| Exam Board | CAIE |
|---|---|
| Module | FP2 (Further Pure Mathematics 2) |
| Year | 2015 |
| Session | November |
| Marks | 10 |
| Paper | Download PDF ↗ |
| Mark scheme | Download PDF ↗ |
| Topic | Chi-squared goodness of fit |
| Type | Chi-squared goodness of fit: Poisson |
| Difficulty | Standard +0.8 This is a standard chi-squared goodness of fit test for a Poisson distribution requiring multiple steps: calculating the sample mean to estimate λ, computing expected frequencies using Poisson probabilities, pooling cells to ensure expected frequencies ≥5, calculating the χ² test statistic, determining degrees of freedom (accounting for estimated parameter and pooled cells), and comparing to critical value. While methodical, it's a routine Further Maths statistics procedure with no novel insight required, placing it moderately above average difficulty. |
| Spec | 5.06b Fit prescribed distribution: chi-squared test |
| Number of goals | 0 | 1 | 2 | 3 | 4 | 5 | 6 or more |
| Frequency | 12 | 16 | 31 | 25 | 13 | 3 | 0 |
| Answer | Marks | Guidance |
|---|---|---|
| Answer/Working | Marks | Guidance |
| \(\lambda = 220/100 = 2.2\) | B1 | Find mean of sample data |
| \(H_0\): Poisson distribution fits data *or* \(\lambda = 2.2\) | B1 | State null hypothesis (AEF) |
| Expected values \(100\lambda^r e^{-\lambda}/r!\): \(11.080\quad 24.377\quad 26.814\quad 19.664\quad 10.815\quad 4.759\quad 2.491\) | M1 A1 | (to 1 d.p.; ignore incorrect final value for M1) |
| Combine last two cells: \(O_i: 3\); \(E_i: 7.25\) | M1* | So that exp. value \(\geq 5\) |
| \(\chi^2 = 0.076+2.879+0.653+1.448+0.441+2.491 = 7.99\) (allow \(7.95\) if 1 d.p. exp. values used) | M1 A1 | Calculate \(\chi^2\) to 2 d.p.; A1 dep M1* |
| 5 cells: \(\chi^2_{3,\,0.95} = 7.815\); 6 cells: \(\chi^2_{4,\,0.95} = 9.488\) (correct); 7 cells: \(\chi^2_{5,\,0.95} = 11.07\) | B1 | Correct tabular value to 3 s.f. |
| Accept \(H_0\) if \(\chi^2 <\) tabular value | M1 | State/imply valid method for conclusion |
| Distribution fits *or* \(\lambda = 2.2\) (requires both values correct; not combining cells [\(\chi^2=8.64\)] can earn B1 B1 M1 A1 M0 M1 B1 M1, max 7) | A1 | Total: 10 marks |
## Question 8:
| Answer/Working | Marks | Guidance |
|---|---|---|
| $\lambda = 220/100 = 2.2$ | B1 | Find mean of sample data |
| $H_0$: Poisson distribution fits data *or* $\lambda = 2.2$ | B1 | State null hypothesis (AEF) |
| Expected values $100\lambda^r e^{-\lambda}/r!$: $11.080\quad 24.377\quad 26.814\quad 19.664\quad 10.815\quad 4.759\quad 2.491$ | M1 A1 | (to 1 d.p.; ignore incorrect final value for M1) |
| Combine last two cells: $O_i: 3$; $E_i: 7.25$ | M1* | So that exp. value $\geq 5$ |
| $\chi^2 = 0.076+2.879+0.653+1.448+0.441+2.491 = 7.99$ (allow $7.95$ if 1 d.p. exp. values used) | M1 A1 | Calculate $\chi^2$ to 2 d.p.; A1 dep M1* |
| 5 cells: $\chi^2_{3,\,0.95} = 7.815$; 6 cells: $\chi^2_{4,\,0.95} = 9.488$ (correct); 7 cells: $\chi^2_{5,\,0.95} = 11.07$ | B1 | Correct tabular value to 3 s.f. |
| Accept $H_0$ if $\chi^2 <$ tabular value | M1 | State/imply valid method for conclusion |
| Distribution fits *or* $\lambda = 2.2$ (requires both values correct; not combining cells [$\chi^2=8.64$] can earn B1 B1 M1 A1 M0 M1 B1 M1, max 7) | A1 | Total: 10 marks |
---8 The number of goals scored by a certain football team was recorded for each of 100 matches, and the results are summarised in the following table.
\begin{center}
\begin{tabular}{ | l | c | c | c | c | c | c | c | }
\hline
Number of goals & 0 & 1 & 2 & 3 & 4 & 5 & 6 or more \\
\hline
Frequency & 12 & 16 & 31 & 25 & 13 & 3 & 0 \\
\hline
\end{tabular}
\end{center}
Fit a Poisson distribution to the data, and test its goodness of fit at the 5\% significance level.
\hfill \mbox{\textit{CAIE FP2 2015 Q8 [10]}}