Which statistical test do you use?

Start here

What decides which test you use?

Every test asks the same question: could chance alone explain the difference you see? Which test answers it depends on four facts about your data.

1. Type of outcome: a measurement, ordered grades, or categories.
2. Number of groups: two, or three or more.
3. Independent or paired: different patients in each group, or the same patients measured twice (or matched pairs).
4. Distribution, for measurements: roughly normal, or skewed with outliers.

Two types of tests

Parametric or non-parametric: what is the difference?

A parametric test assumes the data come from a known distribution, usually the normal distribution, described by parameters such as the mean and SD. A non-parametric test does not assume a distribution. It works on ranks or signs instead of the raw values.

Assumptions of parametric tests

1. Normality: the outcome is roughly normal in each group (for paired tests, the differences). Look at a histogram, and use the Shapiro-Wilk test if needed.
2. Equal variances: the groups have similar spread. Check with the Levene test.
3. Independence: one patient's value gives no information about another's, except where pairing is part of the design.
4. Scale: the outcome is a measurement on an interval or ratio scale, so that a mean has a meaning.

Non-parametric tests

How can a test work without the actual values?

It replaces each value with its rank. The smallest value gets rank 1, the next rank 2, and tied values share the average of their ranks.

Interactive

Move one long stay and watch the t-test and the Mann-Whitney test

The t-test uses the actual values, so one long stay raises the mean and the SD of its group. The Mann-Whitney test uses ranks, so a stay longer than all the others counts as rank 16 whatever its length.

Decide

Which test fits your study?

Answer the questions in order. Each answer links to its worked example.

Comparison

Parametric or non-parametric: which one should you use?

Use a parametric test when its assumptions hold, because it uses the actual values. Use a non-parametric test when they do not hold.

The chi-square test is usually grouped with the non-parametric tests because it works on counts and assumes no distribution for a measurement.

Comparison

Paired or unpaired: are the two sets of values linked?

Values are paired when each value in one set belongs with one value in the other: the same patient before and after, two eyes of one patient, or a case matched to a control.

Interactive

Change how much the women differ and watch the paired and unpaired p-values

The paired test works on each patient's own change, so the variation between patients does not enter its standard error. The unpaired test keeps that variation.

The formula

Student's t-test: is the difference between two means larger than chance would produce?

t is the difference between the means divided by its standard error. The larger t is, the less likely it is that chance alone produced the difference.

t = (mean₁ − mean₂) ÷ SE of the difference

Three types of t-test

• One-sample t-test: one sample mean against a reference value. t = (mean − reference) ÷ (SD ÷ √n), df = n − 1.
• Unpaired (independent samples) t-test: two independent groups. df = n₁ + n₂ − 2.
• Paired t-test: two measurements on the same patients. t = mean difference ÷ (SD of the differences ÷ √n), df = number of pairs − 1.

Worked example 1: unpaired t-test in two small groups

Drug A: 12, 15, 9, 14, 11, 16, 10, 13 mmHg. Drug B: 8, 10, 6, 11, 7, 9, 12, 5 mmHg.

1. Means: A = 100 ÷ 8 = 12.5 and B = 68 ÷ 8 = 8.5. Difference = 4.0 mmHg.
2. Squared deviations from the mean in group A: 0.25, 6.25, 12.25, 2.25, 2.25, 12.25, 6.25, 0.25. They add up to 42. Group B also adds up to 42.
3. SD = √(42 ÷ 7) = √6 = 2.45 mmHg in each group.
4. Pooled SD = √[(7 × 6 + 7 × 6) ÷ 14] = 2.45 mmHg.
5. SE of the difference = 2.45 × √(1/8 + 1/8) = 1.225.
6. t = 4.0 ÷ 1.225 = 3.27, with 8 + 8 − 2 = 14 degrees of freedom.
7. The critical t for 14 df at 5% (two-sided) is 2.145. 3.27 is larger, so p < 0.05. The exact p = 0.006.

pooled SD

√{[(n₁ − 1)SD₁² + (n₂ − 1)SD₂²] ÷ (n₁ + n₂ − 2)}

SE

pooled SD × √(1/n₁ + 1/n₂)

df

degrees of freedom, used to read the critical t from the table

t = 4.0 ÷ 1.225 = 3.27, df 14, p = 0.006 → 95% CI for the difference 1.37 to 6.63 mmHg

Worked example 2: paired t-test

Pain scores before: 4, 9, 6, 8, 5, 7, 3, 8. After 1 hour: 3, 7, 5, 6, 3, 6, 2, 7.

1. Differences: 1, 2, 1, 2, 2, 1, 1, 1. Mean difference = 11 ÷ 8 = 1.375.
2. SD of the differences = 0.518. SE = 0.518 ÷ √8 = 0.183.
3. t = 1.375 ÷ 0.183 = 7.51, df = 7, p = 0.0001. 95% CI for the mean fall 0.94 to 1.81 points.
4. The same 16 numbers put through an unpaired t-test give t = 1.35, p = 0.20.

Worked example 3: one-sample t-test

Made-up teaching data. Hb in 10 adolescent girls: 11.2, 10.8, 12.1, 11.5, 10.9, 11.8, 11.0, 12.3, 10.6, 11.4 g/dL. Does the mean differ from 12 g/dL?

t = (11.36 − 12) ÷ (0.568 ÷ √10) = −0.64 ÷ 0.180 = −3.56, df 9 → p = 0.006

Why the t-test matters

• It gives correct p-values for small samples. Methods for small samples were the problem W. S. Gosset worked on as a brewer at Guinness.
• It does not need the population SD, so it can be used at any sample size.
• It is the basis of the confidence interval for a mean and for a difference between two means.
• It is one of the most used statistical tests in medicine: comparing a sample with a standard, two treatments, or values before and after treatment.

Assumptions: a measurement outcome, roughly normal (for the paired test, the differences), independent patients, and for the unpaired test similar variances. If the variances clearly differ, use the t-test for unequal variances (Welch test).

The formula

Three or more groups: why not run a t-test for each pair?

Each t-test carries a 5% chance of a false positive. The more tests you run, the higher the chance that at least one of them is falsely significant.

The principle of ANOVA

Analysis of variance (ANOVA) compares means by comparing two kinds of variation. If all groups have the same true mean, the group means should differ from each other only about as much as individual patients differ within a group. F is the ratio of the two.

F = variance between groups (mean square between) ÷ variance within groups (mean square within)

MS between

Σ nᵢ (group mean − grand mean)² ÷ (k − 1)

MS within

Σ (each value − its group mean)² ÷ (N − k)

k, N

number of groups, total number of patients

If H₀ is true, both mean squares estimate the same patient-to-patient variance, so F is expected to be close to 1. Group means that truly differ make the mean square between groups larger. Only large values of F count against H₀, so the p-value is the area in the upper tail of the F distribution, although the alternative hypothesis (at least one mean differs) has no direction. F cannot be negative. With only two groups, F = t²: in worked example 1 below, t = 3.266 and F = t² = 10.67, with the same p = 0.006.

Worked example 4: one-way ANOVA

Duration of analgesia (hours). Drug alone: 5.2, 6.1, 5.8, 4.9, 6.4, 5.6 (mean 5.67). With adjuvant A: 7.9, 8.4, 7.2, 8.8, 7.6, 8.1 (mean 8.00). With adjuvant B: 8.6, 9.4, 8.1, 9.9, 8.8, 9.2 (mean 9.00). Grand mean 7.56.

F = 17.56 ÷ 0.346 = 50.7. The critical F for (2, 15) df at 5% is 3.68, so at least one mean differs. The Levene test gives p = 0.91, so the variances are similar. Tukey's post hoc test shows that all three pairs differ, including adjuvant A against adjuvant B (difference 1.0 hour, p = 0.026).

Types of ANOVA

• One-way ANOVA: one grouping factor with three or more independent groups. Example: HbA1c with three drugs.
• Two-way (factorial) ANOVA: two factors at once, such as drug and sex. It tests the effect of each factor and their interaction, that is, whether the effect of the drug differs between men and women. In a factorial design every level of one factor is combined with every level of the other (2 drugs × 2 sexes = 4 groups). Main effects, interaction plots and worked two-factor designs are in the Practical Statistics chapter on two-way ANOVA.
• Repeated measures ANOVA: the same patients at three or more time points. It extends the paired t-test and needs sphericity (Mauchly test). If sphericity fails, apply the Greenhouse-Geisser or Huynh-Feldt correction. The analysis of repeated measurements is covered further in the Practical Statistics chapter on repeated measures.
• Mixed (two-way repeated measures) ANOVA: one factor between patients (treatment group) and one within patients (time).
• ANCOVA: ANOVA adjusted for a continuous covariate, such as the baseline value.

Assumptions of ANOVA

1. Observations are independent (except the planned repetition in repeated measures designs).
2. The outcome is roughly normal in each group.
3. Variances are similar across groups (Levene test). If not, use a post hoc test for unequal variances, such as Games-Howell.
4. For repeated measures, sphericity holds (Mauchly test).
5. The outcome is a measurement. If normality fails, use the Kruskal-Wallis or Friedman test.

How to check normality and equal variances, and what to do when they fail, is in the Practical Statistics chapter on checking assumptions.

After a significant F: post hoc tests

A significant F shows that at least one mean differs, not which one. Post hoc tests compare pairs while controlling the false-positive rate: Tukey, Scheffé, Holm, and Bonferroni, which divides α by the number of comparisons (0.05 ÷ 3 = 0.017). Dunnett's test compares each treatment group with one control group, instead of comparing every pair. How to choose among these tests is covered in the Practical Statistics chapter on multiple comparisons.

Interactive

Move the group means apart or spread the patients out, and watch F

F compares how far the group means lie from the grand mean with how far patients lie from their own group mean. The same distance between the means can give a large or a small F, depending on the spread within the groups.

Non-parametric tests

Mann-Whitney, Wilcoxon, Kruskal-Wallis and Friedman: what does each one do?

Each rank test takes the place of a parametric test when the data are skewed, have outliers, or are ordinal.

The same teaching data, analysed with rank tests

• Mann-Whitney U on length of stay: medians 3 and 5.5 days, U = 8.5, p = 0.015. The t-test on the same data gave p = 0.75.
• Wilcoxon signed-rank on the pain scores: all 8 differences are positive, p = 0.008.
• Kruskal-Wallis on the analgesia data: H = 13.69, df = 2, p = 0.001. One-way ANOVA on the same data gave p < 0.001.

Why non-parametric tests are useful

• They do not need the normality assumption, so they are less likely to give a wrong conclusion when that assumption fails.
• Outliers have little effect, because an extreme value is only the highest or lowest rank.
• They are the correct choice for ordinal data: pain grades, Likert items, Apgar scores.
• They can be used in small samples, where normality cannot be checked.
• When data are normal they lose little: the asymptotic relative efficiency of the Mann-Whitney test against the t-test is 0.955.

Limitations: ranks use less of the information in the data, confidence intervals are harder to calculate, and there are fewer methods for complex designs. The result describes ranks or medians, not a difference in means.

A significant Mann-Whitney test does not by itself show that the medians differ. The test compares the ranks of the two groups, so groups with similar medians but different spread can also give a small p-value. Hart made this point in a 2001 BMJ statistics note, whose title says that differences in spread can be important for this test. Report the median and IQR of each group, so that readers can see which one differs.

The formula

Comparing counts: the chi-square test

Chi-square compares the counts you observed with the counts you would expect if there were no difference between the groups.

χ² = Σ (O − E)² ÷ E

O

observed count in a cell

E

expected count = row total × column total ÷ grand total

df

(rows − 1) × (columns − 1)

Applications of the chi-square test

1. Test of association (independence): are two categorical variables related? Smoking and wound infection.
2. Comparison of proportions (homogeneity): cure rates in two or more treatment groups.
3. Goodness of fit: do observed counts match a stated distribution? Admissions per weekday, or a genetic ratio.
4. Trend: chi-square for trend across ordered groups, such as dose levels.
5. Paired proportions: the related McNemar test, for the same patients classified twice.

Worked example 5: test of association in a 2 × 2 table

1. Expected infections in each row = 100 × 26 ÷ 200 = 13. Expected non-infected = 100 × 174 ÷ 200 = 87.
2. Infected cells: (8 − 13)² ÷ 13 = 1.923 and (18 − 13)² ÷ 13 = 1.923. Non-infected cells: (92 − 87)² ÷ 87 = 0.287 each.
3. χ² = 1.923 + 1.923 + 0.287 + 0.287 = 4.42, with (2 − 1) × (2 − 1) = 1 df.
4. The critical χ² for 1 df at 5% is 3.84. 4.42 is larger: p = 0.036. Infection rates were 8% and 18%.

Yates' continuity correction

Counts are whole numbers, but the chi-square distribution is continuous. For 2 × 2 tables, Yates proposed subtracting 0.5 from each |O − E| before squaring.

χ²(Yates) = Σ (|O − E| − 0.5)² ÷ E

In example 5: (5 − 0.5)² ÷ 13 = 1.558 for each infected cell and 0.233 for each non-infected cell, so χ² = 3.58 and p = 0.058. The correction moved p above 0.05. The Yates-corrected test is overly conservative, and many authors advise against it. State in the protocol which version you will use, before you see the data.

When to use Fisher's exact test

The chi-square p-value is an approximation. It needs adequate expected counts: no more than 20% of cells with an expected count below 5, and no cell below 1. When a table fails this rule, use Fisher's exact test, which calculates the exact probability and is valid at any sample size.

Made-up teaching data. Expected complications = 10 × 6 ÷ 20 = 3 in each row, so 2 of the 4 cells (50%) are below 5. Fisher's exact test gives p = 0.14. The uncorrected chi-square would have given χ² = 3.81, p = 0.051.

Worked example 6: goodness of fit

Made-up teaching data: 140 admissions for acute myocardial infarction by weekday. Monday 32, Tuesday 21, Wednesday 18, Thursday 17, Friday 19, Saturday 16, Sunday 17. If admissions were spread evenly, each day would expect 140 ÷ 7 = 20.

1. (O − E)² ÷ E: Monday (32 − 20)² ÷ 20 = 7.2. Tuesday 0.05, Wednesday 0.2, Thursday 0.45, Friday 0.05, Saturday 0.8, Sunday 0.45.
2. χ² = 9.2, with 7 − 1 = 6 df. The critical χ² for 6 df at 5% is 12.59.
3. p = 0.16. These 140 admissions give no evidence that admissions differ by weekday, despite the high Monday count.

Worked example 7: an R × C table

Made-up teaching data: anaemia in adolescent girls by family diet, 80 girls in each group.

1. Expected anaemic in each row = 80 × 66 ÷ 240 = 22. Expected not anaemic = 80 × 174 ÷ 240 = 58. Every expected count is well above 5.
2. (O − E)² ÷ E: vegetarian (30 − 22)² ÷ 22 = 2.9091 and (50 − 58)² ÷ 58 = 1.1034. Mixed 0 and 0. Non-vegetarian 2.9091 and 1.1034.
3. χ² = 2 × (2.9091 + 1.1034) = 8.03, with (3 − 1) × (2 − 1) = 2 df. The critical χ² for 2 df at 5% is 5.99, and p = 0.018. Anaemia was 37.5%, 27.5% and 17.5%.
4. A significant R × C test shows only that the proportions are not all equal. To find which groups differ, compare pairs with a correction for multiple testing. With Bonferroni (0.05 ÷ 3 = 0.017), only vegetarian against non-vegetarian differs (p = 0.005). Vegetarian against mixed gives p = 0.18, and mixed against non-vegetarian p = 0.13.

Interactive

Edit the 2 × 2 table and check the expected counts before choosing the test

The chi-square p-value needs adequate expected counts. In a 2 × 2 table, one expected count below 5 is already more than 20% of the cells, so Fisher's exact test is used instead.

Comparing proportions

Which test compares proportions: z-test, McNemar test or binomial test?

Yes-or-no outcomes are compared in three designs: two independent groups, the same patients classified twice, or one group against a reference value. Each design has its own test.

Two independent groups: the z-test for two proportions

In worked example 5, infection was 8% with preparation A and 18% with preparation B. Under H₀ both groups share one infection rate, estimated from all 200 patients as 26 ÷ 200 = 0.13.

z = (p₂ − p₁) ÷ √[p(1 − p)(1/n₁ + 1/n₂)]

z = 0.10 ÷ √(0.13 × 0.87 × 0.02) = 0.10 ÷ 0.0476 = 2.103, p = 0.036

z² = 2.103² = 4.42, the same value as the chi-square statistic for this 2 × 2 table, with the same p-value, so the two are one test written in two ways. The z form can be used one-tailed, and it leads directly to a confidence interval for the difference: 10 percentage points, 95% CI 0.8 to 19.2 (calculated with the unpooled SE, 0.047).

The same patients classified twice: the McNemar test

McNemar χ² = (b − c)² ÷ (b + c)

χ² = (22 − 6)² ÷ 28 = 256 ÷ 28 = 9.14, df 1, p = 0.002

b

pairs positive on the first occasion only (anaemic before, not after): 22

c

pairs positive on the second occasion only (anaemic after, not before): 6

a, d

pairs with the same result both times: not used

With the continuity correction, χ² = (|b − c| − 1)² ÷ (b + c) = 8.04, p = 0.005. When b + c is small, use the exact binomial version, which here gives p = 0.004. Analysing the two occasions as independent groups (65% against 45% anaemic, chi-square p = 0.011) is wrong, because the same 80 women appear in both groups.

One group against a reference value: the binomial test

Made-up teaching data: 7 of 20 patients (35%) had a surgical site infection after a new procedure, against an expected rate of 15%. The expected number is only 20 × 0.15 = 3, too small for the normal approximation. The binomial test calculates the probability directly. If the true rate were 15%, the chance of 7 or more infections in 20 patients is 0.022, and the two-tailed p-value is 0.022. The z approximation gives z = 2.50 and a smaller p-value, 0.012, because it is inaccurate with an expected count of 3. The sign test for paired measurements is a binomial test with a reference proportion of 0.5.

Decide

The complete test map

Find your outcome in the left column and your design across the top.

One group against a reference value: one-sample t-test (normal measurement), Wilcoxon signed-rank test (skewed measurement), chi-square goodness of fit (counts in several categories), or the binomial test (one yes-or-no proportion). Correlation and regression are covered in the correlation and regression chapter, and survival methods in the risk and survival chapter.

Software

Which software can you use for the analysis?

Any of these packages will run the tests in this chapter, but none of them chooses the right test for you. Name the software and its version in your methods section.

Qualitative data (interview and focus group transcripts) needs different software: NVivo (Lumivero), ATLAS.ti (paid licences, free trial), MAXQDA (VERBI Software, Berlin), and Taguette (free and open source). Qualitative methods are covered in the qualitative research chapter.

Types of analysis

What types of data analysis are there?

Analyses are classified by purpose, by the number of variables, by the type of data, and by whether they were planned in advance.

By specialty

Choosing a test by specialty

Common myths

Common mistaken beliefs

• ✗Myth: Three t-tests are fine for three groups.

  ✓Each test has a 5% false-positive rate. Three t-tests give a 14.3% chance of at least one false positive if the tests were independent. Use one-way ANOVA, then a post hoc test.

• ✗Myth: Non-parametric tests are weak, so avoid them.

  ✓When data are normal, the Mann-Whitney and Kruskal-Wallis tests have an asymptotic relative efficiency of 0.955 against the t-test and ANOVA. With heavily skewed data they have more power than the parametric tests.

• ✗Myth: The t-test is only for samples smaller than 30.

  ✓When the population SD is unknown, which is almost always, the t-test can be used at any sample size.

• ✗Myth: A significant ANOVA shows which groups differ.

  ✓It shows only that at least one mean differs. A post hoc test (Tukey, Bonferroni) shows which pairs differ.

• ✗Myth: Always apply Yates' correction to a 2 × 2 table.

  ✓The Yates-corrected test is overly conservative. Decide in the protocol, and use Fisher's exact test when expected counts are small.

• ✗Myth: Analysing paired data with an unpaired test loses only a little.

  ✓In worked example 2, the same 16 numbers gave p = 0.20 unpaired and p = 0.0001 paired.

Why this matters to you

Five consequences of choosing the wrong test

1. Protocol sent back

   ICH E9 expects the principal features of the statistical analysis in the protocol. "Appropriate statistical tests will be applied" does not meet that.

2. A wrong conclusion in the thesis

   Analysing paired data with an unpaired test, or skewed data with a t-test, can make a real effect appear as "no significant difference".

3. Questions you cannot answer at the viva

   "Why this test?" needs an answer in terms of outcome type, number of groups, pairing and distribution. "The software gave it" is not an answer.

4. Misreading papers

   Applying parametric tests to non-normal data is one of the most common statistical errors found in journals. Check also for repeated t-tests across several groups.

5. Harm to patients

   A false-positive result can bring an ineffective treatment into practice. A missed real effect can keep a useful one out.

History

History of the common statistical tests

1. 1900

   Karl Pearson

   Published the chi-square criterion for judging whether observed frequencies could have arisen by random sampling from an expected distribution: the chi-square goodness-of-fit test.

2. 1908

   "Student" (William Sealy Gosset)

   A brewer at Arthur Guinness in Dublin from 1899, he published "The probable error of a mean" in Biometrika, the paper behind the t-test for small samples. Guinness allowed him to publish only under a pseudonym, so he signed it "Student".

3. 1925

   Ronald A. Fisher

   Developed analysis of variance in the 1920s at Rothamsted Experimental Station. His first ANOVA paper, with W. A. Mackenzie, appeared in 1923. His 1925 book Statistical Methods for Research Workers included chapters on the analysis of variance.

4. 1934

   Frank Yates

   Published "Contingency tables involving small numbers and the χ² test", which introduced the continuity correction for 2 × 2 tables.

5. 1942

   Jacob Wolfowitz

   Used the term "non-parametric" for methods that do not assume a distribution.

6. 1945

   Frank Wilcoxon

   Introduced tests based on ranks in "Individual comparisons by ranking methods".

7. 1947

   Henry Mann and Donald Whitney

   Extended Wilcoxon's rank method to two groups of different sizes.

8. 1952

   William Kruskal and W. Allen Wallis

   Published "Use of ranks in one-criterion variance analysis", the rank test for three or more groups.

Guidelines

Statistical analysis in guidelines

Steps

Choosing and running a test, step by step

1. Write the primary outcome and name its type: measurement, ordinal or categorical.
2. Count the groups, and decide whether they are independent or paired.
3. For a measurement, plan how you will check the distribution (histogram, Shapiro-Wilk test) and what you will do if it is skewed.
4. Choose the test from the map, with a post hoc test or correction if there are three or more groups.
5. Write the analysis plan in the protocol: tests, α, software and version.
6. After data entry, check the assumptions, run the planned test, and report the effect size with its 95% confidence interval, not only the p-value.
7. Label every analysis you did not plan as exploratory.

Reading a paper

Five checks on the statistical tests in any paper

• Does the test match the outcome: means for normal measurements, ranks for skewed or ordinal data, counts for categories?
• Were paired measurements (before and after, matched pairs) analysed with a paired test?
• With three or more groups, did the authors use ANOVA or Kruskal-Wallis with a named post hoc test, instead of repeated t-tests?
• For a chi-square test, were the expected counts large enough, or was Fisher's exact test used?
• Are the tests, the software and its version named, and are unplanned analyses labelled as exploratory?

Why is it called that?

Word surgery

Exam answers

Model answer outlines

Past papers

Every TNMGRMU question on choosing a test

Each one is answered in this chapter. Tap a sitting to open the university's own question paper.

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