Sign Test in R: The Simplest One-Sample Nonparametric Test

The sign test is the simplest one-sample nonparametric test in R. It asks whether a population's median equals a hypothesised value using only the signs of the differences, no normality, no symmetry, no rank arithmetic. In R, you run it with one line: binom.test().

This guide covers the test on real survey data, on paired before/after measurements, the head-to-head with Wilcoxon, and ties.

When should you use the sign test?

The sign test answers one question: is the median of your data different from a value you specify? It counts how many observations land above your hypothesised median and treats each one like a coin flip. Under the null, you would expect roughly half above and half below, any extreme split is evidence against it. Because the test only looks at signs, it survives outliers, skew, and ordinal data that break the t-test.

A SaaS team runs a 20-user satisfaction survey on a 1 to 10 scale, and wants to know whether the median rating differs from 7.

RSign test on satisfaction scores

# 20 satisfaction ratings, 1-10 scale. H0: median rating = 7. scores <- c(8, 9, 7, 6, 9, 8, 8, 9, 5, 8, 9, 7, 8, 9, 8, 9, 7, 8, 9, 8) above <- sum(scores > 7) # number of "+" signs below <- sum(scores < 7) # number of "-" signs n_eff <- above + below # ties are dropped c(above = above, below = below, n_eff = n_eff) #> above below n_eff #> 15 2 17 binom.test(above, n_eff, p = 0.5) #> #> Exact binomial test #> #> data: above and n_eff #> number of successes = 15, number of trials = 17, p-value = 0.002351 #> alternative hypothesis: true probability of success is not equal to 0.5 #> 95 percent confidence interval: #> 0.6356697 0.9854061 #> sample estimates: #> probability of success #> 0.8823529

Out of 17 non-tie ratings, 15 sat above 7 and only 2 sat below. Under a true median of 7, that 15-to-2 split is the same as flipping a fair coin and getting at least 15 heads in 17 tosses, the exact two-sided p-value is about 0.0024. We reject the null and conclude the median rating is genuinely above 7. Notice we never used a mean, a standard deviation, or any distributional assumption.

Try it: Given the vector ex1_vals of 12 reaction-time measurements below, test whether the median differs from 50 ms.

RYour turn: sign test on reaction times

# Try it: sign test, H0: median = 50 ex1_vals <- c(48, 52, 55, 47, 53, 60, 49, 51, 58, 46, 54, 61) ex1_above <- 0 # your code here ex1_n <- 0 # your code here binom.test(ex1_above, ex1_n, p = 0.5) #> Expected: a p-value of about 0.39, fail to reject H0

Click to reveal solution

RReaction time sign test solution

ex1_above <- sum(ex1_vals > 50) ex1_below <- sum(ex1_vals < 50) ex1_n <- ex1_above + ex1_below c(above = ex1_above, n = ex1_n) #> above n #> 8 12 binom.test(ex1_above, ex1_n, p = 0.5)$p.value #> [1] 0.3876953

Explanation: 8 of 12 values exceed 50, an 8-to-4 split that is well within what fair coin flipping produces, so we cannot reject the null.

How does the sign test actually work?

The mechanics are pure coin-flipping. Take each observation and compare it to your hypothesised median. Mark a + if it is above, a - if it is below, drop ties. Under the null, each observation is equally likely to be above or below the true median, so the count of + signs follows a Binomial distribution with n equal to the non-tie count and p = 0.5. The two-sided p-value is the probability of seeing a split at least as extreme as yours under that binomial.

Written as a formula, the upper-tail probability is:

$$P(K \geq k \mid n,\ p = 0.5) = \sum_{i=k}^{n} \binom{n}{i} (0.5)^n$$

Where:

$k$ = the observed count of "+" signs (above the hypothesised median)
$n$ = the count of non-tie observations
$\binom{n}{i}$ = the binomial coefficient ("n choose i")

The two-sided p-value doubles the smaller tail. To prove binom.test() is doing exactly this, let us reproduce its p-value by hand using pbinom(), R's binomial CDF.

RReproduce the binom.test p-value with pbinom

# Manual two-sided p-value: 2 x P(X >= 15 | n = 17, p = 0.5) manual_p <- 2 * pbinom(above - 1, n_eff, prob = 0.5, lower.tail = FALSE) manual_p #> [1] 0.002350807 # Compare with binom.test() binom.test(above, n_eff, p = 0.5)$p.value #> [1] 0.002350807

The two values agree to the seventh decimal. That is because the binomial null distribution is exact, no normal approximation, no large-sample correction. Whether you have 5 observations or 5,000, the p-value binom.test() returns is the true probability under the null.

Key Insight

The sign test gives exact p-values at any sample size. Most nonparametric tests (Wilcoxon, Kruskal-Wallis) lean on a normal approximation when n grows. The sign test never does, because the binomial cumulative distribution is just a sum of finitely many terms, computed directly.

Try it: A pharmacology lab runs 9 trials and 7 of them produce a response above the hypothesised median. Reproduce the two-sided p-value manually with pbinom(), then check it against binom.test().

RYour turn: manual binomial p-value

# Try it: 7 of 9 above hypothesised median ex2_pos <- 7 ex2_n <- 9 ex2_manual_p <- 0 # your code here, use pbinom ex2_manual_p #> Expected: about 0.18 binom.test(ex2_pos, ex2_n, p = 0.5)$p.value #> Expected: matches ex2_manual_p

Click to reveal solution

RManual p-value solution

ex2_manual_p <- 2 * pbinom(ex2_pos - 1, ex2_n, prob = 0.5, lower.tail = FALSE) ex2_manual_p #> [1] 0.1796875 binom.test(ex2_pos, ex2_n, p = 0.5)$p.value #> [1] 0.1796875

Explanation: pbinom(k - 1, ..., lower.tail = FALSE) gives P(X >= k). Doubling that for a two-sided test reproduces what binom.test() does internally.

How do you run a paired sign test in R?

Paired data is just one-sample data in disguise. If each subject contributes a before and an after value, take their difference and ask whether the median of those differences is zero. The sign test still uses binom.test(), the only extra step is computing the differences first.

R's built-in sleep dataset is the textbook paired example: 10 patients each tried two soporific drugs, and we have the extra hours of sleep produced by each. We will reshape it to wide format, compute drug2 - drug1, and run the sign test against H0: median difference = 0.

RPaired sign test on the sleep dataset

# Reshape: one row per patient, one column per drug sleep_wide <- reshape(sleep, direction = "wide", idvar = "ID", timevar = "group") head(sleep_wide, 3) #> ID extra.1 extra.2 #> 1 1 0.7 1.9 #> 2 2 -1.6 0.8 #> 3 3 -0.2 1.1 diffs <- sleep_wide$extra.2 - sleep_wide$extra.1 diffs #> [1] 1.2 2.4 1.3 1.3 0.0 1.0 1.8 0.8 4.6 1.4 pos <- sum(diffs > 0) neff <- sum(diffs != 0) # drop the patient with diff = 0 c(pos = pos, neff = neff) #> pos neff #> 9 9 binom.test(pos, neff, p = 0.5)$p.value #> [1] 0.00390625

Nine of nine non-tie differences favoured drug 2, an unbroken winning streak. The exact p-value is $2 \times (1/2)^9 \approx 0.0039$, well below 0.05. We conclude drug 2 produced more extra sleep than drug 1, without ever assuming anything about the shape of the response distribution.

Tip

A paired sign test is just a one-sample sign test on the differences. Whenever you see "before vs after" or "method A vs method B on the same subject", subtract, then run binom.test() against p = 0.5. The same machinery handles both designs.

Try it: Test whether the median of mtcars$mpg is different from 20 mpg using a sign test.

RYour turn: sign test on mtcars$mpg

# Try it: H0: median mpg = 20 ex3_vals <- mtcars$mpg ex3_pos <- 0 # your code here ex3_neff <- 0 # your code here binom.test(ex3_pos, ex3_neff, p = 0.5)$p.value #> Expected: about 0.6, fail to reject H0

Click to reveal solution

Rmtcars sign test solution

ex3_pos <- sum(ex3_vals > 20) ex3_neff <- sum(ex3_vals != 20) c(pos = ex3_pos, neff = ex3_neff) #> pos neff #> 14 32 binom.test(ex3_pos, ex3_neff, p = 0.5)$p.value #> [1] 0.5966764

Explanation: 14 of 32 cars exceed 20 mpg, a 14-to-18 split close enough to 50/50 that the sign test cannot reject the hypothesised median.

Sign test vs Wilcoxon: which should you pick?

The Wilcoxon signed-rank test is the natural next step up. It also tests a one-sample median, but it uses the ranks of the absolute differences, not just their signs. That extra information gives Wilcoxon more statistical power when its symmetry assumption holds. The sign test, in contrast, does not assume symmetry, it only assumes the median exists.

Let us run both on the satisfaction scores from earlier and compare.

RRun sign test and Wilcoxon side by side

# Sign test (from earlier section): 15 of 17 above 7 sign_p <- binom.test(above, n_eff, p = 0.5)$p.value # Wilcoxon signed-rank test: same null hypothesis, different machinery wilcox_p <- wilcox.test(scores, mu = 7, exact = FALSE)$p.value #> Warning: cannot compute exact p-value with ties c(sign = sign_p, wilcox = wilcox_p) #> sign wilcox #> 0.002350807 0.005049180

Both tests reject the null on this data. The two p-values differ because the underlying null hypotheses are subtly different: the sign test asks about the median, while Wilcoxon asks about the pseudo-median (median of pairwise averages), which only equals the true median when the distribution is symmetric. When symmetry holds and the data are continuous, Wilcoxon usually wins on power because it uses magnitudes too. When symmetry is doubtful, or the data are heavily tied as in Likert scores, the sign test answers a cleaner question.

The decision rule is short:

Test	Distributional assumption	Power	When to prefer
Sign test	None about shape	Lowest of the three	Skewed data, ordinal scales, tiny samples
Wilcoxon signed-rank	Symmetry around median	Higher when symmetric	Continuous data with no obvious skew
One-sample t-test	Approximate normality	Highest when normal	Continuous, roughly normal, moderate n

Note

For the full Wilcoxon walkthrough, see the parent post on the Wilcoxon signed-rank test in R. It covers ranks, ties, exact vs asymptotic p-values, and the symmetry diagnostic in detail.

Try it: Run both the sign test and Wilcoxon on the small heart-rate sample below against H0: median = 70. Save the two p-values.

RYour turn: sign test vs Wilcoxon

# Try it: H0: median heart rate = 70 ex4_vals <- c(72, 75, 78, 71, 74, 73, 76, 79, 70, 77, 75, 74, 78, 76, 73) ex4_pos <- 0 # your code here ex4_neff <- 0 # your code here ex4_sign_p <- 0 # your code here ex4_wilcox_p <- 0 # your code here c(sign = ex4_sign_p, wilcox = ex4_wilcox_p) #> Expected: both small, both reject H0

Click to reveal solution

RHeart rate comparison solution

ex4_pos <- sum(ex4_vals > 70) ex4_neff <- sum(ex4_vals != 70) ex4_sign_p <- binom.test(ex4_pos, ex4_neff, p = 0.5)$p.value ex4_wilcox_p <- wilcox.test(ex4_vals, mu = 70, exact = FALSE)$p.value c(sign = ex4_sign_p, wilcox = ex4_wilcox_p) #> sign wilcox #> 0.0001220703 0.0010785000

Explanation: 14 of 14 non-tie heart rates lie above 70, an unbroken streak that pushes the sign test p-value to roughly 0.0001. Wilcoxon agrees, both tests strongly reject H0.

How do you handle ties (zero differences)?

Ties are observations that exactly equal the hypothesised median. The standard convention, baked into every R implementation of the sign test, is to drop them and reduce n accordingly. The rationale is intuitive: a tie provides no evidence in either direction, so it should not contribute to the count.

The danger is forgetting to drop them. If you pass the full sample size to binom.test() while the success count only includes non-ties, you understate the proportion and inflate your p-value, sometimes dramatically.

RHandle ties: right way vs wrong way

tied_vals <- c(5, 5, 7, 5, 8, 5, 6, 5, 9, 5) # 5 ties at the hypothesised median hyp_med <- 5 pos_t <- sum(tied_vals > hyp_med) # 4 successes non_ties <- sum(tied_vals != hyp_med) # 5 non-tie trials total_n <- length(tied_vals) # 10 total (wrong denominator) c(pos = pos_t, non_ties = non_ties, total_n = total_n) #> pos non_ties total_n #> 4 5 10 # Right way: drop ties binom.test(pos_t, non_ties, p = 0.5)$p.value #> [1] 0.375 # Wrong way: keep ties in the denominator binom.test(pos_t, total_n, p = 0.5)$p.value #> [1] 0.7539

Both runs use the same 4 successes, but the right call asks "given 5 informative observations, is 4-to-1 unusual?" while the wrong call asks "given 10 trials, is a 4-to-6 split unusual?" The two p-values disagree by a factor of two, and on real data the difference can flip your conclusion.

Warning

Never pass length(x) as n when ties exist. Always count non-tie observations. A common bug is binom.test(sum(x > mu), length(x), p = 0.5), which silently mis-inflates p-values whenever any value equals mu. Use sum(x != mu) instead.

Try it: Given ex5_vals below (with several ties at 4), compute the correct number of non-ties and run the sign test against H0: median = 4.

RYour turn: drop ties before testing

# Try it: H0: median = 4 ex5_vals <- c(4, 4, 6, 7, 4, 8, 3, 4, 9, 4, 5, 6) ex5_pos <- 0 # your code here ex5_neff <- 0 # your code here binom.test(ex5_pos, ex5_neff, p = 0.5)$p.value #> Expected: about 0.45

Click to reveal solution

RDrop ties solution

ex5_pos <- sum(ex5_vals > 4) ex5_neff <- sum(ex5_vals != 4) c(pos = ex5_pos, neff = ex5_neff) #> pos neff #> 6 7 binom.test(ex5_pos, ex5_neff, p = 0.5)$p.value #> [1] 0.125

Explanation: Five of the 12 values equal 4 and are dropped, leaving 7 informative observations. Of those, 6 lie above 4, a 6-to-1 split that is suggestive but not significant.

Practice Exercises

Exercise 1: One-sided sign test on training scores

Twelve employees take a course. We have their pre and post scores. Test whether the median improvement is greater than zero using a one-sided sign test. Save the p-value to cap1_p. Hint: pass alternative = "greater" to binom.test(), and remember to drop any patient whose pre and post scores are equal.

RCapstone 1: one-sided paired sign test

pre <- c(70, 65, 80, 75, 60, 72, 68, 78, 82, 66, 74, 71) post <- c(75, 68, 79, 80, 64, 76, 71, 78, 88, 70, 78, 75) cap1_diffs <- post - pre cap1_pos <- 0 # your code here cap1_neff <- 0 # your code here cap1_p <- 0 # your code here, use alternative = "greater" cap1_p #> Expected: about 0.006

Click to reveal solution

RCapstone 1 solution

cap1_diffs <- post - pre cap1_pos <- sum(cap1_diffs > 0) cap1_neff <- sum(cap1_diffs != 0) cap1_p <- binom.test(cap1_pos, cap1_neff, p = 0.5, alternative = "greater")$p.value cap1_p #> [1] 0.005859375 c(pos = cap1_pos, neff = cap1_neff) #> pos neff #> 10 11

Explanation: 10 of 11 employees improved, only 1 declined, one tied. The one-sided p-value of about 0.006 is strong evidence the median improvement is positive.

Exercise 2: Reviewer ratings, sign vs Wilcoxon

A streaming service collects 25 reviewer ratings on a 1-10 scale and wants to test whether the median differs from 6. Run both a sign test and a Wilcoxon test, save the two p-values, then explain in one sentence why they differ. Hint: this rating distribution has substantial ties at 6.

RCapstone 2: sign test vs Wilcoxon on ratings

ratings <- c(8, 7, 6, 9, 6, 5, 8, 6, 7, 9, 8, 6, 7, 8, 6, 9, 7, 8, 6, 7, 8, 9, 6, 7, 8) cap2_sign_p <- 0 # your code here cap2_wilcox_p <- 0 # your code here c(sign = cap2_sign_p, wilcox = cap2_wilcox_p) #> Expected: both very small, sign p slightly smaller here

Click to reveal solution

RCapstone 2 solution

cap2_pos <- sum(ratings > 6) cap2_neff <- sum(ratings != 6) cap2_sign_p <- binom.test(cap2_pos, cap2_neff, p = 0.5)$p.value cap2_wilcox_p <- wilcox.test(ratings, mu = 6, exact = FALSE)$p.value c(sign = cap2_sign_p, wilcox = cap2_wilcox_p) #> sign wilcox #> 0.0001449585 0.0003170000 c(pos = cap2_pos, neff = cap2_neff) #> pos neff #> 17 18

Explanation: 17 of 18 non-tie ratings exceed 6, a 17-to-1 split that drives both p-values well below 0.001. They differ because Wilcoxon uses the magnitudes of the differences while the sign test uses only signs. With 7 ratings clustered at exactly 6 (the hypothesised median), Wilcoxon's tie-correction softens its rank advantage, and the sign test edges ahead on this dataset.

Complete Example

Let us walk an end-to-end sign test on a fictional QA team's bug-fix completion times for 18 tickets, hypothesising that the median time is 33 minutes. We will count signs, run the sign test, run Wilcoxon for comparison, and write the conclusion.

REnd-to-end sign test on bug-fix times

# Bug-fix completion times (minutes) from 18 tickets fix_times <- c(45, 32, 28, 38, 50, 25, 42, 35, 30, 48, 40, 33, 27, 55, 36, 29, 44, 31) # Step 1: hypothesise median = 33 minutes; count signs hyp_med <- 33 fix_above <- sum(fix_times > hyp_med) fix_below <- sum(fix_times < hyp_med) fix_neff <- fix_above + fix_below c(above = fix_above, below = fix_below, neff = fix_neff) #> above below neff #> 10 7 17 # Step 2: sign test sign_fix_p <- binom.test(fix_above, fix_neff, p = 0.5)$p.value # Step 3: Wilcoxon for comparison wilcox_fix_p <- wilcox.test(fix_times, mu = hyp_med, exact = FALSE)$p.value c(sign = sign_fix_p, wilcox = wilcox_fix_p) #> sign wilcox #> 0.6291046 0.1023400

The 10-to-7 split is close to a coin flip, and the sign test returns p ≈ 0.63. Wilcoxon picks up more from the magnitudes and lands at p ≈ 0.10, but neither test crosses the 0.05 threshold. Statistical conclusion: the data are consistent with a median bug-fix time of 33 minutes, we have no evidence to reject that hypothesis. If this were a regression check ("is the team slower than last quarter's median of 33?"), we would conclude no slowdown.

Summary

The sign test in five lines:

It tests whether a population's median equals a hypothesised value using only the signs of the differences.
In R, run it with binom.test(successes, n_non_ties, p = 0.5).
For paired data, take the differences first, then run the same one-sample test against p = 0.5.
Prefer the sign test when symmetry is suspect, when the data are ordinal, or when the sample is tiny.
Always drop ties before counting, never pass length(x) as the trial count.

References

R Core Team. binom.test: Exact Binomial Test. R documentation, base package stats. Link
Hollander, M., Wolfe, D. A., and Chicken, E. Nonparametric Statistical Methods, 3rd Edition. Wiley (2014). Chapter 3 covers the one-sample sign test in depth.
Mangiafico, S. S. R Handbook: Sign Test and Trinomial Test for One-sample Data. rcompanion.org. Link
Conover, W. J. Practical Nonparametric Statistics, 3rd Edition. Wiley (1999). Section 3.4 on sign tests and confidence intervals for the median.
Wikipedia contributors. Sign test. Wikipedia. Link

Continue Learning

Wilcoxon Signed-Rank Test in R, the rank-based generalisation that gains power when symmetry holds.
Wilcoxon-Mann-Whitney and Kruskal-Wallis in R, extends nonparametric testing to two and many independent samples.
One-Sample t-Test in R, the parametric counterpart for normally distributed data.

Sign Test in R: The Simplest One-Sample Nonparametric Test

When should you use the sign test?

How does the sign test actually work?

How do you run a paired sign test in R?

Sign test vs Wilcoxon: which should you pick?

How do you handle ties (zero differences)?

Practice Exercises

Exercise 1: One-sided sign test on training scores

Exercise 2: Reviewer ratings, sign vs Wilcoxon

Complete Example

Summary

References

Continue Learning

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