Sampling Distribution of a Proportion

Part 1 · Hook

Why this matters

A seed company prints "85% sprout" on every packet. A garden store plants 200 seeds and only 160 come up: 80%. Is the company exaggerating, or is that just the luck of the draw? To answer, you need to know how much a sample proportion of 200 seeds bounces around when the truth really is 85%.

Part 2 · Before you start

What this builds on

Part 3 · Prerequisite check

Quick check before you start

1. For a normal distribution with mean 0.85 and standard deviation 0.025, what is the z-score of 0.80?

$-2$
$2$
$-0.05$
$-0.5$

Show the answer

$z$ $= \frac{0.80 - 0.85}{0.025}$ $= \frac{-0.05}{0.025}$ $= -2$ .

Correct: $-2$ :
$2$ :
$-0.05$ :
$-0.5$ :

2. As the size of a random sample grows, the sampling distribution of $\hat{p}$

stays centered at $p$ and becomes less spread out
moves closer to 0.5 and becomes more spread out
stays centered at $p$ and keeps the same spread
becomes centered at the sample's own proportion

Show the answer

$\hat{p}$ from a random sample is unbiased at every sample size, and larger samples give less variability.

Correct: stays centered at $p$ and becomes less spread out:
moves closer to 0.5 and becomes more spread out:
stays centered at $p$ and keeps the same spread:
becomes centered at the sample's own proportion:

Part 4 · See it

See it first

Two probability histograms of the sample proportion of seeds that sprout when the true proportion is p = 0.85. Top, samples of 200: the bars form a symmetric mound centered at 0.85 that matches the normal curve with mean 0.85 and standard deviation 0.025 drawn over it; large counts hold because 200(0.85) = 170 and 200(0.15) = 30 are both at least 10. Bottom, samples of 20: the bars are skewed left, piled up near 0.85 to 1 with a long tail toward lower values, and the normal curve does not fit; 20(0.15) = 3 is less than 10. — Exact sampling distributions of the proportion of seeds that sprout when p = 0.85. With 200 seeds the normal curve fits; with 20 seeds it does not. LevlPrep original diagram.

Part 5 · Step by step

How it works, step by step

Each sampled seed sprouts (1) or not (0), with a 0-or-1 spread of $\sqrt{p(1 - p)}$ averaging $n$ of them gives $\sigma_{\hat{p}}$ $= \sqrt{\frac{p(1 - p)}{n}}$ , centered at $\mu_{\hat{p}} = p$
The sample is drawn without replacementthe formula is accurate only when $n$ is at most 10% of the population: the 10% condition
At least 10 successes and 10 failures are expected ( $np \ge 10$ , $n(1 - p) \ge 10$ )the sampling distribution is approximately normal: the large counts condition
Both conditions holdstandardize with $z$ $= \frac{\hat{p} - p}{\sigma_{\hat{p}}}$ and find probabilities about $\hat{p}$ from the normal distribution

Part 6 · Key ideas

Key ideas

$\mu_{\hat{p}} = p$ and $\sigma_{\hat{p}}$ $= \sqrt{\frac{p(1 - p)}{n}}$ , using the true $p$ .
10% condition: $n$ at most 10% of the population, so the standard deviation formula holds when sampling without replacement.
Large counts condition: $np \ge 10$ and $n(1 - p) \ge 10$ , so the shape is approximately normal. Never " $n \ge 30$ " for proportions.
When large counts fails, the distribution is skewed: right when $p$ is near 0, left when $p$ is near 1.
Four times the sample size halves $\sigma_{\hat{p}}$ .

Part 7 · Misconception

A common mistake

The wrong idea: The 10% condition is what makes the distribution of $\hat{p}$ normal, so a sample under 10% of the population can always use a normal model.

What actually happens: The 10% condition protects the standard deviation formula (independence). The normal shape comes from the large counts condition, $np \ge 10$ and $n(1 - p) \ge 10$ . Check both, for different reasons.

Part 8 · Check yourself

Check yourself

Exam-style questions. Anything you miss goes into your review queue.

Table

Testing a seed company's sprouting rate

A seed company says that 85% of its tomato seeds sprout. A garden store plants a random sample of 200 seeds from a shipment of 50,000 seeds and records the proportion that sprout. Suppose the company's figure is correct, so $p = 0.85$ .

The setting for the sampling distribution of the sample proportion
Quantity	Value
Population	50,000 seeds in the shipment
True proportion that sprout, $p$	0.85
Sample size, $n$	200 seeds, chosen at random

1. What are the mean and standard deviation of the sampling distribution of $\hat{p}$ , the proportion of the 200 seeds that sprout?

Mean 0.85, standard deviation 0.0252
Mean 0.85, standard deviation 0.000638
Mean 0.85, standard deviation 0.357
Mean 170, standard deviation 5.05

Show the answer

$\mu_{\hat{p}} = p = 0.85$ and $\sigma_{\hat{p}}$ $= \sqrt{\frac{p(1 - p)}{n}}$ $= \sqrt{\frac{0.85(0.15)}{200}}$ $= 0.0252$ .

Correct: Mean 0.85, standard deviation 0.0252: Right: the mean is $p$ and the standard deviation is $\sqrt{p(1 - p)/n}$ .
Mean 0.85, standard deviation 0.000638: That is $\frac{p(1 - p)}{n}$ without the square root: the variance, not the standard deviation.
Mean 0.85, standard deviation 0.357: That is $\sqrt{p(1 - p)}$ , the spread for a single seed. Divide by $n$ inside the square root.
Mean 170, standard deviation 5.05: Those describe the count of seeds that sprout, $np$ and $\sqrt{np(1 - p)}$ , not the proportion.

2. Calculate the standard deviation of the sampling distribution of $\hat{p}$ . Give four decimal places.

Type a number. A table answer and a calculator answer are both accepted.

Show the answer

$\sigma_{\hat{p}}$ $= \sqrt{\frac{p(1 - p)}{n}}$ $= \sqrt{\frac{0.85(0.15)}{200}}$ $= \sqrt{0.0006375}$ $= 0.0252$ . The 10% condition holds (200 is less than 10% of 50,000), so this formula applies.

Answer: 0.0252

3. Why is it reasonable to use a normal model for the sampling distribution of $\hat{p}$ here?

Because $n = 200$ is at least 30
Because 200 is less than 10% of the 50,000 seeds in the shipment
Both $200(0.85) = 170$ and $200(0.15) = 30$ are at least 10
Because the seeds were chosen at random

Show the answer

The large counts condition: $np = 200(0.85) = 170$ and $n(1 - p)$ $= 200(0.15)$ $= 30$ are both at least 10, so the distribution of $\hat{p}$ is approximately normal.

Because $n = 200$ is at least 30: The rule of 30 is for sample means. For proportions, check the expected numbers of successes and failures.
Because 200 is less than 10% of the 50,000 seeds in the shipment: That is the 10% condition. It justifies the standard deviation formula, not the normal shape.
Correct: Both $200(0.85) = 170$ and $200(0.15) = 30$ are at least 10: Right: the large counts condition, with both counts at least 10.
Because the seeds were chosen at random: Random selection is needed, but it does not make the shape normal. The shape depends on the expected numbers of successes and failures.

4. Assuming the company's figure is correct, find the probability that 80% or fewer of the 200 seeds sprout.

Type a number. A table answer and a calculator answer are both accepted.

Show the answer

$\hat{p}$ is approximately normal with mean 0.85 and standard deviation 0.0252. $z$ $= \frac{0.80 - 0.85}{0.0252}$ $= -1.98$ , so $P(\hat{p} \le 0.80)$ $= P(Z \le -1.98)$ $= 0.024$ . Technology (normal, mean 0.85, SD 0.0252, below 0.80) gives 0.0239.

Answer: 0.0238

5. A pollster keeps everything else the same but takes a random sample 4 times as large. By what factor does the standard deviation of $\hat{p}$ change?

It is divided by 4
It is divided by 2
It is multiplied by 2
It does not change, because $p$ is the same

Show the answer

$\sigma_{\hat{p}}$ $= \sqrt{\frac{p(1 - p)}{n}}$ has $n$ under a square root. Replacing $n$ with $4n$ divides it by $\sqrt{4} = 2$ .

It is divided by 4: The sample size is inside a square root, so the effect is $\sqrt{4}$ , not 4.
Correct: It is divided by 2: Right: $\sqrt{4n}$ is twice $\sqrt{n}$ .
It is multiplied by 2: A larger sample gives less spread, not more.
It does not change, because $p$ is the same: The formula depends on $n$ as well as $p$ .

6. About 3% of a factory's light bulbs are defective. An inspector takes a random sample of 250 bulbs from a large batch. What can you say about the shape of the sampling distribution of the proportion defective?

Approximately normal, because 250 is a large sample
Skewed left, because most bulbs are not defective
Approximately normal, because $n(1 - p) = 242.5$
Skewed right, because $250(0.03) = 7.5$ is less than 10

Show the answer

$np = 250(0.03) = 7.5$ , which is less than 10. Large counts fails, so the distribution is skewed right (it piles up near 0 and has a longer tail to the right).

Approximately normal, because 250 is a large sample: Size alone is not enough: the expected number of defective bulbs is only 7.5.
Skewed left, because most bulbs are not defective: With $p$ near 0, the proportion piles up near 0 and stretches to the right: skewed right.
Approximately normal, because $n(1 - p) = 242.5$ : Both counts must be at least 10, and $np = 7.5$ is not.
Correct: Skewed right, because $250(0.03) = 7.5$ is less than 10: Right: large counts fails for the defective bulbs.

7. The garden store plants a random sample of 200 seeds and only 78% sprout. Assuming 85% of all the company's seeds sprout, $P(\hat{p} \le 0.78) = 0.0028$ . Build a sentence that uses this probability to judge the company's claim.

Write the conclusion in context.

Show the answer

A probability is useful only when you tie it back to the claim. If the claim were true, 78% or fewer sprouting would happen in about 3 of every 1,000 samples, so it would be very unusual. That casts doubt on the claim.

A full-credit answer: Assuming 85% of all the company's seeds sprout, a sample proportion of 0.78 or less in a random sample of 200 seeds would be very unusual (probability 0.0028), so this result gives reason to doubt the company's claim.

Part 9 · Summary

Summary

For random samples,

\hat{p}

is centered at

p

with a spread that shrinks as

n

grows. It has mean

p

and standard deviation

\sqrt{p(1 - p)/n}

(accurate when the sample is at most 10% of the population) and is approximately normal when

np \ge 10

and

n(1 - p) \ge 10

. Then

z

= \frac{\hat{p} - p}{\sigma_{\hat{p}}}

turns a question about a sample proportion into a normal probability, which you can tie back to a claim about

p

.

Part 10 · Up next

What comes next

Part 11 · Connections