(a) Find the most powerful test of size α(= 0·05) for testing H₀: μ = 0 vs. H₁: μ = 1, given a random sample of size 25 from N(μ…

(a) Most Powerful Test of Size 0.05 (Neyman–Pearson Lemma)

Both hypotheses are simple, so the Neyman–Pearson lemma gives the most powerful test. X₁, …, X₂₅ are iid N(μ, 16), so σ² = 16 is known and X̄ ~ N(μ, 16/25) = N(μ, 0.64), with standard error σ/√n = 4/5 = 0.8.

The likelihood is L(μ) = (2π·16)^(−n/2) exp[−(1/32) Σ(xᵢ − μ)²]. The likelihood ratio is

λ = L(0)/L(1) = exp[−(1/32){Σxᵢ² − Σ(xᵢ − 1)²}].

Since Σ(xᵢ − 1)² = Σxᵢ² − 2Σxᵢ + n, the bracket equals 2Σxᵢ − n = n(2X̄ − 1), so

λ = exp[−(25/32)(2X̄ − 1)].

The lemma rejects H₀ when λ ≤ k. Because λ is a decreasing function of X̄, this is equivalent to rejecting H₀ for large X̄: the critical region is X̄ > c.

Size condition: under H₀, Z = X̄/0.8 ~ N(0, 1), and P(Z > c/0.8) = 0.05 gives c/0.8 = 1.645 (table area 0.4500 at z = 1.645), so c = 0.8 × 1.645 = 1.316.

Most powerful test: reject H₀ if X̄ > 1.316 (equivalently Σxᵢ > 32.9).

Power at μ = 1: P(X̄ > 1.316 | μ = 1) = P(Z > (1.316 − 1)/0.8) = P(Z > 0.395) ≈ 0.5 − 0.1535 = 0.35. Because the critical region does not depend on the particular value μ₁ > 0, the same test is uniformly most powerful for H₁: μ > 0.

(b) Estimating θ by MLE, Minimum χ² and Modified Minimum χ²

With n = 25, the observed frequencies are O₁ = 6 (defective, p₁ = θ) and O₂ = 19 (non-defective, p₂ = 1 − θ); the expected frequencies are E₁ = 25θ and E₂ = 25(1 − θ).

(i) MLE method. L(θ) ∝ θ⁶(1 − θ)¹⁹, so ln L = 6 ln θ + 19 ln(1 − θ). Setting d(ln L)/dθ = 6/θ − 19/(1 − θ) = 0 gives 6(1 − θ) = 19θ, i.e. θ̂_MLE = 6/25 = 0.24. The second derivative −6/θ² − 19/(1 − θ)² < 0 confirms a maximum.

(ii) Minimum χ² method. Minimise χ² = Σ(Oᵢ − Eᵢ)²/Eᵢ = (6 − 25θ)²/(25θ) + (19 − 25(1 − θ))²/(25(1 − θ)). Since 19 − 25(1 − θ) = 25θ − 6, both numerators equal (25θ − 6)², so χ² = (25θ − 6)²/(25θ(1 − θ)). This is non-negative on 0 < θ < 1 and equals zero only when 25θ = 6, so θ̂_χ² = 0.24.

(iii) Modified minimum χ² method. The observed frequencies replace the expected ones in the denominator: χ²ₘ = (6 − 25θ)²/6 + (25θ − 6)²/19 = (25θ − 6)²(1/6 + 1/19), which vanishes only at θ̂ = 6/25 = 0.24.

All three methods give θ̂ = 0.24: the lot is estimated to be 24% defective. They coincide here because with only two classes and one parameter the observed proportions can be fitted exactly; with more classes the three estimators generally differ, though they are asymptotically equivalent (all BAN estimators).

(c) Mann–Whitney U-Test versus Wilcoxon Test; APGAR Scores

Differences between the two tests:

• Design: Mann–Whitney compares two independent samples; the Wilcoxon test (sign or signed-rank) is for a single sample or matched pairs.
• What is ranked: Mann–Whitney ranks all observations of both samples together; the Wilcoxon signed-rank test ranks the absolute within-pair differences (the plain sign test uses only the signs of the differences).
• Null hypothesis: Mann–Whitney tests that the two populations have the same distribution (location); Wilcoxon tests that the median of the differences is zero.
• Suitability here: programmes A and B involve different women with unequal sample sizes (8 and 7), so pairing is impossible and the Mann–Whitney U-test is the correct choice.

H₀: APGAR score distributions are the same under both programmes; H₁: they differ.

Ranking the 15 pooled scores (tied values get the mean rank): 2→1, 3→2, 5→3, 6→4.5 (two values), 7→7 (three values), 8→10 (three values), 9→13 (three values), 10→15.

• Ranks of A (8, 7, 6, 2, 5, 8, 7, 3): 10, 7, 4.5, 1, 3, 10, 7, 2, so R_A = 44.5.
• Ranks of B (9, 9, 7, 8, 10, 9, 6): 13, 13, 7, 10, 15, 13, 4.5, so R_B = 75.5 (check: 44.5 + 75.5 = 120 = 15 × 16/2).

U_A = R_A − n₁(n₁ + 1)/2 = 44.5 − 36 = 8.5 and U_B = R_B − n₂(n₂ + 1)/2 = 75.5 − 28 = 47.5 (check: U_A + U_B = 56 = n₁n₂). The test statistic is U = min(U_A, U_B) = 8.5.

Conclusion: since U = 8.5 < U₍₀.₀₅₎ = 10, H₀ is rejected at the 5% level. There is a significant difference in APGAR scores between the two programmes, with programme B showing higher scores.