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Choose a route or read from the beginning. All teaching chapters stay open; the grouped index below is expandable.
Define the summaries
Chapters 1–2. Check reported precision, group sizes and variance definitions.
Build the reconstruction
Chapters 3–5. Recover total spread and distinguish it from within-group pooling.
Bound the missing digits
Chapters 6–7. Translate rounded means into a separation interval and variance bounds.
Test the information model
Chapters 8–10. Change rounding precision, dependence and uncertainty assumptions.
Practise and continue
Chapters 11–12. Work a fresh interval and choose related statistics guides.
Full chapter index · Practice and worked checks · Statistics subject root
Rounded group means can make a reconstructed variance look more precise than its inputs allow. The usual combining formula is exact for exact group summaries, but it cannot recover digits that were removed from a report. A small uncertainty in the distance between group means can have a noticeable effect on the combined spread.
This guide first establishes the combining identity, then derives bounds for fictional rounded summaries. The bounds describe missing numerical information, not a confidence interval or a claim about measurement accuracy. Keeping those distinctions visible is the main learning goal.
Chapter index
Define the summaries · Chapters 1–2
1. Ask whether the summaries are exact
2. Define the sample quantities
Build the reconstruction · Chapters 3–5
3. Split one deviation before squaring
4. Work through the six readings
5. Separate pooling from combining
Bound the missing digits · Chapters 6–7
6. Turn mean-rounding into a separation interval
7. Calculate bounds rather than extra digits
Test the information model · Chapters 8–10
8. Change the rounding model before changing the answer
9. Keep rounding bounds separate from statistical uncertainty
10. Questions readers often ask
Practise and continue · Chapters 11–12
11. Practise with a changed separation interval
12. Choose the next reading route
Imagine two groups reported with means 10 and 20, three observations per group and sample variance 4 in each group. If the means are exact, the combined sample variance is 33.2. If they were rounded to the nearest unit, 33.2 is only the result obtained by treating the displayed values as exact.
The true group centres could be closer together or further apart. Their own within-group spreads can remain unchanged while their separation changes.
We will distinguish arithmetic from information. A calculator can give many digits from rounded inputs, but those digits do not establish the omitted group means. Ask for the reporting convention before interpreting the reconstructed result.
For group i, let nᵢ be its number of observations, mᵢ its arithmetic mean and sᵢ² its ordinary sample variance. The sample variance divides the sum of squared deviations from that group’s mean by nᵢ − 1. Here each group has at least two observations.
Let N be the total number of observations and G the number of groups. The combined mean m is the size-weighted mean: m = Σ(nᵢmᵢ)/N. This is an ordinary unweighted dataset partitioned into groups, not a survey with unequal inclusion weights.
For an exact-summary baseline, use fictional readings A = 8, 10, 12 and B = 18, 20, 22. The means are 10 and 20. Each group’s squared deviations are 4, 0 and 4, so each has sample variance 8/2 = 4. Their standard deviations are 2 in the original measurement unit.
Take one reading x in group i. Its deviation from the combined mean can be written x − m = (x − mᵢ) + (mᵢ − m). The first part locates the reading relative to its own group. The second locates the group relative to the whole dataset.
Square this expression. It produces a within-group square, a between-centre square and a cross term. Within each group, Σ(x − mᵢ) = 0, so the cross terms cancel when all readings are summed.
Define W = Σ(nᵢ − 1)sᵢ² and B = Σnᵢ(mᵢ − m)². The exact identity is total sum of squares = W + B. Therefore the combined sample variance is s² = (W + B)/(N − 1).
This algebra does not require normally distributed observations or equal population variances. Those assumptions become relevant to particular inferential procedures, rather than to this descriptive identity.
The combined mean is 15. Within-group squared deviations contribute W = 8 + 8 = 16. The two centres are each five units from 15, so B = 3 × 25 + 3 × 25 = 150.
The combined sample variance is (16 + 150)/5 = 33.2. Its square root is approximately 5.76. Averaging the two group standard deviations would give 2, which misses most of the combined spread.
| Quantity | Calculation | Result |
|---|---|---|
| Group A sample variance | (4 + 0 + 4)/2 | 4 |
| Group B sample variance | (4 + 0 + 4)/2 | 4 |
| Within sum W | 8 + 8 | 16 |
| Between sum B | 3 × 25 + 3 × 25 | 150 |
| Combined sample variance | 166/5 | 33.2 |
Check directly against the combined mean: squared deviations of 8, 10, 12, 18, 20 and 22 from 15 are 49, 25, 9, 9, 25 and 49. They sum to 166. The reconstructed answer agrees with the raw-data answer.
The usual pooled within-group variance is W/(N − G). In this example it is 16/4 = 4. It describes dispersion around the separate group means; it is not the variance of all six readings around one combined mean.
Do not average standard deviations and then square the answer. Even averaging sample variances needs the appropriate degrees-of-freedom weights when the target is this pooled within-group quantity.
Nor should you claim that the combined sample variance must always exceed the pooled variance. They use different denominators. If both groups instead contain 8, 10 and 12, then B is zero: the combined sample variance is 16/5 = 3.2 while the pooled within-group variance remains 4.
The larger combined spread in the first example comes from the substantial mean separation, not from a universal inequality between the two estimators.
CHAPTER 6 OF 12 · Bound the missing digits
6. Turn mean-rounding into a separation interval
Suppose the reporting information guarantees that each unrounded mean lies within 0.5 unit of its displayed value. Use inclusive bounds as a conservative enclosure; the exact treatment of half-unit ties may make some endpoints unattainable.
Then the first mean lies between 9.5 and 10.5, while the second lies between 19.5 and 20.5. Their difference d lies between 9 and 11. We are allowing each mean to vary within its own reported range, without imposing another relationship.
For any two groups, B = [n₁n₂/(n₁ + n₂)]d². This follows by substituting the size-weighted combined mean into the two between-group terms.
With n₁ = n₂ = 3, B = 1.5d². Keeping the sample variances exactly 4 gives W = 16. Thus the combined sample variance is s² = 3.2 + 0.3d².
CHAPTER 7 OF 12 · Bound the missing digits
7. Calculate bounds rather than extra digits
For d between 9 and 11, the squared separation lies between 81 and 121. Therefore s² lies between 3.2 + 0.3 × 81 = 27.5 and 3.2 + 0.3 × 121 = 39.5. These are conservative bounds under the stated reporting assumptions.
The nominal displayed-means calculation, 33.2, sits inside that enclosure. Its apparent precision does not narrow the enclosure.
| Assumed mean separation | Within sum W | Between sum B | Combined sample variance |
|---|---|---|---|
| Lower enclosure d = 9 | 16 | 121.5 | 27.5 |
| Displayed separation d = 10 | 16 | 150 | 33.2 |
| Upper enclosure d = 11 | 16 | 181.5 | 39.5 |
For a concrete compatible case, use readings 7.6, 9.6, 11.6 and 18.4, 20.4, 22.4. The group means are 9.6 and 20.4, each rounds to the displayed integer, and both sample variances remain 4. Their combined sample variance is 3.2 + 0.3 × 10.8² = 38.192.
A direct calculation from those six readings gives the same answer. This confirms that the discrepancy is lost input information, rather than a failure of the combining identity.
CHAPTER 8 OF 12 · Test the information model
8. Change the rounding model before changing the answer
If the means are instead reported within 0.05 unit, their difference lies between 9.9 and 10.1. The same model gives bounds 32.603 and 33.803. Better retained input precision narrows this particular source of uncertainty.
Now return to the half-unit mean bounds, but suppose each reported sample variance 4 is itself known only to lie between 3.5 and 4.5. Then W lies between 14 and 18. Combining the allowed extremes gives a conservative total-variance enclosure from (14 + 121.5)/5 = 27.1 to (18 + 181.5)/5 = 39.9.
That broader range is conditional on the stated admissible summaries. Raw observations or additional constraints may rule out some combinations. Do not silently assume that every independently listed endpoint can occur together.
A different dependence matters too. If both exact means share one identical unknown additive offset, their difference remains 10. With exact within-group variances, this common shift leaves the combined variance 33.2. Independent rounding ranges and one shared calibration offset are different information models.
CHAPTER 9 OF 12 · Test the information model
9. Keep rounding bounds separate from statistical uncertainty
These bounds describe what can be reconstructed from incomplete reported digits. They are not a sampling confidence interval, a probability distribution for the missing digits or an assessment of instrument bias.
No normality or independence assumption is needed for the descriptive sums-of-squares identity. But statistical inference from the observations requires a suitable data-generating model. A rounding enclosure cannot replace that model.
Check that summaries concern a non-overlapping partition, compatible units, the same variable and known sample sizes. Overlapping groups, unequal observation weights or incompatible variance denominators require another calculation.
If the raw readings are available, calculate from them and preserve sufficient precision in the analysis record. If only rounded summaries survive, report the assumptions and the sensitivity rather than presenting the nominal reconstruction as exact.
Is the combining formula wrong when rounded inputs give another result?
No. It is being supplied approximate summaries. Exact algebra cannot reconstruct the missing input digits.
Are 27.5 and 39.5 confidence limits?
No. They enclose the possible numerical reconstructions under the stated half-unit mean bounds and exact variance assumptions. They have no assigned probability coverage.
Does an unknown common offset create the same problem?
Not for variance when it shifts every observation equally. The mean separation and centred deviations remain unchanged. Separate reporting errors can change the separation.
Can I average standard deviations to avoid the problem?
No. That changes the quantity and discards the between-group contribution. Keep the proper formula and make its input precision visible.
CHAPTER 11 OF 12 · Practise and continue
11. Practise with a changed separation interval
Use two groups of two observations each. Their exact sample variances are 2, but their reported means 5 and 11 each have bounds of plus or minus 0.1 unit. Work out the range of the true mean difference before calculating the variance.
The difference lies between 5.8 and 6.2. Here W = 2 + 2 = 4 and n₁n₂/(n₁ + n₂) = 1, so s² = (4 + d²)/3. The resulting enclosure is 12.5466 to 14.1467, with values rounded outward for reporting.
Treating the displayed means as exact gives 40/3, approximately 13.33. That nominal value is compatible with the bounds, but it does not erase them.
Now add the same unknown constant to both groups. Explain why the enclosure based on their separation is unchanged. Finally, identify what extra records would let you replace the enclosure with an exact descriptive reconstruction: the unrounded compatible means, sizes and variances, or the original observations.
For the broader map of measurement, distributions, study design and uncertainty, return to The Gold Standard of Statistics. For teaching the subject through questions and evidence, continue with Teaching Statistics: questions, variation and valid inference.
For foundational and computational routes, read How Variance Works and Welford’s algorithm for running means and online variance. Those guides explain variance and stable accumulation; this article tests the information lost when group summaries have already been rounded.
For unequal-variance mean comparisons, use Alexander–Govern and precision weights and the companion Alexander–Govern mean-comparison guide. These comparison methods address a different target from reconstructing one dataset’s total spread.
For the primary framework, read NIST’s one-way ANOVA explanation and NIST’s repeatability and pooled-variance discussion. The examples and derivation here are original teaching work.
Connect the calculation to The Gold Standard of Education: learning transfers into life, or explore the How X Works hub. A useful final explanation names the spread being described before presenting its number.
Sources checked 21 October 2026. Worked examples are educational models with the assumptions stated above.