A biscuit manufacturer claims that, on average, 1 in 3 packets of biscuits contain a prize offer. Gerry suspects that the proportion of packets containing the prize offer is less than 1 in 3. In order to test the manufacturer's claim, he buys 20 randomly selected packets. He finds that exactly 2 of these packets contain the prize offer.

Mark scheme

Marking points from the official mark scheme (Oct/Nov 2020). The mark or mark type is shown in brackets.

1. [B1] H₀: P(contains offer) = 1/3, H₁: P(contains offer) < 1/3 (Allow p for P(contains offer) but not just proportion)
2. [M1] Calculate P(X ≤ 2 | H₀) where X ~ B(20, 1/3) = P(0) + P(1) + P(2)
3. [A1] = 0.0176 (3sf)
4. [M1] Compare p-value ('0.0176') < 0.1 (For valid comparison. SC comparison of 0.982(4) > 0.9 scores M1 and recovers the previous M1 A1)
5. [A1 FT] Reject H₀. No evidence (at 10% level) to support manufacturers claim (In context. Not definite. No contradictions. (Accept critical region method: M1 A1 for correctly finding critical region of < 4; 2 in critical region M1; A1 conclusion))
6. [M1] M1 for 1 – (one term omitted or extra or incorrect) or omit '1–'
7. [A1] for all correct expression
8. [A1] = 0.318 (3sf) As final answer.
9. [B1] Concluding that prop is 1 in 3 when it is actually less(1 in 7) (OE, in context.)