Sampling distributions of proportions

Polls, quality-control checks and medical studies all report a sample proportion. To judge whether a sample proportion is surprising, you need to know how much $\hat{p}$ naturally varies from sample to sample. This lesson gives you the center, spread and shape of the sampling distribution of $\hat{p}$, and shows how to use it to find probabilities.

Where the formulas come from

Suppose a population has a proportion $p$ of "successes," and you take a random sample of size $n$. Let $X$ be the number of successes in the sample. If the sample is taken with replacement (or the population is so large that it hardly matters), $X$ is a binomial random variable, which you studied in Unit 4:

$$\mu_X = np, \qquad \sigma_X = \sqrt{np(1-p)}.$$

The sample proportion is $\hat{p} = \dfrac{X}{n}$, which is just $X$ multiplied by the constant $\dfrac{1}{n}$. Multiplying a random variable by a constant multiplies its mean and its standard deviation by that constant, so

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

The two conditions

Each formula comes with a condition, and AP graders expect you to check them by name, with numbers.

The 10% condition justifies the standard deviation formula. When you sample without replacement, the draws are not quite independent: removing one success slightly changes the chance that the next person is a success. If the sample is no more than 10% of the population, this effect is tiny and the formula $\sqrt{p(1-p)/n}$ is accurate. For example, a sample of 150 students from a school of 2,400 is fine because $150 \le 0.10(2400) = 240$.

The Large Counts condition justifies using a normal model. A binomial distribution is skewed when the expected number of successes or failures is small. When both $np$ and $n(1-p)$ are at least 10, the distribution of $\hat{p}$ is close enough to normal to use normal calculations.

Finding probabilities about $\hat{p}$

Once the conditions are met, you find a probability the same way you did for any normal distribution: standardize, then use a table or technology.

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

Here is an excerpt of the standard normal table (area to the left of $z$) with the values used in this lesson.

$z$	$-2.00$	$-1.50$	$-1.25$	$-1.00$	$1.00$	$1.25$	$1.50$	$2.00$
Area to the left	$0.0228$	$0.0668$	$0.1056$	$0.1587$	$0.8413$	$0.8944$	$0.9332$	$0.9772$

Sample size and spread

Because $n$ sits under a square root in the denominator, the spread of $\hat{p}$ shrinks slowly as the sample grows. Quadrupling $n$ cuts $\sigma_{\hat{p}}$ in half. For a fixed $n$, the spread is largest when $p = 0.5$, because $p(1-p)$ is maximized there. That is why pollsters who don't know $p$ plan their sample sizes using $p = 0.5$: it gives the most conservative (largest) standard deviation.

Practice

Use the table excerpt above where needed. Give probabilities to four decimal places.

Practice 1

In a large city, 25% of households have a dog. A random sample of 300 households is selected. Find the standard deviation of the sampling distribution of $\hat{p}$, the proportion of sampled households with a dog.

Your answer

Enter a number. Fractions like 3/4 and sqrt(2) are OK.

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Practice 2

A factory knows that 4% of the light bulbs it produces are defective. An inspector selects a random sample of 200 bulbs from a day's production of 50,000. Can the sampling distribution of $\hat{p}$ be modeled as approximately normal?

AYes, because $200 \le 0.10(50{,}000)$.BNo, because the expected number of defective bulbs, $200(0.04) = 8$, is less than 10.CYes, because the sample size of 200 is greater than 30.DNo, because the population of bulbs is not normally distributed.

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Practice 3

A small college has 1,200 students, and 45% of them live on campus. A researcher plans to select a random sample of 300 students without replacement. Which statement about the formula $\sigma_{\hat{p}} = \sqrt{p(1-p)/n}$ is correct?

AIt is appropriate because $np = 135$ and $n(1-p) = 165$ are both at least 10.BIt is appropriate because the sample was selected at random.CIt is not appropriate because $p$ is less than 0.5.DIt is not appropriate because the sample is more than 10% of the population.

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Practice 4

In a large state, 20% of adults have a library card. A random sample of 400 adults is selected. Find the probability that more than 24% of the sample have a library card.

Your answer

Enter a number. Fractions like 3/4 and sqrt(2) are OK.

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Practice 5

Return to the high school where 60% of the 3,000 students get less than 8 hours of sleep, with random samples of 150 students. Find the probability that the sample proportion is between 0.55 and 0.65.

Your answer

Enter a number. Fractions like 3/4 and sqrt(2) are OK.

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Practice 6

A pollster wants the standard deviation of $\hat{p}$ to be at most 0.02. She does not know $p$, so she uses the conservative value $p = 0.5$. What is the smallest sample size that achieves her goal?

Your answer

Enter a number. Fractions like 3/4 and sqrt(2) are OK.

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Practice 7

A company claims that 80% of its customers are satisfied. For random samples of 100 customers, $\hat{p}$ is approximately normal with mean 0.80 and standard deviation 0.04. A consultant surveys 100 randomly selected customers and finds that 76% are satisfied. Which conclusion is most reasonable?

AThe result is very unusual, because 0.76 is less than 0.80.BThe result proves the claim is false, because the probability of getting exactly 0.76 is very small.CThe result is not unusual, because a sample proportion of 0.76 or lower happens about 16% of the time when $p = 0.80$.DThe result is not unusual, because a sample proportion of 0.76 or lower happens about 84% of the time when $p = 0.80$.

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