Actuarial Statistics · Confidence intervals and prediction intervals

Confidence Intervals for Normal Mean and Variance

Q: When should I use the t distribution instead of the normal for a confidence interval?

Use t when you are estimating the mean of a normal population and σ is unknown, so you use s instead. The degrees of freedom are n − 1. Use the normal distribution only when σ is known, or in a large-sample approximation where the question allows it.

Q: How do I find a confidence interval for the variance of a normal population?

Compute (n − 1)s², then divide by the upper and lower chi-square points with n − 1 degrees of freedom. For 95%, use the 97.5% point for the lower limit and the 2.5% point for the upper limit. The interval is not symmetric around s².

Q: Why is the chi-square interval not symmetric?

The chi-square distribution is skewed to the right. The two percentage points are not equally far from the centre. Dividing a fixed number by each of them gives limits at unequal distances from s².

Q: Does the confidence interval for the mean need normality?

For small samples, yes. The t interval relies on the data being normal. For large samples, the central limit theorem makes the interval approximately valid even if the data are not normal. The variance interval is much more sensitive to non-normality.

Q: What happens to the interval if I increase the confidence level?

It becomes wider. A higher confidence level needs a larger critical value, so the margin grows. A larger sample size makes the interval narrower for the same confidence level.

Updated 11 October 2026 · Fact-checked

A confidence interval gives a range that captures the true parameter with a stated probability over repeated samples. For a normal mean, use z if σ is known and t with n − 1 degrees of freedom if σ is unknown. For the variance, use the chi-square distribution with n − 1 degrees of freedom.

Understand Confidence Intervals for Normal Mean and Variance

A confidence interval is a range calculated from sample data. It is built so that, over many repeated samples, a stated percentage of such intervals (say 95%) contain the true parameter. It does not say there is a 95% probability that the true value lies in your one calculated interval. The parameter is fixed. The interval is random before you take the sample.

Intervals come from a pivotal quantity. This is a function of the data and the parameter whose distribution does not depend on any unknown parameter. For a normal sample, (X̄ − μ) ÷ (σ/√n) follows N(0,1) when σ is known. You find the middle 1 − α of that distribution and rearrange to isolate μ.

When σ is unknown, you replace it with the sample standard deviation s. That adds extra uncertainty, so the pivot (X̄ − μ) ÷ (S/√n) follows a t distribution with n − 1 degrees of freedom. The t distribution has heavier tails than the normal, so the interval is wider. As n grows, t approaches the normal. This result needs the underlying data to be normal.

For the variance, the pivot is (n − 1)S² ÷ σ², which follows a chi-square distribution with n − 1 degrees of freedom. This distribution is not symmetric. So the interval is not 'estimate ± margin'. You use two different percentage points, and the larger one gives the lower limit.

You can get an interval for the standard deviation by taking square roots of the limits for the variance. You cannot do the same shortcut for the mean and variance together. Each interval is separate.

Key rules to remember

Mean, σ known: x̄ ± z(1−α/2) × σ ÷ √n; Use when the population is normal (or n is large) and σ² is known. For 95%, z = 1.96. For 99%, z = 2.576. For 90%, z = 1.645.
Mean, σ unknown: x̄ ± t(n−1, 1−α/2) × s ÷ √n; Use when the population is normal and σ² is estimated by s². Degrees of freedom = n − 1. Here s² = Σ(x − x̄)² ÷ (n − 1).
Variance: ( (n − 1)s² ÷ χ²(n−1, 1−α/2) , (n − 1)s² ÷ χ²(n−1, α/2) ); Here χ²(n−1, p) is the point with probability p below it. The larger value goes under the lower limit. The population must be normal.
Standard deviation: ( √lower limit for σ² , √upper limit for σ² ); Take square roots of the variance interval limits.
Pivotal quantities: (X̄ − μ) ÷ (σ/√n) ~ N(0,1); (X̄ − μ) ÷ (S/√n) ~ t(n−1); (n − 1)S² ÷ σ² ~ χ²(n−1); These are the starting points of each interval. They hold for a random sample from a normal distribution.
Sample size for a mean, σ known: n ≥ ( z(1−α/2) × σ ÷ E )²; E is the required half-width. Always round n up to the next whole number.

How to solve Confidence Intervals for Normal Mean and Variance questions

Use this method for any question asking for an interval for a normal mean or variance.

1Identify the parameter: mean μ, variance σ², or standard deviation σ. Check the data are described as normal, or that the sample is large enough to justify it for the mean.
2Note n and the sample statistics. If you are given raw data or sums, compute x̄ = Σx ÷ n and s² = (Σx² − n x̄²) ÷ (n − 1).
3Decide on the distribution. Mean with σ known: standard normal. Mean with σ unknown: t with n − 1 degrees of freedom. Variance: chi-square with n − 1 degrees of freedom.
4Convert the confidence level to α. For 95%, α = 0.05, so you need the 0.975 point (and the 0.025 point for chi-square).
5Read the percentage points from the Tables. For chi-square, find both the lower and upper points at n − 1 degrees of freedom.
6Substitute into the correct formula. For variance, divide (n − 1)s² by the upper point to get the lower limit, and by the lower point to get the upper limit.
7State the interval with sensible rounding and units, and add a one-line interpretation. Mention the normality assumption.
8Check: the mean interval must be centred on x̄, and the variance interval must contain s² but not be centred on it.

Quickest way: Three-way choice and one-line calculation

When to use it: Use under time pressure for MCQs and for the calculation part of written questions.

Ask two questions: 'mean or variance?' and 'is σ given?'. This tells you z, t or chi-square at once.
Write n − 1 immediately. It is the degrees of freedom for t and chi-square.
Get the table value first, then compute the margin: critical value × s ÷ √n for the mean, or (n − 1)s² ÷ each chi-square point for the variance.
Sense check: the t value must be larger than the matching z value. For the variance, the lower limit must be below s² and the upper limit above it.
In MCQs, eliminate options that are symmetric around s² for a variance interval, or that use z when σ is unknown.

Common mistakes in Confidence Intervals for Normal Mean and Variance

Using z instead of t when σ is estimated from the sample.
Students remember the 1.96 value and apply it automatically.
Fix: Check whether σ is given as a known value. If only s is available, use t with n − 1 degrees of freedom.
Using n instead of n − 1 as the degrees of freedom.
The sample size is the first number in the question, so students use it without thinking.
Fix: Write 'df = n − 1' as soon as you start. One degree of freedom is used up in estimating the mean.
Writing a variance interval as s² ± margin.
Students copy the structure of the mean interval.
Fix: The chi-square distribution is skewed. Use the formula with two different percentage points. The interval is not centred on s².
Swapping the chi-square points so the lower limit is bigger than the upper limit.
Students match 'lower' with the lower table value.
Fix: Divide by the larger point to get the smaller limit. Check that lower < s² < upper.
Confusing the variance interval with the standard deviation interval, or forgetting to take square roots when asked for σ.
Questions switch between σ and σ² and students lose track.
Fix: Finish the variance interval first. Then take square roots of both limits if σ is asked for.
Interpreting a 95% interval as '95% probability that μ is in this interval'.
The wording sounds natural but treats μ as random.
Fix: Say that 95% of intervals constructed this way would contain μ. State the method, not a probability for this one interval.

Worked examples

Example 1

A random sample of 16 claim-processing times (in minutes) from a normal population has sample mean 52.4 and sample standard deviation 6.0. Find a 95% confidence interval for the population mean. Use t(15) 97.5% point = 2.131.

Show the solution

The variance is unknown, so use the t distribution with n − 1 = 15 degrees of freedom.
Standard error = s ÷ √n = 6.0 ÷ √16 = 6.0 ÷ 4 = 1.5.
Margin = 2.131 × 1.5 = 3.1965.
Lower limit = 52.4 − 3.1965 = 49.2035. Upper limit = 52.4 + 3.1965 = 55.5965.

Answer: The 95% confidence interval for μ is approximately (49.20, 55.60) minutes, assuming the times are normally distributed.

Example 2

Using the same sample (n = 16, s = 6.0, so s² = 36), find a 95% confidence interval for σ² and for σ. Use χ²(15) points: 2.5% point = 6.262 and 97.5% point = 27.488.

Show the solution

The pivot is (n − 1)S² ÷ σ² ~ χ²(15). Compute (n − 1)s² = 15 × 36 = 540.
Lower limit for σ² = 540 ÷ 27.488 = 19.64 (dividing by the larger point).
Upper limit for σ² = 540 ÷ 6.262 = 86.23 (dividing by the smaller point).
Check: 19.64 < 36 < 86.23, and the interval is not centred on 36.
For σ, take square roots: √19.64 = 4.43 and √86.23 = 9.29.

Answer: The 95% confidence interval for σ² is approximately (19.64, 86.23), and for σ approximately (4.43, 9.29), assuming the data are normal.

Exam tips

Always state the assumption: the sample comes from a normal distribution. Examiners award marks for it, and variance intervals depend on it heavily.
Show the pivotal quantity and its distribution before substituting. In written answers, the method marks are for this step.
Write the table values you use, with the degrees of freedom. This lets you earn marks even if you misread a later number.
When the question gives Σx and Σx², compute s² with n − 1 in the denominator. A wrong s² carries through every later step.
In Paper B (computer-based), you can use R. qt(0.975, n−1) and qchisq(c(0.025, 0.975), n−1) give the critical values, but still write the formula and interpretation.

Practice questions from Confidence intervals and prediction intervals