Experiment 1
Change only the new-method easy share to 90%. Predict both observed rates.
Check the prediction
New-method success becomes 84%; old stays 74%. The within-difficulty advantage was always ten percentage points.
Causality and scientific reasoning
Compare like populations before interventions.
Explain a reversal in aggregate rates and state the assumptions needed to interpret standardization causally.
Start with: Conditional percentages and weighted averages. A confounder influences both intervention choice and outcome.
01 · Commit to a prediction
What do you expect?
Yes. With 10% easy jobs in the new-method group and 90% in the old, their overall success rates are 36% and 74%. Different job mixes reverse the comparison.
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02 · Change an assumption
Predict. Change. Explain.
Before changing a control, say what should move and why. Start with the experiments below. Reset restores the starting model; it preserves your written responses.
Experiment 2
Reset. Change only the target easy share from 50% to 0%.
Check the prediction
The standardized rates become 30% and 20%. The observed group rates stay 36% and 74%. Here the target changes the rates but not the 10-point contrast, because the advantage is 10 points in both strata. Set hard-job success to 20% and move the target again. Now the answer depends on which population you ask about.
0 to 100 · step 1
0 to 100 · step 1
0 to 100 · step 1
0 to 100 · step 1 · old method: 20
Calculated model output. The table gives the same values. Displayed values are rounded; calculations keep full precision.
Read the plotted data as a table
03 · Connect the mechanism
From the picture to the quantities.
Difficulty changes the probability of success and can influence which method is chosen. Pooling unlike job mixes combines a method difference with a composition difference. Standardization computes both methods on the same target mixture. That repairs this constructed confounding pattern, but arithmetic cannot certify that every confounder was measured.
Observed new = 0.9 a + 0.3(1−a)
Observed old = 0.8 b + 0.2(1−b)
Standardized new = 0.9 q + 0.3(1−q)
Standardized old = 0.8 q + 0.2(1−q)
Observed old = 0.8 b + 0.2(1−b)
Standardized new = 0.9 q + 0.3(1−q)
Standardized old = 0.8 q + 0.2(1−q)
a and b are the fractions of easy jobs within their respective method groups. q is the easy-job fraction in a common target population. All lie in [0,1]; controls display percent. A percentage-point difference subtracts two percentages. Rates here are specified population probabilities, not noisy sample estimates.
Worked example
For 1,000 new-method jobs, imagine 100 easy and 900 hard: expected successes are 90 + 270 = 360. For 1,000 old-method jobs with 900 easy, successes are 720 + 20 = 740. Now evaluate both on 500 easy and 500 hard: new gives 450 + 150 = 600, old gives 400 + 100 = 500. The common-population contrast is +10 percentage points.
Assumptions and limits
A causal interpretation requires consistency (each method is well defined), conditional exchangeability (no remaining confounding within difficulty), positivity (both methods possible in every relevant stratum), and an appropriate target. In this synthetic model all stratum probabilities are supplied even if a group mix is 0% or 100%. Real data then lack overlap and cannot identify those missing rates without extra assumptions. Adjusting for a collider or a consequence of treatment can introduce bias.
The annotated sources distinguish established results from this lesson’s original examples.
04 · Follow the structure
Where else does this apply?
Policy evaluation
Compare programs using the same pre-intervention case mix.
Boundary: Controlling every available variable is not sufficient and can be harmful.
Ecological intervention
Habitat quality may influence both restoration choice and survival.
Boundary: Spatial spillovers violate a simple assumption that each unit is unaffected by others.
05 · Retrieve without hints
Close the explanation. Try a new case.
Write an answer before opening its feedback. Later, return directly here without rereading above. Recognition, explanation and transfer are separate outcomes. No page action or answer reveal measures mastery.
Recognition and explanation
Reveal reasoning and rubric
Only the weights. Both methods receive the same target mixture while conditional outcome rates remain fixed.
Self-check: Name the conditioned variable and explain the different denominators.
Calculation
Reveal reasoning and rubric
New: 0.25 × 0.9 + 0.75 × 0.3 = 0.45. Old: 0.25 × 0.8 + 0.75 × 0.2 = 0.35. Difference: ten percentage points.
Self-check: Show both weights sum to one and retain hard jobs.
Novel transfer
Reveal reasoning and rubric
No. Conditional exchangeability fails. Randomization, better measurements or an explicit justified identification strategy is needed. The calculation still describes the supplied rates but does not establish the intervention effect.
Self-check: Separate the computed association from the unsupported causal interpretation.
On a later day, try again and record actual evidence in the curriculum. A later unaided explanation and a fresh transfer problem give stronger evidence than immediate familiarity. No reminder is scheduled.
Sources & scope
What supports the lesson?
- Hernán and Robins, Causal Inference: What If, chapters 2 and 3. Exchangeability, positivity, consistency and standardization.
Original teaching examples. Reference links need a connection; the lesson itself does not. Built 2026-10-11. Learner understanding is not assessed.