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You are reading the 2026 curriculum (exams up to November 2026). Sitting in February 2027 or later? Go to the 2027 Quantitative Methods guide.

By Daniil Tarakanov Updated Oct 16, 20262026 curriculum

Is a correlation significant? The t-test of correlation

Test H₀: ρ = 0 with t = r√(n − 2) ÷ √(1 − r²), which has n − 2 degrees of freedom. If |t| exceeds the critical value, the correlation is statistically significant. The same correlation becomes more significant as the sample grows, so always look at both r and n.

6 min cram, 28 min full read. Based on the 2026 CFA® Program Level I curriculum, Quantitative Methods, Learning Module 9. This is the 2026 version of the lesson. Sitting in February 2027 or later? Read the updated 2027 version.

Tailor this to you: take the self-assessment

Before this: Hypothesis Testing, Statistical Measures of Asset Returns

On the exam

Explain parametric and nonparametric tests of the hypothesis that the population correlation coefficient equals zero, and determine whether the hypothesis is rejected at a given level of significance

Explain the parametric test of a correlation and determine whether H₀: ρ = 0 is rejected at a given level. Expect to compute t from r and n, then compare it with a critical value the question supplies. One option usually uses n − 1 or leaves out the √(1 − r²) term.

How deeply to learn each part

Know the idea: Understand it in general. It frames other topics, but you will not calculate with it.
Use it: Understand it well enough to solve a question with it. Recognize when it applies.
Memorize: Learn it by heart. The exam will not give it to you.

You know this one. The full lesson is folded away. The takeaways, formulas and checkpoint below are your quick refresher, and the checkpoint is a fair test of whether you still do.

You tested strong here. The full lesson is folded away. The takeaways, formulas and checkpoint below are your quick refresher, and the checkpoint is a fair test of whether you still know it.

What does the t-test of a correlation test?

It tests whether the true population correlation, $\rho$ , differs from zero, given only the sample correlation coefficient $r$ you measured. A sample of paired returns will almost never show exactly zero correlation, even for unrelated series. The question is whether the $r$ you see is too far from zero to be luck.

This is a parametric test of correlation : it assumes both variables are approximately normally distributed. The usual hypotheses are two-tailed.

H_0: \rho = 0 \qquad H_a: \rho \neq 0

How do you compute the t-statistic?

Scale the correlation by the square root of the degrees of freedom and divide by the square root of what the correlation leaves unexplained.

t-statistic for a correlation Memorize

t = \frac{r\sqrt{n-2}}{\sqrt{1-r^2}}, \qquad df = n - 2

The t-statistic for a correlation follows a t-distribution with $n-2$ degrees of freedom . If $|t|$ is larger than the critical value, reject $H_0$ and call the correlation statistically significant .

Worked example Airline shares and jet fuel

Monthly returns of an airline stock and changes in jet-fuel prices over 40 months show $r = 0.35$ . The two-tailed 5% critical value for 38 degrees of freedom is 2.024.

Numerator: $0.35 \times \sqrt{38} = 0.35 \times 6.164 = 2.158$ .
Denominator: $\sqrt{1 - 0.35^2} = \sqrt{0.8775} = 0.937$ .
$t = 2.158 \div 0.937 =$ 2.30.
2.30 > 2.024, so reject $H_0$ : the correlation is statistically significant.

On your calculator

0.35×38√x÷(1−0.35x²)√x= 2.3032

√x square root · x² square

0.35ENTER38g√x× 2.1575
1ENTER0.35ENTER×−g√x÷ 2.3032

ENTER enter (on the HP, puts the number on the stack) · g blue shift key · √x square root

Questions also turn the formula around: they give $t$ and $n$ and ask for $r$ , or give $r$ and $t$ and ask how many observations were used.

Rearrange itt-test of a correlation: solve for any one

The question asks for

A sample of 40 observations has a correlation of 0.35. The t-statistic for H₀: ρ = 0 is closest to:

Sample correlation, rSample size, n

t = \frac{r n - 2}{1 - r ^{2}}

0.35 × √38 ÷ √(1 − 0.1225) = 2.3032

Test statistic, t2.3032

On your calculator

0.35×38√x÷(1−0.35x²)√x=

√x square root · x² square

0.35ENTER38g√x×1ENTER0.35ENTER×−g√x÷

ENTER enter (on the HP, puts the number on the stack) · g blue shift key · √x square root

InteractiveIs this correlation significant? Move r and the sample size

Try this 0 of 3

Keep r at 0.35 and α at 5%. Find the smallest sample in which it is still significant.
Make a weak correlation (|r| of 0.15 or less, not zero) significant at 5%.
Load the worked example (n 40, r 0.35) and switch α to 1%.

Your sample: 40 months, r = 0.35

Smallest significant |r| at 5%

Smallest significant |r|
Your |r|: significant

Correlation r0.35

t-statistic2.30

df = n − 238

Critical t (5%)±2.024

Smallest significant |r|0.312

DecisionSignificant: reject H₀

Sample

Sample size n40

Sample correlation r0.35

Significance level α (two-tailed)

Show working

H₀: ρ = 0 (no correlation in the population); Hₐ: ρ ≠ 0. Two-tailed test.

t = r√(n − 2) ÷ √(1 − r²) = 0.35 × √38 ÷ √(1 − 0.1225) = 2.1575 ÷ 0.9367 = 2.3032.

Degrees of freedom = n − 2 = 40 − 2 = 38. Critical value at 5%, two-tailed: ±2.024.

|t| = 2.30 > 2.024, so reject H₀: the correlation is statistically significant (p = 0.0268).

Same test, turned around: the smallest significant |r| at n = 40 is t_c ÷ √(t_c² + n − 2) = 2.024 ÷ √(4.0982 + 38) = 0.312.

A significant correlation still says nothing about cause: both series may simply respond to a third factor.

t-statistic on your calculator

0.35×38√x÷(1−0.35x²)√x=t = 2.3032

√x square root · x² square

0.35ENTER38g√x×1ENTER0.35ENTER×−g√x÷t = 2.3032

ENTER enter (on the HP, puts the number on the stack) · g blue shift key · √x square root

Display: 2.3032

With enough data, even a weak correlation is statistically significant; whether it matters economically is a separate question.

Why does sample size matter so much?

Because $n$ sits inside the numerator. Keep $r$ fixed and every extra observation makes $t$ larger, so a weak correlation that is noise in a small sample becomes significant in a big one.

Reading the t-test of a correlation

Compute t from r and nH₀: ρ = 0, df = n − 2

|t| greater than the critical value Memorize a large t is far from zero
- Reject H₀ the correlation is statistically significant
|t| at or below the critical value Memorize the sample could easily be noise
- Fail to reject H₀ not significant, which is not the same as proving ρ = 0

Exam tip Failing to reject is not proof that the correlation is zero. It only means this sample is not strong enough evidence. Never write “accept H₀”.

Go deeper Statistical significance is not economic significance

With 500 monthly observations, a correlation of just 0.12 gives $t = 0.12 \times \sqrt{498} \div \sqrt{1 - 0.0144} = 2.70$ , comfortably significant. Yet $r^2 = 1.4\%$ means the other variable explains almost none of the variation. A significant correlation can still be too weak to build a trade or a hedge on. Conversely, a large correlation in a tiny sample can fail the test. Report both the size of $r$ and the evidence behind it. And remember the test says nothing about cause: two series can move together because of a third factor or by coincidence.

Go deeper Link to simple regression

If you regress one variable on the other with a single explanatory variable, the t-statistic of the slope coefficient is exactly the same number as the t-statistic for the correlation, with the same $n-2$ degrees of freedom. Testing “the slope is zero” and “the correlation is zero” are the same hypothesis seen from two sides. You meet the regression version in the next module.

Memorize this

MemorizeThe correlation t-test

Item	Value
Null hypothesis	$H_0: \rho = 0$ (two-tailed, $H_a: \rho \neq 0$ )
Statistic	$t = r\sqrt{n-2} \,/\, \sqrt{1-r^2}$
Degrees of freedom	$n - 2$
Decision	$\|t\| >$ critical value → reject $H_0$

Memory hook Same df as a one-variable regression, which estimates two coefficients (intercept and slope): n − 2. The root of 1 − r² is the "noise" under the line.

Know it cold Reject only when |t| beats the critical value the question gives. Never "accept" H₀.

Reading the t-test of a correlation

Compute t from r and nH₀: ρ = 0, df = n − 2

|t| greater than the critical value Memorize a large t is far from zero
- Reject H₀ the correlation is statistically significant
|t| at or below the critical value Memorize the sample could easily be noise
- Fail to reject H₀ not significant, which is not the same as proving ρ = 0

t-statistic for a correlation

t = \dfrac{r\sqrt{n-2}}{\sqrt{1-r^2}},\quad df = n - 2

Quick example

A correlation of 0.52 from 27 observations. t = 0.52 × √25 ÷ √(1 − 0.2704) = 2.60 ÷ 0.854 = 3.04, with 25 degrees of freedom.

On your calculator

0.52×25√x÷(1−0.52x²)√x= 3.0439

√x square root · x² square

0.52ENTER25g√x×1ENTER0.52ENTER×−g√x÷ 3.0439

ENTER enter (on the HP, puts the number on the stack) · g blue shift key · √x square root

How it is asked

Compute t, then decide

r = 0.35 from 40 observations, critical value 2.024. Conclusion?

t = 0.35 × √38 ÷ √(1 − 0.1225) = 2.30 > 2.024: reject H₀. Option using n − 1 gives 2.33, a near-miss trap.
Same r, different n

The correlation is unchanged but the sample doubles. What happens to t?

√(n − 2) grows, so t rises and the correlation is more likely to be significant.
Degrees of freedom

Correlation of two return series, 62 observations. Degrees of freedom?

n − 2 = 60. The trap is 61 (n − 1).

Exam traps

Using n − 1 (or n) instead of n − 2 degrees of freedom.
Forgetting the √(1 − r²) in the denominator.
Concluding that a significant correlation means one variable causes the other.
Treating a small correlation from a huge sample as economically important: significant is not the same as large.

Checkpoint

4 exam-style questions. About 90 seconds each.

1A sample of 62 observations has a correlation of 0.28 between two return series. The t-statistic for H₀: ρ = 0 is closest to:
2An analyst tests H₀: ρ = 0 with 62 observations and gets a t-statistic of 2.26. The two-tailed 5% critical value is 2.000. The analyst should:
3Holding the sample correlation constant, increasing the sample size will most likely:
4An analyst finds a correlation of 0.35 between two series with 20 observations. With a two-tailed 5% critical value of 2.101, the t-statistic and the conclusion are closest to:

Key takeaways

t = r√(n − 2) ÷ √(1 − r²), with n − 2 degrees of freedom.
Bigger samples make the same r more significant.
Significance is not causation, and not size.

t-statistic for a correlation

t = \dfrac{r\sqrt{n-2}}{\sqrt{1-r^2}},\quad df = n - 2

Quick example

A correlation of 0.52 from 27 observations. t = 0.52 × √25 ÷ √(1 − 0.2704) = 2.60 ÷ 0.854 = 3.04, with 25 degrees of freedom.

On your calculator

0.52×25√x÷(1−0.52x²)√x= 3.0439

√x square root · x² square

0.52ENTER25g√x×1ENTER0.52ENTER×−g√x÷ 3.0439

ENTER enter (on the HP, puts the number on the stack) · g blue shift key · √x square root

Next lesson Spearman rank correlation: the nonparametric test

Questions about this lesson?

Where this comes back

Simple Linear Regression Quantitative Methods
Portfolio Risk and Return: Part I Portfolio Construction
Market Efficiency Equity Investments

Questions candidates ask

How do you test if a correlation is significant?

State H₀: ρ = 0, compute t = r√(n − 2) ÷ √(1 − r²) with n − 2 degrees of freedom, and compare |t| with the critical value. If |t| is larger, reject H₀ and call the correlation statistically significant.

Why does the correlation t-test use n − 2 degrees of freedom?

Two quantities are estimated from the data before the test can be run, which uses up two degrees of freedom and leaves n − 2. It is the same count as in simple regression, where the intercept and the slope are estimated, and the correlation test gives the same t as the slope test.

Can a small correlation be statistically significant?

Yes. With a large sample, even r = 0.12 gives a big t because √(n − 2) is large. Significance says the correlation is unlikely to be zero, not that it is large or useful. Judge economic importance separately.

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