Grade 12 Advanced Statistics Practice — Medium

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Question 1 of 39: The variance of B(n, p) is:

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Answer key for parents & teachers (39 questions)
  1. The variance of B(n, p) is:np(1−p). Var(X) = np(1−p). Standard deviation = √(np(1−p)).
  2. A Poisson distribution is used when:Only the average rate λ is known. Poisson: use when you know the average rate λ of rare events per interval, but not n and p separately.
  3. For Poisson, P(X=k) = λᵏe^(−λ)/k!. The mean and variance both equal:λ. For a Poisson distribution, μ = σ² = λ. Mean equals variance.
  4. 10 coin flips with p=0.5. Using binomial, P(exactly 6 heads) = ?All three are equivalent. All three expressions equal C(10,6)·(0.5)¹⁰ = 210/1024 ≈ 0.205. They are the same formula.
  5. The normal approximation to the binomial is valid when:np ≥ 10 and n(1−p) ≥ 10. The rule of thumb: np ≥ 10 AND n(1−p) ≥ 10 ensures the binomial is approximately normal.
  6. Poisson approximates binomial when n is large and p is:Small (close to 0). Use Poisson when n is large and p is small, with λ = np. Classic examples: rare defects, accidents.
  7. For B(20, 0.4): mean = ?, standard deviation = ?μ=8, σ=√4.8≈2.19. μ = np = 20·0.4 = 8. σ² = np(1−p) = 20·0.4·0.6 = 4.8. σ = √4.8 ≈ 2.19.
  8. The 68-95-99.7 rule for the normal distribution states:68% of data is within σ, 95% within 2σ, 99.7% within 3σ. Empirical rule: 68% within ±1σ, 95% within ±2σ, 99.7% within ±3σ of the mean.
  9. For Binomial, which additional condition must be satisfied?Trials must be independent. Binomial requires: (1) fixed n, (2) each trial independent, (3) exactly 2 outcomes, (4) constant p.
  10. If X ~ B(10, 0.3): P(X = 0) = ?(0.7)¹⁰. P(X=0) = C(10,0)·(0.3)⁰·(0.7)¹⁰ = (0.7)¹⁰ ≈ 0.028.
  11. Which distribution models the number of email spam per day if average is 5 per day?Poisson with λ=5. Rare events with a known average rate → Poisson. Spam emails per day with λ = 5.
  12. For X ~ N(100, 225) (μ=100, σ=15): What z-score corresponds to X = 130?z = 2. z = (x − μ)/σ = (130 − 100)/15 = 30/15 = 2.
  13. The significance level α represents:The probability of rejecting H₀ when it is true (Type I error rate). α = P(Type I error) = P(rejecting H₀ | H₀ is true). Common values: 0.05 or 0.01.
  14. A p-value is defined as:P(observing data this extreme or more extreme, given H₀ is true). p-value = P(data as extreme as observed | H₀). Small p-value → data unlikely under H₀ → evidence against H₀.
  15. Type I error is:Rejecting H₀ when H₀ is true (false positive). Type I error = false positive. Probability = α. We incorrectly reject a true null hypothesis.
  16. Type II error is:Failing to reject H₀ when H₀ is false (false negative). Type II error = false negative. Probability = β. We fail to detect a real effect.
  17. The power of a test is:1 − β. Power = P(reject H₀ | Hₐ is true) = 1 − β. Higher power → better at detecting true effects.
  18. The z-test statistic z = (x̄ − μ₀)/(σ/√n). What does σ/√n represent?Standard error of the mean. σ/√n is the standard error of the mean (SEM) — how much x̄ varies from sample to sample.
  19. A t-test is used instead of z-test when:The population σ is unknown (we use s). Use t-test when σ is unknown and you must estimate it with the sample standard deviation s.
  20. Degrees of freedom for a one-sample t-test with n observations:n − 1. df = n − 1 for a one-sample t-test. We lose one degree of freedom for estimating μ.
  21. If p-value = 0.12 and α = 0.05, we:Fail to reject H₀ (but do not prove H₀). 0.12 > 0.05, so we fail to reject H₀. We never "accept" H₀ — we just don't have enough evidence against it.
  22. "Statistical significance" at α = 0.05 means:There is less than a 5% chance the results are due to random variation if H₀ is true. Statistical significance says only that the observed result is unlikely under H₀ (p < α). It says nothing about practical importance.
  23. A two-tailed test is used when Hₐ states:μ ≠ μ₀. A two-tailed test (Hₐ: μ ≠ μ₀) rejects H₀ for deviations in either direction.
  24. A one-tailed test is preferred when:The researcher predicts a specific direction (greater or less). One-tailed: Hₐ: μ > μ₀ or Hₐ: μ < μ₀. Use when theory/prior evidence specifies the direction.
  25. The chi-square test for independence tests whether:Two categorical variables are related in a contingency table. Chi-square test of independence: tests if two categorical variables in a contingency table are associated or independent.
  26. A z-test for population mean: sample n=100, x̄=52, σ=10, H₀: μ=50. z = ?z = 2. z = (52−50)/(10/√100) = 2/1 = 2.
  27. We "never accept H₀" because:Failing to reject doesn't prove H₀ is true — the test just lacks evidence against it. Absence of evidence ≠ evidence of absence. Failing to reject H₀ means insufficient evidence — not proof that H₀ is true.
  28. A 95% confidence interval means:95% of the time, intervals constructed this way will contain the true parameter. Correct interpretation: if we repeated the study many times, 95% of all intervals would contain the true parameter. For THIS interval, the parameter is either in it or not.
  29. The critical value z* for a 90% CI is approximately:1.645. z* ≈ 1.645 for a 90% CI.
  30. The critical value z* for a 99% CI is approximately:2.576. z* ≈ 2.576 for a 99% CI.
  31. The margin of error E = z*·(σ/√n). To halve E, multiply n by:4. E ∝ 1/√n. To halve E: 1/√(4n) = (1/2)·(1/√n). So multiply n by 4.
  32. Sample n=36, x̄=52, σ=12. 95% CI:(52−3.92, 52+3.92) = (48.08, 55.92). ME = 1.96·(12/6) = 1.96·2 = 3.92. CI = (52 − 3.92, 52 + 3.92) = (48.08, 55.92).
  33. The formula to find required sample size for margin of error E is:n = (z*σ/E)². E = z*·σ/√n → √n = z*σ/E → n = (z*σ/E)². Always round UP.
  34. Convenience sampling (surveying whoever is nearby) leads to:Convenience bias / non-random sample. Convenience sampling is non-random, leading to biased estimates that may not represent the population.
  35. A voluntary response sample (people who choose to respond) is biased because:It overrepresents people with strong opinions. Voluntary response attracts people with strong opinions (often negative), biasing the results.
  36. A confounding variable is one that:Is related to both the explanatory and response variable, creating spurious associations. A confound is associated with both the treatment and outcome, making it hard to isolate the true effect.
  37. If σ doubles and n stays the same, the margin of error:Doubles. ME = z*σ/√n. If σ → 2σ, then ME → 2ME. Margin of error doubles.
  38. If n quadruples and σ stays the same, the margin of error:Halves. ME = z*σ/√n. √(4n) = 2√n. So ME is halved.
  39. Which study design best establishes causation?Randomized controlled experiment. Randomized experiments (with random assignment to treatment/control) best establish causation by controlling confounders.