A researcher read a magazine article which stated that boys aged 1 to 3 prefer green to orange. It claimed that, when offered a green cube and an orange cube to play with, a boy is more likely to choose the green one. The researcher disagrees with this claim. She believes that boys of this age are equally likely to choose either colour. In order to test her belief, the researcher carried out a hypothesis test at the 5% significance level. She offered a green cube and an orange cube to each of 10 randomly chosen boys aged 1 to 3, and recorded the number, X, of boys who chose the green cube. Out of the 10 boys, 8 boys chose the green cube.

Mark scheme

Marking points from the official mark scheme (Oct/Nov 2023). The mark or mark type is shown in brackets.

1. [B1] oe (e.g. invalid because it should be a tail probability compared with 0.05). Need to find P(X ≥ 8)
2. [B1] Allow p = 0.5. Allow p > 0.5. H0: P(green) = 0.5, H1: P(green) > 0.5
3. [M1] Attempt 0.0439 + P(X=9) + P(X= 10). Must see Binomial expressions B(10,0.5). P(X ≥ 8) = 0.0439 + 10C9 × (0.5)¹ × (0.5)⁹ + 0.5¹⁰
4. [A1] SC B1 0.0547 or 0.0546 with no working. = 0.0547 or 0.0546 (3 sf)
5. [M1] Valid comparison of tail probability with 0.05. 0.0547 > 0.05
6. [A1FT] In context, not definite. No contradictions. Allow 'There is insufficient evidence to reject the hypothesis that boys like green and orange equally'. Not definite, e.g. not 'They don't prefer green' or 'Researchers claim true' 'Magazine's claim untrue'. Any mention of ‘claim' must be clear which claim it is. [Do not reject H0] ‘There is insufficient evidence [at the 5% level] to accept the hypothesis that boys prefer green.' Or 'There is sufficient evidence to support the researcher's claim.'
7. [B1] Mark independently. H0 was not rejected
8. [M1] Finding P(9,10) using B(10,0.5). Could be seen in (a)(ii). 10C9×(0.5)¹×(0.5)⁹ + 0.5¹⁰ or "0.0547" – 0.0439
9. [A1] P(Type I error) = 11/1024 or 0.0107 (3 sf) or 0.0108