Exercise 2: Required Sample Size for a 5% to 6% Effect — Possible Solution
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GIVEN
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p1 = 0.05, p2 = 0.06, 80% power, alpha = 0.05 (two-tailed)
z_alpha/2 = 1.96, z_beta = 0.84

STEP 1: THE AVERAGE PROPORTION
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p-bar = (p1 + p2) / 2 = (0.05 + 0.06) / 2 = 0.055

STEP 2: APPLYING THE SAMPLE SIZE FORMULA
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n = 2(z_alpha/2 + z_beta)^2 x p-bar(1 - p-bar) / (p2 - p1)^2
  = 2(1.96 + 0.84)^2 x 0.055 x 0.945 / (0.01)^2
  = 2(2.8)^2 x 0.051975 / 0.0001
  = 2 x 7.84 x 0.051975 / 0.0001
  ~= 8,150 per group

So approximately 8,150 users would need to be assigned to EACH group
(control and treatment) - roughly 16,300 total - to have an 80%
chance of detecting a genuine 1-percentage-point improvement from 5%
to 6%, at the standard alpha=0.05 significance level.

WHY THIS WORKS AS AN ANSWER
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The sample size is computed by applying this chapter's own formula
directly with the given effect size and the standard published z_beta
and z_alpha/2 constants, showing each substitution explicitly rather
than presenting only the final number.