Answer

Why does a t test use the t distribution and ANOVA the F distribution?

Inspired by a question on Cross Validated · October 21, 2026

hypothesis testingdistributionsanovat test

The short answer

A hypothesis test needs to know how its statistic behaves when nothing is going on, so that it can say whether the observed value is unusual. That behaviour is called the null distribution (or reference distribution) of the statistic. It is not chosen; it is derived. You write down the statistic, write down the assumptions (for example, independent observations from normal distributions with equal variances), assume the null is true, and use probability theory to find the distribution of the statistic. The t, F and chi-square distributions are simply the names of the answers that keep coming up.

So the t test uses a t distribution because its statistic is built in a way that produces one, and the ANOVA uses an F distribution because its statistic is a ratio of two variance estimates. When the maths has no exact answer, statisticians fall back on large-sample approximations or on simulating the null distribution directly.

Where the common distributions come from

Almost every classical test is built from a few pieces, and each piece has a known distribution when the data are normal:

• Standard normal (z). A sample mean minus its null value, divided by its true standard error, is exactly standard normal. This only works if you know the population standard deviation, which you almost never do.
• Chi-square. Add up the squares of df independent standard normal variables and you get a chi-square variable with df degrees of freedom. A sample variance is one of these in disguise: (n − 1)s²/σ² follows a chi-square distribution with n − 1 degrees of freedom.
• t. A standard normal divided by the square root of an independent chi-square over its degrees of freedom follows a t distribution. That is exactly what you get when the true standard error is replaced by one estimated from the sample. William Gosset ("Student") worked this out in 1908.
• Why t has heavier tails. The estimated standard error is itself uncertain, which spreads the statistic out, most noticeably when there are few degrees of freedom.
• F. The ratio of two independent chi-square variables, each divided by its degrees of freedom, follows an F distribution, named by George Snedecor in honour of Ronald Fisher.
• Why ANOVA uses F. When the null is true, the between-groups and within-groups mean squares are two independent estimates of the same variance, so their ratio is F with k − 1 and N − k degrees of freedom.

These pieces are connected. A t statistic squared is an F statistic with 1 numerator degree of freedom, which is why a two-group ANOVA and a pooled two-sample t test always give the same p-value (see t test vs ANOVA with two groups). And the degrees of freedom are not decoration: they tell you which member of the family to use, because each one describes how much information went into a variance estimate (see what degrees of freedom are).

See it in R and Python

The code checks each derivation by brute force. It generates 10,000 data sets in which the null hypothesis is true and compares the simulated statistics with the theoretical distribution: a pooled two-sample t test with 5 per group (8 degrees of freedom), a one-way ANOVA with 3 groups of 5 (2 and 12 degrees of freedom), the t² = F identity, a chi-square goodness-of-fit test for 60 rolls of a fair die, and finally the variance-ratio F test applied to data that break its normality assumption.

set.seed(70820)
reps <- 10000

# 1. Two-sample t statistic (5 per group) when the null is true
t_stat <- replicate(reps, t.test(rnorm(5), rnorm(5), var.equal = TRUE)$statistic)
probs <- c(0.90, 0.95, 0.975, 0.995)
round(rbind(simulated = quantile(t_stat, probs),
            t_8 = qt(probs, df = 8), normal = qnorm(probs)), 3)
c(reject_with_normal = mean(abs(t_stat) > qnorm(0.975)),
  reject_with_t8 = mean(abs(t_stat) > qt(0.975, df = 8)))

# 2. One-way ANOVA F statistic (3 groups of 5) when the null is true
g <- factor(rep(1:3, each = 5))
F_stat <- replicate(reps, oneway.test(rnorm(15) ~ g, var.equal = TRUE)$statistic)
round(rbind(simulated = quantile(F_stat, c(0.90, 0.95, 0.99)),
            F_2_12 = qf(c(0.90, 0.95, 0.99), 2, 12)), 3)
mean(F_stat > qf(0.95, 2, 12))

# 3. With two groups, F is exactly t squared
y <- rnorm(10); g2 <- factor(rep(1:2, each = 5))
c(t_squared = unname(t.test(y ~ g2, var.equal = TRUE)$statistic^2),
  F = unname(oneway.test(y ~ g2, var.equal = TRUE)$statistic))

# 4. Pearson chi-square for 60 rolls of a fair die: chi-square(5) is an approximation
chi_stat <- replicate(reps, {
  obs <- tabulate(sample(6, 60, replace = TRUE), 6)
  sum((obs - 10)^2 / 10)
})
c(reject_rate = mean(chi_stat > qchisq(0.95, 5)))

# 5. Same statistic, broken assumption: variance-ratio F test, 20 per group
p_normal <- replicate(reps, var.test(rnorm(20), rnorm(20))$p.value)
p_skewed <- replicate(reps, var.test(rexp(20), rexp(20))$p.value)
c(normal_data = mean(p_normal < 0.05), exponential_data = mean(p_skewed < 0.05))

import numpy as np
from scipy import stats

rng = np.random.default_rng(70820)
reps = 10000

# 1. Two-sample t statistic (5 per group) when the null is true
t_stat = stats.ttest_ind(rng.normal(size=(reps, 5)), rng.normal(size=(reps, 5)), axis=1).statistic
probs = [0.90, 0.95, 0.975, 0.995]
print("simulated", np.round(np.quantile(t_stat, probs), 3))
print("t_8      ", np.round(stats.t.ppf(probs, 8), 3))
print("normal   ", np.round(stats.norm.ppf(probs), 3))
print("reject_with_normal", np.mean(np.abs(t_stat) > stats.norm.ppf(0.975)),
      "reject_with_t8", np.mean(np.abs(t_stat) > stats.t.ppf(0.975, 8)))

# 2. One-way ANOVA F statistic (3 groups of 5) when the null is true
y = rng.normal(size=(reps, 3, 5))
F_stat = stats.f_oneway(y[:, 0], y[:, 1], y[:, 2], axis=1).statistic
print("simulated", np.round(np.quantile(F_stat, [0.90, 0.95, 0.99]), 3))
print("F_2_12   ", np.round(stats.f.ppf([0.90, 0.95, 0.99], 2, 12), 3))
print("reject", np.mean(F_stat > stats.f.ppf(0.95, 2, 12)))

# 3. With two groups, F is exactly t squared
y2 = rng.normal(size=10)
print("t_squared", stats.ttest_ind(y2[:5], y2[5:]).statistic ** 2,
      "F", stats.f_oneway(y2[:5], y2[5:]).statistic)

# 4. Pearson chi-square for 60 rolls of a fair die: chi-square(5) is an approximation
obs = np.array([np.bincount(r, minlength=6) for r in rng.integers(0, 6, size=(reps, 60))])
chi_stat = ((obs - 10) ** 2 / 10).sum(axis=1)
print("reject_rate", np.mean(chi_stat > stats.chi2.ppf(0.95, 5)))

# 5. Same statistic, broken assumption: variance-ratio F test, 20 per group
def var_test_p(a, b):
    ratio = a.var(axis=1, ddof=1) / b.var(axis=1, ddof=1)
    p = stats.f.cdf(ratio, 19, 19)
    return np.minimum(1, 2 * np.minimum(p, 1 - p))   # two-sided, as in R's var.test

p_normal = var_test_p(rng.normal(size=(reps, 20)), rng.normal(size=(reps, 20)))
p_skewed = var_test_p(rng.exponential(size=(reps, 20)), rng.exponential(size=(reps, 20)))
print("normal_data", np.mean(p_normal < 0.05), "exponential_data", np.mean(p_skewed < 0.05))

            90%   95% 97.5% 99.5%
simulated 1.387 1.844 2.299 3.370
t_8       1.397 1.860 2.306 3.355
normal    1.282 1.645 1.960 2.576
reject_with_normal     reject_with_t8 
            0.0807             0.0484 
            90%   95%   99%
simulated 2.792 3.840 6.964
F_2_12    2.807 3.885 6.927
[1] 0.0485
t_squared         F 
0.6648932 0.6648932 
reject_rate 
     0.0477 
     normal_data exponential_data 
          0.0549           0.2720

All figures come from one seeded run of the R code. The theoretical quantiles involve no randomness, and Python reproduces them exactly. Python's random numbers differ from R's, so its simulated values differ slightly (for example 8.5% false positives with the normal cutoff instead of 8.1%, and 27.4% instead of 27.2% for the variance test on skewed data), but they show the same pattern.

What the numbers show

• The derived distribution matches reality. The simulated t statistics had a 97.5th percentile of 2.299, against 2.306 for the t distribution with 8 degrees of freedom. The simulated F statistics had a 95th percentile of 3.840, against 3.885 for F(2, 12). The small gaps are simulation noise.
• The wrong member of the family costs you. The normal distribution's 97.5th percentile is 1.960, well below 2.306. Using it as the cutoff with only 5 per group gave a false-positive rate of 8.1% instead of 5%; the t cutoff gave 4.8%. The difference between t and normal shrinks as the degrees of freedom grow.
• The families are linked. With two groups, the squared t statistic and the F statistic were identical (0.6649 for both).
• Some reference distributions are approximations. The chi-square goodness-of-fit statistic is only approximately chi-square, with the approximation improving as expected counts grow. With 10 expected rolls per face it worked well: 4.8% false positives at the 5% level.
• The derivation is only as good as its assumptions. The variance-ratio F test rejected 5.5% of the time with normal data, but 27.2% of the time with exponential data, even though the null (equal variances) was true in both cases. Its F distribution relies heavily on normality, unlike the ANOVA F test, which is fairly robust to non-normal data when group sizes are equal.

When there is no exact formula

Exact results like Gosset's are the exception. For most modern methods statisticians use one of three routes:

• Large-sample theory. The central limit theorem makes many statistics approximately normal in large samples, which is where z tests for proportions and Wald tests in regression come from.
• Likelihood ratios. Wilks's theorem shows that, under regularity conditions, twice the log of a likelihood ratio is approximately chi-square, with degrees of freedom equal to the number of parameters the null fixes. That is why likelihood ratio tests in logistic regression report chi-square statistics.
• Permutation tests. If the null says group labels do not matter, you can shuffle the labels many times, recompute the statistic each time, and use the shuffled values as the reference distribution. This needs no distributional formula at all.
• Simulation or the bootstrap. For awkward statistics you can simulate data under the null, or resample the data you have (see why bootstrapping works), and read the cutoffs off the simulated values, exactly as the R code above does.

Whichever route a test takes, the practical lesson is the same: the reference distribution is part of the test's assumptions. When the assumptions clearly fail, choose a test whose null distribution does not depend on them. Our test chooser and the DASS blog can help you pick one.

How to report t and F statistics in APA style (7th edition)

The degrees of freedom in parentheses tell the reader which reference distribution you used, so always include them, along with the exact p-value and an effect size. With placeholders for your own results:

"Scores were higher in the treatment group (M = x.xx, SD = x.xx) than in the control group (M = x.xx, SD = x.xx), t(8) = x.xx, p = .xxx, d = x.xx."

"Mean scores differed across the three conditions, F(2, 12) = x.xx, p = .xxx, η² = .xx." If you used a permutation or bootstrap test instead, say so in the method section and give the number of resamples, for example "p values were obtained from 10,000 random permutations of the group labels".

Related tools and guides

• Check a reported p-value
• ANOVA power and sample size calculator
• APA 7 formatter for ANOVA
• How to interpret the F statistic and p value in a one-way ANOVA
• What do p values and t values mean?
• What are degrees of freedom in statistics?
• t test vs ANOVA with two groups
• Which statistical test should I use?

More answered questions

• Kolmogorov-Smirnov, t test or Mann-Whitney: which detects a change from a baseline?
• Significant ANOVA but no significant pairwise comparisons: how can that happen?
• t tests vs the F test in regression: what is the difference?

Written by AskStats with AI assistance. This is general information, not advice for your specific data or study. When the results matter, check your approach with a qualified statistician.