A machine is supposed to produce random digits. Bob thinks that the machine is not fair and that the probability of it producing the digit 0 is less than 1/10. In order to test his suspicion he notes the number of times the digit 0 occurs in 30 digits produced by the machine. He carries out a test at the 10% significance level.

Mark scheme

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

1. [B1] Ho: P(0) = 1/10 H1: P(0) < 1/10 Accept p.
2. [M1] For B(30,0.1) Used not just stated.
3. [M1] P(X= 0) = 0.9^30 [= 0.0424] [<0.1]
4. [B1] P(X = 0 or 1) = 0.9^30 + 30×0.9^29×0.1=0.184 [>0.1] Accept 0.184 or 0.183.
5. [A1] Rejection region is 0 zeros Dependent on M1 M1 and at least one comparison, no errors seen. SC One unsupported correct answer 0.0424/0.184(or 0.183) and correct rejection region scores B1; with comparison with 0.1 scores B2. Two unsupported correct answers 0.0424 and 0.184(or 0.183) and correct rejection region scores B2 or if with one comparison with 0.1 scores B3.
6. [B1] 0.0424 FT their (b) must have a critical region (only follow though Binomial), dependent on answer < 0.1.
7. [B1] Bin(30, 1/40) SOI
8. [M1] 1-0.975^30 FT their rr and with Bin(30, 1/40)).
9. [A1] 0.532 (3dp) SC Unsupported correct answer scores B2 only.
10. [B1] Not concluding that the probability is less than 1/10, when in fact it is. In context.