Random Sample from a School

A school has 1,800 students. A counselor selects a simple random sample of 120 students without replacement to estimate the proportion who ride the bus to school. Suppose the population proportion who ride the bus is \(p=0.30\). Check whether the usual standard deviation formula is appropriate, then calculate the standard deviation of \(\hat{p}\).

1
State.
Let \(\hat{p}\) be the proportion of sampled students who ride the bus. The population size is \(N=1{,}800\), the sample size is \(n=120\), and \(p=0.30\).
2
Plan and check conditions.
The sample is described as a simple random sample, and its size is fixed. The sampling is without replacement, so check the 10% condition: \(0.10(1{,}800)=180\), and \(120\leq180\). The condition is met, so the dependence from sampling without replacement is small enough to use the usual standard deviation formula.
3
Do.
Substitute \(p=0.30\), \(1-p=0.70\), and \(n=120\):
$$ \sigma_{\hat{p}} =\sqrt{\frac{(0.30)(0.70)}{120}} =\sqrt{\frac{0.21}{120}} =\sqrt{0.00175} \approx0.0418 $$

Check the rounded result by squaring it: \(0.0418^2=0.00174724\), which is close to \(0.00175\). The small difference is due to rounding.

4
Conclude in context.
Across repeated simple random samples of 120 students drawn without replacement from this school, the sample proportion who ride the bus typically varies about 0.0418, or 4.18 percentage points, from the population proportion of 0.30.

The 10% condition justifies using the standard deviation formula; it does not guarantee that the sample proportion in any one sample will be within 0.0418 of 0.30.

Worked Example: A City Population

Worked Example: A City Population

A city has 42,000 residents. A planning group takes a simple random sample of 1,500 residents without replacement and studies whether they support adding protected bike lanes. Suppose \(p=0.54\) of all city residents support the proposal. Check the 10% condition and calculate the standard deviation of the sample proportion.

Check the sampling conditions. The sample is random and has a fixed size. Because sampling is without replacement, compare \(n\) with 10% of \(N\):

$$ 0.10N=0.10(42{,}000)=4{,}200 \qquad\text{and}\qquad 1{,}500\leq4{,}200 $$

The sample is less than 10% of the city population, so the condition is met. The usual standard deviation formula is appropriate.

Calculate. Here, \(1-p=0.46\), so \(p(1-p)=(0.54)(0.46)=0.2484\):

$$ \sigma_{\hat{p}} =\sqrt{\frac{(0.54)(0.46)}{1{,}500}} =\sqrt{\frac{0.2484}{1{,}500}} =\sqrt{0.0001656} \approx0.0129 $$

As a check, \(0.0129^2=0.00016641\), close to \(0.0001656\); the difference comes from rounding the standard deviation. In context, sample proportions of residents who support the proposal typically vary about 0.0129, or 1.29 percentage points, from 0.54 across repeated random samples of 1,500 residents.

What If the Sample Is More Than 10%?

Suppose a school has 720 students and a survey samples 90 students without replacement. Ten percent of the school is \(0.10(720)=72\), and \(90>72\). The 10% condition is not met. Under the standard AP approach, you should not claim that the usual standard deviation formula is justified by this condition. The issue is not that a calculation with the formula is impossible; it is that the approximation based on treating observations as independent is not supported by the stated rule.

The reason can be seen by considering what happens as a sample gets closer to a census. If all 720 students were selected, every student would be included, so the sample proportion would equal the population proportion with no sampling variation. But the usual formula \(\sqrt{p(1-p)/n}\) would still give a positive value for \(0<p<1\). For a large fraction of the population, the fact that selected individuals are not replaced matters more; the usual formula no longer reflects that reduced variation well.

For additional perspective, the exact standard deviation for a simple random sample without replacement from a known finite population includes a finite-population adjustment:

$$ \sigma_{\hat{p}} =\sqrt{\frac{p(1-p)}{n}}\sqrt{\frac{N-n}{N-1}} $$

The second factor is less than 1 and accounts for the reduced variation caused by sampling a substantial fraction of the population. This adjustment is not needed for the routine AP calculation when the 10% condition holds. It helps explain why the condition matters: with a small sampling fraction, the adjustment is close to 1; as the sample approaches the full population, it gets closer to 0.

For the school example, suppose \(p=0.40\). The usual formula gives \(\sqrt{(0.40)(0.60)/90}=\sqrt{0.0026667}\approx0.0516\). The finite-population adjustment is \(\sqrt{(720-90)/(720-1)}=\sqrt{630/719}\approx0.9361\). Multiplying gives an exact standard deviation of about \((0.0516)(0.9361)=0.0483\). Squaring this rounded value gives \(0.0483^2=0.00233289\), close to the unrounded variance \(0.0026667(630/719)\approx0.0023366\). The adjustment makes the exact spread smaller than the usual formula suggests. In an AP response, the key point is that the 10% condition fails, so the standard independence-based formula is not justified by the usual rule.

What the 10% Condition Does—and Does Not—Establish

The 10% condition is about dependence created by sampling without replacement. It does not say that the sample was randomly selected, and it cannot make a biased sampling method representative. If a survey asks only students who visit the library, for example, a small sample compared with the whole school does not remove the selection bias.

The condition also does not determine whether the sampling distribution of \(\hat{p}\) is approximately normal. That is a separate question. As covered in Normal Models for Sample Means and Proportions, the Large Counts condition for a normal model of \(\hat{p}\) checks whether \(np\geq10\) and \(n(1-p)\geq10\). The 10% condition concerns the sampling fraction; the Large Counts condition concerns the expected numbers of successes and failures.

Keep the conditions separate: Use the 10% condition to support treating observations as approximately independent when sampling without replacement. Use the Large Counts condition to assess whether a normal model for \(\hat{p}\) is appropriate. One condition does not replace the other.

Common Mistakes and AP Exam Communication

  • Using the wrong population size. Compare the sample with the population actually being sampled. A school survey uses the school’s student population, not the city’s total population.
  • Checking the wrong direction. The requirement is \(n\leq0.10N\). If \(n\) is greater than 10% of \(N\), the condition fails.
  • Calling selections independent without qualification. Without replacement, selections are dependent. When the condition holds, say they can be treated as approximately independent for the standard deviation calculation.
  • Treating the condition as proof of a representative sample. The 10% check does not address undercoverage, voluntary response, or other sources of bias.
  • Confusing the 10% condition with Large Counts. A sample can meet the 10% condition but fail one of the Large Counts checks, or vice versa. They answer different questions.
  • Using the standard formula automatically when the condition fails. State that the usual formula is not justified by the 10% rule, rather than silently assuming independence.
AP Exam Tip: Write the condition with the actual values: “The population has \(N=\ldots\), so 10% is \(\ldots\). Since \(n=\ldots\) is no greater than that value, the 10% condition is met.” Then connect it to the method: “Because the sample is drawn without replacement but is no more than 10% of the population, we can treat the observations as approximately independent and use the standard deviation formula.”

Key Takeaway

Sampling without replacement makes selections dependent because each selected individual changes the pool that remains. When the sample is no more than 10% of the population, that change is small enough for AP Statistics to treat the observations as approximately independent when calculating the standard deviation of \(\hat{p}\).

Key takeaway: For a random sample of fixed size drawn without replacement, check \(n\leq0.10N\). If the condition is met, use \(\sigma_{\hat{p}}=\sqrt{p(1-p)/n}\) for the standard deviation. The condition supports the formula; it does not establish random selection, eliminate bias, or check normality.

Check Your Understanding

For each situation, identify the population size and sample size, then apply the 10% condition where appropriate.

  1. A school has 2,400 students. A random sample of 180 students is selected without replacement. Is the 10% condition met? Show the comparison.
  2. A city has 15,000 residents. A survey takes a random sample of 1,800 residents without replacement. Is the usual standard deviation formula justified by the 10% condition? Explain.
  3. Why are selections dependent when sampling without replacement, and why does a small sampling fraction make that dependence less important?
  4. A sample meets the 10% condition but has \(np=8\). What does the 10% condition establish, and what separate issue still needs checking before using a normal model?
  5. A student says, “The sample is less than 10% of the population, so it must be representative.” Explain why that conclusion is not justified.

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