Exam Pattern: Two-Sample Z-Test Power and Sample Size

This note consolidates the normal midterm derivation question and the matching makeup question where the same formula is reused for direct power computation.

Source Questions

  • STA305 Midterm W2026, Question 3
  • STA305 Makeup W2026, Question 3(b)
  • STA305 Makeup W2026, Question 3(c)

Setup

Assume two independent normal groups with means $\mu_1, \mu_2$ and common variance $\sigma^2$. Let

$$ \theta = \mu_1 - \mu_2. $$

For the one-sided test

$$ H_0: \theta = 0 \quad \text{versus} \quad H_1: \theta > 0, $$

the usual $z$-statistic is

$$ Z = \frac{\bar{Y}_1 - \bar{Y}_2}{\sigma \sqrt{1/n_1 + 1/n_2}}. $$

Reject $H_0$ when

$$ Z > z_{\alpha}. $$

Power Formula

Under the alternative $\theta > 0$,

$$ Z \sim \mathcal{N}\left( \frac{\theta}{\sigma \sqrt{1/n_1 + 1/n_2}}, 1 \right). $$

Therefore,

$$ 1-\beta

1 - \Phi\left( z_{\alpha}

\frac{\theta}{\sigma \sqrt{1/n_1 + 1/n_2}} \right). $$

Equal Allocation Sample Size

If $n_1 = n_2 = n/2$, then

$$ 1/n_1 + 1/n_2 = \frac{4}{n}, $$

so the power becomes

$$ 1-\beta

1 - \Phi\left( z_{\alpha} - \frac{\theta \sqrt{n}}{2\sigma} \right). $$

Solving for the total sample size gives

$$ n = \frac{4 \sigma^2 (z_{\alpha} + z_{\beta})^2}{\theta^2}. $$

Direct Power Computation

If $n_1$ and $n_2$ are given, plug them directly into

$$ 1-\beta

1 - \Phi\left( z_{\alpha}

\frac{\theta}{\sigma \sqrt{1/n_1 + 1/n_2}} \right). $$

That is the key move in the makeup version where each arm size is specified.

R Template

alpha <- 0.05
theta <- 1
sigma <- 3
n1 <- 80
n2 <- 80

1 - pnorm(qnorm(1 - alpha) - theta / (sigma * sqrt(1 / n1 + 1 / n2)))

Exam Checklist

  • Write the rejection rule using $z_{\alpha}$
  • Standardize under the alternative, not only under $H_0$
  • Keep the common variance factor $\sigma$ in the denominator
  • For equal allocation, replace $n_1 = n_2 = n/2$
  • Distinguish direct power calculation from solving for $n$