CFA Level I Exam · Estimation and Hypothesis Testing

Tests of Means, Variances and Correlation for CFA Level I

Updated 7 October 2026 · Fact-checked

These tests compare a sample statistic with a hypothesised value. Use a t-test for a mean and for correlation, with df of n − 1 or n − 2. Use a paired t-test on the differences when samples are dependent. Use chi-square for one variance and F for two variances. Compute the statistic, compare it with the critical value, then decide.

Understand Tests of Means, Variances and Correlation

Every test here follows the same frame: state H0 and Ha, compute a test statistic, compare it with a critical value, and decide whether to reject H0. What changes is the statistic and its distribution. Pick the right pair and the rest is arithmetic.

For a mean, the statistic is (sample mean − hypothesised mean) ÷ standard error. If the population variance is known, use a z-test. If it is unknown, which is the usual case, use a t-test with n − 1 degrees of freedom. The t-distribution has fatter tails than the normal, so its critical values are larger. With a large sample the two give nearly the same answer. For a nonnormal population with a small sample, the test is not reliable.

When you compare two means, ask one question: are the samples independent or dependent? If they are independent, such as two different funds, you use a difference-in-means test. If they are dependent, such as the same fund before and after a change, you use a paired comparison. Compute the difference for each pair, then run a one-sample t-test on those differences. Pairing removes the shared noise and often gives a more powerful test.

For variances, the distributions are not symmetric. A test of one variance against a value uses the chi-square distribution with n − 1 df. A test of whether two variances are equal uses the F-distribution, the ratio of two sample variances, with two df values. Both are sensitive to nonnormality, and chi-square and F values are never negative.

For correlation, you test whether the population correlation is zero. Convert the sample correlation r into a t-statistic with n − 2 df. A small r can be significant with a large n, and a large r can be insignificant with a small n. Significance does not mean the relationship is strong or causal.

Key formulas to remember

t-test for a single mean: t = (x̄ − μ0) ÷ (s ÷ √n), df = n − 1; Use when σ is unknown. If σ is known, use z = (x̄ − μ0) ÷ (σ ÷ √n).
Paired comparison: t = (d̄ − μd0) ÷ (sd ÷ √n), df = n − 1; d̄ is the mean of the paired differences and sd their standard deviation. n is the number of pairs. μd0 is usually 0.
Difference in means, independent samples, equal variances assumed: sp² = [(n1 − 1)s1² + (n2 − 1)s2²] ÷ (n1 + n2 − 2); t = [(x̄1 − x̄2) − (μ1 − μ2)0] ÷ √(sp²/n1 + sp²/n2); df = n1 + n2 − 2; Both populations assumed normal. If variances are not assumed equal, use √(s1²/n1 + s2²/n2) in the denominator, with df approximated.
Chi-square test of one variance: χ² = (n − 1)s² ÷ σ0², df = n − 1; σ0² is the hypothesised variance. The distribution is skewed, so two-tailed tests use two different critical values.
F-test of two variances: F = s1² ÷ s2², df = n1 − 1 and n2 − 1; Put the larger sample variance on top, so F ≥ 1. For a two-tailed test at significance α, use the α/2 table value.
Test of correlation (H0: ρ = 0): t = r√(n − 2) ÷ √(1 − r²), df = n − 2; Reject H0 if |t| exceeds the critical value. Assumes the two variables are approximately bivariate normal.

How to solve Tests of Means, Variances and Correlation questions

Use this sequence for any question on tests of means, variances or correlation.

1Identify the parameter: one mean, two means, one variance, two variances, or correlation.
2For two means, decide whether the samples are independent or dependent. Dependent pairs mean a paired t-test on the differences.
3Write H0 and Ha, and note whether the test is one-tailed or two-tailed from the wording.
4Choose the statistic: t or z for means, chi-square for one variance, F for two variances, t with n − 2 df for correlation.
5Compute the statistic carefully. Check which n is used: pairs, sample size, or n − 2.
6Find df and the critical value for the stated significance level and tails.
7Compare: reject H0 if the statistic falls beyond the critical value, otherwise fail to reject.
8State the conclusion in words, and do not say that you accept H0.

Quickest way: Match the test, then compare the statistic

When to use it: Use this for most multiple-choice items, where you have about 90 seconds and the options are three numbers or three conclusions.

Name the test from the data wording: before and after, or same stocks measured twice, means paired; variance means chi-square or F; correlation means t with n − 2.
Compute the statistic once. On the TI BA II Plus, for correlation, work the denominator first. With r = 0.3, key 0.3 [x²] [+/−] [+] 1 [=] to get 0.91, then press [√x] to get 0.9539, and store it with [STO] 1. Next key 36 [√x] [×] 0.3 [=] to get 1.8 (here n = 38, so n − 2 = 36). Then press [÷] [RCL] 1 [=] to get 1.887.
Eliminate options that use the wrong df, wrong n, or a missing square root. These are common wrong options.
If the question only asks for the decision, compare |statistic| with the critical value given. Do not look up extra values you do not need.
Check the direction: a one-tailed test uses the full α in one tail, while a two-tailed test splits it.

Common mistakes in Tests of Means, Variances and Correlation

Using an independent-samples test on paired data.
You see two sets of numbers and assume they are separate groups.
Fix: If the same units are measured twice, or the observations are matched, work with the differences and use n pairs and n − 1 df.
Using n instead of n − 2 for the correlation test.
Other tests use n − 1, so the pattern feels familiar.
Fix: Correlation uses n − 2 in both the formula and the df, because two sample means are estimated, one for each variable, so df = n − 2.
Putting the smaller variance in the F numerator.
You follow the order the samples are listed in the question.
Fix: Always put the larger sample variance on top so F ≥ 1, and match the numerator df to that sample.
Forgetting the square root of n when computing the standard error.
You rush and divide s by n or leave s alone.
Fix: Standard error is s ÷ √n. Check it: a standard error should be smaller than s but not tiny.
Using a z-test when σ is not given.
The sample is large and you think z is always fine.
Fix: If only the sample standard deviation is given, use t. Use z only when the question gives the population σ.
Reading a significant correlation as strong or causal.
A small p-value feels like proof of a major link.
Fix: Significance only says ρ is unlikely to be zero. With a large n, even a small r can be significant, and correlation does not show causation.

Worked examples

Example 1

An analyst compares the returns of 16 stocks in the month before and the month after a rule change. The mean of the paired differences (after − before) is 0.8% and the standard deviation of the differences is 2.0%. The test of H0: μd = 0 has a t-statistic closest to: A. 0.40, B. 1.60, C. 6.40.

Show the solution

The same stocks are measured twice, so the samples are dependent. Use a paired t-test on the differences.
n = 16 pairs, so the standard error is sd ÷ √n = 2.0 ÷ 4 = 0.5.
t = (d̄ − 0) ÷ standard error = 0.8 ÷ 0.5 = 1.60.
Check the wrong options: 0.40 comes from dividing d̄ by sd (0.8 ÷ 2.0) and forgetting to divide sd by √n, and 6.40 comes from dividing by 2.0 ÷ 16 instead of 2.0 ÷ √16.
With df = 15, the two-tailed 5% critical value is 2.131, so 1.60 does not reject H0.

Answer: B. t = 1.60 with 15 df. At the 5% level, two-tailed, you fail to reject H0 because 1.60 is below 2.131.

Example 2

Using 38 monthly observations, the sample correlation between two funds' returns is 0.30. Testing H0: ρ = 0, the test statistic is closest to: A. 1.80, B. 1.89, C. 1.94.

Show the solution

Use t = r√(n − 2) ÷ √(1 − r²) with n − 2 = 36 degrees of freedom.
Numerator: 0.30 × √36 = 0.30 × 6 = 1.80.
Denominator: √(1 − 0.09) = √0.91 = 0.9539.
t = 1.80 ÷ 0.9539 = 1.887, which rounds to 1.89.
Option A forgets the denominator. Option C wrongly uses √38 instead of √36.
The two-tailed 5% critical value for 36 df is about 2.03, so 1.89 is not beyond it.

Answer: B. t ≈ 1.89 with 36 df. At the 5% level, two-tailed, you fail to reject H0, so the correlation is not significantly different from zero.

Exam tips

Look for the dependence clue. Words like same portfolio, before and after, or matched pairs point to a paired test, and the question often gives only d̄ and sd.
Memorise the df rule for each test: n − 1 for a mean, a paired test, and one variance; n1 + n2 − 2 for pooled means; n − 2 for correlation; two values for F.
Wrong options are usually the result of a classic slip: a missing √n, wrong df, or an inverted F ratio. Compute once, then check which option matches a slip.
For chi-square and F questions, check whether the test is one-tailed or two-tailed before reading the table. A two-tailed F-test splits α across both tails.
There is no penalty for a wrong answer, so always choose one. If you are short on time, eliminate options that give the wrong direction or an impossible value, such as a negative F.

Practice questions from Estimation and Hypothesis Testing

Tests of Means, Variances and Correlation in other exams

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

Tests of Means, Variances and Correlation: frequently asked questions

What is the difference between a z-test and a t-test?

Use a z-test when the population standard deviation is known. Use a t-test when you only have the sample standard deviation, which is the usual case. The t-distribution has fatter tails and depends on degrees of freedom, and it approaches the normal as n grows.

How do I do a paired comparisons test in CFA Level I?

Subtract each pair to get the differences, then find their mean d̄ and standard deviation sd. Compute t = (d̄ − 0) ÷ (sd ÷ √n) with n − 1 df, where n is the number of pairs. Compare it with the critical t-value and decide.

What is the chi-square test statistic for a variance?

The statistic is χ² = (n − 1)s² ÷ σ0², with n − 1 degrees of freedom. Here s² is the sample variance and σ0² the hypothesised variance. The distribution is skewed, so a two-tailed test uses a lower and an upper critical value.

How do I set up the F-test for equality of variances?

Divide the larger sample variance by the smaller, so F is at least 1. The df are n − 1 for each sample, with the numerator df tied to the larger variance. For a two-tailed test at significance α, use the critical value for α/2 in the upper tail.

What is the formula for testing whether a correlation is significant?

Use t = r√(n − 2) ÷ √(1 − r²) with n − 2 df, testing H0: ρ = 0. If |t| exceeds the critical value, reject H0. Significance does not mean the relationship is strong or causal.