Statistics · S.IC.B.4

Sampling distribution of a sample proportion

A county says that 35% of its households compost. A random sample of 150 households finds 40% composting. Is that gap a surprise, or the ordinary wobble of a sample?

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Shape of a binomial and the normal approximationStatistics

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Conditions for one proportionStatistics Inference for proportionsGrades 11 to 12 Sampling distribution of a differenceStatistics

Where you are 7 in Sampling distributions

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▴ 2 above Simulating a sampling distribution and a margin of error The ten percent condition Sampling distribution of a sample mean

Sampling distribution of a sample proportion

Sampling distribution of a difference

Why this exists

Kadri works in a county waste office. The county’s records say 35\% of its households compost food scraps (a made-up figure, as is every number in this lesson). Kadri surveys a random sample of 150 households and finds 60 that compost, a share of 60/150 = 0.40. Has composting grown, or did her sample happen to catch a few extra composters?

One sample cannot answer that by itself. What answers it is the pattern the sample share makes over all the samples Kadri could have drawn: where it centers, how far it typically strays, and what shape it takes. That pattern is the sampling distribution of a sample proportion, the proportion case of Sampling distributions.

The lesson takes two days. Day 1 finds the center and spread of the sample share from the binomial count, and the condition the spread needs. Day 2 adds the condition for a normal shape and uses the normal model to find chances. Tiit and Maarja, who also work in the office, run the other surveys.

Definition

Each household in a sample either composts (a success) or does not. If X is the number of successes in a sample of size n, the sample proportion is

\hat{p} = \frac{X}{n}

The population proportion p is the share of successes in the whole population; here p = 0.35. The sampling distribution of \hat{p} is the distribution of the values \hat{p} takes over all possible random samples of size n from the population. Its mean is \mu_{\hat{p}} and its standard deviation \sigma_{\hat{p}}.

distribution of	share	spread
all households	p	\displaystyle \sqrt{p(1-p)}
one sample	\hat{p}
\hat{p}, all samples	\mu_{\hat{p}}	\sigma_{\hat{p}}

The population’s spread treats each household as a score of 1 (yes) or 0 (no); “Why it is true” below shows where \displaystyle \sqrt{p(1-p)} comes from.

Intuition

Before dividing by n, the count X of composting households in a random sample is a binomial count, as long as the households can be treated as independent. The sample share is that same count on a new scale. Figure 1 draws the bars once, with both scales underneath: 60 households is the share 0.40, 45 is 0.30. Dividing by n squeezes the axis evenly, so the center, the spread and the shape of \hat{p} are those of X, each divided by n.

Figure 1Samples of 150 households when 35\% compost (made-up share). Each bar is one possible count and, on the lower scale, the same sample’s share; dividing by 150 relabels the axis and changes nothing else. Shaded bars are the samples with \hat{p} > 0.40.

A larger sample makes the count spread more in households but less as a share, because the count’s spread grows like \displaystyle \sqrt{n} while the division is by n itself. The pile of shares tightens around p as n grows. Simulating a sampling distribution and a margin of error watched this happen in a simulated pile of sample proportions; this lesson proves the formulas behind it and puts a normal model on top.

Results

Day 1: center and spread

For a random sample of size n from a population with proportion p:

\mu_{\hat{p}} = p, \qquad \sigma_{\hat{p}} = \sqrt{\frac{p(1-p)}{n}}

The center holds for every random sample, so \hat{p} is an Bias and variability of an estimator of p. The spread needs independent observations: sampling with replacement, or without replacement with n at most 10\% of the population (the The ten percent condition).

Day 2: the normal shape

When there are at least ten expected successes and ten expected failures,

np \ge 10 \quad \text{and} \quad n(1-p) \ge 10

(the large-counts condition), the sampling distribution of \hat{p} is approximately normal, \displaystyle N\!\left(p, \sqrt{p(1-p)/n}\right). Chances for \hat{p} then come from

z = \frac{\hat{p} - p}{\sqrt{p(1-p)/n}}

and the standard normal areas of Normal calculations: areas and percentiles. The formulas of Day 1 hold whether or not this condition holds; the condition is only about shape.

Why it is true

Center and spread. With independent observations, X is binomial with mean np and standard deviation \displaystyle \sqrt{np(1-p)} (Mean and standard deviation of a binomial). Dividing a random variable by the constant n divides its mean and its standard deviation by n:

\begin{aligned} \mu_{\hat{p}} &= \frac{np}{n} = p \\ \sigma_{\hat{p}} &= \frac{\sqrt{np(1-p)}}{n} = \sqrt{\frac{p(1-p)}{n}} \end{aligned}

The last step writes \displaystyle n = \sqrt{n^2} and cancels one n under the root. The center needs no independence, since the mean of a sum is always the sum of the means; the spread uses the binomial standard deviation, which does.

A mean in disguise. Score each household 1 for yes and 0 for no. The population of scores has mean p, and its variance is p(1-p)^2 + (1-p)(0-p)^2 = p(1-p). The sample mean of the scores is X/n = \hat{p}, so the Sampling distribution of a sample mean gives the same answer, \displaystyle \sigma/\sqrt{n} = \sqrt{p(1-p)}/\sqrt{n}.

Shape. Dividing by n relabels the axis without changing the shape of the bars. The Shape of a binomial and the normal approximation showed that a binomial count with np \ge 10 and n(1-p) \ge 10 is close to normal, so \hat{p} = X/n is too, under the same condition.

Worked examples

Normal areas use z rounded to two decimal places and \Phi(z), the area to the left of z; for a negative z, \Phi(-z) = 1 - \Phi(z).

z	\Phi(z)	z	\Phi(z)
1.05	0.8531	1.58	0.9429
1.18	0.8810	1.66	0.9515
1.28	0.8997	2.50	0.9938
1.41	0.9207	2.57	0.9949

Cutoffs: \Phi(1.645) = 0.950 and \Phi(1.960) = 0.975.

Example 1

Day 1: center and spread for 150 households

The county has 52{,}000 households. Find the mean and standard deviation of \hat{p} for Kadri’s samples of 150, check the condition the spread needs, and interpret both numbers.

Solution

The count first. X has mean 150(0.35) = 52.5 households and standard deviation \displaystyle \sqrt{150(0.35)(0.65)} = \sqrt{34.125} \approx 5.842 households. Dividing by 150:

\begin{aligned} \mu_{\hat{p}} &= 0.35 \\ \sigma_{\hat{p}} &= \sqrt{\frac{0.35(0.65)}{150}} \approx 0.0389 \end{aligned}

which matches 5.842/150 \approx 0.0389. The households are drawn without replacement, and 150 \le 0.10(52{,}000) = 5{,}200, so the ten percent condition holds and the formula applies.

In context: over all random samples of 150 households, the sample shares average 0.35, and a typical sample share misses 0.35 by about 0.039, close to 4 percentage points. Kadri’s 0.40 is about 1.3 such steps above the center.

Example 2

Day 1: a bigger sample, a tighter share

Tiit plans samples of 600 households. Find \sigma_{\hat{p}}, and find the smallest sample for which \sigma_{\hat{p}} is at most 0.02.

Solution

\sigma_{\hat{p}} = \sqrt{\frac{0.2275}{600}} \approx 0.0195

Four times the households, half the spread, since \displaystyle \sqrt{4} = 2 sits in the denominator. For a spread of at most 0.02, square both sides of \displaystyle \sqrt{0.2275/n} \le 0.02 (both sides are positive, so the direction holds):

n \ge \frac{0.2275}{0.02^2} = 568.75

A sample size is a whole number, so the smallest is n = 569. Check: \displaystyle \sqrt{0.2275/569} \approx 0.019996, while n = 568 gives about 0.020013, just over. Both 600 and 569 are far under 5{,}200.

Example 3

Day 1: when the town is small

Maarja studies one town of 1{,}200 households, 35\% of which compost, and wants a sample of 150 drawn without replacement. Does \displaystyle \sigma_{\hat{p}} = \sqrt{p(1-p)/n} apply? What is the largest sample for which it does, and what is \sigma_{\hat{p}} there?

Solution

The ten percent condition needs n \le 0.10(1{,}200) = 120. A sample of 150 is 12.5\% of the town, so the condition fails. Each household drawn changes the mix of those left, the draws are not close to independent, and the formula’s 0.0389 overstates the spread; for this town the true standard deviation of \hat{p} works out to about 0.0364. The center is still 0.35, since unbiasedness never needed independence.

The largest sample that meets the condition is 120, with

\sigma_{\hat{p}} = \sqrt{\frac{0.2275}{120}} \approx 0.0435

Example 4

Day 2: is Kadri’s 40% unusual?

If p = 0.35, how likely is a sample of 150 households with \hat{p} > 0.40? Check the conditions first.

Solution

Conditions. Random sample; 150 \le 5{,}200 (Example 1); and the expected counts are

150(0.35) = 52.5, \qquad 150(0.65) = 97.5

both at least 10, so the normal model N(0.35, 0.0389) applies. Figure 1 shows it following the bars.

The chance.

z = \frac{0.40 - 0.35}{0.03894} \approx 1.28

so P(\hat{p} > 0.40) \approx 1 - 0.8997 = 0.1003. About one sample in ten finds a share this high when the true share is 35\%. Kadri’s result is not strong evidence that composting has grown.

The exact binomial chance, from technology, is P(X \ge 61) \approx 0.0864. The normal area is a little larger because the cut at 0.40 runs through the middle of the bar for 60 households and keeps its right half, the effect met in Shape of a binomial and the normal approximation. With 150 households, a difference of one household is one bar, so the two answers agree to about a percentage point.

Example 5

Day 2: the middle 95% of shares

For samples of 150 households, find P(0.30 < \hat{p} < 0.40), and find the interval around 0.35 that holds the middle 95\% of sample shares.

Solution

The cuts 0.30 and 0.40 sit 0.05 either side of 0.35, so z \approx \pm 1.28 and

P(0.30 < \hat{p} < 0.40) \approx 0.8997 - 0.1003 = 0.7994

For the middle 95\%, leave 2.5\% in each tail, so z = \pm 1.960:

\begin{aligned} 0.35 \pm 1.960(0.03894) &= 0.35 \pm 0.0763 \\ &\approx 0.2737 \text{ to } 0.4263 \end{aligned}

Ninety-five samples in a hundred find between about 27\% and 43\% composting, between about 41 and 64 households out of 150. A survey of 150 pins a share down only to within about 8 percentage points, which is why Tiit wants 600.

Example 6

Day 2: when the large-counts condition fails

Only 4\% of households keep a worm bin. Kadri’s sample of 150 finds 12 with one, \hat{p} = 0.08. Is a normal model fair for P(\hat{p} \ge 0.08)? Compare it with the exact chance P(X \ge 12) \approx 0.0179 from technology.

Solution

The expected count of successes is 150(0.04) = 6, under 10, so the condition fails. Figure 2 shows why it matters: the bars are pressed against the wall at 0 and trail off to the right.

Figure 2Samples of 150 households when 4\% keep a worm bin (made-up share). Only 6 successes are expected, the bars lean right against the wall at 0, and the normal curve with the same center and spread fits badly at both ends.

Using the normal model anyway, \displaystyle \sigma_{\hat{p}} = \sqrt{0.04(0.96)/150} = \sqrt{0.000256} = 0.016 and

z = \frac{0.08 - 0.04}{0.016} = 2.5

so the normal chance is 1 - 0.9938 = 0.0062. The exact chance is 0.0179, about three times as large: the symmetric curve is too thin in the long right tail. At the other end, the curve puts the same 0.0062 below \hat{p} = 0, a share no sample can have. The mean and spread are still right; the shape is wrong. For 6 expected successes to grow to 10, Kadri would need 0.04n \ge 10, that is n \ge 250 households.

Example 7

Day 2: the same share, a bigger sample

Maarja surveys a random sample of 600 households across the county and finds 240 that compost, again \hat{p} = 0.40. How unusual is that if p = 0.35?

Solution

Conditions. 600 \le 5{,}200; expected counts 600(0.35) = 210 and 600(0.65) = 390, both at least 10. From Example 2, \sigma_{\hat{p}} \approx 0.0195.

z = \frac{0.40 - 0.35}{0.01947} \approx 2.57

so P(\hat{p} \ge 0.40) \approx 1 - 0.9949 = 0.0051 (exact binomial, 0.0061). This time the normal area is the smaller one, the mirror of Example 4: the event \hat{p} \ge 0.40 counts the whole bar for 240 households, but the cut at 0.40 runs through its middle and the normal area keeps only its right half. The same five-point gap that happens one sample in ten at n = 150 happens about one sample in two hundred at n = 600. Maarja’s result is strong evidence that the share is now above 35\%; Kadri’s was not. The gap did not change; the yardstick \sigma_{\hat{p}} did.

Boundaries

Independent observations. \sigma_{\hat{p}} needs independent households. Asking 150 households on one street, where neighbors share one curbside program, links the answers, and \hat{p} can vary far more than the formula says.

The ten percent condition. Without replacement from a small population, \hat{p} varies less than \displaystyle \sqrt{p(1-p)/n} (Example 3). Only the spread is affected, never the center.

The large-counts condition. A share near 0 or 1 with too few expected successes or failures gives a skewed, lumpy \hat{p}, and normal chances can be off by a factor of several in the tail (Example 6). Ten is a guide, not a sharp line: the fit improves gradually as both counts grow.

A known p. Every calculation here starts from the population share. In practice p is what a survey is trying to find, and \sigma_{\hat{p}} is estimated by putting the sample’s \hat{p} in place of p; that estimate is the Standard error, and it changes how the conditions are checked.

Discreteness. \hat{p} takes only the values 0, 1/n, 2/n, \dots, 1. A normal model is a smooth stand-in, so a chance at a cut that lands on a possible value, such as \hat{p} > 0.40 against \hat{p} \ge 0.40, can differ from the exact answer by about one bar (Example 4).

Where this goes next

The conditions of this lesson, random sample, ten percent and large counts, return as the Conditions for one proportion, with the sample’s own counts standing in for np and n(1-p). With p unknown and \hat{p} in its place, the normal model of Day 2 becomes the confidence intervals and tests of Inference for proportions. Two shares compared, say this year’s survey against last year’s, need the gap between two independent statistics: Sampling distribution of a difference builds that gap for means, and Interval for a difference of proportions uses it for shares. Since \hat{p} is the sample mean of 0 and 1 scores, its normal shape on Day 2 is one case of the The central limit theorem.

Exercises

Unless told otherwise, p = 0.35, samples are random from the county (52{,}000 households), and the \Phi table above applies. Two other made-up shares appear: 60\% of households in a neighboring county recycle glass, and 5\% of households here keep chickens.

Day 1: center and spread

Find \mu_{\hat{p}} and \sigma_{\hat{p}} for samples of 100 households.
For samples of 350 households, find the mean and standard deviation of the count X of composting households, and then \sigma_{\hat{p}} from them.
How many households must a sample have for \sigma_{\hat{p}} to be exactly 0.025?
Find \mu_{\hat{p}} and \sigma_{\hat{p}} for samples of 240 households from the glass county.
For samples of 150, find \sigma_{\hat{p}} when p = 0.5 and when p = 0.9, and compare both with 0.0389 for p = 0.35.
A town has 2{,}600 households. What is the largest sample, drawn without replacement, that meets the ten percent condition? Does a sample of 150? Of 300?
Tiit doubles a sample from 150 to 300 households. Find the new \sigma_{\hat{p}} and the factor by which it shrank.

Day 2: the large-counts condition

Check the large-counts condition: (a) n = 25, p = 0.35; (b) n = 30, p = 0.35; (c) n = 200, p = 0.03; (d) n = 200, p = 0.95.
Find the smallest n that meets the condition when p = 0.35, and when p = 0.06.
Maarja wants the chance that a sample of 120 households finds at least 10\% keeping chickens. Is a normal model justified? What is the smallest sample for which it would be?

Day 2: normal calculations

For samples of 100, find P(\hat{p} < 0.30).
For samples of 250, find \sigma_{\hat{p}} and P(\hat{p} > 0.40).
For samples of 350, find P(0.32 < \hat{p} < 0.38).
For samples of 100, find the interval around 0.35 that holds the middle 95\% of sample shares.
For samples of 250, find the share that \hat{p} exceeds only 5\% of the time.
For samples of 240 from the glass county, find P(\hat{p} \le 0.55).
Kadri surveys 500 households and finds 190 that compost. Find \hat{p} and P(\hat{p} \ge \text{that value}) if p = 0.35. Is this strong evidence that the share has grown?
Find the smallest sample size for which P(\hat{p} > 0.40) \le 0.025 when p = 0.35.

Reasoning

Interpret \sigma_{\hat{p}} \approx 0.0389 for samples of 150 in one sentence about households and sample shares.
Show that p(1-p) = 0.25 - (p - 0.5)^2, and explain what it says about which p gives the most spread in \hat{p}.
Explain why the ten percent condition is needed for \sigma_{\hat{p}} but not for \mu_{\hat{p}}.
Tiit says: “150 households tell us less about a county of 52{,}000 than about a town of 2{,}000.” Respond, assuming both have p = 0.35.
Explain why the large-counts condition asks for ten expected failures as well as ten expected successes.
Explain why \sigma_{\hat{p}} falls as n grows while the standard deviation of the count X rises.

Answers

\mu_{\hat{p}} = 0.35; \displaystyle \sigma_{\hat{p}} = \sqrt{0.2275/100} \approx 0.0477.
Mean 350(0.35) = 122.5 households; standard deviation \displaystyle \sqrt{350(0.2275)} = \sqrt{79.625} \approx 8.923; \sigma_{\hat{p}} = 8.923/350 \approx 0.0255.
364 households: n = 0.2275/0.025^2 = 364.
\mu_{\hat{p}} = 0.60; \displaystyle \sigma_{\hat{p}} = \sqrt{0.60(0.40)/240} = \sqrt{0.001} \approx 0.0316.
p = 0.5: \displaystyle \sqrt{0.25/150} \approx 0.0408, a little more than 0.0389. p = 0.9: \displaystyle \sqrt{0.09/150} \approx 0.0245, much less.
260 households, since 0.10(2{,}600) = 260. A sample of 150 meets it; 300 does not.
\displaystyle \sqrt{0.2275/300} \approx 0.0275; it shrank by the factor \displaystyle 1/\sqrt{2} \approx 0.707.
(a) 8.75 and 16.25: fails. (b) 10.5 and 19.5: holds. (c) 6 and 194: fails. (d) 190 and 10: holds, just.
p = 0.35: n \ge 10/0.35 \approx 28.6, so 29. p = 0.06: n \ge 10/0.06 \approx 166.7, so 167 (at 166, np = 9.96).
No: 120(0.05) = 6 expected successes, under 10, so \hat{p} is skewed right. The smallest sample is 10/0.05 = 200.
0.1469: z = -0.05/0.0477 \approx -1.05, and 1 - 0.8531 = 0.1469.
\displaystyle \sigma_{\hat{p}} = \sqrt{0.2275/250} \approx 0.0302; z = 0.05/0.0302 \approx 1.66, so 1 - 0.9515 = 0.0485.
0.7620: z = \pm 0.03/0.0255 \approx \pm 1.18, and 0.8810 - 0.1190 = 0.7620.
From 0.2565 to 0.4435: 0.35 \pm 1.960(0.0477) = 0.35 \pm 0.0935.
About 0.3996: 0.35 + 1.645(0.03017) \approx 0.35 + 0.0496.
0.0571: z = -0.05/0.0316 \approx -1.58, and 1 - 0.9429 = 0.0571.
\hat{p} = 190/500 = 0.38; \displaystyle \sigma_{\hat{p}} = \sqrt{0.2275/500} \approx 0.0213, z = 0.03/0.0213 \approx 1.41, so 1 - 0.9207 = 0.0793. Not strong evidence: about 8 samples in 100 do this well by chance.
350 households: \displaystyle \sqrt{n} \ge 1.960\sqrt{0.2275}/0.05 \approx 18.70, so n \ge 349.6. At n = 349, z \approx 1.958, just short.
Over all random samples of 150 households, the share that composts typically misses the county’s 35\% by about 3.9 percentage points.
0.25 - (p^2 - p + 0.25) = p - p^2 = p(1-p). A square is never negative, so p(1-p) is largest, 0.25, at p = 0.5, and falls toward 0 as p nears 0 or 1. An even split gives the most spread in \hat{p}.
The mean of a sum is the sum of the means whether or not the draws are linked, so \mu_{\hat{p}} = p always. The binomial standard deviation assumes independent trials, and draws without replacement are close to independent only when the sample is a small part of the population.
Both samples are under 10\% of their population, so both have \displaystyle \sigma_{\hat{p}} = \sqrt{0.2275/150} \approx 0.0389. The spread depends on n and p, not on the population’s size, so the two samples are about equally precise. If anything the town’s sample is slightly more precise, since 150 is 7.5\% of the town and drawing without replacement trims the spread a little (Example 3).
Successes and failures are two sides of one count: the number of failures is n - X. A share near 1, such as 0.95, piles the bars against the wall at \hat{p} = 1, and it is the expected failures, n(1-p), that measure how far the center is from that wall.
The count’s standard deviation is \displaystyle \sqrt{np(1-p)}, growing like \displaystyle \sqrt{n}. The share divides it by n, which grows faster, leaving \displaystyle \sqrt{p(1-p)}/\sqrt{n}, which shrinks like \displaystyle 1/\sqrt{n}.