A-Level Mathematics October/November 2024 Q7(a): The number of accidents per year on a certain road has the distribution Po(λ). In the p…

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  Let $X$ be the number of accidents in two years. The original rate is $\lambda = 3.3$ per year.

  The council wishes to test if the rate has decreased. Hypotheses: $H_0: \lambda = 3.3$ $H_1: \lambda < 3.3$

  Under $H_0$, the number of accidents in two years, $X$, follows a Poisson distribution with parameter $2 \times 3.3 = 6.6$. So, $X \sim \text{Po}(6.6)$.

  The test is carried out at the 5% significance level. We need to find the critical region, which is of the form $X \le c$, such that $P(X \le c) < 0.05$.

  We test values for $c$: $P(X \le 2) = P(X=0) + P(X=1) + P(X=2)$ $= e^{-6.6} \left( \frac{6.6^0}{0!} + \frac{6.6^1}{1!} + \frac{6.6^2}{2!} \right)$ $= e^{-6.6} (1 + 6.6 + 21.78)$ $= e^{-6.6} (29.38) = 0.04004...$ Since $0.04004... < 0.05$, the value $c=2$ is in the critical region.

  Now we check the next value, $c=3$: $P(X \le 3) = P(X \le 2) + P(X=3)$ $= 0.04004... + e^{-6.6} \frac{6.6^3}{3!}$ $= 0.04004... + 0.06518... = 0.1052...$ Since $0.1052... > 0.05$, the critical region is $X \le 2$.

  A Type I error occurs when $H_0$ is rejected but $H_0$ is true. This means the result falls into the critical region ($X \le 2$) when the distribution is indeed $\text{Po}(6.6)$.

  Probability of a Type I error = $P(X \le 2)$ under $H_0$.

  $P(\text{Type I error}) = 0.04004...$

  $P(\text{Type I error}) = 0.0400$ (3 s.f.)

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  How the marks are awarded

  • B1 — Stating the correct parameter for the Poisson distribution over a two-year period, $\lambda = 6.6$.
  • M1 — Attempting to find the critical region by calculating a Poisson probability and comparing it to 0.05. The expression for $P(X \le 2)$ must be seen, i.e., $e^{-6.6}(1 + 6.6 + 6.6^2/2)$. Using $\lambda=3.3$ is acceptable for this method mark.
  • B1 — Calculating $P(X \le 3)$ and showing it is greater than 0.05 to confirm the boundary of the critical region. An unsupported value of 0.105 is sufficient.
  • A1 — Stating the final correct probability of a Type I error, which is the calculated $P(X \le 2)$, rounded correctly to 3 significant figures as 0.0400.

  Common mistakes

  • Using the wrong parameter for the Poisson distribution, i.e., using $\lambda = 3.3$ for the two-year period instead of $\lambda = 6.6$.
  • Stating that the probability of a Type I error is exactly the significance level (0.05), forgetting that for a discrete distribution it is the actual probability of the rejection region.
  • Incorrectly identifying the critical region, for example by choosing $X \le 3$ because $P(X \le 3)$ is the closest probability to 0.05, rather than the first one below 0.05.
  • Calculating the probability for a single value, e.g., $P(X=2)$, instead of the cumulative probability for the region, $P(X \le 2)$.

  Examiner tip: For hypothesis tests with discrete distributions, remember that the probability of a Type I error is the actual probability of the calculated rejection region (P(X ≤ c)), not simply the stated significance level.

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