Do I need a nonparametric test if my data are not normally distributed?

Not automatically. The t test and ANOVA assume the sampling distribution of the mean is approximately normal, not that the raw values are. With reasonable sample sizes the central limit theorem makes that hold even when the data are skewed. A rank test is the better choice when skew is strong and the sample is small.

Do parametric and nonparametric tests answer the same question?

No. A t test compares means; a Mann-Whitney test asks whether one distribution tends to produce larger values than the other, which is a different null hypothesis. In skewed data they can point in different directions, and that difference is informative rather than an error.

Is a nonparametric test always safer?

No. Rank tests discard the magnitude of the values, so when the data are genuinely well behaved they have less power than the parametric alternative. Choosing a rank test defensively costs you the ability to detect real effects.

What is the best way to decide?

Look at the shape and the sample size together, and report the assumption check either way. If the two tests agree, the choice does not matter and saying so is enough. If they disagree, that disagreement belongs in the write-up.

Parametric or nonparametric?

"My data are not normally distributed, so I need a nonparametric test" is the most common wrong inference in applied statistics. The t test does not assume your data are normal. It assumes the mean would be normal if you repeated the study — and that holds far more often than a histogram suggests.

The short answer

Judge the departure and the sample size together. Strong skew with n = 10: use a rank test. Mild skew with n = 80: the t test is fine. When the two agree, report that they agree. When they disagree, you have learned something about your data and it belongs in the write-up.

What "parametric" is actually claiming

A t test estimates a mean and relies on a known sampling distribution to convert the estimate into a p value. That distribution is approximately normal by the central limit theorem, regardless of the shape of the raw data. The assumption is about the sampling distribution, not the histogram.

A rank test makes a weaker claim: it asks whether one group tends to produce larger values than the other, using only the ordering. Fewer assumptions, but also less information used — the magnitude of each value is discarded.

Where they disagree, on real numbers

Two groups drawn from a strongly right-skewed distribution, eighty per group — the shape you get with cost data, reaction times, or anything bounded at zero.

Test	Statistic	p
Welch independent-samples t test	t(145.61) = −0.35	.723
Mann–Whitney U	U = 2513, z = −2.34	.019

Same data. One test finds nothing, the other finds a significant shift. Neither is broken.

The t test is asking about the difference in means, and in a heavily skewed distribution the mean is dominated by a few extreme values, so its standard error is large. The rank test is asking whether values from one group tend to sit higher, which the skew does not obscure.

This is the case where the choice is the answer. If your research question is about typical values — "do these patients usually recover faster" — the rank test is closer to what you meant. If it is about the total or the average, the mean is the quantity of interest and the t test is measuring the right thing, even when it fails to reach significance.

The small-sample case

The opposite situation: the same shape, but ten per group. There the t test and the rank test both come out non-significant, with p values far apart and neither meaningful. That is not a contradiction, it is underpower. With ten observations from a skewed distribution there is simply not enough information, and the right response is to collect more data rather than to choose the test whose p value you prefer.

Three things that decide it in practice

Sample size. Below about 30 per group, departures from normality matter more; above that, the sampling distribution is doing the work.
Direction of the skew. Mild skew is tolerable. Extreme skew, or heavy tails driven by outliers, is where means become unstable.
What the question is about. Means and totals belong to parametric tests; "typically higher" belongs to ranks. This is the part people skip.

Never do this

Do not run both tests and report whichever gives the smaller p value. That is not robustness checking, it is choosing your answer after seeing it, and it inflates the error rate in a way no correction can recover.

The defensible version: decide from the design and the shape before looking at the comparison, state the decision, and — if a reviewer might object — report the alternative in a footnote.

What the rank tests are called

Parametric	Rank-based counterpart	What the rank test asks
Independent t test	Mann–Whitney U	Does one group tend to score higher?
Paired t test	Wilcoxon signed-rank	Do the differences tend to be positive?
One-way ANOVA	Kruskal–Wallis	Does any group tend to score higher?
Repeated-measures ANOVA	Friedman	Does any condition tend to score higher?

Frequently asked

Do I need a nonparametric test if my data are not normal?

Not automatically. The assumption is about the sampling distribution of the mean, which is far more forgiving than the histogram. Strong skew with a small sample is the real warning sign.

Do the two kinds of test answer the same question?

No. One compares means, the other asks about the ordering of values. That is why they can disagree on the same data without either being wrong.

Is nonparametric always safer?

It is more robust but less powerful. On well-behaved data it can miss effects the t test would have found.

What if a reviewer insists on normality?

Report the check, report the effect size and confidence interval, and show that the conclusion does not depend on the choice — either because both tests agree, or because you can say why the mean is the quantity of interest.

Which test should I use? · t test vs ANOVA · Run a t test · Method notes