Statistical tables can give three different answers to a question about two failures in ten trials. Do you want exactly two failures, at most two, or at least two? At a failure rate of 20%, those answers are about 30%, 68% and 62%. The calculation starts with the wording of the question.
Use the calculator to find a probability or critical value, then follow the worked example for your table. For a printed reference, download the PDF. The Excel workbook contains the full lookup grids and preserves the leading zeros in the random numbers.
Statistical tables calculator
Choose a table and its inputs. Each selector shows the available grid points, and the result updates immediately. For example, choose Student’s t, 9 degrees of freedom and 0.05 to see both the one-tail and two-tail cutoffs.
Which table answers your question?
Probability, critical value, or observed statistic?
A probability lookup starts with an outcome and returns an area. For example, a z score of 1.96 gives an upper area near 0.025. A critical-value lookup reverses the question. Start with a tail area and find the boundary that leaves that much probability beyond it. Finally, an observed statistic comes from the data. Compare it with a cutoff only after matching the model, tail and degrees of freedom.
Binomial distribution table for a fixed number of trials
A supplier ships ten components. Each has a 20% chance of failing an inspection, and the outcomes are independent. How often should an inspector expect exactly two failures? First, choose n = 10 and p = 0.20. Then read across k = 2. The point probability is 0.301990, or about 30.2%. Although two is the expected count, the inspector will see some other count on roughly seven out of ten shipments.
| Trials n | Success probability p | Count k | P(X = k) | P(X ≤ k) | P(X ≥ k) |
|---|---|---|---|---|---|
| 10 | 0.2 | 0 | 0.107374 | 0.107374 | 1 |
| 10 | 0.2 | 1 | 0.268435 | 0.37581 | 0.892626 |
| 10 | 0.2 | 2 | 0.30199 | 0.6778 | 0.62419 |
| 10 | 0.2 | 3 | 0.201327 | 0.879126 | 0.3222 |
| 10 | 0.2 | 4 | 0.08808 | 0.967207 | 0.120874 |
| 10 | 0.2 | 5 | 0.026424 | 0.993631 | 0.032793 |
| 10 | 0.2 | 6 | 0.005505 | 0.999136 | 0.006369 |
| 10 | 0.2 | 7 | 0.000786 | 0.999922 | 0.000864 |
| 10 | 0.2 | 8 | 7.3728e-05 | 0.999996 | 7.79264e-05 |
| 10 | 0.2 | 9 | 4.096e-06 | 1 | 4.1984e-06 |
| 10 | 0.2 | 10 | 1.024e-07 | 1 | 1.024e-07 |
Exactly, at most, or at least?
For an acceptance rule that permits up to two failures, use the cumulative column instead. It gives 0.677800. By contrast, an investigation that opens at two or more failures has probability 0.624190. Those two probabilities add to more than one because both include exactly two failures. In fact, their sum equals one plus the point probability. That identity gives you a quick way to audit all three columns.
For sampling without replacement, ask how much each draw changes the pool. For example, drawing ten invoices from a batch of twelve with a known number of errors changes the odds substantially. A hypergeometric model accounts for the shrinking pool. The binomial calculation assumes the same success probability on each trial.
Source · SciPy binomial functions
Poisson distribution table for arrivals and event counts
Suppose a help desk receives an average of two calls per half hour. Under a homogeneous Poisson process, the count in that interval has mean λ = 2. Consequently, a quiet half hour with zero calls has probability about 0.135335. More surprisingly, exactly one call and exactly two calls have equal probabilities of about 0.270671 each.
| Mean λ | Count k | P(X = k) | P(X ≤ k) | P(X ≥ k) |
|---|---|---|---|---|
| 2 | 0 | 0.135335 | 0.135335 | 1 |
| 2 | 1 | 0.270671 | 0.406006 | 0.864665 |
| 2 | 2 | 0.270671 | 0.676676 | 0.593994 |
| 2 | 3 | 0.180447 | 0.857123 | 0.323324 |
| 2 | 4 | 0.090224 | 0.947347 | 0.142877 |
| 2 | 5 | 0.036089 | 0.983436 | 0.052653 |
| 2 | 6 | 0.01203 | 0.995466 | 0.016564 |
| 2 | 7 | 0.003437 | 0.998903 | 0.004534 |
| 2 | 8 | 0.000859 | 0.999763 | 0.001097 |
| 2 | 9 | 0.000191 | 0.999954 | 0.000237 |
| 2 | 10 | 3.81899e-05 | 0.999992 | 4.64981e-05 |
Match the rate to the window
For a full hour at the same rate, switch to λ = 4. The probability of no calls falls to about 0.018316. Alternatively, multiply the two half-hour zero-call probabilities because the model gives independent counts in disjoint intervals. Both routes produce the same answer.
A staffing manager may care about overload instead. For example, if the desk can handle three calls, overload begins at four. Use P(X ≥ 4) with λ = 2 to get 0.142877. Before applying that result to Monday mornings, check whether calls arrive in bursts or whether the rate changes across shifts. Either feature can make a single average misleading.
Source · SciPy Poisson functions
Standard normal distribution table for areas under the curve
A normal table can show the area below z, above z, or between zero and z. First, inspect the row for zero. A left-area table returns 0.500000 there, whereas a mean-to-score table returns zero. Next, inspect z = 1.00. A left-area table increases to 0.841345, while a right-area table decreases to 0.158655. Together, these two checks identify the convention.
| Score z | Left area | Right area | Two-tail area |z| |
|---|---|---|---|
| -3 | 0.00135 | 0.99865 | 0.0027 |
| -2.58 | 0.00494 | 0.99506 | 0.00988 |
| -1.96 | 0.024998 | 0.975002 | 0.049996 |
| -1 | 0.158655 | 0.841345 | 0.317311 |
| 0 | 0.5 | 0.5 | 1 |
| 1 | 0.841345 | 0.158655 | 0.317311 |
| 1.64 | 0.949497 | 0.050503 | 0.101005 |
| 1.96 | 0.975002 | 0.024998 | 0.049996 |
| 2.58 | 0.99506 | 0.00494 | 0.00988 |
| 3 | 0.99865 | 0.00135 | 0.0027 |
Translate a measurement into a z score
Suppose a filling machine produces normally distributed weights with mean 500 g and standard deviation 4 g. A 508 g package sits two standard deviations above the mean because (508 − 500) / 4 = 2. Therefore, about 2.28% of packages exceed 508 g. If the customer also flags packages under 492 g, the two-tail probability rises to about 4.55%.
For a sample mean, standardize using its standard error. For instance, averaging four independent packages divides the standard deviation of the mean by two. The same physical threshold now represents a different z score. Record whether the question concerns one item or an average before consulting the table.
Source · SciPy normal functions
Table of random numbers for reproducible sampling
A random-number table gives you a sequence of digits. Each five-digit group here preserves zeros on the left. To reproduce the sequence in Python, initialize random.Random(417) once, draw randrange(100000) for each cell, and format each integer to five digits. Then move across ten columns before starting the next row.
| Row | 1 | 2 | 3 | 4 | 5 | 6 | 7 | 8 | 9 | 10 |
|---|---|---|---|---|---|---|---|---|---|---|
| 1 | 78315 | 94770 | 10479 | 05247 | 97566 | 50844 | 37481 | 86854 | 06005 | 45824 |
| 2 | 00835 | 85626 | 60581 | 00042 | 50817 | 10300 | 04063 | 41682 | 50645 | 00681 |
| 3 | 34440 | 10722 | 46211 | 33355 | 01421 | 43566 | 54902 | 14010 | 61604 | 64934 |
| 4 | 03460 | 79522 | 75812 | 53547 | 92094 | 14555 | 67145 | 56786 | 32903 | 93880 |
| 5 | 47940 | 19367 | 16532 | 54734 | 95634 | 68626 | 55463 | 79141 | 36608 | 85949 |
| 6 | 46779 | 17579 | 42803 | 37355 | 20923 | 45013 | 25169 | 12252 | 78715 | 04050 |
| 7 | 67820 | 96749 | 16336 | 66689 | 52348 | 34813 | 54846 | 03709 | 65336 | 48885 |
| 8 | 93811 | 89206 | 86403 | 27470 | 77301 | 43983 | 35477 | 89162 | 90233 | 95327 |
| 9 | 97540 | 30585 | 41060 | 04240 | 13271 | 02977 | 88978 | 80176 | 39251 | 30624 |
| 10 | 51118 | 53412 | 15268 | 06614 | 26706 | 89920 | 15460 | 60261 | 83217 | 22084 |
| 11 | 09681 | 82943 | 35096 | 10540 | 44218 | 60678 | 52372 | 91406 | 29979 | 70013 |
| 12 | 22481 | 96064 | 23795 | 79566 | 06055 | 09945 | 54350 | 18092 | 09240 | 79942 |
| 13 | 28091 | 35966 | 93619 | 59188 | 92694 | 49196 | 65359 | 37195 | 48989 | 60429 |
| 14 | 68227 | 35883 | 66060 | 13586 | 65196 | 18240 | 11154 | 73475 | 05654 | 94322 |
| 15 | 54940 | 77507 | 42798 | 05936 | 05924 | 48691 | 77189 | 40150 | 84249 | 34332 |
| 16 | 27332 | 24984 | 58134 | 09775 | 07417 | 47723 | 31979 | 70365 | 93124 | 85142 |
| 17 | 34621 | 21664 | 12367 | 15379 | 28053 | 85907 | 49440 | 52441 | 12745 | 10017 |
| 18 | 37088 | 21843 | 11540 | 14880 | 98100 | 00971 | 47991 | 51511 | 99841 | 46328 |
| 19 | 29181 | 76552 | 93085 | 27163 | 51560 | 17660 | 36246 | 92124 | 76650 | 04657 |
| 20 | 08046 | 20795 | 83932 | 36245 | 64339 | 77353 | 97429 | 15942 | 05152 | 93130 |
Draw invoice IDs without favoring low numbers
Imagine a population of 73 invoices labeled 01 through 73. Before looking at the digits, choose the first row, start in column one, and take the first two digits of each group. Accept 01 through 73. For a sample without replacement, skip 00, values above 73, and any ID already selected. Continue in the same direction until the sample reaches its planned size.
In the first row, 78 and 94 fall outside the roster. The next two groups select invoices 10 and 05. After skipping 97, the next valid ID is 50. Why skip instead of wrapping 74 back to 01? With 100 possible two-digit strings and 73 IDs, a wraparound rule gives some IDs two chances and others only one. By contrast, rejecting out-of-range values gives every eligible ID one route to selection. A fixed seed lets a reader audit the classroom example. For a real audit, record the roster order, starting rule and selection procedure before drawing the sample.
Source · Python random number generator
Student’s t distribution table for an estimated mean
Suppose ten independent observations from a normal population give a mean of 12 and a sample standard deviation of 3. The standard error is 3 / √10, or about 0.949. Because the mean calculation uses one degree of freedom, select df = 9. For a two-sided 95% confidence interval, read the 0.05 row and the two-tail column to get 2.262157.
| df | α | Upper one-tail | Two-tail ± cutoff |
|---|---|---|---|
| 1 | 0.05 | 6.313752 | 12.706205 |
| 2 | 0.05 | 2.919986 | 4.302653 |
| 5 | 0.05 | 2.015048 | 2.570582 |
| 9 | 0.05 | 1.833113 | 2.262157 |
| 10 | 0.05 | 1.812461 | 2.228139 |
| 15 | 0.05 | 1.75305 | 2.13145 |
| 20 | 0.05 | 1.724718 | 2.085963 |
| 25 | 0.05 | 1.708141 | 2.059539 |
| 30 | 0.05 | 1.697261 | 2.042272 |
| 60 | 0.05 | 1.670649 | 2.000298 |
| 120 | 0.05 | 1.657651 | 1.97993 |
Build the interval and check the tail
Multiplying 2.262157 by the standard error gives a margin of about 2.146. Therefore, the interval runs from 9.854 to 14.146. If the question asks only for a 95% upper confidence bound, the multiplier changes to 1.833113. The upper bound then becomes about 13.739.
The difference matters most in small samples. For example, replacing 2.262 with 1.960 shrinks the two-sided margin by roughly 13%. As the degrees of freedom increase, the t cutoffs approach their normal counterparts. In a regression, use the residual degrees of freedom from the fitted model. A simple n − 1 rule applies to the one-sample mean example.
Source · SciPy t distribution
Chi-square distribution table for variance and counts
A sample of ten independent normal observations has sample variance 4. To estimate population variance, start with (n − 1)s² = 36. A two-sided 95% interval leaves 0.025 in each tail. At nine degrees of freedom, the lower and upper chi-square cutoffs are about 2.7004 and 19.0228.
| df | Tail area α | Lower cutoff | Upper cutoff |
|---|---|---|---|
| 1 | 0.025 | 0.000982 | 5.023886 |
| 2 | 0.025 | 0.050636 | 7.377759 |
| 5 | 0.025 | 0.831212 | 12.832502 |
| 9 | 0.025 | 2.700389 | 19.022768 |
| 10 | 0.025 | 3.246973 | 20.483177 |
| 15 | 0.025 | 6.262138 | 27.488393 |
| 20 | 0.025 | 9.590777 | 34.169607 |
| 25 | 0.025 | 13.11972 | 40.646469 |
| 30 | 0.025 | 16.790772 | 46.979242 |
| 60 | 0.025 | 40.481748 | 83.297675 |
| 120 | 0.025 | 91.572642 | 152.211403 |
Why the larger cutoff gives the smaller endpoint
First divide 36 by 19.0228 to obtain 1.8925. Next divide 36 by 2.7004 to obtain about 13.33. Thus, the interval for variance stretches much farther upward than downward from the estimate of 4. To report standard deviation in the original measurement units, take the square root of each endpoint. That gives about 1.38 to 3.65.
For a contingency table, the lookup answers a different question. A standard independence test with three rows and four columns uses (3 − 1)(4 − 1) = 6 degrees of freedom. At an upper-tail area of 0.05, its cutoff is about 12.592. Check the expected cell counts and sampling design before interpreting the large-sample approximation.
Source · SciPy chi-square distribution
F distribution table for joint restrictions and variance ratios
In a classical linear regression, an F test can ask whether several slope coefficients are jointly zero. Suppose a restricted model has residual sum of squares 160 and an unrestricted model has 120. The additional regressors use two degrees of freedom, and the unrestricted model retains 20 residual degrees of freedom. Consequently, the observed statistic is [(160 − 120) / 2] / [120 / 20] = 3.3333.
| Numerator df | Denominator df | Upper area α | Upper cutoff |
|---|---|---|---|
| 1 | 20 | 0.05 | 4.351244 |
| 2 | 20 | 0.05 | 3.492828 |
| 3 | 20 | 0.05 | 3.098391 |
| 4 | 20 | 0.05 | 2.866081 |
| 5 | 20 | 0.05 | 2.71089 |
| 6 | 20 | 0.05 | 2.598978 |
| 8 | 20 | 0.05 | 2.447064 |
| 10 | 20 | 0.05 | 2.347878 |
| 12 | 20 | 0.05 | 2.277581 |
| 20 | 20 | 0.05 | 2.124155 |
Keep the numerator and denominator in order
At α = 0.05, the F cutoff for (2, 20) is about 3.4928. Since 3.3333 falls below that threshold, the test fails to reject the joint restriction. For this comparison, fit both models to exactly the same observations. Otherwise, a change in the estimation sample also changes the residual sums of squares.
One identity makes an excellent spreadsheet check. With one numerator degree of freedom, the upper 5% F cutoff equals the square of the two-sided 5% t cutoff at the same denominator degrees of freedom. For example, at 20 residual degrees of freedom, 2.085963 squared gives about 4.3512. Compare those two entries in the workbook to catch a reversed tail or a misplaced df.
Source · SciPy F distribution
Durbin–Watson statistic and its two critical bounds
Order regression residuals by time before calculating d. In the numerator, sum the squared changes between neighboring residuals. In the denominator, sum all squared residuals. For example, the sequence 1, 1, −1, −1 gives d = 4 / 4 = 1. Reordering the same values as 1, −1, 1, −1 gives d = 12 / 4 = 3. The sums of squares match, yet the timing changes the diagnostic.
| Observations n | Slopes k | Lower bound dL | Upper bound dU |
|---|---|---|---|
| 15 | 2 | 0.945 | 1.543 |
| 20 | 2 | 1.1 | 1.537 |
| 25 | 2 | 1.206 | 1.55 |
| 30 | 2 | 1.284 | 1.567 |
| 40 | 2 | 1.391 | 1.6 |
| 50 | 2 | 1.462 | 1.628 |
| 60 | 2 | 1.514 | 1.652 |
| 80 | 2 | 1.586 | 1.688 |
| 100 | 2 | 1.634 | 1.715 |
Read the inconclusive region
Suppose a regression includes an intercept, two slopes and 20 observations. For a one-sided 5% test of positive first-order autocorrelation, the bounds are 1.100 and 1.537. A statistic of 0.95 falls below the lower bound, so reject the no-autocorrelation null. By contrast, 1.30 lies between the bounds and leaves the test inconclusive. A value of 1.80 exceeds the upper bound, so this positive-direction test fails to reject.
For negative autocorrelation, apply the same one-sided rule to 4 − d. Treat a value equal to a printed bound as a boundary case because the source rounds the entries. A lagged dependent variable in the regression changes the validity of the classical bounds procedure. Also, the four-observation example above illustrates arithmetic only. Use the actual regression sample size for inference.
Source · Penn State critical bounds
Wilcoxon W table for paired differences
Here W means the smaller of the positive and negative rank sums in a two-sided signed-rank test. Start with independent pairs and compute one difference per pair using a consistent direction. Under the null, a continuous symmetric difference distribution centers on zero. Rank the absolute differences from smallest to largest, then restore each sign.
| Nonzero pairs n | Two-sided α | Reject at W ≤ | Attained size |
|---|---|---|---|
| 4 | 0.05 | None | 0 |
| 5 | 0.05 | None | 0 |
| 6 | 0.05 | 0 | 0.03125 |
| 7 | 0.05 | 2 | 0.046875 |
| 8 | 0.05 | 3 | 0.039062 |
| 9 | 0.05 | 5 | 0.039062 |
| 10 | 0.05 | 8 | 0.048828 |
| 11 | 0.05 | 10 | 0.041992 |
| 12 | 0.05 | 13 | 0.04248 |
| 13 | 0.05 | 17 | 0.047852 |
| 14 | 0.05 | 21 | 0.049438 |
| 15 | 0.05 | 25 | 0.047913 |
Why a nominal 5% test can reject only 3.125% of the time
Consider six differences of +1, +2, +3, +4, +5 and +6. Their positive ranks sum to 21 and their negative ranks sum to zero. Therefore W = 0. Under the null, six independent signs produce 64 equally likely patterns. Only the all-positive and all-negative patterns have W = 0, giving a two-sided probability of 2 / 64 = 0.03125.
For five pairs, even complete agreement gives 2 / 32 = 0.0625. As a result, the 5% table shows “None” for that sample size. A zero cutoff would incorrectly allow rejection. If zero differences or ties in the absolute differences occur, use software that handles them explicitly. For independent samples, use a rank-sum procedure with its own reference distribution.
Source · SciPy signed-rank test
Kolmogorov–Smirnov critical values for a distribution check
The one-sample, two-sided KS statistic D measures the largest vertical gap between the empirical cumulative distribution and a fully specified continuous reference distribution. For example, an analyst could compare draws from a simulation with a Uniform(0,1) model fixed before examining those draws. The lookup uses the finite-sample distribution of D for the selected n.
| Sample size n | Two-sided α | Critical D |
|---|---|---|
| 5 | 0.05 | 0.563275 |
| 10 | 0.05 | 0.409246 |
| 15 | 0.05 | 0.337596 |
| 20 | 0.05 | 0.294075 |
| 25 | 0.05 | 0.264041 |
| 30 | 0.05 | 0.241703 |
| 40 | 0.05 | 0.210115 |
| 50 | 0.05 | 0.188406 |
| 75 | 0.05 | 0.15442 |
| 100 | 0.05 | 0.134028 |
| 200 | 0.05 | 0.095158 |
Check both sides of every step
Suppose the ordered observations are 0.10, 0.20, 0.60 and 0.90. Immediately after the second observation, the empirical curve reaches 2 / 4 = 0.50, while the uniform curve equals 0.20. The gap is 0.30. Checking the other observations and the left limits at each step gives no larger gap, so D = 0.30.
At n = 20, the 5% cutoff is approximately 0.2941. A statistic of 0.30 would exceed that cutoff for twenty observations. However, the four-observation example must use the much larger cutoff for n = 4. Likewise, if you estimate the mean and standard deviation of a normal reference from the tested data, the ordinary KS calibration no longer applies. Use a fitted-parameter procedure or a parametric bootstrap that repeats the fitting step.
Source · SciPy finite-sample KS distribution
ADF critical values for a unit-root test
The augmented Dickey–Fuller regression relates a series change to its lagged level, optional deterministic terms and lagged changes. Under the unit-root null, the coefficient on the lagged level in this regression equals zero. Although the statistic looks like a t ratio, its null distribution depends on the deterministic terms. More negative values provide evidence against the unit-root null.
| Effective T | Regression | Left area α | Critical tau |
|---|---|---|---|
| 25 | Constant | 0.05 | -2.986489 |
| 25 | Constant + trend | 0.05 | -3.603468 |
| 50 | Constant | 0.05 | -2.92136 |
| 50 | Constant + trend | 0.05 | -3.502275 |
| 100 | Constant | 0.05 | -2.890906 |
| 100 | Constant + trend | 0.05 | -3.455343 |
| 250 | Constant | 0.05 | -2.873172 |
| 250 | Constant + trend | 0.05 | -3.428199 |
| 500 | Constant | 0.05 | -2.867338 |
| 500 | Constant + trend | 0.05 | -3.419307 |
How a trend changes the decision
Consider an ADF statistic of −3.10 with 100 usable regression observations. At 5%, the constant-only cutoff is about −2.8912. The statistic falls below it, so that specification rejects the unit-root null. With a constant and linear trend, the cutoff is about −3.4553. The same numerical statistic would now fail to reject. In an actual analysis, refitting with a trend will usually change the statistic as well.
Choose the deterministic terms from the economic question and the series behavior before comparing results. Then count the observations that actually entered the fitted regression. For example, a series of 104 complete consecutive levels with three lagged differences leaves 100 usable observations after differencing and lagging. These entries use MacKinnon’s single-series response surfaces. An Engle–Granger test on estimated residuals requires a different calibration.
Source · MacKinnon, 2010, Tables 2 and 3
Three checks before you report a result
Read the inequality aloud. For example, “at least four” includes four. Consequently, subtract the cumulative probability through three from one. In Excel, the Poisson example uses =1-POISSON.DIST(3,2,TRUE).
Check the units of the answer. A t cutoff of 2.262 measures standard errors. By contrast, a KS cutoff of 0.2941 measures a vertical difference between two cumulative probabilities. Neither number, on its own, gives the probability that the null hypothesis is true.
Keep precision until the decision. The printed tables round their entries, and the DW source publishes rounded bounds. Therefore, when a statistic sits very close to a boundary, record more digits or consult the original procedure. For discrete tests, also check whether the rule includes equality. Wilcoxon W at its cutoff can still fall inside the rejection region.
How to document a statistical table lookup
Give your reader enough detail to reproduce one cell. For example, write “The joint test gave F(2, 20) = 3.333. The 5% upper cutoff was 3.493, so we failed to reject the two restrictions.” For a paired test, add the definition of W and the treatment of ties. For a unit-root test, name the deterministic terms, lag count and effective sample size. Those details let another reader check the conclusion even when their software uses a different table layout.
For further practice, see 12 solved regression diagnostics exercises and the NIST distribution gallery.
