Two Rates, Two Denominators, One Question

Suppose 18 of 30 people in Group A experience an outcome, compared with 12 of 30 in Group B.

The observed proportions are .60 and .40. The difference is 20 percentage points, not a 20 percent relative change.

Those distinctions matter enormously when communicating research.

An estimator, not yet an inference

For independent groups with sample sizes $n_A,n_B$, define

$$\hat\Delta=\hat p_A-\hat p_B.$$

An approximate unpooled standard error for the difference is

$$\widehat{\operatorname{SE}}(\hat\Delta) =\sqrt{\frac{\hat p_A(1-\hat p_A)}{n_A}+ \frac{\hat p_B(1-\hat p_B)}{n_B}}.$$

Here that is $\sqrt{.6(.4)/30+.4(.6)/30}=\sqrt{.016}\approx .12649$.

The naive unpooled 95% normal interval is approximately

$$.20\pm1.96(.12649)=[-.048,.448].$$

The interval includes zero. It also relies on approximations whose performance varies with sample size and underlying proportions.

Score-based or other better-calibrated intervals deserve consideration instead of treating this formula as the universal default.

Why some tests pool and intervals don’t

Under the null hypothesis $H_0:p_A=p_B$, the pooled proportion is

$$\hat p_0=\frac{x_A+x_B}{n_A+n_B}.$$

For our example it is $(18+12)/(30+30)=.5$.

Using a shared estimate to evaluate a null of equality can make sense for certain score tests. It does not mean that an effect-size confidence interval should always use the same pooled standard error.

Risk difference, relative risk, and odds ratio

• Risk difference: $\hat p_A-\hat p_B=.20$.
• Risk ratio: $\hat p_A/\hat p_B=1.5$.
• Odds ratio: $(.6/.4)/(.4/.6)=2.25$.

These are different mathematical measures. An odds ratio of 2.25 does not mean the event is 2.25 times as probable.

And none of them, by itself, establishes a causal effect of treatment. That interpretation depends on the design.

Check your understanding

1. Find the risk difference and risk ratio for 12/20 versus 9/30.
2. Why does a null-equality test sometimes use a pooled estimate?
3. Why should a reported effect include both its units and uncertainty?

Worked responses

1. Proportions .60 and .30, difference .30 and risk ratio 2.

2. The null asserts a common probability across groups.

3. Magnitude without context or sampling uncertainty can exaggerate the evidence.

Work through the entire comparison

Suppose two independent synthetic groups have 60 successes among 100 observations and 40 among 100. The estimated risk difference is $\widehat\Delta=0.6-0.4=0.2$. For a large-sample interval, an unpooled estimated standard error is

$$SE_{CI}=\sqrt{\frac{\hat p_1(1-\hat p_1)}{n_1}+\frac{\hat p_2(1-\hat p_2)}{n_2}} =\sqrt{0.0048}\approx0.069282.$$

A normal 95% interval is approximately $0.2\pm1.96(0.069282)=[0.06421,0.33579]$. This is a large-sample Wald interval; small or extreme samples call for better-calibrated score-based or exact procedures. An interval formula that can exceed the parameter’s possible range should not be silently clipped and presented as having unchanged coverage.

For the equality null, the common proportion estimate is $\hat p=(60+40)/200=0.5$. The null standard error is $\sqrt{\hat p(1-\hat p)(1/n_1+1/n_2)}=\sqrt{0.005}$. Thus $z=0.2/\sqrt{0.005}\approx2.82843$, with a two-sided normal p-value about $0.00468$. The pooled test estimates variance under equality, whereas the unpooled interval estimates uncertainty without imposing equality. Software continuity corrections may yield a different displayed approximation; record the setting.

The same data give risk ratio $0.6/0.4=1.5$ and odds ratio $(0.6/0.4)/(0.4/0.6)=2.25$. Report the chosen effect scale and the event definition. Reversing “success” and “failure” reverses a risk difference and reciprocates an odds ratio, but does not generally reciprocate the risk ratio for the original event.

Additional practice

1. Compare 30/50 with 20/50. Calculate the difference, risk ratio, odds ratio, and unpooled standard error.
2. Why is this independent-groups formula inappropriate for the same people measured twice?
3. If one group has no observed successes, does that prove its population risk is zero?

Worked additional solutions

1. Rates remain $0.6,0.4$, giving $0.2$, $1.5$, and $2.25$. The standard error is $\sqrt{0.0096}\approx0.09798$, larger because the group sizes are halved.
2. Paired binary outcomes have dependence; the numbers changing in opposite directions carry the relevant information. Independent variance addition omits the covariance.
3. No. Zero is a sample count. A suitable one-sided or two-sided interval still permits positive population risks, depending on the procedure and confidence level.