AQA A-Level Mathematics Paper 3, June 2025: Question 18

📐 Step-by-Step Calculation

1. Find standard normal critical value (z):
   For area = 0.25 in the lower tail:
   z = -0.6745 (or z = ±0.6745)
2. Calculate lower quartile (a):
   (a - 5.7) / 1.2 = -0.6745
   a = 5.7 - 0.6745(1.2) = 4.89 mmol/l
3. Calculate upper quartile (b):
   By symmetry: b = 5.7 + 0.6745(1.2) = 6.51 mmol/l
4. Compute IQR:
   IQR = b - a = 6.5094 - 4.8906 = 1.62 mmol/l
   Alternative method: 2 × (5.7 - a) = 2 × 0.8094 = 1.62

✅ Correct Answer & Mark Scheme

• M1 (AO 3.4): States z = (±) 0.67(45) for inverse normal, or implied by seeing 4.89 to 4.9 or 6.5 to 6.51 .
• A1 (AO 1.1b): Obtains correct lower quartile in range [4.89, 4.9] OR correct upper quartile in range [6.5, 6.51] .
• M1 (AO 3.4): Uses a complete method for IQR: e.g. b - a , 2(b - 5.7) , or 2(5.7 - a) .
• A1 (AO 1.1b): Obtains final answer in range [1.6, 1.62] . (Condone missing units).

💡 Key Knowledge

The quartiles divide continuous probability into four equal parts of 0.25:

• Lower quartile a = Q₁ has P(X < a) = 0.25 .
• Upper quartile b = Q₃ has P(X > b) = 0.25 , meaning P(X < b) = 0.75 .
• Due to the symmetry of the Normal distribution around the mean μ:
  IQR = 2 × (μ - a) = 2 × 0.6745σ ≈ 1.349σ .

❌ Common Errors

• Premature Rounding: Rounding z to 0.67 too early yields IQR = 1.608 , which is acceptable, but rounding intermediate values to 1 d.p. (e.g. 4.9 and 6.5) can lead to inaccurate answers outside acceptable boundaries.
• Variance vs Standard Deviation: Dividing by 1.2² = 1.44 instead of σ = 1.2 when standardising.

📐 Full Step-by-Step Solution

1. State Hypotheses:
   H₀: μ = 5.7
   H₁: μ < 5.7 (one-tailed test for reduction)
2. Define Sample Distribution under H₀:
   X̄ ~ N(5.7, 1.2² / 160)
   Standard error: σ / √n = 1.2 / √160 ≈ 0.09487
3. Calculate p-value (Method 1 - Recommended):
   Using calculator: P(X̄ < 5.6) = 0.146 (to 3 s.f.)
   Alternative (z-score method):
   z = (5.6 - 5.7) / (1.2 / √160) = -1.054
   Critical value at 5% one-tailed is z = -1.645
4. Compare & Make Statistical Inference:
   0.146 > 0.05 (or -1.054 > -1.645 )
   Therefore, do not reject H₀ (accept H₀).
5. Contextual Conclusion:
   There is insufficient evidence to suggest that the mean cholesterol level has reduced after taking the dietary supplement.

✅ Mark Scheme Breakdown

• B1 (AO 2.5): Both hypotheses correct: H₀: μ = 5.7 and H₁: μ < 5.7 (must use parameter μ).
• M1 (AO 1.1a): Uses normal distribution with mean 5.7 and variance 1.2² / 160 (or standard error 1.2 / √160 ≈ 0.095).
• A1 (AO 1.1b): Obtains p-value in range [0.145, 0.147] , test statistic z in range [-1.1, -1.05] , or critical sample mean in range [5.54, 5.55] .
• M1 (AO 3.5a): Correctly compares their p-value with 0.05 (e.g. 0.146 > 0.05 ) or test statistic with -1.645 .
• A1F (AO 2.2b): Correct inference stated: "Do not reject H₀" / "Accept H₀". Follows through their comparison.
• R1 (AO 3.2a): Fully correct conclusion written in context, stating non-definitively that there is insufficient evidence to suggest the mean cholesterol level has reduced.

🧠 Exam Technique & Examiner Insights

• Parameter Notation: Always use μ (population mean) in hypotheses. Using x̄ loses the B1 mark immediately.
• Two-part Conclusion: Always give both:
  1. Mathematical statement: "Do not reject H₀".
  2. Real-world context: "Insufficient evidence to suggest...".
• Avoid Definite Language: Never write "The supplement does not work" or "This proves mean cholesterol is 5.7". Examiners require non-definitive wording like "suggest", "indicate", or "support".

❌ Common Errors to Avoid

• Forgetting √n: Using standard deviation 1.2 instead of 1.2 / √160 for the sample mean. Remember: individuals vary by σ, but sample averages vary by σ / √n !
• Wrong Tail Direction: The supplement aims to reduce cholesterol, so this is strictly lower tail: H₁: μ < 5.7 and P(X̄ ≤ 5.6) .
• Two-Tailed Rejection Threshold: Using critical value z = -1.96 (2.5%) instead of the correct 1-tailed 5% critical value z = -1.645 .