| Exam Board | OCR |
|---|---|
| Module | S2 (Statistics 2) |
| Year | 2007 |
| Session | June |
| Marks | 7 |
| Paper | Download PDF ↗ |
| Mark scheme | Download PDF ↗ |
| Topic | Hypothesis test of a Poisson distribution |
| Type | Find actual significance level |
| Difficulty | Standard +0.3 This is a straightforward hypothesis testing question requiring students to identify a one-tailed alternative hypothesis, calculate significance level using Poisson tables for λ=15 (6×2.5), and find the critical value by checking tables. While it involves multiple parts and requires careful table reading, the concepts are standard S2 material with no novel problem-solving or proof required. |
| Spec | 2.05a Hypothesis testing language: null, alternative, p-value, significance2.05c Significance levels: one-tail and two-tail5.02j Poisson formula: P(X=x) = e^(-lambda)*lambda^x/x!5.02k Calculate Poisson probabilities |
| Answer | Marks | Guidance |
|---|---|---|
| (i) \(H_1: \lambda > 2.5\) or \(15\) | B1 | \(\lambda > 2.5\) or \(15\), allow \(\mu\), don't need "\(H_1\)" |
| Use parameter \(15\) | M1 | |
| \(P(> 23)\) | M1 | |
| \(1 - 0.9805 = 0.0195\) or \(1.95\%\) | A1 | Answer, \(1.95\%\) or \(2\%\) or \(0.0195\) or \(0.02\) [SR: 2-tailed, 3.9% gets 3/3 here] |
| (ii) \(P(\leq 23 \mid \lambda = 17) = 0.9367\); \(P(\leq 23 \mid \lambda = 18) = 0.8989\); Parameter = \(17\) | M1 | One of these, or their complement; 9367, 8989, 0.9047, 0.8551, 9317, 8933, 9907, 9805 |
| A1 | Parameter \(17\) [17.1076], needs P(≤ 23), two [SR: if insufficient evidence can give B1 for 17] | |
| \(\lambda = 17/6\) or \(2.83\) | M1 | Their parameter ÷ 6 [2.85] |
| A1 | ||
| (iii) \(H_0: p = 0.19, H_1: p < 0.19\) where \(p\) is population proportion | B2 | |
| \(0.81^{20} + 20 \times 0.81^{19} \times 0.19 = 0.0841\); Compare 0.1 | M1 | Binomial probabilities, allow 1 term only |
| A1 | Correct expression [0.0148 + 0.0693] | |
| A1 | Probability, a.r.t. 0.084 | |
| \(0.1\) | B1 | Explicit comparison of "like with like" |
| A1 | [P(≤ 2) = 0.239] | |
| (ii) Letters not independent | B1 | Correct modelling assumption, stated in context; Allow "random", "depends on message", etc |
(i) $H_1: \lambda > 2.5$ or $15$ | B1 | $\lambda > 2.5$ or $15$, allow $\mu$, don't need "$H_1$"
Use parameter $15$ | M1 |
$P(> 23)$ | M1 |
$1 - 0.9805 = 0.0195$ or $1.95\%$ | A1 | Answer, $1.95\%$ or $2\%$ or $0.0195$ or $0.02$ [SR: 2-tailed, 3.9% gets 3/3 here]
(ii) $P(\leq 23 \mid \lambda = 17) = 0.9367$; $P(\leq 23 \mid \lambda = 18) = 0.8989$; Parameter = $17$ | M1 | One of these, or their complement; 9367, 8989, 0.9047, 0.8551, 9317, 8933, 9907, 9805
| A1 | Parameter $17$ [17.1076], needs P(≤ 23), two [SR: if insufficient evidence can give B1 for 17]
$\lambda = 17/6$ or $2.83$ | M1 | Their parameter ÷ 6 [2.85]
| A1 |
(iii) $H_0: p = 0.19, H_1: p < 0.19$ where $p$ is population proportion | B2 |
$0.81^{20} + 20 \times 0.81^{19} \times 0.19 = 0.0841$; Compare 0.1 | M1 | Binomial probabilities, allow 1 term only
| A1 | Correct expression [0.0148 + 0.0693]
| A1 | Probability, a.r.t. 0.084
$0.1$ | B1 | Explicit comparison of "like with like"
| A1 | [P(≤ 2) = 0.239]
(ii) Letters not independent | B1 | Correct modelling assumption, stated in context; Allow "random", "depends on message", etc5 The number of system failures per month in a large network is a random variable with the distribution $\operatorname { Po } ( \lambda )$. A significance test of the null hypothesis $\mathrm { H } _ { 0 } : \lambda = 2.5$ is carried out by counting $R$, the number of system failures in a period of 6 months. The result of the test is that $\mathrm { H } _ { 0 }$ is rejected if $R > 23$ but is not rejected if $R \leqslant 23$.\\
(i) State the alternative hypothesis.\\
(ii) Find the significance level of the test.\\
(iii) Given that $\mathrm { P } ( R > 23 ) < 0.1$, use tables to find the largest possible actual value of $\lambda$. You should show the values of any relevant probabilities.
\hfill \mbox{\textit{OCR S2 2007 Q5 [7]}}