Solution :-

Part (a): Probability of Type I Error

Null hypothesis \(H_0: \lambda = 3.3\)

Alternative hypothesis \(H_1: \lambda < 3.3\)

Significance level \(\alpha = 5\% = 0.05\)

Two years: \(\lambda = 3.3 \times 2 = 6.6\)

\(X \sim Po(6.6)\)

We need to find the critical region. Let \(c\) be the critical value.

\(P(X \le c) \le 0.05\)

Using Poisson distribution tables or calculator:

\(P(X \le 2) = 0.0496\)

\(P(X \le 3) = 0.1026\)

Critical region: \(X \le 2\)

Probability of Type I error is \(P(X \le 2)\) when \(\lambda = 6.6\).

\(P(X \le 2) = P(X=0) + P(X=1) + P(X=2)\)

\(P(X=0) = e^{-6.6} \approx 0.001357\)

\(P(X=1) = 6.6e^{-6.6} \approx 0.008956\)

\(P(X=2) = \frac{6.6^2 e^{-6.6}}{2} \approx 0.02955\)

\(P(X \le 2) = 0.001357 + 0.008956 + 0.02955 = 0.039863\)

Probability of Type I error is approximately 0.0399.

\(\lambda = 6.6\)

\(P(X \le 2) = e^{-6.6} (1 + 6.6 + \frac{6.6^2}{2}) [= 0.0400] [< 0.05]\) (Expression must be seen. No end errors. Allow use of 3.3 here.)

or \(e^{-6.6} (1 + 6.6 + 21.78)\)

or \(0.001360 + 0.008978 + 0.02963\)

\(P(X \le 3) = e^{-6.6} (1 + 6.6 + \frac{6.6^2}{2} + \frac{6.6^3}{6})\) or \(0.0400 + … = 0.105 [> 0.05]\) (Condone unsupported 0.105.)

\(P(\text{Type I error}) = 0.0400 \text{ (3 sf)}\) (Allow 0.040 or 0.04 AWRT. SC unsupported ans of 0.0400 can score max B1B1B1.)

Part (b): Carry out the test

Given \(X = 2\), which is in the critical region \(X \le 2\).

Therefore, we reject \(H_0\).

Conclusion: There is sufficient evidence at the 5% significance level to conclude that the value of \(\lambda\) has decreased.

\(H_0: \lambda = 6.6, H_1: \lambda < 6.6\) (May be seen in part (a) and award B1 mark here. Accept \(\mu\) or \(\lambda\). Accept 3.3 or 6.6.)

\([P(X \le 2) = 0.0400] 0.04 < 0.05\) (For comparing their \(P(X \le 2)\) any \(\lambda\) with 0.05.)

[\(\text{Reject } H_0\)] There is evidence to suggest that mean number of accidents has decreased. (In context, not definite. No contradictions. CWO.)

Part (c): Probability of Type II Error

\(H_0: \lambda = 3.3\)

\(H_1: \lambda < 3.3\)

Critical region: \(X \le 2\) when \(\lambda = 6.6\) (from part a).

True value of \(\lambda = 0.6\), so for two years, \(\lambda = 0.6 \times 2 = 1.2\).

\(X \sim Po(1.2)\)

Type II error is failing to reject \(H_0\) when \(H_1\) is true.

We fail to reject \(H_0\) when \(X > 2\).

Probability of Type II error is \(P(X > 2)\) when \(\lambda = 1.2\).

\(P(X > 2) = 1 – P(X \le 2)\)

\(P(X \le 2) = P(X=0) + P(X=1) + P(X=2)\)

\(P(X=0) = e^{-1.2} \approx 0.3012\)

\(P(X=1) = 1.2e^{-1.2} \approx 0.3614\)

\(P(X=2) = \frac{1.2^2 e^{-1.2}}{2} \approx 0.2169\)

\(P(X \le 2) = 0.3012 + 0.3614 + 0.2169 = 0.8795\)

\(P(X > 2) = 1 – 0.8795 = 0.1205\)

Probability of Type II error is approximately 0.1205.

\(P(X > 2)\) attempted, with any \(\lambda\).

\(P(X > 2) = 1 – e^{-1.2} (1 + 1.2 + \frac{1.2^2}{2})\) (Expression must be seen. Correct \(\lambda\). No end errors.)

or \(= 1 – e^{-1.2} (1 + 1.2 + 0.72)\)

or \(= 1 – (0.3012 + 0.3614 + 0.2169)\)

\(= 0.121 \text{ (3 sf)} \text{ or } 0.120\) (SC unsupported answer scores B2.)

Part (d): Normal Approximation

\(\lambda = 0.6\) per year, so for 50 years, \(\lambda = 0.6 \times 50 = 30\).

\(X \sim Po(30)\)

For large \(\lambda\), Poisson can be approximated by Normal distribution.

\(X \sim N(\mu, \sigma^2)\), where \(\mu = \lambda = 30\) and \(\sigma^2 = \lambda = 30\).

\(\sigma = \sqrt{30} \approx 5.477\)

We want \(P(X > 19)\). Using continuity correction, \(P(X > 19.5)\).

\(Z = \frac{X – \mu}{\sigma} = \frac{19.5 – 30}{5.477} = \frac{-10.5}{5.477} \approx -1.917\)

\(P(Z > -1.917) = 1 – P(Z < -1.917)\)

From standard normal tables, \(P(Z < -1.917) \approx 0.0276\)

\(P(Z > -1.917) = 1 – 0.0276 = 0.9724\)

Probability is approximately 0.9724.

\(N(18, 18)\) seen or implied.

\(\frac{10.5 – 18}{\sqrt{18}} [= -1.768]\) (Allow with no or incorrect continuity correction. Their 18.)

\(P(X > -1.768) = \Phi(1.768)\) (ft their standardised value. Area consistent with their values.)

\(= 0.961 \text{ or } 0.962 \text{ (3 sf)}\)