| Exam Board | Edexcel |
|---|---|
| Module | S3 (Statistics 3) |
| Year | 2005 |
| Session | June |
| Paper | Download PDF ↗ |
| Mark scheme | Download PDF ↗ |
| Topic | Hypothesis test of a Poisson distribution |
| Type | Comment on test validity or assumptions |
| Difficulty | Standard +0.3 This is a standard chi-squared goodness-of-fit test for a Poisson distribution, requiring students to estimate λ from data, calculate expected frequencies, combine cells appropriately, compute the test statistic, and compare to critical values. While it involves multiple steps and careful bookkeeping (12 marks reflects this), it follows a well-rehearsed procedure taught explicitly in S3 with no novel problem-solving required, making it slightly easier than average. |
| Spec | 5.06c Fit other distributions: discrete and continuous |
| Number of restarts | Frequency |
| 0 | 99 |
| 1 | 65 |
| 2 | 22 |
| 3 | 12 |
| 4 | 2 |
| Answer | Marks | Guidance |
|---|---|---|
| Content | Marks | Guidance |
| Hypotheses: \(H_0\): Poisson distribution is a suitable model \(H_1\): Poisson distribution is not a suitable model | B1 | |
| Parameter estimate: \(\hat{\lambda} = \frac{(0 \times 99) + (1 \times 65) + \cdots + (4 \times 2)}{200} = \frac{153}{200} = 0.765\) | M1, A1 | |
| **Using \(P(X=x) = 0.765 \frac{e^{-0.765}}{x!}\) where \(X\) represents the number of defects. 200r P(X=x) | M1 | |
| 2oox P(X=x) | ||
| Expected frequency table: | ||
| \(X\) | Observed frequency | Expected frequency |
| 0 | 99 | 93.06678... |
| 1 | 65 | 71.19604... |
| 2 | 22 | 27.2350... |
| 3 | 12 | 6.94482... |
| \(\geq 4\) | 2 | 1.56040... |
| Degrees of freedom: \(\nu = 4 - 1 - 1 = 2\); Critical region: \(\chi^2 > 5.991\) from Poisson | B1, B1√ | |
| Chi-squared statistic: \(\sum \frac{(O-E)^2}{E} = 5.47360...\) | M1, A1 | |
| Conclusion: \(5.47\) is not in the critical region. Number of computer failures per day can be modelled by a Poisson distribution | A1√ (A) |
| Content | Marks | Guidance |
|---------|-------|----------|
| **Hypotheses:** $H_0$: Poisson distribution is a suitable model $H_1$: Poisson distribution is not a suitable model | B1 | |
| **Parameter estimate:** $\hat{\lambda} = \frac{(0 \times 99) + (1 \times 65) + \cdots + (4 \times 2)}{200} = \frac{153}{200} = 0.765$ | M1, A1 | |
| **Using $P(X=x) = 0.765 \frac{e^{-0.765}}{x!}$ where $X$ represents the number of defects. 200r P(X=x) | M1 | |
| | | 2oox P(X=x) |
| **Expected frequency table:** | | |
| $X$ | Observed frequency | Expected frequency | M1, A1 | |
| 0 | 99 | 93.06678... | |
| 1 | 65 | 71.19604... | 0,2 | A1, A1 |
| 2 | 22 | 27.2350... | | |
| 3 | 12 | 6.94482... | | |
| $\geq 4$ | 2 | 1.56040... | 8.5046? | A1 |
| **Degrees of freedom:** $\nu = 4 - 1 - 1 = 2$; **Critical region:** $\chi^2 > 5.991$ from Poisson | B1, B1√ | |
| **Chi-squared statistic:** $\sum \frac{(O-E)^2}{E} = 5.47360...$ | M1, A1 | |
| **Conclusion:** $5.47$ is not in the critical region. Number of computer failures per day can be modelled by a Poisson distribution | A1√ (A) | |
---The number of times per day a computer fails and has to be restarted is recorded for 200 days. The results are summarised in the table.
\begin{center}
\begin{tabular}{|c|c|}
\hline
Number of restarts & Frequency \\
\hline
0 & 99 \\
\hline
1 & 65 \\
\hline
2 & 22 \\
\hline
3 & 12 \\
\hline
4 & 2 \\
\hline
\end{tabular}
\end{center}
Test whether or not a Poisson model is suitable to represent the number of restarts per day. Use a 5\% level of significance and state your hypothesis clearly.
(Total 12 marks)
\hfill \mbox{\textit{Edexcel S3 2005 Q5}}