Step 11: Confidence Intervals
- A single estimate hides how wrong it might be. A confidence interval gives a range of believable values for a parameter, built as estimate ± critical value × standard error.
- For a mean, use \( \bar{x} \pm t^* \cdot s/\sqrt{n} \) with \( df = n – 1 \). For a proportion, use \( \hat{p} \pm z^* \sqrt{\hat{p}(1-\hat{p})/n} \).
- “95% confident” describes the method: across many samples, about 95% of the intervals it produces contain the true value. It is not a 95% chance that one particular interval is right.
- Exam point: higher confidence gives a wider interval; a bigger sample gives a narrower one. Always state the conditions and interpret the interval in context.
From one number to a range
A polling firm asks 1,000 random voters about a proposal, and 430 say they support it. The easy headline is “Support stands at 43%”. But a different 1,000 voters might have given 41% or 45%. Step 9 showed that sample results always bounce around. Reporting 43% as if it were the exact truth hides that.
The honest version is “43%, give or take about 3 percentage points”, so the true support is probably somewhere between 40% and 46%. That range is a confidence interval.
Nearly every confidence interval in this book has the same structure:
The estimate sits in the middle. The standard error measures how much the estimate jumps between samples. The critical value decides how many standard errors wide to make the interval, and that is where the confidence level comes in. Everything after the ± sign is the margin of error. A narrow interval means you know a lot; a wide one means you do not.
What “95% confident” really means
This is the most misunderstood idea in introductory statistics. The confidence level describes the procedure, not any single interval. Imagine repeating the study many times: a new random sample each time, a new interval each time. About 95% of those intervals would contain the true parameter, and about 5% would miss it.
In real life you only get one interval, and you cannot tell whether it is one of the hits or one of the misses. So the safe wording is: “We are 95% confident that the true mean lies between A and B.” Here “confident” is short for “produced by a method that works 95% of the time”.
The t-distribution and an interval for a mean
To build an interval for \( \mu \), you need the standard error \( \sigma/\sqrt{n} \). But if you do not know the population mean, you almost certainly do not know \( \sigma \) either. So you use the sample standard deviation \( s \) instead, giving \( s/\sqrt{n} \).
That swap adds extra uncertainty, because \( s \) also changes from sample to sample. To allow for it, you use the t-distribution instead of the normal. It looks like the normal (symmetric, bell-shaped, centred on 0) but has heavier tails, which makes intervals a little wider. Its exact shape depends on the degrees of freedom, \( df = n – 1 \). As \( df \) grows, the t-distribution gets closer and closer to the normal.
| df | 8 | 9 | 14 | 15 | 19 | 24 | 29 | 49 | ∞ (z) |
|---|---|---|---|---|---|---|---|---|---|
| 90% | 1.860 | 1.833 | 1.761 | 1.753 | 1.729 | 1.711 | 1.699 | 1.677 | 1.645 |
| 95% | 2.306 | 2.262 | 2.145 | 2.131 | 2.093 | 2.064 | 2.045 | 2.010 | 1.960 |
| 99% | 3.355 | 3.250 | 2.977 | 2.947 | 2.861 | 2.797 | 2.756 | 2.680 | 2.576 |
Use this when the data is a random sample and either the population is roughly normal or \( n \ge 30 \).
A random sample of 25 students studies a mean of \( \bar{x} = 68 \) minutes a day, with \( s = 10 \) minutes. Build a 95% confidence interval for the mean study time of all students.
We are 95% confident that the true mean daily study time for all students is between about 63.9 and 72.1 minutes.
For a 99% interval with the same data, only \( t^* \) changes, to 2.797. The margin becomes \( 2.797 \times 2.0 \approx 5.59 \), giving (62.41, 73.59). The interval is wider: 11.2 minutes compared with 8.3. More confidence costs you precision. The only way to get both is a larger sample, which shrinks the standard error.
An interval for a proportion
For a proportion the structure is the same, with two changes. The standard error uses \( \hat{p} \), and you use \( z^* \) from the normal instead of \( t^* \). There is no separate \( \sigma \) to estimate, because a proportion’s spread is set by the proportion itself.
You need a random sample with at least 10 successes and 10 failures.
Of 1,000 randomly sampled voters, 430 support a proposal. Build a 95% confidence interval for the true level of support.
We are 95% confident that true support is between about 39.9% and 46.1%. That is the “43%, give or take 3 points” from the start. The whole interval is below 50%, so the data suggests the proposal lacks majority support.
Margin of error, sample size and judging claims
The margin of error is the ± part that newspapers quote. Four things could affect it:
| Factor | Effect on margin of error | Under your control? |
|---|---|---|
| Sample size \( n \) | Larger \( n \) gives a smaller margin (via \( \sqrt{n} \)) | Yes, the main lever |
| Confidence level | Higher level gives a larger margin | Yes, but at a cost |
| Variability (\( s \) or \( \hat{p} \)) | More variable data gives a larger margin | Mostly no |
| Population size \( N \) | No effect | Irrelevant |
Choosing a sample size. Decide the margin you need, \( E \), and solve for \( n \). Always round up to the next whole person.
If you have no earlier estimate of the proportion, use \( p^* = 0.5 \). It gives the largest, and so safest, sample size. For a 3-point margin at 95%: \( n = (1.960/0.03)^2 \times 0.25 \approx 1067.1 \), so 1,068 people. That is why national polls so often sample about a thousand people.
Judging a claim. Any value inside the interval is believable given the data; a value outside it is not well supported. A maker says its batteries last 500 hours. A sample of 16 gives \( \bar{x} = 486 \), \( s = 20 \). With \( SE = 5 \) and \( t^* = 2.131 \) (df = 15), the 95% interval is \( 486 \pm 10.66 = (475.3,\ 496.7) \). The claimed 500 hours lies outside, so the data does not support the claim. This is the same question a hypothesis test asks, which is Step 11.
Practise: 6 questions
Try each one on paper first, then check the worked answer. Use the t-table above.
EasyQ1. Find \( t^* \) for a 95% interval with \( n = 20 \), and \( z^* \) for a 99% interval for a proportion. Show worked answer
For 99%, \( z^* = 2.576 \).
EasyQ2. A 95% interval for a mean is (18, 26). What are the point estimate and the margin of error? Show worked answer
The margin of error is half the width: \( (26 – 18)/2 = 4 \).
MediumQ3. \( \bar{x} = 100 \), \( s = 15 \), \( n = 9 \), and the population is roughly normal. Build a 95% interval for \( \mu \). Show worked answer
\( df = 8 \), so \( t^* = 2.306 \).
Margin \( = 2.306 \times 5 = 11.53 \).
Interval: \( 100 \pm 11.53 = (88.47,\ 111.53) \).
MediumQ4. In a random sample of 400 people, 240 said yes. Build a 95% interval for the true proportion. Show worked answer
\( SE = \sqrt{0.6 \times 0.4/400} = \sqrt{0.0006} \approx 0.0245 \).
Margin \( = 1.960 \times 0.0245 \approx 0.048 \).
Interval: \( 0.60 \pm 0.048 = (0.552,\ 0.648) \).
MediumQ5. You want a 95% interval for a proportion with a margin of error of 0.05, and you have no earlier estimate. What sample size do you need? Show worked answer
\( n = (1.960/0.05)^2 \times 0.5 \times 0.5 = 39.2^2 \times 0.25 = 1536.64 \times 0.25 = 384.16 \).
Round up: \( n = 385 \).
ExamQ6. A hospital samples 25 patients at random. Their recovery times have \( \bar{x} = 14 \) days and \( s = 5 \) days, and the data is roughly symmetric. (a) State and check the conditions. [2] (b) Build a 95% confidence interval. [3] (c) Interpret it in one sentence. [1] (d) Managers claim the true mean is 18 days. Does your interval support this? [2] Show worked answer
(b) \( SE = 5/\sqrt{25} = 1 \) [1]; \( df = 24 \), \( t^* = 2.064 \) [1]; margin \( = 2.064 \times 1 \approx 2.06 \), so the interval is \( (11.94,\ 16.06) \) days [1].
(c) We are 95% confident that the true mean recovery time for all patients is between about 11.9 and 16.1 days [1].
(d) No [1]: 18 days lies outside the interval, so it is not a believable value for the true mean given this data [1].
Check yourself
1. Which is the correct meaning of a 95% confidence level?
2. You change from a 95% interval to a 99% interval with the same data. What happens?
3. Why do we use \( t^* \) rather than \( z^* \) in an interval for a mean?
Your score: 0 out of 3
The short version
- A point estimate is one best guess; a confidence interval adds its uncertainty, in the form estimate ± critical value × standard error.
- “95% confidence” means the method captures the true value in about 95% of samples. It is not a probability about one interval or a statement about the data.
- Mean: \( \bar{x} \pm t^* \cdot s/\sqrt{n} \) with \( df = n – 1 \). Proportion: \( \hat{p} \pm z^* \sqrt{\hat{p}(1-\hat{p})/n} \), needing 10 successes and 10 failures.
- Higher confidence widens the interval; a larger \( n \) narrows it; population size does not matter. Plan with \( n = (z^*\sigma/E)^2 \) or \( (z^*/E)^2 p^*(1-p^*) \), using \( p^* = 0.5 \) if unsure, and round up.
- The full chapter in Statistics Made Simple has many more worked examples and a full set of tiered practice questions with a complete answer key.
Further reading on EconTutorials
Older in-depth articles on this topic from our blog: