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The jackknife is a resampling method that re-computes a statistic many times, each time leaving out one observation, and uses the spread and average of those leave-one-out results to estimate the statistic’s bias and standard error. It needs no distributional formula for the statistic, only the ability to recompute it on a smaller sample. This guide shows every step with a five-number example you can check by hand, says which statistics the method handles and which it does not, and gives Excel, R and Python code that reproduces the numbers.
What is the jackknife? Definition in one paragraph
The jackknife (also called leave-one-out resampling) takes a sample of n observations and an estimator, then builds n “jackknife samples”, each one the original sample with a single observation removed. The estimator is recomputed on every jackknife sample, giving n replicates. The average of the replicates yields a bias estimate, and their variation yields a standard error. It was proposed by M. H. Quenouille for bias correction and later refined and named by John Tukey, who used it for confidence limits (McIntosh, 2016). Unlike the bootstrap, the jackknife samples without replacement and always produces exactly n samples, so its output is deterministic: run it twice on the same data and you get the same answer. For how the two methods relate, see bootstrapping in statistics.
Jackknife formulas: replicates, bias and standard error
Let the sample be x = (x1, …, xn) and let the estimator be θ̂ = s(x). Every symbol below is defined once:
- x(i): the sample with the i-th observation left out (size n − 1).
- θ̂(i) = s(x(i)): the i-th jackknife replicate, the estimator computed on that reduced sample.
- θ̂(·) = (1/n) Σ θ̂(i): the average of the n replicates.
- Standard error: SEjack = √[ ((n − 1)/n) Σ (θ̂(i) − θ̂(·))² ]. The factor (n − 1)/n is needed because jackknife samples are all very similar to the original, so their raw spread is far too small (Berkeley Stat 152 handout).
- Bias: biasjack = (n − 1)(θ̂(·) − θ̂).
- Bias-corrected estimate: θ̂* = θ̂ − biasjack = nθ̂ − (n − 1)θ̂(·).
Jackknife example: worked numbers
Take the five observations 3, 5, 7, 11, 14 (n = 5) and the plug-in variance, which divides by n and is therefore a biased estimator. That choice lets the bias correction visibly do something.
Step 1, full-sample estimate. The mean is (3 + 5 + 7 + 11 + 14)/5 = 8. The squared deviations are 25, 9, 1, 9, 36, which sum to 80, so θ̂ = 80/5 = 16.0.
Step 2, leave-one-out replicates. Recompute the divide-by-n variance on each four-number sample:
| Observation left out | Remaining values | Replicate θ̂(i) |
|---|---|---|
| 3 | 5, 7, 11, 14 | 12.1875 |
| 5 | 3, 7, 11, 14 | 17.1875 |
| 7 | 3, 5, 11, 14 | 19.6875 |
| 11 | 3, 5, 7, 14 | 17.1875 |
| 14 | 3, 5, 7, 11 | 8.75 |
Step 3, average. θ̂(·) = (12.1875 + 17.1875 + 19.6875 + 17.1875 + 8.75)/5 = 75/5 = 15.0.
Step 4, bias and correction. biasjack = (5 − 1)(15.0 − 16.0) = −4.0, so θ̂* = 16.0 − (−4.0) = 20.0. This equals the familiar n − 1 sample variance, 80/4 = 20, which is a useful sanity check: the jackknife has recovered the textbook correction without being told the formula.
Step 5, standard error. The squared deviations of the replicates from 15.0 are 7.9102, 4.7852, 21.9727, 4.7852 and 39.0625, which sum to 78.5156. Then SEjack = √(4/5 × 78.5156) = √62.8125 ≈ 7.93.
Second check, the mean. For the sample mean the five replicates are 9.25, 8.75, 8.25, 7.25 and 6.5; their average is 8.0, so the bias is exactly 0 and the SE is 2.0, which equals s/√n = 4.472/2.236. For the mean, the jackknife variance reduces to the usual unbiased variance of the mean, (1/(n(n − 1))) Σ(xi − x̄)² (McIntosh, 2016).
Reading the jackknife results
| Output | What it tells you | In the example |
|---|---|---|
| Bias near 0 | Dropping one observation does not systematically shift the estimate on average; the mean always lands here. | Mean: bias 0 |
| Bias clearly non-zero | The estimator is biased; θ̂* = nθ̂ − (n − 1)θ̂(·) is the corrected value. | Variance: bias −4.0, corrected 20.0 |
| Jackknife SE | The estimated standard deviation of the estimator across samples of this size. For large samples it is roughly equivalent to the delta-method SE (McIntosh, 2016). | Mean: 2.0; variance: 7.93 |
Assumptions, limits and common mistakes
When the jackknife works and when it fails
The Berkeley Stat 152 handout is direct about the limit: the jackknife only works well for linear statistics such as the mean, and it fails to give accurate estimates for non-smooth statistics such as the median and for non-linear ones such as the correlation coefficient.
| Statistic type | Jackknife | Why |
|---|---|---|
| Mean (linear) | Works well | Reproduces the unbiased variance of the mean |
| Median (non-smooth) | Inaccurate with delete-1 | Non-smooth statistic (Berkeley Stat 152 handout) |
| Correlation coefficient (non-linear) | Inaccurate | Stated in the handout |
Delete-d jackknife
The delete-d jackknife leaves out d observations at a time, giving samples of size n − d and C(n, d) of them. For the median it is consistent when √n/d → 0 and n − d → ∞ (Berkeley Stat 152 handout).
Common mistakes
- Forgetting the (n − 1)/n factor. Using the plain variance of the replicates understates the SE badly.
- Applying delete-1 to a median or quantile and trusting the SE.
- Confusing it with the bootstrap. The jackknife samples without replacement and gives n samples; the bootstrap samples with replacement and gives as many as you choose.
Jackknife vs bootstrap in brief
The jackknife needs only n recomputations and gives the same answer every time, so it is cheap to compute; the Berkeley Stat 152 handout describes it as requiring less computational power than more recent techniques such as the bootstrap. The two are often used together: the bias-corrected and accelerated (BCa) bootstrap interval uses a jackknife-based acceleration factor, covered in bootstrapping in statistics. A related idea appears in regression, where the HC3 robust standard error approximates a jackknife; see robust standard errors HC0 to HC3.
How to report a jackknife result (APA-style example)
State the statistic, the sample size, the jackknife variant and the quantity you report. Example wording:
The sample mean was M = 8.00 (delete-one jackknife SE = 2.00, n = 5). The delete-one jackknife bias-corrected variance was 20.00 (jackknife SE = 7.93).
Related reporting terms: confidence interval and effect size.
How to calculate the jackknife in Excel, R or Python
Excel
Put the data in A2:A6. In B2 enter =VAR.P(A3:A6), in B3 =VAR.P(A2,A4:A6), in B4 =VAR.P(A2:A3,A5:A6), in B5 =VAR.P(A2:A4,A6) and in B6 =VAR.P(A2:A5). Then:
Full estimate: =VAR.P(A2:A6) -> 16 Replicate mean: =AVERAGE(B2:B6) -> 15 Bias: =(5-1)*(AVERAGE(B2:B6)-VAR.P(A2:A6)) -> -4 Corrected: =VAR.P(A2:A6)-(5-1)*(AVERAGE(B2:B6)-VAR.P(A2:A6)) -> 20 SE: =SQRT((5-1)/5*DEVSQ(B2:B6)) -> 7.925
R
x <- c(3, 5, 7, 11, 14); n <- length(x) v <- function(a) mean((a - mean(a))^2) # divide-by-n variance theta <- v(x) reps <- sapply(seq_len(n), function(i) v(x[-i])) bias <- (n - 1) * (mean(reps) - theta) # -4 corrected <- theta - bias # 20 se <- sqrt((n - 1) / n * sum((reps - mean(reps))^2)) # 7.925
Python
import numpy as np x = np.array([3, 5, 7, 11, 14]); n = len(x) theta = np.var(x) # divide-by-n variance reps = np.array([np.var(np.delete(x, i)) for i in range(n)]) bias = (n - 1) * (reps.mean() - theta) # -4.0 corrected = theta - bias # 20.0 se = np.sqrt((n - 1) / n * np.sum((reps - reps.mean())**2)) # 7.925
Frequently asked questions
What is jackknife?
The jackknife is a leave-one-out resampling method: recompute a statistic on each sample with one observation removed, then use the replicates to estimate the statistic’s bias and standard error.
What are some jackknife examples?
The worked example above uses the variance of 3, 5, 7, 11, 14: the jackknife finds a bias of −4.0, corrects 16.0 to 20.0, and gives an SE of about 7.93. For the mean of the same data it gives bias 0 and SE 2.0.
Who invented the jackknife?
M. H. Quenouille proposed it in 1949 for bias correction; John Tukey later refined it, gave it its current name and used it for confidence limits (McIntosh, 2016).
Is the jackknife the same as the bootstrap?
No. The jackknife samples without replacement and produces exactly n deterministic samples; the bootstrap samples with replacement and produces as many as the analyst chooses.
Can I use the jackknife for the median?
Not with the delete-one version, which is inaccurate for non-smooth statistics. The delete-d version can be consistent under the condition given above.
Sources
- Berkeley Statistics, Stat 152 course handout “Jackknife”: https://www.stat.berkeley.edu/~hhuang/STAT152/Jackknife-Bootstrap.pdf. Checked 2026-10-29.
- McIntosh, A. I. (2016). The Jackknife Estimation Method. arXiv:1606.00497 [stat.ME]: https://arxiv.org/pdf/1606.00497. Checked 2026-10-29.
- Shao, J., and Tu, D. (1995). The Jackknife and Bootstrap. Springer. Further reading on the theory; no figures on this page are taken from it.
