Confidence Intervals for the Mean (AI HL)

A confidence interval for the mean gives you a range that is very likely to contain the true population mean, built from a single sample. The tricky part is not the arithmetic - your GDC does that - it's choosing the right method and reading the output correctly. This page focuses on that choice, with worked examples and the mistakes that lose marks. It's part of the broader Confidence Intervals topic.

11 questions on this sub-topic.

Practise confidence intervals for the mean → Try exam-style questions

Choosing your interval

Covered under IB syllabus reference AHL4.16: confidence intervals for the mean of a normal population, using the normal distribution when \(\sigma\) is known and the \(t\)-distribution when \(\sigma\) is unknown - regardless of sample size. You should also be able to interpret what the resulting interval means in context.

z-interval vs t-interval

\(\bar{x} \pm z^* \dfrac{\sigma}{\sqrt{n}}\)  or  \(\bar{x} \pm t^* \dfrac{s}{\sqrt{n}}\)

These aren't given as algebra in the formula booklet - your GDC builds them directly from raw or summary data, so the real skill is spotting which routine to use. If the question states the population standard deviation \(\sigma\), use a \(z\)-interval; if you only have a sample standard deviation \(s\) (however large \(n\) is), use a \(t\)-interval.

Confidence level

90% → 95% → 99%

A higher confidence level needs a larger critical value, which widens the interval.

Need the full syllabus wording and GDC walkthrough? See Confidence Intervals.

Worked examples

1
Medium
GDC
[3 marks]

A sample of \(n=25\) has mean \(\bar{x}=50\) and standard deviation \(s=4.\) Find a 95% confidence interval for the population mean.

Worked solution

Use a \(t\)-interval with \(n=25\); technology gives M1
\((48.3,\ 51.7).\) A1
\((48.3,\ 51.7).\) A1

M1 T-interval, df = 24 A1 Lower 48.3 A1 Upper 51.7
2
Hard
GDC
[5 marks]

A population has known standard deviation \(\sigma = 10\). A sample of \(n = 64\) gives \(\bar{x} = 75\). Construct a 95% CI using the \(z\)-interval formula \(\bar{x} \pm z^* \dfrac{\sigma}{\sqrt{n}}\), where \(z^* = 1.96.\)

Worked solution

\(\text{SE} = \dfrac{10}{\sqrt{64}}\) M1
\(= 1.25.\) A1
Margin \(= 1.96 \times 1.25\) M1
\(= 2.45.\) A1
CI: \((72.55,\ 77.45).\) A1

M1 Standard error A1 SE = 1.25 M1 \(z^* \times \text{SE}\) A1 Correct answer of 2.45 A1 Interval

Common mistakes

Ready to practise properly?

10 confidence-interval questions on the mean, marked instantly like the real exam.

Quick answers

When do I use a z-interval instead of a t-interval for the mean?

Use a \(z\)-interval only when the population standard deviation \(\sigma\) is given directly. Use a \(t\)-interval whenever you only have the sample standard deviation \(s\), regardless of how large the sample is.

Why does a 99% confidence interval come out wider than a 95% one?

A higher confidence level needs a larger critical value, which widens the margin of error and therefore the interval - you are trading precision for a higher chance of capturing the true mean.

← Back to Applications & Interpretation HL topics