Pool only under the equality null

Try

An original synthetic comparison has 72 successes among 120 observations in one independent group and 48 among 120 in the other. The question is whether the population success probabilities differ. State H0:p1=p2 and Ha:p1≠p2 before looking at the observed 0.2 difference.

Predict

Predict why an equality-null test can combine the success counts to estimate one common rate. Then contrast that with an interval for p1−p2, which should allow different unknown rates. Pooling is a consequence of a particular null, not a universal instruction whenever there are two groups.

Experiment

Under the equality null, p̂pool=(72+48)/(120+120)=0.5. The null standard error is √[0.5·0.5(1/120+1/120)]=√(1/240)≈0.064550. Thus z=0.2/0.064550≈3.098. A supplied two-sided normal reference gives p≈0.00195. At α=0.05, reject equality under the assumptions.

Expected null successes and failures are 60 in each group. Compare them with observed counts 72,48 and 48,72. The null creates expected values; it does not rewrite the observed data. A causal conclusion still requires an appropriate randomized treatment design.

Understand

The pooled estimate is total successes divided by total observations, not generally the unweighted average of group proportions. With unequal sample sizes, the larger group contributes more information about the common null probability. For example, 8 successes in 10 and 36 in 90 pool to 44/100=0.44, not the average of 0.8 and 0.4.

For the conventional equality-null test, z=[(p̂1−p̂2)−0]/√[p̂pool(1−p̂pool)(1/n1+1/n2)]. The numerator is the observed departure from zero difference; the denominator estimates its spread under the fitted common-rate null. A nonzero difference null requires additional care and is not handled by blindly reusing this pooled formula.

Conditions include an appropriate sampling or randomized design, independent groups, suitable within-group independence, and adequate expected counts under the null. A common check examines n1p̂pool, n1(1−p̂pool), n2p̂pool and n2(1−p̂pool). Label the expected-count check clearly rather than mixing it with the observed-count check for the unpooled interval.

Choose the tail from the alternative. A two-sided test uses deviations in both directions; an upper-tailed test concerns p1>p2. Reversing group labels changes the sign and corresponding alternative but should not change a properly matched substantive conclusion. A supplied table gives an approximate p-value; rounded statistics should not be presented as yielding an invented exact probability.

A significant result indicates evidence against equality under the specified model. It does not quantify all practical consequences, establish causality from observational groups or prove every individual responds differently. Report a suitable difference estimate and interval alongside the test. The interval’s unpooled uncertainty and the equality test’s pooled-null uncertainty have different purposes and need not produce perfectly identical approximate boundary decisions.

Apply

Explain the pooled denominator to someone who thinks pooling means averaging percentages equally. Use an unequal-size example and reconstruct successes. Then write a conclusion for the worked comparison that distinguishes a representative observational sample from a randomized experiment: the arithmetic can be identical while the permissible causal statement differs.

Continue

Next extend categorical comparisons to tables with more categories. A chi-square statistic adds discrepancies across cells rather than assigning an arbitrary numerical scale to categories.