Actuarial Statistics · Hypothesis testing and goodness of fit

Tests for Means and Variances: t, Chi-Square and F Tests

Q: What is the difference between a paired t-test and a two-sample t-test?

A paired t-test is for linked observations, such as before and after readings on the same people. You take the differences and test whether their mean is zero, with n − 1 degrees of freedom. A two-sample t-test is for two independent groups, and uses a standard error built from both samples.

Q: How do I do an F-test for equality of variances in CS1?

State H0: σX² = σY². Compute F = SX² ÷ SY² with degrees of freedom (n − 1, m − 1). For a two-tailed test, put the larger variance on top and compare with the upper α ÷ 2 critical value. Reject H0 if F exceeds it.

Q: How do I do a chi-square test for the variance of a normal distribution?

Compute (n − 1)S² ÷ σ0² and compare it with the χ²(n − 1) distribution. For H1: σ² > σ0², reject if the value exceeds the upper critical point. For H1: σ² < σ0², reject if it is below the lower point. The data must be normal.

Q: When do I use a z-test instead of a t-test?

Use z when the population standard deviation is known, with normal data or a large sample. Use t when you estimate it from the sample. In the exam, a given σ is the signal for z.

Q: Do I always pool the variances in a two-sample t-test?

No. Pool only if the question tells you to assume equal variances, or if an F-test supports it. Otherwise use the unpooled standard error.

Updated 11 October 2026 · Fact-checked

These tests check a claim about a population mean or variance using sample data. Use a z-test or t-test for means, a chi-square test for one variance, and an F-test to compare two variances. Pick the test from what is known, compute the statistic, compare it with the table value, then conclude.

Understand Tests for Means and Variances

A hypothesis test starts with a claim about a population, called the null hypothesis (H0). You collect a sample and ask: if H0 were true, how surprising is this sample? If it is very surprising, you reject H0. The test statistic measures that surprise, and a critical value or p-value turns it into a decision.

For a mean, the sample mean X̄ is centred on the true mean. If the data are normal and the population variance σ² is known, the standardised mean follows a standard normal, so you use a z-test. If σ² is unknown, you replace σ with the sample standard deviation S. That extra uncertainty makes the statistic follow a t distribution with n − 1 degrees of freedom. The t distribution has fatter tails than the normal, and it approaches the normal as n grows.

For a variance, the key fact is that for a normal sample, (n − 1)S² ÷ σ² follows a chi-square distribution with n − 1 degrees of freedom. Chi-square is skewed and only takes positive values, so the upper and lower critical values are different numbers. Compare the statistic with the right one for your alternative hypothesis.

For two samples, you compare means or variances. To compare means with independent samples, you need to know whether the two variances can be taken as equal. The F-test checks that: the ratio of two sample variances follows an F distribution when the population variances are equal. If you then accept equal variances, you pool them and use the pooled t-test.

Paired samples are different. Each observation in one sample is linked to one in the other, such as the same policyholder before and after. You take the differences and run a one-sample t-test on them. This removes the person-to-person variation, so it is usually more powerful than a two-sample test on the same data. All these tests assume normal data, or a large enough sample for the Central Limit Theorem to help with means. The variance tests are much more sensitive to non-normality.

Key rules to remember

One sample mean, σ known: Z = (X̄ − μ0) ÷ (σ ÷ √n) ~ N(0, 1); Use for normal data with known σ. For large n with unknown σ, a z approximation is common but state it.
One sample mean, σ unknown: T = (X̄ − μ0) ÷ (S ÷ √n) ~ t(n − 1); S² = Σ(xi − x̄)² ÷ (n − 1). Degrees of freedom are n − 1.
One sample variance: χ² = (n − 1)S² ÷ σ0² ~ χ²(n − 1); Needs normal data. Upper-tail test for H1: σ² > σ0², lower-tail for H1: σ² < σ0².
Two independent means, equal variances: T = (X̄ − Ȳ − δ0) ÷ (Sp × √(1/n + 1/m)) ~ t(n + m − 2); δ0 is usually 0. Pooled variance Sp² = [(n − 1)SX² + (m − 1)SY²] ÷ (n + m − 2).
Two independent means, variances known: Z = (X̄ − Ȳ − δ0) ÷ √(σX²/n + σY²/m) ~ N(0, 1); Use when both population variances are given.
Two independent means, variances unequal: T = (X̄ − Ȳ − δ0) ÷ √(SX²/n + SY²/m); Welch approach: the distribution is only approximately t. Use it only if the question tells you to or the F-test rejects equal variances.
F-test for variance ratio: F = (SX² ÷ σX²) ÷ (SY² ÷ σY²) ~ F(n − 1, m − 1); Under H0: σX² = σY², F = SX² ÷ SY². The first degrees of freedom belong to the numerator.
Lower F critical value: F(1 − α; a, b) = 1 ÷ F(α; b, a); Tables often give only upper points. Use this to get the lower point. Here F(α; a, b) is the upper α point.
Paired t-test: Di = Xi − Yi; T = (D̄ − δ0) ÷ (SD ÷ √n) ~ t(n − 1); n is the number of pairs. Differences must be roughly normal.

How to solve Tests for Means and Variances questions

Use this order for any question on means or variances. It stops you picking the wrong test.

1Identify what is being tested: one mean, one variance, two means or two variances. Check whether the samples are independent or paired.
2State H0 and H1 in symbols, and decide one-tailed or two-tailed from the wording. For example, 'has increased' means a one-tailed upper test.
3Check what is known. Known σ gives z. Unknown σ gives t. A variance claim gives chi-square. A comparison of variances gives F. State the normality assumption.
4For two independent means with unknown variances, test equality of variances with the F-test first, or use the assumption the question gives. Then pool or use Welch.
5Compute the test statistic with the correct degrees of freedom. Keep sample sizes and the n − 1 divisors straight.
6Find the critical value at the given significance level, or the p-value. For two-tailed tests, split α between the tails.
7Compare and decide: reject H0 if the statistic falls in the critical region. Do not say 'accept H0'; say 'there is insufficient evidence to reject H0'.
8Write a conclusion in context, in one sentence that uses the problem's words.

Quickest way: Test-choice shortcut and quick check

When to use it: Use this in MCQs and in the first minute of a written question, when you must choose the test fast.

Ask: mean or variance? Variance means chi-square for one sample and F for two.
For means, ask: is σ given? If yes, z. If no, t.
Ask: are the data linked pairs? If yes, subtract and do one-sample t on the differences. Degrees of freedom are pairs − 1.
Two independent samples with unknown variances: assume equal and pool unless told otherwise. Degrees of freedom are n + m − 2.
Put the larger sample variance on top in the F-test for a two-tailed test, so you only need the upper critical value. Use a significance level of α ÷ 2 for the upper point, and the degrees of freedom of the larger variance first.
Sanity check: the t statistic should be roughly (difference) ÷ (standard error). If it is huge or tiny, recheck the standard error.

Common mistakes in Tests for Means and Variances

Using a two-sample t-test on paired data.
Both involve two columns of numbers and the pairing is easy to miss.
Fix: Ask whether each value in one sample belongs with one specific value in the other. If so, take differences and do a one-sample t-test with n − 1 degrees of freedom.
Using n instead of n − 1 in the sample variance or the degrees of freedom.
Students mix up the population variance divisor with the sample variance divisor.
Fix: Sample variance S² divides by n − 1. A one-sample t or chi-square has n − 1 degrees of freedom. A pooled t has n + m − 2.
Using only the upper critical value of chi-square or F for a two-tailed test.
Z and t are symmetric, so students expect the same for skewed distributions.
Fix: Chi-square and F need two different critical values for a two-tailed test, at α ÷ 2 in each tail. For F, put the larger variance on top so only the upper point is needed.
Writing F degrees of freedom in the wrong order.
The two numbers look interchangeable, but the table is not symmetric.
Fix: Numerator degrees of freedom come first. If the numerator is the sample with n observations, use F(n − 1, m − 1).
Pooling variances without checking they are plausibly equal.
The pooled formula is the one most often practised.
Fix: Run the F-test for equal variances, or use the assumption stated in the question. If variances differ clearly, use the unpooled standard error.
Concluding 'H0 is accepted' or giving a bare 'reject'.
Students stop once the comparison is done.
Fix: Say 'reject H0' or 'do not reject H0', then add a sentence in context, for example that there is evidence the mean claim amount differs.

Worked examples

Example 1

Two independent normal samples of claim processing times (in minutes) are taken. Sample X: n = 10, x̄ = 52, sX² = 16. Sample Y: m = 12, ȳ = 48, sY² = 25. (a) Test at the 5% level whether the population variances are equal. (b) Assuming equal variances, test at the 5% level whether the population means differ.

Show the solution

(a) H0: σX² = σY². H1: σX² ≠ σY². Two-tailed.
Put the larger variance on top: F = 25 ÷ 16 = 1.5625, with degrees of freedom (11, 9).
The upper 2.5% point of F(11, 9) is about 3.9. Since 1.5625 < 3.9, do not reject H0. Equal variances are reasonable.
(b) H0: μX = μY. H1: μX ≠ μY. Two-tailed.
Pooled variance: sp² = (9 × 16 + 11 × 25) ÷ 20 = (144 + 275) ÷ 20 = 419 ÷ 20 = 20.95.
Standard error: √[20.95 × (1/10 + 1/12)] = √(20.95 × 0.18333) = √3.8408 = 1.9598.
T = (52 − 48) ÷ 1.9598 = 2.041, with 10 + 12 − 2 = 20 degrees of freedom.
The critical value for t(20) at 2.5% in each tail is 2.086. Since 2.041 < 2.086, do not reject H0.

Answer: The variance test does not reject equal variances (F = 1.5625). The t statistic is 2.041, below 2.086, so at the 5% level there is insufficient evidence that the mean processing times differ. The result is close to the boundary, so say so.

Example 2

A random sample of 15 observations from a normal population has sample variance 31.5. Test at the 5% level H0: σ² = 20 against H1: σ² > 20.

Show the solution

The statistic is χ² = (n − 1)s² ÷ σ0². Under H0 it follows χ²(14).
Compute: χ² = 14 × 31.5 ÷ 20 = 441 ÷ 20 = 22.05.
The test is one-tailed upper. The 5% upper critical value of χ²(14) is 23.685.
Since 22.05 < 23.685, the statistic is not in the critical region.
Do not reject H0.

Answer: The statistic is 22.05 against a critical value of 23.685, so there is insufficient evidence at the 5% level that the population variance exceeds 20.

Exam tips

Write H0, H1, the statistic's distribution and its degrees of freedom every time. Method marks are awarded for each, even if your arithmetic slips.
State your assumptions explicitly: normality, independence, and equal variances for pooling. Examiners look for them.
In computer-based questions, show the R call or Excel formula you used, such as t.test with paired = TRUE or var.test, and read off the statistic, degrees of freedom and p-value. Then write the conclusion in words.
Know how to read a table that gives only upper points. Use the reciprocal rule for lower F points and the symmetry of t.
MCQs often hinge on choosing the test or the degrees of freedom. Check pairing, whether σ is known, and n − 1 versus n + m − 2 before you calculate.

Practice questions from Hypothesis testing and goodness of fit

Tests for Means and Variances in other exams

The same ground in other exams, if you are preparing for more than one or want another angle on it.

Tests for Means and Variances: frequently asked questions

What is the difference between a paired t-test and a two-sample t-test?

A paired t-test is for linked observations, such as before and after readings on the same people. You take the differences and test whether their mean is zero, with n − 1 degrees of freedom. A two-sample t-test is for two independent groups, and uses a standard error built from both samples.

How do I do an F-test for equality of variances in CS1?

State H0: σX² = σY². Compute F = SX² ÷ SY² with degrees of freedom (n − 1, m − 1). For a two-tailed test, put the larger variance on top and compare with the upper α ÷ 2 critical value. Reject H0 if F exceeds it.

How do I do a chi-square test for the variance of a normal distribution?

Compute (n − 1)S² ÷ σ0² and compare it with the χ²(n − 1) distribution. For H1: σ² > σ0², reject if the value exceeds the upper critical point. For H1: σ² < σ0², reject if it is below the lower point. The data must be normal.

When do I use a z-test instead of a t-test?

Use z when the population standard deviation is known, with normal data or a large sample. Use t when you estimate it from the sample. In the exam, a given σ is the signal for z.

Do I always pool the variances in a two-sample t-test?

No. Pool only if the question tells you to assume equal variances, or if an F-test supports it. Otherwise use the unpooled standard error.