Two group averages do not automatically get equal votes

Question

NIST’s Dataplot manual defines a weighted mean as the sum of each value multiplied by its weight, divided by the sum of the weights. It identifies frequency data as one use.

That is the source rule for this workshop; the cases and calculations below are Gretna Junction’s original, explicitly hypothetical teaching examples.

Imagine being handed two group summaries and asked for one combined result. Before adding anything, decide whether each group should count equally or each underlying observation should count equally.

Those are different questions. This workshop concerns invented numerical records, not the performance of any Southside school, employer, service or community.

No local observations were collected.

Counts

In the first invented dataset, Group A contains 10 observations whose exact average is 20 points. Group B contains 90 observations whose exact average is 40 points.

Adding the two averages and dividing by two gives 30. That calculation gives the small group half the influence even though it contains only one tenth of the observations.

To give each observation equal influence, recover each group’s total points from its exact average and count. Group A contributes 10 × 20 = 200 points.

Group B contributes 90 × 40 = 3,600 points. Together they contribute 3,800 points across 100 observations.

Dividing 3,800 by 100 gives 38 points. The result is closer to 40 because 90 of the 100 observations belong to that group.

Neither 30 nor 38 is a calculator malfunction. Thirty answers an equal-group question; 38 answers an equal-observation question.

A useful caption should identify which question was answered. Simply calling either result the average leaves that decision hidden.

Mixture

Now create a second invented dataset. Give Group A 90 observations at an exact group average of 20, and Group B 10 observations at an exact group average of 40.

The two group averages have not changed. Their equal-group average is still 30.

But the combined total is now 1,800 + 400 = 2,200 points across 100 observations, so the equal-observation result is 22.

Comparing 38 in the first dataset with 22 in the second might look like a change in underlying group performance. In this constructed example it is entirely a change in the mixture: the low-average group has become the large group.

This does not explain any real trend. It demonstrates why an overall change alone cannot tell a reader whether group results changed, group sizes changed, or both happened.

Subdivision

Return to the first dataset and divide Group B into two subgroups. Suppose one has 30 observations averaging exactly 40, and the other has 60 observations also averaging exactly 40.

Nothing about the original 100 observations has changed. The totals are now 200, 1,200 and 2,400 points, still adding to 3,800.

The combined result remains 38.

A simple average of the three displayed subgroup averages would instead be (20 + 40 + 40) ÷ 3, or about 33.33. Merely redrawing the group boundaries would have changed that summary from 30 to about 33.33.

This is a practical diagnostic for the intended equal-observation calculation: check whether dividing a group into smaller reporting units changes its influence without adding observations.

That diagnostic does not forbid an equal-group analysis. It asks the author to defend why the reporting units, rather than their members, receive equal weight.

An analysis of typical groups may deliberately use equal group weights; it should not quietly be described as the combined result for all observations.

Missing counts

Suppose the 20-point and 40-point averages are supplied but their counts are missing. The combined equal-observation result cannot be recovered uniquely.

Our two datasets already prove that: identical displayed group averages produced either 38 or 22 once the counts were specified. Choosing 30 would add an equal-group assumption, not retrieve missing information.

A blank count also should not be entered as zero merely to make a spreadsheet formula run. Zero observations means no observations contribute from that group; an unknown count means the contribution has not been established.

In this exercise the correct response is to retain the two summaries separately until the intended weights are available, or to state the equal-group assumption openly.

Rounding

So far the invented group averages were exact. Suppose instead that 20 and 40 were rounded to the nearest whole point, while the counts 10 and 90 were exact.

Ignoring the convention for exact halfway ties, the first unrounded average could lie between 19.5 and 20.5, and the second between 39.5 and 40.5.

Using the lower endpoints gives (10 × 19.5 + 90 × 39.5) ÷ 100 = 37.5. Using the upper endpoints gives 38.5.

Thus the displayed inputs do not justify pretending that 38.000 is an exact recovered result. This interval is original arithmetic about rounding bounds, not a confidence interval, a measured error estimate or a prediction about real data.

Readback

For this first invented dataset, a transparent result would say: 100 observations, group counts 10 and 90, group averages treated as exact, combined average 38 points. That sentence carries enough information for someone to reconstruct the numerator and denominator.

If the averages were rounded, the sentence should say that instead of adding extra decimal places.

Before using a combined figure in a local discussion, ask for the grouping rule, the intended weights and the precision of the inputs. Then check the arithmetic separately from the interpretation.

Getting 38 correctly does not establish that the fictional points measure anything valuable; it establishes only that this stated equal-observation calculation was performed consistently.