How Many Decimal Places in Statistics?

By Jude Wallis · Published

Keep every digit your calculator gives you through the intermediate steps, and round only the final answer. Report test statistics to 2 or 3 decimals and p-values to 3 or 4. Rounding a standard error or a z-score mid-calculation can move the p-value enough to flip a decision.

Precision and rounding belong to the "Carrying Out a Test" steps of the Fall 2026 AP Statistics course, Unit 3 topics 3.7 and 3.15 and Unit 4 topic 4.5, where the test statistic and p-value are computed and reported.

The rule: round once, at the end

There are two separate questions here, and mixing them up is the whole problem.

  • How much precision do you carry while calculating? All of it. Every digit the calculator holds, at every intermediate step.
  • How much precision do you report? Enough for a reader to follow and check your work, and no more. Usually two to four decimals.

Rounding early is not a small approximation that washes out. The rounded value goes into a division, then that quotient goes into a tail probability, and each step can magnify the error rather than shrink it. A standard error rounded from 1.7527 to 1.8 is a change of under 3%; by the time it reaches the p-value it has become a 12% change, which is enough to move a result from one side of α\alpha to the other.

The practical version of the rule: never retype a number you just computed. Chain the calculation, or store the intermediate value and recall it. Retyping a rounded readout is the single most common way students introduce this error.

How many digits to report for each quantity

Use this as the default, and follow any instruction in the problem that overrides it.

QuantityReport toNotes
Sample mean xˉ\bar{x}, standard deviation ss3 or 4 significant digitskeep full precision internally
Sample proportion p^\hat{p}3 or 4 decimalsnever round to 2 before using it
Standard error4 decimalsthis is the value most often rounded too soon
Test statistic zz, tt, or χ2\chi^22 or 3 decimals2.66 or 2.661, not 2.7
p-value3 or 4 decimalswrite p<0.0001p < 0.0001 rather than p=0p = 0
Confidence interval endpoints3 or 4 decimals, or the units of the problem(0.552, 0.648), or 6.07 to 13.93 hours
Expected counts in a chi-square tablekeep the decimals24.6 is a valid expected count, do not round to 25
Correlation rr and r2r^23 or 4 decimals
Degrees of freedominteger for one-sample procedurestwo-sample tt gives a decimal dfdf; keep it

Two entries deserve a comment. Expected counts genuinely are not whole numbers, and rounding them changes every term of the chi-square statistic, so leave them alone. And the decimal degrees of freedom from a two-sample t procedure are correct, not a display glitch. Why df is a decimal in a two-sample t-test explains where it comes from.

One formatting habit worth keeping: report a proportion as a proportion (0.308) or as a percent (30.8%), and say which. Do not report 0.308% for a proportion of 0.308.

Why rounding a z or t moves the p-value

A p-value is a tail area, and out in the tail a small shift in the test statistic is a large fraction of an already small area. That is why the digits you keep in zz or tt show up in the p-value you report.

Watch what a two-decimal round does to a two-sided z-test:

zzTwo-sided p-value
2.1349 (full precision)0.032769
2.13 (rounded to 2 decimals)0.033171
2.1 (rounded to 1 decimal)0.035729

Rounding to two decimals moved the p-value in the third decimal place, from 0.0328 to 0.0332. Rounding to one decimal moved that same third decimal place more than seven times as far, from 0.033 to 0.036. Neither changed the decision here, but neither was harmless either, and reporting 0.036 when the answer is 0.0328 is simply wrong.

Rounding earlier than the test statistic is worse, because the error gets divided into a small number. Suppose 37 of 120 sampled households recycle and you test H0:p=0.25H_0: p = 0.25. The sample proportion is p^=37/120=0.308333\hat{p} = 37/120 = 0.308333, and the standard error under the null is 0.25(0.75)/120=0.039528\sqrt{0.25(0.75)/120} = 0.039528. Full precision gives z=1.4757z = 1.4757 and p=0.1400p = 0.1400. Round p^\hat{p} to 0.31 first and you get z=1.5179z = 1.5179 and p=0.1290p = 0.1290, an error of about 8% in the reported p-value, from a rounding that looked completely innocent.

The rounding that flips a decision

Here is the case that should end the habit for good.

A random sample of 30 measurements has xˉ=53.6\bar{x} = 53.6 and s=9.6s = 9.6. Test H0:μ=50H_0: \mu = 50 against Ha:μ50H_a: \mu \neq 50 at α=0.05\alpha = 0.05, with df=29df = 29.

Full precision. SE=9.630=9.65.477226=1.752712SE = \frac{9.6}{\sqrt{30}} = \frac{9.6}{5.477226} = 1.752712, so t=3.61.752712=2.0540t = \frac{3.6}{1.752712} = 2.0540 and p=0.0491p = 0.0491. Since 0.04910.050.0491 \leq 0.05, you reject H0H_0.

Standard error rounded to 1.8 first. Then t=3.61.8=2.0000t = \frac{3.6}{1.8} = 2.0000 and p=0.0549p = 0.0549. Since 0.0549>0.050.0549 > 0.05, you fail to reject H0H_0.

Same data, opposite conclusions, and the only difference is that someone wrote down 1.8 instead of 1.752712. The critical value here is t=2.045t^* = 2.045 for df=29df = 29, and the true statistic 2.0540 clears it by less than one hundredth. Rounding the standard error to one decimal is enough to lose that. Rounding it to four decimals is not, and neither is rounding the test statistic itself to two: 3.61.7527=2.0540\frac{3.6}{1.7527} = 2.0540 still clears tt^*, and so does a reported t=2.05t = 2.05.

The honest reading of this example is not that the rounding was the difference between right and wrong science. A p-value of 0.0491 and a p-value of 0.0549 are nearly the same evidence, which is the point of what is a good p-value. It is that your arithmetic should not be what decides the sentence you write. Compute it correctly, report 0.0491, and note that the result is marginal.

Keeping full precision on a calculator

Three habits on a TI-84 remove almost all of this risk.

  • Use `Ans`. The last computed value is available as `Ans` (press `2nd` then `(-)`), and it carries every digit the calculator has, not the ones on screen. Compute the standard error, then divide directly into it.
  • Store, do not retype. `STO` (the arrow key above `ON`) writes a value into a letter variable. Store the standard error in a variable and use that variable in the next line.
  • Let the built-in test do the whole thing. `STAT` then `TESTS` runs the one-sample t-test, the one-proportion z-test, and the chi-square tests end to end at full precision, so no intermediate value is ever rounded. How to run a t-test on the TI-84 walks through the menus, and reading calculator output covers what the display is telling you.

After `1-Var Stats`, the summary values live under `VARS` then `5:Statistics`, so you can pull xˉ\bar{x} and ss into a later calculation without copying digits by hand.

The same principle applies to any tool. On this site, the p-value calculator and the one-sample t-test calculator compute from the inputs you type at full precision, so type the raw numbers rather than a rounded statistic you worked out yourself.

Showing enough digits in written work

On free-response questions your solution is read as an argument, not just checked as a final number, so the digits you show are part of the communication. A reader should be able to take your written values and reproduce your next line.

That gives a simple standard: show the quantity that goes into the next step at enough precision that redoing the step lands on your answer. If you write SE=1.75SE = 1.75 and then t=2.054t = 2.054, the two lines do not agree, and a reader has to guess whether you rounded or made an error. Writing SE=1.7527SE = 1.7527 removes the doubt.

A few conventions worth adopting.

  • Keep units on every reported quantity: 6.07 hours, 3.2 mm Hg, 30.8%.
  • Round confidence interval endpoints to the same number of decimals as each other, and to a precision the context can support.
  • Write p<0.0001p < 0.0001 when the p-value underflows the display. Writing p=0p = 0 claims something impossible.
  • Do not round a value and then use the rounded value later in the same problem. Report the rounded version; keep computing with the full one.
  • If a problem says "round to three decimal places," that instruction applies to the final answer, not to the working.

The AP Statistics free-response guide covers how the rest of a written answer should be laid out.

How three innocent-looking roundings change the same p-value

Twelve reaction times, in seconds, are 0.42, 0.38, 0.51, 0.47, 0.35, 0.44, 0.49, 0.41, 0.53, 0.39, 0.46, 0.45. Test H0:μ=0.40H_0: \mu = 0.40 against Ha:μ0.40H_a: \mu \neq 0.40 at α=0.05\alpha = 0.05. Compute the p-value at full precision, then recompute it after rounding the sample mean to two decimals, and again after rounding the sample standard deviation to two decimals.

  1. Sum the data: 0.42+0.38+0.51+0.47+0.35+0.44+0.49+0.41+0.53+0.39+0.46+0.45=5.300.42 + 0.38 + 0.51 + 0.47 + 0.35 + 0.44 + 0.49 + 0.41 + 0.53 + 0.39 + 0.46 + 0.45 = 5.30, with n=12n = 12.

  2. Sample mean: xˉ=5.3012=0.4416667\bar{x} = \frac{5.30}{12} = 0.4416667 seconds.

  3. Sum of squared deviations from the mean is 0.0323667, so the sample variance is 0.032366711=0.0029424\frac{0.0323667}{11} = 0.0029424 and s=0.0029424=0.0542441s = \sqrt{0.0029424} = 0.0542441.

  4. Check conditions: treat the 12 trials as a random sample, and with n=12n = 12 the population must be plausibly normal. The values are tightly clustered with no outlier, so proceed.

  5. Full precision. SE=0.054244112=0.05424413.4641016=0.0156589SE = \frac{0.0542441}{\sqrt{12}} = \frac{0.0542441}{3.4641016} = 0.0156589, so t=0.44166670.400.0156589=2.6609t = \frac{0.4416667 - 0.40}{0.0156589} = 2.6609 with df=11df = 11, and the two-sided p-value is p=0.0222p = 0.0222.

  6. Now round the mean to two decimals first, as the data themselves are given to two decimals. Using xˉ=0.44\bar{x} = 0.44: t=0.440.400.0156589=2.5545t = \frac{0.44 - 0.40}{0.0156589} = 2.5545, and p=0.0268p = 0.0268.

  7. Compare: 0.0222 against 0.0268. The third decimal place of the p-value changed, from 0.022 to 0.027, and the reported p-value is off by 21%.

  8. Now instead round the standard deviation to two decimals. Using s=0.05s = 0.05: SE=0.053.4641016=0.0144338SE = \frac{0.05}{3.4641016} = 0.0144338, so t=0.04166670.0144338=2.8868t = \frac{0.0416667}{0.0144338} = 2.8868 and p=0.0148p = 0.0148.

  9. Compare: 0.0222 against 0.0148, an error of a third of the p-value, and this time in the direction that overstates the evidence.

  10. For reference, rounding ss to three decimals instead (s=0.054s = 0.054) gives t=2.6729t = 2.6729 and p=0.0217p = 0.0217, still wrong in the fourth decimal even though the rounding looks harmless.

At full precision, t=2.661t = 2.661 with df=11df = 11 and p=0.0222p = 0.0222, so you reject H0H_0 at α=0.05\alpha = 0.05. Rounding xˉ\bar{x} to 0.44 first gives p=0.0268p = 0.0268; rounding ss to 0.05 first gives p=0.0148p = 0.0148. All three decisions agree here, but only one p-value is correct. Report t=2.661t = 2.661 and p=0.0222p = 0.0222, and carry the unrounded values through every step to get there.

Frequently asked questions

How many decimal places should a p-value have?

Three or four, for example p=0.0222p = 0.0222 or p=0.022p = 0.022. If the value is smaller than 0.0001, write p<0.0001p < 0.0001 instead of a long string of zeros, and never write p=0p = 0, since a tail area is never exactly zero.

How many decimals for a z-score or t-statistic?

Two or three, such as z=2.13z = 2.13 or t=2.661t = 2.661. That is precise enough for a reader to reproduce the p-value from a table or a calculator. Do not round to one decimal, since it can shift the p-value by several units in the third decimal place: z=2.1349z = 2.1349 gives p=0.0328p = 0.0328, and z=2.1z = 2.1 gives p=0.0357p = 0.0357.

Does rounding really change the answer?

It can change the decision. In the example on this page, a standard error of 1.752712 rounded to 1.8 moved a p-value from 0.0491 to 0.0549, which crosses α=0.05\alpha = 0.05 and reverses the written conclusion. The fix costs nothing: chain the calculation instead of retyping.

Should I round expected counts in a chi-square test?

No. Expected counts are not required to be whole numbers, and a value like 24.6 goes into the statistic exactly as it is. Rounding them to integers changes every term of the sum and can also change whether the large counts condition appears to be met.

What if the problem tells me to round to a certain number of places?

Follow it for the final reported answer only. Keep full precision in every intermediate step, then round once at the end. An instruction about how to present a result is not an instruction about how to compute it.