Solution
How Many Per Arm
Show the problem again
Two groups are compared with a two-sided z-test at significance level 0.05, using equal group sizes and a known common standard deviation σ = 10. The test should have power 80% to detect a true difference in means of 4. Use z = 1.96 (for the 2.5% tail) and z = 0.84 (for 80% power), and the formula n = 2 σ² (z_a + z_b)² / δ² per group. What is the required sample size per group? Give an integer.
Worked solution
The formula gives n = 2 × 100 × (1.96 + 0.84)² / 16 = 2 × 100 × 7.84 / 16 = 98 exactly. Since this is already an integer there is no rounding up. The key structure: the variance of the difference of means is 2σ²/n, and rejection with 80% power requires the true difference to sit (z_a + z_b) standard errors from zero.
Source: Standard two-sample sample-size formula from clinical-trial design. Statement written for AxiomIQ.