What a confidence interval does and does not tell you
In Topic 3.3 you built intervals of the form p̂ ± z* √[p̂(1 − p̂)/n]. Now you need to say what an interval means, and use it to make a decision.
Start with one fact. The population proportion p is a fixed number. The interval changes from sample to sample, because p̂ does. So any one interval you calculate may or may not contain p. You never find out which, because you do not know p.
That is why statistics uses the word confidence, not probability, for a single interval. Confidence describes how reliable the method is, not one particular result.
Interpreting the interval
An interpretation of a C% confidence interval for p has three parts:
- the confidence level, C%;
- the interval, from the lower limit a to the upper limit b;
- the parameter in context: the proportion, the response and the population.
A reliable template:
We are C% confident that the interval from a to b captures the proportion of all [population] who [response].
For example: “We are 95% confident that the interval from 0.538 to 0.602 captures the proportion of all adult residents of Brennock who support the new park.”
Avoid “the interval captures p̂” or “the proportion of the 900 residents”. The sample proportion is the centre of the interval, so the interval always contains it. The interpretation is about the population.
Interpreting the confidence level
The confidence level answers the question: “How often does this method work?”
If we took many random samples of the same size from this population and built a C% interval from each one, about C% of those intervals would capture the true proportion.
Figure 1 shows this with a simulation: 20 random samples of 150 people from a population in which p = 0.40 is known, with a 95% interval built from each.
Two things to notice:
- 19 of the 20 intervals (95%) capture p here. In a different run you might see 18 or 20. The 95% is a long-run rate.
- Interval 16 looks just like the others. If it were your only sample, nothing would tell you it had missed. That is the “may or may not contain p” idea in action.
Using an interval to justify a claim
An interval gives a range of plausible values for p. Compare a claimed value with the interval:
| Where the claimed value lies | What you can conclude |
|---|---|
| Outside the interval | The claimed value is not plausible. The interval gives convincing evidence against the claim that p equals it. |
| Inside the interval | The claimed value is plausible. There is no convincing evidence against the claim. This does not prove the claim, because other values are plausible too. |
Many claims are about a direction, such as “a majority” (p > 0.5) or “less than a quarter” (p < 0.25). Then check where the whole interval lies:
- every plausible value satisfies the claim → convincing evidence for the claim;
- the interval straddles the boundary → not convincing evidence either way;
- every plausible value contradicts the claim → convincing evidence against it.
Always finish with a sentence in context that refers to the interval.
How confidence level and sample size change the interval
The margin of error is z* × √[p̂(1 − p̂)/n]. Change one piece and the others follow.
| Change (all else the same) | z* | Standard error | Margin of error and width |
|---|---|---|---|
| Increase the confidence level | increases | no change | increase |
| Decrease the confidence level | decreases | no change | decrease |
| Increase the sample size n | no change | decreases | decrease |
The width is roughly proportional to 1/√n. Multiplying n by 4 roughly halves the width; multiplying n by 9 cuts it to about a third. To get more confidence and a narrow interval, you need a bigger sample.
Worked example 1: interpreting an interval and judging three claims
Context. Brennock, a fictional town of about 12,000 adults, plans a new park. The council asked a random sample of 900 adult residents; 513 said they support the park. The conditions for a z-interval are met: the sample is random, 900 ≤ 10% of 12,000 = 1,200, and 513 successes and 387 failures are both at least 10.
(a) Construct and interpret a 95% interval.
- p̂ = 513 ÷ 900 = 0.57.
- SE = √(0.57 × 0.43 ÷ 900) = 0.0165.
- MOE = 1.960 × 0.0165 = 0.0323.
- Interval: 0.57 ± 0.0323 = (0.538, 0.602).
Interpretation. We are 95% confident that the interval from 0.538 to 0.602 captures the proportion of all adult residents of Brennock who support the new park.
(b) Interpret the confidence level. If the council took many random samples of 900 adults and built a 95% interval from each, about 95% of those intervals would capture the true proportion of Brennock adults who support the park.
(c) Judge three claims.
- “A majority of adults support the park.” Every value in (0.538, 0.602) is above 0.5. The interval gives convincing evidence that more than half of Brennock’s adults support the park.
- “Exactly 60% support the park.” 0.60 is inside the interval, so it is plausible. There is no convincing evidence against this claim. But values such as 0.55 are plausible too, so the claim is not proved.
- “At least two-thirds support the park.” Two-thirds is about 0.667, above the whole interval. The interval gives convincing evidence against this claim.
Check with Figure 2. At 90% the interval (0.543, 0.597) excludes 0.60, so claim 2 would be judged differently at a lower confidence level. Claims 1 and 3 reach the same conclusion at all three levels.
Worked example 2: predicting the effect of changes
Context. A fictional survey of 400 randomly chosen adults found p̂ = 0.30. The 95% interval is 0.30 ± 0.0449, that is (0.255, 0.345), with standard error 0.02291.
(a) Same data, 99% confidence. z* rises from 1.960 to 2.576. The standard error is unchanged. MOE = 2.576 × 0.02291 = 0.0590, so the interval (0.241, 0.359) is wider. The margin of error grows by a factor of 2.576 ÷ 1.960 ≈ 1.31.
(b) Same data, 90% confidence. z* = 1.645. MOE = 1.645 × 0.02291 = 0.0377, giving (0.262, 0.338). Lower confidence, narrower interval.
(c) 95% confidence, but n = 1,600 with the same p̂. n is 4 times larger, so the standard error is divided by √4 = 2: SE = 0.01146. MOE = 1.960 × 0.01146 = 0.0225, half of 0.0449. The interval is (0.278, 0.322).
(d) What sample size cuts the 95% margin of error to about a third? Divide the width by 3 by multiplying n by 3² = 9: n = 3,600. Check: MOE = 1.960 × √(0.3 × 0.7 ÷ 3,600) = 0.0150, which is one third of 0.0449.
Lesson. You cannot get higher confidence for free. With a fixed sample, more confidence means a wider, less precise interval. Only a larger sample gives both.
Worked example 3: judging four written interpretations
Context. A fictional sports league selected 240 of its registered players at random; 30% said they had had a minor injury this season. The 90% interval for the proportion of all registered players with a minor injury this season is (0.251, 0.349). Four students wrote these sentences. Which are correct?
- “We are 90% confident that the interval from 0.251 to 0.349 captures the proportion of all registered players in the league who had a minor injury this season.” Correct. It names the level, both endpoints and the population proportion in context.
- “There is a 90% chance that between 25.1% and 34.9% of players were injured.” Wrong. It treats the fixed population proportion as random. The 90% belongs to the method.
- “90% of the 240 players had injury rates between 0.251 and 0.349.” Wrong. It describes individuals, and the sample instead of the population.
- “In many random samples of 240 players, about 90% of the 90% intervals built would capture the true proportion of injured players.” Correct. This interprets the confidence level, not the interval, and it does so properly.
Then judge a claim. The league’s insurer assumes that 35% of players are injured each season. 0.35 is just above the upper limit, 0.349, so at 90% confidence it is not plausible: the interval gives convincing evidence that the true proportion is lower than 35%. At 95% the interval is (0.242, 0.358), which does include 0.35, so the conclusion is borderline. State the confidence level with every conclusion.
Conditions still matter
Everything on this page assumes the interval was built correctly. If the sample was not random, the confidence level has no meaning, because the long-run capture rate relies on random sampling. If the large-counts condition fails, the true capture rate can be well below the stated C%. And the margin of error only allows for random sampling variation. It cannot rescue a biased sample, a leading question or a high non-response rate.
Common misconceptions
- “There is a 95% probability that p is in this interval.” After you calculate it, the interval either contains p or it does not. Say “95% confident”, and explain the 95% as the long-run capture rate of the method.
- “95% of the population (or sample) have values between a and b.” The interval estimates one number, the population proportion. It says nothing about individuals.
- “95% of sample proportions fall in this interval.” The capture rate is about intervals capturing p, not about where other p̂ values land.
- “The claimed value is in the interval, so the claim is true.” Inside means plausible, not proved.
- “A higher confidence level gives a more precise interval.” It gives a wider, less precise interval.
- “Doubling the sample size halves the margin of error.” Doubling n divides the margin of error by √2 ≈ 1.41. You need 4 times the sample to halve it.
- “Increasing the confidence level changes the standard error.” The standard error depends only on p̂ and n. The confidence level changes z*.
- Interpreting the interval without context. Name the proportion, the response and the population every time.
Where this leads
Next, Topic 3.5 sets up a significance test for a population proportion, which asks a related question in a different way: Setting Up a Test for a Population Proportion. Try the practice questions now, then use the revision notes and the checklist to consolidate. To review how the interval is built, return to Topic 3.3.
Quick questions
Can I say there is a 95% probability that p is in my interval?
No. Once the interval is calculated, p is either inside it or not; nothing is random any more. The 95% describes the method: about 95% of intervals made this way, from many random samples, capture p.
If the claimed value is inside the interval, has the claim been proved?
No. A value inside the interval is plausible, so the data give no convincing evidence against the claim. But many other values are plausible too, so the claim is not proved.