Conditions for one proportion
A sample gave a share. Before that share is stretched into a claim about the whole population, what must be true of how the sample was taken?
Where you are 5 in Inference for proportions
Why this exists
A school district has 4000 high school students (made-up data throughout this lesson). Adeola, who works in the district office, wants p, the share of them with a part-time job. She takes a Simple random samples and the sampling frame of 200 students and finds 54 with a job, so \displaystyle \hat{p} = \frac{54}{200} = 0.27.
The last few lessons turn one number like this into a range of plausible values for p: the Standard error, the Critical value z^*, the Point estimates and the margin of error and the Confidence level and interpreting an interval it promises. All of them rest on the Sampling distribution of a sample proportion: \hat{p} is centered at p, has standard deviation \displaystyle \sqrt{\frac{p(1-p)}{n}}, and is close to normal in shape. Those three facts are not automatic. Each holds only for data gathered in a certain way, and the arithmetic of an interval runs just as smoothly when they fail. The conditions for one proportion say whether the arithmetic means anything.
Definition
Inference about a population proportion p from a sample proportion \hat{p}, by the normal methods of this course, needs three conditions.
- Random. The data come from a random sample of the population, or from a randomized experiment, one that uses Random assignment and confounding of subjects to treatments.
- Ten percent. When the sample is drawn without replacement, it is at most a tenth of the population: n \le 0.10\,N.
- Large counts. The sample holds at least 10 successes and at least 10 failures. For a confidence interval this is checked with the sample’s own counts:
Since \hat{p} is the number of successes divided by n, the number n\hat{p} is simply the count of successes, and n(1-\hat{p}) the count of failures.
A significance test, the subject of One sample test for a proportion in the next unit, checks large counts with the claimed value p_0 of its null hypothesis instead: np_0 \ge 10 and n(1-p_0) \ge 10.
Intuition
Each condition guards one of the three facts about \hat{p}.
Random guards the center. A random method favors no one, so \hat{p} is an Bias and variability of an estimator of p: its sampling distribution is centered on p. A Convenience and voluntary response samples or a group of volunteers can sit off center by an amount nobody knows. A margin of error measures only chance variation, never bias.
Ten percent guards the spread. The formula \displaystyle \sqrt{\frac{p(1-p)}{n}} assumes independent draws. Draws without replacement are not quite independent; the The ten percent condition lesson showed that the true standard deviation is the formula times \displaystyle \sqrt{\frac{N-n}{N-1}}, a factor of at least about 0.95 when n \le 0.10\,N.
Large counts guards the shape. The critical value z^* comes from a normal curve. When a sample expects only a few successes, \hat{p} piles up against 0, the smallest value it can take, with a longer tail to the right, and the normal curve fits it poorly.
In Figure 1, left, samples of 30 from a population with share 0.1 expect 30(0.1) = 3 successes. The bars lean right, and the normal curve spills below \hat{p} = 0, onto values no sample can produce. On the right, samples of 200 with share 0.27 expect 54 successes, and the bars follow the curve closely.
Results
The three conditions and what each protects.
| Condition | Check | Protects |
|---|---|---|
| Random | random sample, or random assignment | \hat{p} centered on p |
| Ten percent | n \le 0.10\,N, when sampling without replacement | the spread \displaystyle \sqrt{\frac{p(1-p)}{n}} |
| Large counts | at least 10 successes and 10 failures | a nearly normal shape |
Interval or test. A confidence interval checks large counts with n\hat{p} and n(1-\hat{p}). A significance test checks them with np_0 and n(1-p_0).
When one fails. No calculation repairs a failed random condition: the data describe the people measured and nobody else. A failed ten percent condition makes the usual standard error too large, so the interval is wider than it needs to be; it can be reported as such, or corrected with the factor \displaystyle \sqrt{\frac{N-n}{N-1}}. A failed large counts condition calls for a larger sample, or for a method that does not lean on the normal curve.
Why it is true
Why at least 10. Under the normal model \hat{p} has mean p and standard deviation \displaystyle \sqrt{\frac{p(1-p)}{n}}. The smallest value \hat{p} can take is 0, and it lies this many standard deviations below the mean:
The inequality holds because 1 - p \le 1. If np \ge 10, zero sits at least \displaystyle \sqrt{10} \approx 3.16 standard deviations below the center, and a normal curve has less than 0.001 of its area that far out on one side. The wall at 0 then barely touches the distribution, and it has room to be nearly symmetric. The same argument at the wall \hat{p} = 1 uses n(1-p) \ge 10. The cutoff 10 is a convention, like the ten percent line: nothing breaks suddenly at 9.
Why \hat{p} for an interval and p_0 for a test. The condition is really about np and n(1-p), the expected counts under the true share. An interval is built because p is unknown, so the sample’s \hat{p} stands in, and n\hat{p} is just the observed count. A test asks how \hat{p} would behave if p were p_0; every probability in it is computed in that world, so the distribution that must be nearly normal is the one centered on p_0, and its expected counts are np_0 and n(1-p_0).
Worked examples
Adeola’s sample
Adeola’s simple random sample of 200 of the district’s 4000 students finds 54 with a part-time job. Check the conditions for a confidence interval for p.
SolutionRandom. A simple random sample is a random method, so \hat{p} has no bias from the way students were chosen.
Ten percent. She samples without replacement (no student is asked twice), so the condition applies: \displaystyle \frac{200}{4000} = 0.05 \le 0.10. Met.
Large counts. The sample has 54 successes and 200 - 54 = 146 failures, both at least 10. Met.
All three hold, so the interval of One sample interval for a proportion and sample size may be built from \hat{p} = 0.27, and its stated confidence level can be trusted.
Volunteers at one club
Babatunde wants the same share. He hands out a survey at a meeting of the career club, and the 40 students who fill it in include 22 with a part-time job. Check the conditions.
SolutionTen percent: 40 \le 400, met. Large counts: 22 successes and 18 failures, met.
Random: fails. These are members of one club who chose to answer, and a career club may well draw students who already work. The share \displaystyle \frac{22}{40} = 0.55 could be far from the district’s p, and nothing in the data says how far. Passing the other two conditions does not help: they describe the spread and shape of chance variation, and the problem here is not chance.
What Babatunde can say is a description: 22 of the 40 students who answered have a part-time job. To estimate p he needs a random sample of the district.
One school of 1,500
Njeri takes a simple random sample of 600 of the 1500 students at one high school in the district and finds 192 with a part-time job. Check the conditions, and say what follows.
SolutionRandom: met, but for the population that was sampled, which is this one school. The result says nothing directly about the district’s 4000 students.
Large counts: 192 and 408, met.
Ten percent: \displaystyle \frac{600}{1500} = 0.40 > 0.10. Fails.
Here \displaystyle \hat{p} = \frac{192}{600} = 0.32. The usual standard error is \displaystyle \sqrt{\frac{0.32 \times 0.68}{600}} \approx 0.0190. For a sample this large a share of its population, the correction factor is \displaystyle \sqrt{\frac{900}{1499}} \approx 0.7749, so the spread is really the usual value times this factor, \approx 0.0148 (computed before rounding). The usual value is about \displaystyle \frac{1}{0.7749} \approx 1.29 times too large, and an interval built on it would be about 29\% wider than needed.
So the failure errs on the safe side: such intervals capture p more often than the stated confidence level, but they undersell the precision. Njeri can report the usual interval with that note, or use the corrected standard error.
Only seven successes
Before her main survey, Adeola ran a pilot: a random sample of 80 students, asked whether they work more than 20 hours a week. 7 said yes. Check the conditions, and say what follows.
SolutionRandom: met. Ten percent: 80 \le 400, met. Large counts: 7 successes and 73 failures; 7 < 10, so it fails.
Here \displaystyle \hat{p} = \frac{7}{80} = 0.0875 and the standard error is \displaystyle \sqrt{\frac{0.0875 \times 0.9125}{80}} \approx 0.0316. Zero, the smallest possible \hat{p}, is only \displaystyle \frac{0.0875}{0.0316} \approx 2.77 standard errors below the estimate. Figure 1, left, shows the same effect, more strongly, for 3 expected successes: bunched against 0, longer on the right. A normal interval treats the two sides as mirror images, so its real capture rate need not match the stated confidence level.
The plain remedy is a larger sample. If the share is near 0.0875, ten expected successes need \displaystyle n \ge \frac{10}{0.0875} \approx 114.3, so at least 115 students; this is a guide only, since a new sample’s \hat{p} will differ. Methods built from exact binomial probabilities also exist, beyond this course.
The same data, interval and test
Njeri takes a random sample of 150 district students and finds 12 who work more than 20 hours a week. A report from last year claimed 5\%. Check large counts (a) for a confidence interval and (b) for a test of the claim p_0 = 0.05.
Solution(a) The interval uses the sample’s counts: 12 successes and 138 failures, both at least 10. Met.
(b) The test asks what samples would look like if p were 0.05, so it uses the expected counts under that claim: 150(0.05) = 7.5 and 150(0.95) = 142.5. Since 7.5 < 10, it fails.
The same data pass for one procedure and fail for the other. Here \hat{p} = 0.08 sits above the claim, and samples drawn from a population with share 0.05 would expect too few successes for a normal model.
An experiment with random assignment
Babatunde recruits 100 student volunteers and randomly assigns 50 to a job-search workshop and 50 to no workshop. A month later, 21 of the workshop group and 12 of the others have found a part-time job. Which conditions hold?
SolutionRandom: met by random assignment. It makes the groups alike apart from chance, so a difference in job shares can be credited to the workshop. It does not make the volunteers a random sample of the district, so the conclusion reaches only students like these volunteers.
Ten percent: not needed. No sample is drawn without replacement from a population; the randomness is in the assignment.
Large counts, group by group: 21 and 29 in the workshop group, 12 and 38 in the other, all at least 10. Met. The comparison itself belongs to Test for a difference of proportions.
Boundaries
- The conditions concern how the data were gathered and the shape of chance variation. Meeting them does not make an estimate right: a 95\% interval built correctly still misses p for about 5\% of samples.
- A random sample licenses inference only to the population actually sampled. A random sample of one school speaks for that school.
- Ten percent applies only to sampling without replacement. Random assignment and random processes such as spinning a spinner do not need it.
- The cutoffs 10\% and 10 are conventions placed where the normal methods work well; they are not sharp walls.
- A sample in which every member is a success, or none is, gives a standard error of \displaystyle \sqrt{\frac{0 \times 1}{n}} = 0 and an interval of width zero. Large counts fails there, which is exactly the warning wanted.
Where this goes next
With the conditions checked, One sample interval for a proportion and sample size builds the interval \hat{p} \pm z^*\,\text{SE} and its interpretation. The One sample test for a proportion uses the same three conditions with p_0 in the large counts check. Comparing two shares, in Interval for a difference of proportions and its test, checks each condition for each group. Inference for a mean has its own list, Conditions for one mean, with the same random and ten percent conditions and a shape check suited to means.
Exercises
Checking the three conditions
- Adeola takes a new simple random sample of 300 of the district’s 4000 students and finds 84 with a part-time job. Check the three conditions for a confidence interval.
- Njeri takes a simple random sample of 500 of the 4000 students and finds 140 with a job. Check the conditions.
- What is the largest simple random sample from the 4000 students that meets the ten percent condition?
- A random sample of 60 students finds 9 who work at a restaurant. Is large counts met for the share who work at a restaurant?
- A random sample of 90 students finds 83 who own a phone and 7 who do not. Check large counts for the share who own a phone.
- Adeola plans a new survey and expects the share with a job to be near 0.27. About how small could her sample be and still be expected to meet large counts?
When a condition fails
- Babatunde asks the first 50 students he meets outside the cafeteria; 19 have a job. Which condition fails, and what can he report?
- The district posts an online poll; 312 students answer, and 171 say they have a job. Check the conditions.
- Njeri takes a simple random sample of 250 of the 900 students at a small high school. Check ten percent, and find by what factor the usual standard error overstates the true one.
- A random sample of 40 seniors finds 4 who hold two jobs. Check large counts, and estimate the sample size needed if the share is near 0.1.
- A state agency takes a simple random sample of 2000 students from all high school students in the state, several hundred thousand of them. Does the ten percent condition hold?
Intervals and tests
- Babatunde will test the claim p_0 = 0.30 with a random sample of 200 students. Check large counts for the test.
- A random sample of 120 students finds 15 who work weekends only. Check large counts for an interval, and for a test of p_0 = 0.06.
- What is the smallest sample size for which a test of p_0 = 0.02 meets large counts?
- With n = 200, which claimed values p_0 pass the large counts check for a test?
- Babatunde randomly assigns 60 volunteers, 30 to a resume class and 30 to none. A month later 14 in the class group and 8 in the other have a job. Which conditions hold?
Reasoning
- Babatunde says his club survey in Example 2 passed two of the three conditions, so an interval from it is “mostly fine.” Explain what is wrong.
- Explain in your own words why an interval checks large counts with \hat{p} but a test checks them with p_0.
- Show that, under the normal model, \hat{p} = 1 lies \displaystyle \sqrt{\frac{n(1-p)}{p}} standard deviations above p, and that this is at least \displaystyle \sqrt{10} when n(1-p) \ge 10.
- Njeri says the failed ten percent condition in Example 3 makes her result useless. Explain what the failure does to her interval.
- Explain why the ten percent condition is checked for a random sample but not for a randomized experiment.
Answers
- All three are met. Random: a simple random sample. Ten percent: \displaystyle \frac{300}{4000} = 0.075 \le 0.10. Large counts: 84 and 216, both at least 10.
- Random and large counts (140 and 360) are met; ten percent is not, since \displaystyle \frac{500}{4000} = 0.125 > 0.10. The correction factor is \displaystyle \sqrt{\frac{3500}{3999}} \approx 0.9355, so the usual standard error is about 1.07 times the true one: a mild failure.
- 400, since 0.10 \times 4000 = 400.
- No. The counts are 9 and 51, and 9 < 10.
- It fails on the failures side: 83 \ge 10 but 7 < 10. A share near 1 piles up against the wall at 1 just as a share near 0 piles up against 0.
- About 38 students. Successes: n(0.27) \ge 10 gives n \ge 37.04. Failures: n(0.73) \ge 10 gives n \ge 13.7. The stricter one rounds up to 38.
- Random fails: this is a Convenience and voluntary response samples. Ten percent (50 \le 400) and large counts (19 and 31) pass, but they cannot rescue it. He can report that 19 of the 50 students he asked have a job, with no estimate for the district.
- Random fails: the students chose themselves. Ten percent (312 \le 400) and large counts (171 and 141) pass, but the share \displaystyle \frac{171}{312} \approx 0.548 describes only those who chose to answer.
- It fails: \displaystyle \frac{250}{900} \approx 0.278 > 0.10. The factor is \displaystyle \sqrt{\frac{650}{899}} \approx 0.8503, so the usual standard error is about \displaystyle \frac{1}{0.8503} \approx 1.18 times the true one, and an interval would be about 18\% wider than needed. Any inference reaches only that school.
- It fails: 4 < 10. With a share near 0.1, n(0.1) \ge 10 needs n \ge 100.
- Yes. The condition needs N \ge 10 \times 2000 = 20{,}000, and several hundred thousand is far more.
- 200(0.30) = 60 and 200(0.70) = 140, both at least 10. Met.
- Interval: 15 and 105, met. Test: 120(0.06) = 7.2 < 10, so it fails, although 120(0.94) = 112.8 passes.
- n = 500. Successes: n(0.02) \ge 10 needs n \ge 500; then failures are 500(0.98) = 490.
- 0.05 \le p_0 \le 0.95. From 200p_0 \ge 10, p_0 \ge 0.05; from 200(1 - p_0) \ge 10, p_0 \le 0.95.
- Random: met by random assignment, so a difference can be credited to the class, for volunteers like these. Ten percent: not needed. Large counts: 14 and 16 in the class group pass, but 8 < 10 in the other group fails, so the normal methods are not justified for this comparison.
- The two conditions he passed are about the spread and shape of chance variation. The one he failed is about bias, which no margin of error measures. An interval from volunteers at one club has an unknown error that may be far larger than its stated margin.
- Large counts needs the expected counts under the true share. An interval does not know p, so it uses \hat{p}, and n\hat{p} is the observed count. A test computes everything as if p = p_0, so it needs the distribution centered on p_0 to be nearly normal, and its expected counts are np_0 and n(1-p_0).
- The distance is \displaystyle \frac{1 - p}{\sqrt{p(1-p)/n}} = \sqrt{\frac{n(1-p)^2}{p(1-p)}} = \sqrt{\frac{n(1-p)}{p}}. Since p \le 1, this is at least \displaystyle \sqrt{n(1-p)}, which is at least \displaystyle \sqrt{10} when n(1-p) \ge 10.
- Not useless. The usual standard error is too large by a factor of about 1.29, so her interval is wider than it needs to be and captures p more often than stated. The estimate \hat{p} is still centered on that school’s p. She can note the failure, or divide the width by about 1.29 using the correction factor.
- The condition protects the formula for the spread when people are drawn without replacement from a population. An experiment draws no such sample; its chance variation comes from the random assignment, so there is no population fraction to check.