Math bank

z* and margin of error for a proportion

Module F · freq 50

z∗=Φ−1 ⁣(1+γ2)ME=z∗p^(1−p^)nnormality  ⟺  np^(1−p^)>5z^{*}=\Phi^{-1}\!\left(\tfrac{1+\gamma}{2}\right) \qquad \text{ME}=z^{*}\sqrt{\tfrac{\hat p(1-\hat p)}{n}} \qquad \text{normality}\iff n\hat p(1-\hat p)>5

Signals: This is the single most frequent question type — it shows up in every one of the 50/50 past papers, almost guaranteed to appear · a) "Given Z~N(0,1) ... P(−z∗<Z<z∗)=P(-z^*<Z<z^*)=some number" · b) "In a random sample of n ..., k showed ..." → find p̂ / margin of error / whether normality is reasonable
Knowledge point:Confidence intervals & proportions →
Generating question…
Method
  1. a) Two-sided symmetric → middle γ, each tail (1−γ)/2 → z = stats.norm.ppf((1+γ)/2)
  2. b-i) p̂ = k/n (you can leave it as a fraction)
  3. b-ii) ME = z*·√(p̂(1−p̂)/n), and note z* uses the confidence level from part b) (not necessarily the same as γ in part a))
  4. b-iii) Compute n·p̂(1−p̂); normality is only reasonable if it is >5
Pitfalls where marks are lost
  • a) Using norm.ppf(γ) — two-sided needs (1+γ)/2. γ=0.93 → the 0.965 quantile
  • **b-ii) Reusing z* from part a)** — the two parts may have different confidence levels, so read carefully
  • b-iii) Picking "Yes, because n is large (>30)" — this is a trap option! The normality criterion for a proportion has nothing to do with n>30; what matters is np^(1−p^)>5n\hat p(1-\hat p)>5