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Sampling distributions for proportions · Tutorial 406 of 1000

How Sample Size Changes the Spread of p-hat

Compare the spread of sample proportions at different sample sizes and use the square-root relationship to predict how sample size changes variability.

Intermediate 9 min read

What You'll Learn

  • Calculate and compare the standard deviation of p-hat for sample sizes of 50, 200, and 800.
  • Explain why quadrupling sample size halves the standard deviation when the population proportion stays fixed.
  • Use sample-size ratios to predict how the spread of sample proportions will change.
  • Distinguish changes in sample-size spread from changes caused by a different population proportion.
  • Interpret standard deviations in proportion units and percentage points.

What Changes When the Sample Size Changes?

In Standard Deviation of p-hat Formula, you learned that the standard deviation of the sampling distribution of \(\hat{p}\) is \(\sqrt{p(1-p)/n}\). The population proportion \(p\) determines the center of that distribution, while the sample size \(n\) also affects how widely sample proportions vary around the center. Here we hold \(p\) fixed and focus on what happens to the spread as \(n\) changes.

A larger sample generally produces sample proportions that vary less from sample to sample. But the change is not proportional to the sample size: doubling \(n\) does not halve the standard deviation. The square root in the formula means that sample size has a square-root relationship with spread. In particular, multiplying the sample size by four divides the standard deviation by two.

Key relationship: When the population proportion \(p\) stays the same, the standard deviation of \(\hat{p}\) is inversely proportional to \(\sqrt{n}\). If the sample size is multiplied by \(k\), the standard deviation is multiplied by \(1/\sqrt{k}\).

Comparing \(n=50\), \(n=200\), and \(n=800\)

Suppose the same population proportion applies to each sample size. For instance, let \(p=0.40\). Then \(p(1-p)=(0.40)(0.60)=0.24\), and the standard deviation for each sample size is calculated using that same numerator. Only \(n\), the denominator, changes.

$$ \sigma_{\hat{p}}=\sqrt{\frac{0.24}{n}} $$

The calculations give approximately 0.0693 for \(n=50\), 0.0346 for \(n=200\), and 0.0173 for \(n=800\). The sample size increases by a factor of four from 50 to 200, and again by a factor of four from 200 to 800. Each time, the standard deviation is cut in half.

Sample size \(n\)Standard deviation \(\sigma_{\hat{p}}\)Spread in percentage points
50\(\sqrt{0.24/50}\approx0.0693\)About 6.93
200\(\sqrt{0.24/200}\approx0.0346\)About 3.46
800\(\sqrt{0.24/800}\approx0.0173\)About 1.73

The percentage-point column expresses the same standard deviations on a percentage scale. For example, 0.0346 in proportion units is 3.46 percentage points. These values describe the spread of sample proportions across repeated samples; they are not guaranteed maximum distances from \(p\).

The halving is visible directly in the formula. If \(n\) is replaced by \(4n\), the new standard deviation is:

$$ \sqrt{\frac{p(1-p)}{4n}} =\frac{1}{2}\sqrt{\frac{p(1-p)}{n}} $$

The factor of four inside the square root becomes a factor of two outside it. More generally, if the new sample size is \(kn\), the new standard deviation is the original standard deviation divided by \(\sqrt{k}\). This lets you predict a change in spread without recalculating from scratch.

Useful comparisons: Doubling \(n\) multiplies the standard deviation by \(1/\sqrt{2}\), approximately 0.707. Quadrupling \(n\) halves it. Multiplying \(n\) by nine makes the standard deviation one-third as large.

Worked Example: Comparing Sample Sizes of 50, 200, and 800

Worked Example: Comparing Sample Sizes of 50, 200, and 800

Suppose 40% of the residents of a large town use a community recreation center at least once a month. For each of three sample sizes, a simple random sample is drawn from the town’s 20,000 residents. Compare the standard deviations of the sampling distributions of the sample proportion who use the center monthly.

1
State.
Let \(\hat{p}\) be the proportion in a sample who use the center monthly. The population proportion is \(p=0.40\). We will compare \(\sigma_{\hat{p}}\) for \(n=50\), \(n=200\), and \(n=800\).
2
Plan and check conditions.
Each sample is described as a simple random sample, and each sample size is fixed. The largest sample is 800, and \(0.10(20{,}000)=2{,}000\), so all three sample sizes meet the 10% condition for sampling without replacement. The standard deviation formula is appropriate. The Large Counts condition is about whether a normal model is appropriate; it is not needed just to calculate these standard deviations.
3
Do.
For all three calculations, \(p(1-p)=(0.40)(0.60)=0.24\):
$$ \begin{aligned} n=50:\quad &\sigma_{\hat{p}} =\sqrt{\frac{0.24}{50}} =\sqrt{0.0048} \approx0.0693\\ n=200:\quad &\sigma_{\hat{p}} =\sqrt{\frac{0.24}{200}} =\sqrt{0.0012} \approx0.0346\\ n=800:\quad &\sigma_{\hat{p}} =\sqrt{\frac{0.24}{800}} =\sqrt{0.0003} \approx0.0173 \end{aligned} $$

As a rounding check, \(0.0693^2\approx0.004802\), \(0.0346^2\approx0.001197\), and \(0.0173^2\approx0.000299\). These are close to 0.0048, 0.0012, and 0.0003, respectively; the small differences are due to rounding the standard deviations.

4
Conclude in context.
Across repeated simple random samples of town residents, the sample proportion who use the center monthly has a standard deviation of about 0.0693 for samples of 50, 0.0346 for samples of 200, and 0.0173 for samples of 800. Each fourfold increase in sample size halves the standard deviation.

The mean of each sampling distribution is 0.40, as established in Mean of the Sampling Distribution of p-hat. Increasing the sample size changes the spread, not that center.

Predicting the Change Before Calculating

The ratio of sample sizes gives a quick way to compare standard deviations when \(p\) is fixed. If the sample size changes from \(n_1\) to \(n_2\), divide \(n_2\) by \(n_1\) to find the sample-size multiplier. Then divide the old standard deviation by the square root of that multiplier.

$$ \frac{\sigma_{\hat{p},\,2}}{\sigma_{\hat{p},\,1}} =\sqrt{\frac{n_1}{n_2}} $$

For example, increasing a sample size from 80 to 320 multiplies \(n\) by four, so the standard deviation is multiplied by \(\sqrt{80/320}=\sqrt{1/4}=1/2\). Increasing it from 80 to 720 multiplies \(n\) by nine, so the standard deviation becomes one-third as large. This ratio method is useful for comparing designs, checking a calculation, and explaining the effect of sample size in context.

Worked Example: A Fourfold Increase in a Plant Sample

Worked Example: A Fourfold Increase in a Plant Sample

In a large greenhouse inventory, 25% of the plants show a particular leaf marking. A simple random sample is taken from 5,000 plants. Compare the standard deviations of the sample proportion for sample sizes of 80 and 320.

Check the sampling conditions. Both samples are described as random, and the sample sizes are fixed. For the larger sample, \(0.10(5{,}000)=500\), and \(320\leq500\); therefore both sample sizes satisfy the 10% condition. We can use the standard deviation formula.

Calculate. Here, \(p=0.25\), \(1-p=0.75\), and \(p(1-p)=0.1875\):

$$ \begin{aligned} n=80:\quad \sigma_{\hat{p}} &=\sqrt{\frac{(0.25)(0.75)}{80}} =\sqrt{0.00234375} \approx0.0484\\ n=320:\quad \sigma_{\hat{p}} &=\sqrt{\frac{(0.25)(0.75)}{320}} =\sqrt{0.0005859375} \approx0.0242 \end{aligned} $$

The checks \(0.0484^2\approx0.002343\) and \(0.0242^2\approx0.000586\) agree with the quantities under the square roots, subject to rounding. The sample size increases by a factor of \(320/80=4\), so the predicted standard deviation ratio is \(\sqrt{80/320}=0.5\). The calculated values also give \(0.0242/0.0484=0.5\).

Across repeated random samples, the sample proportion of plants with the marking typically varies about 0.0484 from 0.25 for samples of 80, compared with about 0.0242 for samples of 320. Those standard deviations are about 4.84 and 2.42 percentage points, respectively.

Worked Example: Planning for a Smaller Spread

Worked Example: Planning for a Smaller Spread

A nature center estimates that 36% of the trees in a large preserve belong to a particular species. It plans to estimate this proportion using simple random samples from an inventory of 8,000 trees. Compare the standard deviations for sample sizes of 100 and 400, and explain whether the larger sample achieves half the spread.

Check the sampling conditions. Both sample sizes are fixed, and the samples are simple random samples. For the larger sample, \(0.10(8{,}000)=800\), and \(400\leq800\). Thus, the 10% condition is met for both samples, and the formula is appropriate.

Calculate. With \(p=0.36\), the complementary proportion is \(1-p=0.64\), so \(p(1-p)=0.2304\):

$$ \begin{aligned} n=100:\quad \sigma_{\hat{p}} &=\sqrt{\frac{(0.36)(0.64)}{100}} =\sqrt{0.002304} =0.048\\ n=400:\quad \sigma_{\hat{p}} &=\sqrt{\frac{(0.36)(0.64)}{400}} =\sqrt{0.000576} =0.024 \end{aligned} $$

Squaring the results verifies the calculation: \(0.048^2=0.002304\), and \(0.024^2=0.000576\). The sample size increases by a factor of four, and the standard deviation decreases from 0.048 to 0.024, exactly half. In context, the sample proportion of trees belonging to this species typically varies about 4.8 percentage points from 0.36 for samples of 100, and about 2.4 percentage points for samples of 400.

This comparison illustrates a planning trade-off: a larger sample reduces sampling variability, but a fourfold sample size is needed to cut the standard deviation in half. The calculation describes spread under the stated sampling process; it does not guarantee that any one sample proportion will be close to 0.36.

What the Comparison Does—and Does Not—Say

These comparisons isolate the effect of sample size because \(p\) is held fixed. If both \(p\) and \(n\) change, the standard deviation may change for two reasons. For example, at the same sample size, \(p(1-p)\) is larger when \(p\) is closer to 0.50 than when \(p\) is closer to 0 or 1. To attribute a difference in spread specifically to sample size, compare samples from the same population or otherwise keep \(p\) the same.

A smaller standard deviation means that the sampling distribution is less spread out around its mean. It does not change the population proportion, remove bias from a poor sampling method, or guarantee that a particular sample will be more accurate. The random sampling process still matters. As in the earlier tutorial on the standard deviation formula, use the 10% condition when sampling without replacement from a finite population.

Also keep the units straight. The standard deviation formula returns a value in proportion units: 0.024 means 2.4 percentage points. It is not a count of individuals, and it is not the standard deviation of individual yes-or-no responses. It describes the sample-to-sample variability of \(\hat{p}\).

Common Mistakes and AP Exam Communication

  • Assuming spread decreases in direct proportion to \(n\). Doubling the sample size does not halve the standard deviation. Use the square-root relationship: doubling \(n\) multiplies spread by \(1/\sqrt{2}\).
  • Forgetting that \(p\) must stay fixed for a clean sample-size comparison. If the population proportion differs between settings, a change in standard deviation may reflect both \(p\) and \(n\).
  • Calling the standard deviation a guaranteed error bound. A standard deviation describes spread; it does not say every sample proportion is within that distance of \(p\).
  • Confusing proportions with percentage points. A standard deviation of 0.0346 is 3.46 percentage points, not 0.0346 percentage points.
  • Omitting conditions when the samples are drawn without replacement. Identify the random sampling method and check the 10% condition using the largest sample size being compared.
AP Exam Tip: When asked how spread changes, state both the sample-size multiplier and its square-root effect. For example: “The sample size is multiplied by four, so the standard deviation is divided by \(\sqrt{4}=2\).” Then interpret the comparison in context and in proportion units or percentage points.

Key Takeaway

For a fixed population proportion, \(\sigma_{\hat{p}}=\sqrt{p(1-p)/n}\) decreases as sample size increases, but it decreases according to the square root of \(n\), not in direct proportion to \(n\). When \(n\) is multiplied by four, the standard deviation is halved. For \(p=0.40\), increasing \(n\) from 50 to 200 to 800 changes the standard deviation from about 0.0693 to 0.0346 to 0.0173.

Key takeaway: Compare sample sizes using the square-root rule: multiplying \(n\) by \(k\) multiplies the standard deviation by \(1/\sqrt{k}\), provided \(p\) stays the same and the sampling conditions are appropriate.

Check Your Understanding

Use the square-root relationship and the standard deviation formula to answer each question.

  1. For a population proportion of 0.30, compare the standard deviations for sample sizes of 60 and 240. Show both calculations and describe the relationship.
  2. If a sample size is increased from 100 to 900 while \(p\) stays fixed, by what factor does the standard deviation change?
  3. A student says that doubling a sample size halves the standard deviation of \(\hat{p}\). Explain the error and give the correct multiplier.
  4. Why must \(p\) stay fixed to isolate the effect of sample size when comparing two standard deviations?
  5. For a sample proportion standard deviation of 0.025, express the same spread in percentage points.

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