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Two-Way Tables and Association · Tutorial 215 of 996

Simpson's Paradox in a Worked Example

Work through numerical tables to see how subgroup comparisons and an overall comparison can point in opposite directions.

Beginner 10 min read

What You'll Learn

  • Define Simpson’s paradox using subgroup and overall outcome rates
  • Calculate outcome rates from two-way table counts
  • Verify a reversal by comparing each subgroup with the pooled totals
  • Explain how different subgroup sizes can produce the reversal
  • Avoid treating a pooled comparison as if it described every subgroup

When a Comparison Reverses

A comparison can look clear within every relevant subgroup and still point in the opposite direction when all the observations are combined. This is not an arithmetic contradiction. The subgroup comparisons and the overall comparison use different collections of observations, and the groups being compared may be represented in different proportions across the subgroups.

This reversal is called Simpson’s paradox. The name describes a pattern that can arise in data; it does not mean that a table is incorrect or that a particular explanation has been established. To see the pattern, calculate the rates carefully, first within each subgroup and then from the combined totals.

Definition: Simpson’s paradox is a reversal in the direction of an association when data are combined across subgroups: one group has a higher (or lower) outcome rate than another in every subgroup, but a lower (or higher) rate in the pooled data. A subgroup rate is calculated using the observations in one subgroup; a pooled rate is calculated using the combined observations across subgroups.

For a success or outcome recorded as present or absent, the rate for a group is the number with the outcome divided by the number in that group. Always identify the denominator before comparing rates. A count of successes alone is not a rate: groups of different sizes can have different counts even when their rates are similar.

A Worked Reversal in Two Departments

Suppose, in a hypothetical example, two applicant groups—Redwood and Blue—apply to two departments. The outcome is whether an applicant is accepted. The table reports accepted and not accepted applicants, with the acceptance rate calculated within each group and department.

DepartmentGroupAcceptedNot acceptedApplicantsAcceptance rate
EngineeringRedwood1822018/20 = 90%
EngineeringBlue72189072/90 = 80%
ArtsRedwood832408/40 = 20%
ArtsBlue19101/10 = 10%

Worked Example: Comparing Subgroup Rates With the Pooled Rate

Using the department table, compare Redwood and Blue within each department, then calculate each group’s overall acceptance rate.

Engineering. Redwood’s acceptance rate is \(18/20=0.90\), or 90%. Blue’s is \(72/90=0.80\), or 80%. Redwood’s rate is 10 percentage points higher.

Arts. Redwood’s acceptance rate is \(8/40=0.20\), or 20%. Blue’s is \(1/10=0.10\), or 10%. Redwood’s rate is also higher here, by 10 percentage points.

Pooled totals. Redwood has \(18+8=26\) accepted applicants out of \(20+40=60\), so its overall rate is

$$ \frac{26}{60}=0.4333\ldots \approx 43.3\%. $$

Blue has \(72+1=73\) accepted applicants out of \(90+10=100\), so its overall rate is

$$ \frac{73}{100}=0.73=73.0\%. $$

Redwood has the higher acceptance rate in both departments, but its pooled rate is lower: approximately 43.3% compared with 73.0% for Blue. The direction of the comparison has reversed. The subgroup difference and the pooled difference answer distinct descriptive questions, so both calculations must be reported accurately.

The table also shows a major difference in the groups’ department distributions. Redwood’s 60 applicants include 20 in Engineering and 40 in Arts. Blue’s 100 applicants include 90 in Engineering and 10 in Arts. In this example, Engineering has higher acceptance rates for both groups than Arts does. Blue has many more applicants in Engineering, while a greater share of Redwood’s applicants are in Arts. As a result, combining the counts gives greater influence to the department where acceptance is more common for each group.

This explanation describes the arithmetic pattern in this table. It does not establish why the applicant distributions differ, nor does the table alone show what would happen if individuals or applications were assigned differently. Keep the conclusion tied to the rates actually calculated.

Why the Reversal Is Possible

The short derivation below expresses the essential arithmetic without requiring any special statistical model. Suppose group \(G\) has \(a_1\) outcomes among \(n_1\) observations in subgroup 1, and \(a_2\) outcomes among \(n_2\) observations in subgroup 2. When the observations are pooled, the outcome count is \(a_1+a_2\), and the total number of observations is \(n_1+n_2\). Therefore the pooled rate is

$$ \frac{a_1+a_2}{n_1+n_2}. $$

For the Redwood example, the numerator is \(18+8=26\), and the denominator is \(20+40=60\). For Blue, they are \(72+1=73\) and \(90+10=100\). Thus the pooled comparison depends on how many observations from each subgroup enter each group’s totals, not just on the direction of the two subgroup comparisons.

Here is a direct numerical check using only the totals. The pooled difference, Redwood minus Blue, is

$$ \frac{26}{60}-\frac{73}{100} =0.4333\ldots-0.73 =-0.2967\ldots. $$

The negative result says Redwood’s pooled rate is about 29.7 percentage points lower. Within Engineering, the difference is \(90\%-80\%=10\) percentage points; within Arts, it is \(20\%-10\%=10\) percentage points. The subgroup differences are both positive, while the pooled difference is negative. That sign change is the reversal.

A Second Example With Different Counts

The pattern is not limited to acceptance decisions. In another hypothetical example, two groups complete a task in three settings with different levels of difficulty. The outcome is completing the task successfully.

SettingGroupSuccessfulUnsuccessfulTotalSuccess rate
EasyP911090%
EasyQ821080%
ModerateP641060%
ModerateQ551050%
HardP208010020%
HardQ191010%

Worked Example: Checking a Reversal Across Three Settings

Check whether Group P has a higher success rate in every setting, and compare the result with the pooled rates.

In the Easy setting, the rates are \(9/10=90\%\) for P and \(8/10=80\%\) for Q. In the Moderate setting, they are \(6/10=60\%\) and \(5/10=50\%\). In the Hard setting, they are \(20/100=20\%\) and \(1/10=10\%\). Group P’s rate is higher in all three settings.

For the pooled rate, add the successful counts and the group totals separately. Group P has \(9+6+20=35\) successes among \(10+10+100=120\) attempts:

$$ \frac{35}{120}=0.2917\ldots \approx 29.2\%. $$

Group Q has \(8+5+1=14\) successes among \(10+10+10=30\) attempts:

$$ \frac{14}{30}=0.4667\ldots \approx 46.7\%. $$

So P has the higher success rate within each setting but the lower overall rate: approximately 29.2% compared with 46.7%. P contributes far more attempts from the Hard setting than Q does. Since both groups have lower success rates in that setting, this difference in the composition of their totals can reverse the pooled comparison.

It would be incorrect to report only “P does better” based on the subgroup rates, or only “Q does better” based on the pooled rates, without saying which comparison is being described. Each statement leaves out the level of comparison. A complete description names the outcome, the groups, and whether the rate is within a particular setting or pooled across settings.

Reading a Reversal Carefully

A useful way to read a possible Simpson’s paradox is to keep three questions separate:

1
What is the outcome rate?
For each group, divide the number with the outcome by the number in that same group. Keep the denominator attached to its subgroup.
2
What is the subgroup comparison?
Compare the groups within every listed subgroup. Record the direction of each comparison rather than relying on a single count.
3
What is the pooled comparison?
Add the outcome counts and group totals across the subgroups, calculate the overall rates, and check whether the direction differs from every subgroup comparison.

A reversal should be described only when the rates support it. If one group is higher in some subgroups and lower in others, but the pooled direction differs from only some of them, that is not the defining pattern given here. Likewise, a difference in raw outcome counts is not enough: the relevant quantities are rates with the appropriate denominators.

Common Mistake

  • Wrong reasoning: “Group P has the higher rate in every setting, so P must have the higher overall rate.” This assumes the subgroup comparisons automatically determine the pooled comparison.
  • Correction: Calculate the pooled rate from the combined outcome counts and combined group totals. The subgroup sizes can differ between groups, so the pooled comparison can point in the opposite direction. Report the subgroup and pooled results distinctly.

Another mistake is to describe the reversal as proof that one comparison is the “real” one. The table supports multiple descriptive comparisons, each with its own denominator and scope. Deciding which comparison is relevant depends on the question being asked; the arithmetic alone does not settle that question.

Key takeaway: Simpson’s paradox occurs when a group’s rate is higher in every subgroup but lower after pooling, or vice versa. Check all subgroup rates and the pooled rates from their respective counts and denominators before describing the pattern.

Check Your Understanding

Use rates and denominators in each answer.

  1. In the department example, what are Redwood’s and Blue’s acceptance rates in Engineering, and which is higher?
  2. Using the department example, calculate the pooled acceptance rate for each group from the totals.
  3. In the three-setting example, is Group P’s rate higher or lower in the Hard setting? Show the fractions used.
  4. Explain in one or two sentences how different subgroup sizes can contribute to the reversal in the department example.
  5. A group has a larger number of successful outcomes than another group. Why is that fact alone not enough to conclude that its success rate is higher?