The masses, in grams, of apples from a certain farm have mean µ and standard deviation 5.2. The farmer says that the value of µ is 64.6. A quality control inspector claims that the value of µ is actually less than 64.6. In order to test his claim he chooses a random sample of 100 apples from the farm.

Mark scheme

Marking points from the official mark scheme (Oct/Nov 2021). The mark or mark type is shown in brackets.

1. [B1] Ho: µ = 64.6, H₁: µ < 64.6 (Allow population mean, not just 'mean'.)
2. [M1] (63.5 - 64.6) / (5.2 / √100) (Standardising. Must have √100.)
3. [A1] -2.115 (Accept -2.12 (3sf))
4. [M1] '2.115' > 1.96 or '–2.115' <-1.96 [do not accept Ho] (Valid comparison (0.0172 < 0.025 for area comparison).
5. [A1 FT] There is evidence that µ < 64.6 (Not definite, e.g. not ‘µ < 64.6’. in context. No contradictions. Accept critical value method leading to 63.5 < 63.58 or 64.6 > 64.52.)
6. [M1] m - 64.6 = -1.96 * (5.2 / √100) (Finding the critical value using N(64.6, 5.2/√100) and a z value.)
7. [A1] m = 63.5808
8. [M1] (63.5808 - 62.7) / (5.2 / √100) [= 1.694] (Standardising using N(62.7, 5.2/√100) and a critical value.)
9. [M1] 1 – Φ('1.694') (For area consistent with their values.)
10. [A1] 0.0451 (Accept answers that round to 0.045.)