The masses of cereal boxes filled by a certain machine have mean 510 grams. An adjustment is made to the machine and an inspector wishes to test whether the mean mass of cereal boxes filled by the machine has decreased. After the adjustment is made, he chooses a random sample of 120 cereal boxes. The mean mass of these boxes is found to be 508 grams. Assume that the standard deviation of the masses is 10 grams.

Mark scheme

Marking points from the official mark scheme (May/June 2024). The mark or mark type is shown in brackets.

1. [B1] H₀: Population mean mass = 510 g H₁: Population mean mass < 510 g Allow 'μ' but not just ‘mean'.
2. [M1] (508 - 510) / (10 / √120) Standardising must have √120.
3. [A1] = -2.191 or -2.190
4. [M1] -2.191 < -1.96 or 2.191 > 1.96 Area comparison: 0.0143 or 0.0142 < 0.025 For valid comparison. Inequality sign the wrong way round scores M1 A0.
5. [A1FT] [Reject H₀] There is sufficient evidence to suggest that the [mean] mass has decreased In context (must be ‘decreased' OE, not ‘changed'); not definite. No contradictions. Condone 'there is sufficient evidence to support the inspector's claim'. NB: Accept alternative method using critical value (= 508.21) and comparison with 508. Condone 509.79 compared with 510. Two tail test scores maximum B0 M1 A1 M1 A0; must have comparison with 0.0125 or 2.24/2.241.
6. [M1] cv - 510 / (10 / √120) = -1.96 Standardising to find critical value (must use 510 and 10/√120). Accept ± 1.96.
7. [A1] cv = 508.21 Accept 3 sf if nothing better seen. Note: cv could be found in (a).
8. [M1] z = ± (508.21 - 506) / (10 / √120) [= 2.421] Standardising with their 508.21 and 506 (must use 10/√120).
9. [M1] P(X̄ > 508.21 | μ = 506) = 1 – Ф(‘2.421') For area consistent with their working.
10. [A1] = 0.0077 to 0.0080 (2sf) Note: (510-506) / (10/√120) scores max M0 A0 M1 M1 A0.