FRM Exam Part I · Common Univariate Random Variables

Chi-Squared, Student's t and F Distributions Explained

Updated 11 October 2026 · Fact-checked

These three distributions are built from the normal. Chi-squared (sum of squared standard normals) tests a variance. Student's t (normal divided by a scaled chi-squared root) tests a mean when variance is estimated. F (ratio of two scaled chi-squared variables) compares variances or tests joint restrictions. Find the degrees of freedom, compute the statistic, compare it with the critical value.

Understand Chi-Squared, Student's t and F Distributions

All three distributions come from the normal distribution. You use them because real data rarely give you the true variance. You estimate it from a sample, and that estimate adds extra uncertainty.

The chi-squared distribution with k degrees of freedom is the distribution of the sum of k squared independent standard normal variables. It cannot be negative. It is skewed to the right, and the skew fades as k grows. Its mean is k and its variance is 2k. You use it for inference on a single variance, because (n − 1)s² ÷ σ² follows a chi-squared distribution with n − 1 degrees of freedom when the data are normal.

The Student's t distribution arises when you standardise a sample mean using the sample standard deviation instead of the true one. It is symmetric around zero and bell-shaped like the normal, but it has fatter tails. That gives larger critical values, especially with few degrees of freedom. As degrees of freedom rise, it converges to the standard normal. The mean is 0 (for k > 1) and the variance is k ÷ (k − 2) (for k > 2), which is always above 1.

The F distribution is the ratio of two independent chi-squared variables, each divided by its own degrees of freedom. It has two degrees-of-freedom parameters, numerator and denominator, and their order matters. It is non-negative and right-skewed. You use it to compare two variances and to test joint hypotheses in regression, such as whether all slope coefficients are zero.

Degrees of freedom count the independent pieces of information left after you estimate parameters. A sample of n observations with the mean estimated leaves n − 1. Most exam errors come from using n instead of n − 1.

Key formulas to remember

Chi-squared definition: χ²(k) = Z₁² + Z₂² + … + Z_k², with Z_i independent standard normal; Mean = k, variance = 2k. Values are always ≥ 0 and right-skewed.
Variance test statistic: χ² = (n − 1) s² ÷ σ₀², with n − 1 degrees of freedom; Assumes the data are normally distributed. σ₀² is the variance under the null hypothesis.
Student's t definition: t(k) = Z ÷ √(χ²(k) ÷ k); Z and the chi-squared variable are independent. Mean 0 for k > 1.
t variance: Var(t) = k ÷ (k − 2), for k > 2; Always greater than 1, which is why tails are fatter than the normal. It tends to 1 as k grows.
t statistic for a mean: t = (x̄ − μ₀) ÷ (s ÷ √n), with n − 1 degrees of freedom; Use when the population variance is unknown and estimated by s.
F definition: F(d₁, d₂) = [χ²(d₁) ÷ d₁] ÷ [χ²(d₂) ÷ d₂]; The two chi-squared variables are independent. d₁ is numerator df, d₂ is denominator df.
F test for equal variances: F = s₁² ÷ s₂², with n₁ − 1 and n₂ − 1 degrees of freedom; Convention: put the larger sample variance on top, so F ≥ 1. Assumes both samples are normal and independent.
Link between t and F: [t(k)]² = F(1, k); The square of a t variable with k df is an F variable with 1 and k df.
Useful t critical values (two-tailed, 5%): df 10: 2.228 | df 15: 2.131 | df 30: 2.042 | very large df: 1.960; Learn the pattern: values fall toward the normal value of 1.96 as df rises.

How to solve Chi-Squared, Student's t and F Distributions questions

Use this routine for any question on these distributions. It stops you from picking the wrong distribution or the wrong degrees of freedom.

1Identify what is being tested: one mean (t), one variance (chi-squared), two variances or a joint regression restriction (F).
2Write the null and alternative hypotheses. Note whether the test is one-tailed or two-tailed.
3Compute the sample statistic using the matching formula. Use n − 1 in the variance formula and the standard error s ÷ √n in the t formula.
4Find the degrees of freedom: n − 1 for t and chi-squared, (n₁ − 1, n₂ − 1) for the F test of two variances.
5Look up the critical value from the table given, using the right tail probability. For a two-tailed test, split the significance level across both tails.
6Compare: reject the null if the statistic falls beyond the critical value. For a chi-squared two-tailed test, check both the lower and upper critical values.
7State the conclusion in words, and check that it makes sense, for example that F is not below 1 when you put the larger variance on top.

Quickest way: Pick the distribution, then check df and tails

When to use it: Use this when the exam gives you a statistic and four answer choices and you have little time.

Match keyword to distribution: mean with estimated variance means t, single variance means chi-squared, ratio of variances means F.
Compute the statistic once, using n − 1 for the sample variance. A wrong answer choice often uses n.
Use estimation shortcuts: a t critical value is a little above the normal value (1.96 two-tailed, 1.645 one-tailed at 5%), and the gap shrinks as df rises.
Remember the chi-squared mean is k. If the statistic is far above k, the sample variance is much larger than the null value.
Reject quickly if the statistic is clearly beyond the critical value. Only check the table closely when the numbers are near.

Common mistakes in Chi-Squared, Student's t and F Distributions

Using n instead of n − 1 as the degrees of freedom
Students remember sample size and forget that estimating the mean uses up one degree of freedom.
Fix: For one sample, df = n − 1 for both t and chi-squared tests. Write df before you open the table.
Using the normal critical value 1.96 when variance is estimated from a small sample
The t and normal curves look alike, so students treat them as the same.
Fix: With an estimated variance and a small sample, use the t critical value. For example, 2.131 at 15 df, not 1.96.
Treating the chi-squared distribution as symmetric
Students carry over normal-distribution habits, such as lower critical value equals minus the upper one.
Fix: Chi-squared is right-skewed and non-negative. Look up lower and upper critical values separately for a two-tailed test.
Swapping numerator and denominator degrees of freedom in the F distribution
The notation F(d₁, d₂) looks symmetric, but the two parameters have different roles.
Fix: The variance on top uses numerator df. Match each sample's n − 1 to its place in the ratio.
Saying the t distribution has a variance of 1 or the same tails as the normal
Students remember that t tends to normal and forget the finite-df case.
Fix: Variance is k ÷ (k − 2), which is above 1. Tails are fatter, so critical values are larger.
Applying the variance tests to non-normal data without comment
The formulas work mechanically, so students forget the assumption.
Fix: The chi-squared variance test and the F test of variances are sensitive to non-normality. If a question mentions fat tails, treat the result with caution.

Worked examples

Example 1

A risk analyst has 25 daily returns on a portfolio. The sample standard deviation is 1.4%. She tests H₀: σ = 1.0% against H₁: σ > 1.0% at the 5% significance level. The 5% upper critical value of chi-squared with 24 degrees of freedom is 36.415. Which is correct? A) Statistic 1.96, do not reject. B) Statistic 47.04, reject H₀. C) Statistic 49.00, reject H₀. D) Statistic 24.00, do not reject.

Show the solution

Identify the test: one variance, so use chi-squared with n − 1 = 24 degrees of freedom.
Compute s² = 1.4² = 1.96 and σ₀² = 1.0² = 1.00, both in squared percent.
Statistic = (n − 1) s² ÷ σ₀² = 24 × 1.96 ÷ 1.00 = 47.04.
Compare with the critical value: 47.04 > 36.415, so the statistic is in the rejection region.
Option C uses n = 25 instead of n − 1 = 24, which gives 49.00. That is wrong.

Answer: B. The statistic is 47.04, which exceeds 36.415, so reject H₀. There is evidence that the true volatility is above 1.0%.

Example 2

An analyst has 16 monthly excess returns with a sample mean of 1.0% and a sample standard deviation of 2.0%. She tests H₀: μ = 0 against H₁: μ ≠ 0 at the 5% level. The two-tailed 5% critical value of t with 15 degrees of freedom is 2.131. What is the t statistic and the conclusion?

Show the solution

The population variance is unknown and the sample is small, so use the t test with n − 1 = 15 degrees of freedom.
Standard error = s ÷ √n = 2.0 ÷ √16 = 2.0 ÷ 4 = 0.5%.
t = (x̄ − μ₀) ÷ SE = (1.0 − 0) ÷ 0.5 = 2.00.
Compare with the critical value: |2.00| < 2.131, so the statistic is not in the rejection region.
Note that using the normal value 1.96 would wrongly reject, since 2.00 > 1.96. The fatter t tails change the decision.

Answer: t = 2.00. This is below 2.131, so do not reject H₀. There is not enough evidence at the 5% level that the mean excess return differs from zero.

Exam tips

When a question says the variance is estimated from the sample and n is small, expect the answer to use t rather than the normal.
Always compute degrees of freedom first. Distractor options are often built from the wrong df, such as n instead of n − 1.
Expect conceptual questions on shape: chi-squared is right-skewed and non-negative, t is symmetric with fat tails, F is right-skewed and has two df parameters.
Know the links: t² with k df equals F(1, k), and t tends to normal as df grows. These are quick marks.
In regression questions, remember that the F statistic tests joint restrictions while a t statistic tests a single coefficient.

Practice questions from Common Univariate Random Variables

Chi-Squared, Student's t and F Distributions in other exams

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

CFA Level IStudent's t, Chi-Square and F Distributions

Chi-Squared, Student's t and F Distributions: frequently asked questions

What is the difference between the t distribution and the normal distribution?

Both are symmetric and bell-shaped with mean zero. The t distribution has fatter tails and a variance of k ÷ (k − 2), above 1, because the standard deviation is estimated. As degrees of freedom increase, the t distribution converges to the standard normal.

What are degrees of freedom in a chi-squared distribution?

They are the number of independent squared standard normal variables added together. This also gives the mean (k) and the variance (2k). In a variance test on a sample of n observations, the degrees of freedom are n − 1.

When is the F distribution used in hypothesis testing?

Use it to compare two population variances using the ratio of sample variances. It is also used in regression to test joint hypotheses, such as whether several coefficients are all zero. The statistic has numerator and denominator degrees of freedom, and the order matters.

Do I need to memorise t critical values for FRM Part I?

Not every value. Know the key ones, such as about 1.96 for large samples at 5% two-tailed, and understand that critical values rise as degrees of freedom fall. When a table is needed, the question usually supplies the value.

Why does the chi-squared test for variance assume normal data?

The result that (n − 1)s² ÷ σ² follows a chi-squared distribution depends on the data being normal. With fat-tailed or skewed data the test can mislead, much more so than the t test for a mean.