Predict differences between sample proportions

Try

Two synthetic production processes have known success probabilities 0.8 and 0.6. Independently observe 100 outcomes from A and 150 from B. A success is coded 1 and a failure 0. Before observing anything, predict the difference between the two sample success proportions.

There are three objects to keep separate. A population model gives probabilities for individual outcomes. A sample is one realized list of outcomes. A sampling distribution describes a statistic over repeated samples generated by the same design. The statistic here is D = p̂A − p̂B; a hat marks a sample proportion. Its center need not equal each observed difference.

Predict

Predict whether doubling both sample sizes changes the center, halves the standard deviation, or does something else. Write your prediction before looking at the graph. Also decide whether subtracting two proportions means subtracting their variances. A variance cannot be negative, which is already a useful check on a proposed rule.

Experiment

For independent Bernoulli outcomes, E(p̂) = p and Var(p̂) = p(1−p)/n. Thus E(D) = 0.8−0.6 = 0.2. Independence between the two samples gives

Var(D) = 0.8·0.2/100 + 0.6·0.4/150 = 0.0016 + 0.0016 = 0.0032.

The standard deviation is √0.0032 ≈ 0.05657. Doubling both sizes halves the variance to 0.0016 and changes the standard deviation to 0.04. It divides spread by √2, not by 2.

Design	Center	Variance	Standard deviation
100 from A, 150 from B	0.2	0.0032	about 0.05657
200 from A, 300 from B	0.2	0.0016	0.04

Inspect the shared horizontal axis. Mark the interval from 0.12 to 0.28 and explain why the larger-sample curve assigns more probability to it. These are normal approximation curves. Actual finite-sample proportions and their difference take discrete values; the drawing does not turn them into continuous measurements.

Understand

Why add variances for a difference? After centering U and V, expand the square of (U−EU)−(V−EV). Its expected square is Var(U)+Var(V)−2Cov(U,V). Covariance is the average product of centered deviations. Independence makes that product average zero. The minus sign on V is squared in its own variance term.

The expected success/failure counts are 80/20 and 90/60. All four exceed the usual large-count check of 10, supporting a normal approximation here. Standardizing 0.28 gives z = (0.28−0.2)/√0.0032 = √2. With the supplied normal CDF value Φ(√2) ≈ 0.92135, P(D > 0.28) ≈ 0.07865. Φ gives area to the left; the requested upper tail requires subtraction from 1. This exercise supplies the CDF rather than teaching calculator navigation.

Check design separately from counts. Independent Bernoulli trials give the variance above exactly. Simple random sampling without replacement from a fixed population of size N has variance p(1−p)/n multiplied by (N−n)/(N−1). Here p is the fixed population fraction. Sampling at most 10% supports ignoring this correction approximately; it does not make the draws literally independent. Independently choosing the two groups is a separate requirement.

Apply

Suppose the observed successes are 78 and 93. The realized proportions are 0.78 and 0.62, giving difference 0.16. Compare three calculations before deciding which belongs in a report:

Question	Variance calculation
Known model used above	0.8·0.2/100 + 0.6·0.4/150 = 0.0032
Estimate uncertainty for a difference interval	0.78·0.22/100 + 0.62·0.38/150 ≈ 0.003286667
Test a null of equal probabilities	0.684·0.316·(1/100+1/150) = 0.0036024

The pooled estimate 0.684 is 171 successes out of 250. Pooling here imposes equality under the null. An interval estimates two potentially different probabilities separately. Standard errors are the square roots of these variance estimates. Neither estimated formula replaces stipulated population parameters in a known-model sampling question.

Now consider paired before/after indicators with joint probabilities 0.4 for (1,1), 0.3 for (1,0), 0.1 for (0,1), and 0.2 for (0,0). Derive the marginal probabilities 0.7 and 0.5, then covariance 0.4−0.7·0.5 = 0.05. For n independent people the mean paired difference has variance (0.21+0.25−0.10)/n = 0.36/n. The independent-column answer 0.46/n ignores pairing; subtracting variances would even give a negative answer. The independent unit is a person, not a column entry.

Continue

For each practice response, name the random statistic, its population parameters, and the sampling design before calculating. State whether a probability uses an exact distribution or a normal approximation. Larger samples reduce random spread under the model; they do not repair selection bias or missing pairing information. The next lesson carries these distinctions to quantitative measurements and mean differences.