Moderate -0.3 This is a straightforward one-tailed binomial hypothesis test with clearly stated hypotheses (H₀: p=0.92, H₁: p<0.92), requiring students to state the binomial distribution assumption, calculate P(X≤16) using tables or formula, and compare to 5% significance level. While it requires multiple steps (stating assumption, setting hypotheses, calculating probability, making conclusion), each step follows a standard procedure taught explicitly in S2 with no novel problem-solving required, making it slightly easier than average.
4 A train company claims that \(92 \%\) of trains on a particular line arrive on time. Sanjeep suspects that the true percentage is less than \(92 \%\). He chooses a random sample of 20 trains on this line and finds that exactly 16 of them arrive on time. Making an assumption that should be stated, test at the 5\% significance level whether Sanjeep's suspicion is justified. [0pt]
[6]
OE. No contradictions. Method using normal approximation: If first B1B1 earned: \(\text{CV} - 1.566\left(\text{from } \frac{16.5 - 20 \times 0.92}{\sqrt{20 \times 0.92 \times 0.08}}\text{, with continuity correction}\right)\) or CV \(= 1.978\) (without continuity correction). comp \(z = 1.645\). No evidence that % decreased (1.566) or evidence that % decreased (1.978) is awarded SC2 after B marks
Total: 6
## Question 4:
| Answer | Marks | Guidance |
|--------|-------|----------|
| Trains are independent OR probability of being on time is constant | B1 | Must be in context |
| $H_0$: P(on time) $= 0.92$; $H_1$: P(on time) $< 0.92$ | B1 | Both. Allow '$p$' or $\pi$ |
| $1 - \left({}^{20}C_{17} \times 0.92^{17} \times 0.08^3 + {}^{20}C_{18} \times 0.92^{18} \times 0.08^2 + 20 \times 0.92^{19} \times 0.08 + 0.92^{20}\right)$ | M1 | Allow one end error. Must have $1 - \ldots$ |
| $= 0.0706$ (3 sf) | A1 | |
| Compare with 0.05 | M1 | Valid comparison needed |
| No evidence that percentage less than 92% | A1FT | OE. No contradictions. Method using normal approximation: If first B1B1 earned: $\text{CV} - 1.566\left(\text{from } \frac{16.5 - 20 \times 0.92}{\sqrt{20 \times 0.92 \times 0.08}}\text{, with continuity correction}\right)$ or CV $= 1.978$ (without continuity correction). comp $z = 1.645$. No evidence that % decreased (1.566) or evidence that % decreased (1.978) is awarded **SC2** after B marks |
| **Total: 6** | | |
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4 A train company claims that $92 \%$ of trains on a particular line arrive on time. Sanjeep suspects that the true percentage is less than $92 \%$. He chooses a random sample of 20 trains on this line and finds that exactly 16 of them arrive on time. Making an assumption that should be stated, test at the 5\% significance level whether Sanjeep's suspicion is justified.\\[0pt]
[6]\\
\hfill \mbox{\textit{CAIE S2 2019 Q4 [6]}}