| Exam Board | OCR MEI |
|---|---|
| Module | S3 (Statistics 3) |
| Year | 2008 |
| Session | June |
| Marks | 17 |
| Paper | Download PDF ↗ |
| Mark scheme | Download PDF ↗ |
| Topic | Poisson distribution |
| Type | Goodness-of-fit test for Poisson |
| Difficulty | Standard +0.3 This is a standard S3 question covering routine chi-squared goodness of fit test and confidence interval calculation. Part (a)(i) requires basic mean/variance calculation and comparison (standard Poisson property check), part (a)(ii) is a textbook chi-squared test with expected frequencies, and part (b) is a straightforward t-distribution confidence interval. All techniques are standard bookwork with no novel insight required, though the 10 marks for the chi-squared test reflects the computational steps involved. Slightly above average difficulty due to the multi-step nature and need for careful calculation. |
| Spec | 5.05d Confidence intervals: using normal distribution5.06b Fit prescribed distribution: chi-squared test |
| Number of bees | 0 | 1 | 2 | 3 | 4 | 5 | 6 | 7 | \(\geq 8\) |
| Number of intervals | 6 | 16 | 19 | 18 | 17 | 14 | 6 | 4 | 0 |
| Answer | Marks |
|---|---|
| Evidence could support Poisson since the variance is fairly close to the mean. | B1, B1, E1 |
| Answer | Marks | Guidance |
|---|---|---|
| \(f_o\) | 6 | 16 |
| \(f_e\) | 4.50 | 13.97 |
| Merged | 22 |
| Answer | Marks | Guidance |
|---|---|---|
| Suggests Poisson model does fit... at any reasonable level of significance. | M1, A1, A1, M1, M1, A1, A1, A1, A1 | Calculation of expected frequencies. Last cell correct. All others correct, but ft if wrong. Combining cells. (Condone if not combined as fully as shown above, but require top two cells combined as a minimum.) Calculation of \(\chi^2\). (Condone wrong last cell.) Depends on both of the preceding M marks. Allow correct df (= cells – 2) from wrongly grouped or ungrouped table, and FT. Otherwise, no FT if wrong. ft only c's test statistic. ft only c's test statistic. Or other sensible comment. |
| Answer | Marks | Guidance |
|---|---|---|
| CI is given by \(1.465 \pm \frac{2.262}{\sqrt[2]{}} \times \frac{0.3288}{\sqrt{10}} = 1.465 \pm 0.2352 = (1.2298, 1.7002)\) | M1, B1, B1, A1 | If both 1.465 and \(0.3288/\sqrt{10}\) are correct. If \(t_9\) used. 95% 2-tail point for \(c's\) \(t\) distribution (Independent of previous mark). c.a.o. Must be expressed as an interval. |
## (a)(i)
$\bar{x} = \frac{310}{100} = 3.1$
$s^2 = \frac{1288 - 100 \times 3.1^2}{99} = \frac{327}{99} = 3.303$
Evidence could support Poisson since the variance is fairly close to the mean. | B1, B1, E1 |
## (a)(ii)
| $f_o$ | 6 | 16 | 19 | 18 | 17 | 14 | 6 | 4 | 0 |
|---|---|---|---|---|---|---|---|---|---|
| $f_e$ | 4.50 | 13.97 | 21.65 | 22.37 | 17.33 | 10.75 | 5.55 | 2.46 | 1.42 |
| Merged | | 22 | | | | | | 9.43 |
$\chi^2 = 0.6747 + 0.3244 + 0.8537 + 0.0063 + 0.9826 + 0.0345 = 2.876(2)$
Refer to $\chi^2_i$, e.g. Upper 10% point is 7.779.
Not significant.
Suggests Poisson model does fit... at any reasonable level of significance. | M1, A1, A1, M1, M1, A1, A1, A1, A1 | Calculation of expected frequencies. Last cell correct. All others correct, but ft if wrong. Combining cells. (Condone if not combined as fully as shown above, but require top two cells combined as a minimum.) Calculation of $\chi^2$. (Condone wrong last cell.) Depends on both of the preceding M marks. Allow correct df (= cells – 2) from wrongly grouped or ungrouped table, and FT. Otherwise, no FT if wrong. ft only c's test statistic. ft only c's test statistic. Or other sensible comment.
## (b)
CI is given by $1.465 \pm \frac{2.262}{\sqrt[2]{}} \times \frac{0.3288}{\sqrt{10}} = 1.465 \pm 0.2352 = (1.2298, 1.7002)$ | M1, B1, B1, A1 | If both 1.465 and $0.3288/\sqrt{10}$ are correct. If $t_9$ used. 95% 2-tail point for $c's$ $t$ distribution (Independent of previous mark). c.a.o. Must be expressed as an interval.
---\begin{enumerate}[label=(\alph*)]
\item A researcher is investigating the feeding habits of bees. She sets up a feeding station some distance from a beehive and, over a long period of time, records the numbers of bees arriving each minute. For a random sample of 100 one-minute intervals she obtains the following results.
\begin{center}
\begin{tabular}{|c|c|c|c|c|c|c|c|c|c|}
\hline
Number of bees & 0 & 1 & 2 & 3 & 4 & 5 & 6 & 7 & $\geq 8$ \\
\hline
Number of intervals & 6 & 16 & 19 & 18 & 17 & 14 & 6 & 4 & 0 \\
\hline
\end{tabular}
\end{center}
\begin{enumerate}[label=(\roman*)]
\item Show that the sample mean is 3.1 and find the sample variance. Do these values support the possibility of a Poisson model for the number of bees arriving each minute? Explain your answer. [3]
\item Use the mean in part (i) to carry out a test of the goodness of fit of a Poisson model to the data. [10]
\end{enumerate}
\item The researcher notes the length of time, in minutes, that each bee spends at the feeding station. The times spent are assumed to be Normally distributed. For a random sample of 10 bees, the mean is found to be 1.465 minutes and the standard deviation is 0.3288 minutes. Find a 95\% confidence interval for the overall mean time. [4]
\end{enumerate}
\hfill \mbox{\textit{OCR MEI S3 2008 Q4 [17]}}