| Exam Board | CAIE |
|---|---|
| Module | FP2 (Further Pure Mathematics 2) |
| Year | 2014 |
| Session | November |
| Marks | 9 |
| 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: (1) estimating the parameter λ from data, (2) calculating expected frequencies, (3) combining cells appropriately, (4) computing the test statistic, and (5) comparing to critical values. While methodical, it involves multiple computational steps, parameter estimation reducing degrees of freedom, and careful cell combination—more demanding than typical A-level questions but still a textbook Further Maths procedure without novel insight required. |
| Spec | 5.06b Fit prescribed distribution: chi-squared test |
| Number sold | 0 | 1 | 2 | 3 | 4 | 5 | 6 | 7 | \(\geqslant 8\) |
| Number of Saturdays | 7 | 20 | 39 | 16 | 14 | 2 | 1 | 1 | 0 |
| Answer | Marks | Guidance |
|---|---|---|
| Answer | Marks | Guidance |
| \(\lambda = \frac{225}{100} = 2.25\) | B1 | Find mean of sample data |
| \(H_0\): Poisson distribution fits data | B1 | State null hypothesis; A.E.F. |
| Expected values: \(10.540\ \ 23.715\ \ 26.679\ \ 20.009\ \ 11.255\ \ 5.065\ \ 1.899\ \ 0.6105\ \ 0.2275\) | M1 A1 | Find expected values \(\frac{100\lambda^r e^{-\lambda}}{r!}\) to 1 d.p.; ignore incorrect final value for M1 |
| \(O_i\): \(\ldots\ 16\quad 14\quad 4\); \(E_i\): \(\ldots\ 20.009\quad 11.255\quad 7.802\) | *M1 | Combine last four cells so that expected value \(\geq 5\) |
| \(\chi^2 = 1.189 + 0.582 + 5.690 + 0.803 + 0.6695 + 1.853 = 10.8\) (allow 10.7) | M1 A1 | Calculate \(\chi^2\) to 1 d.p.; A1 dependent on *M1 |
| \(\chi^2_{4,\ 0.975} = 11.14\) (if cells combined) | B1 | State or use consistent tabular value to 1 d.p.; \([\chi^2_{7,\ 0.975} = 16.01,\ \chi^2_{5,\ 0.975} = 12.83]\) |
| \(\chi^2 < 11.1\) so Poisson distribution fits | B1\(\checkmark\) | Consistent conclusion on two \(\chi^2\) values; A.E.F. |
# Question 8:
| Answer | Marks | Guidance |
|--------|-------|----------|
| $\lambda = \frac{225}{100} = 2.25$ | B1 | Find mean of sample data |
| $H_0$: Poisson distribution fits data | B1 | State null hypothesis; A.E.F. |
| Expected values: $10.540\ \ 23.715\ \ 26.679\ \ 20.009\ \ 11.255\ \ 5.065\ \ 1.899\ \ 0.6105\ \ 0.2275$ | M1 A1 | Find expected values $\frac{100\lambda^r e^{-\lambda}}{r!}$ to 1 d.p.; ignore incorrect final value for M1 |
| $O_i$: $\ldots\ 16\quad 14\quad 4$; $E_i$: $\ldots\ 20.009\quad 11.255\quad 7.802$ | *M1 | Combine last four cells so that expected value $\geq 5$ |
| $\chi^2 = 1.189 + 0.582 + 5.690 + 0.803 + 0.6695 + 1.853 = 10.8$ (allow 10.7) | M1 A1 | Calculate $\chi^2$ to 1 d.p.; A1 dependent on *M1 |
| $\chi^2_{4,\ 0.975} = 11.14$ (if cells combined) | B1 | State or use consistent tabular value to 1 d.p.; $[\chi^2_{7,\ 0.975} = 16.01,\ \chi^2_{5,\ 0.975} = 12.83]$ |
| $\chi^2 < 11.1$ so Poisson distribution fits | B1$\checkmark$ | Consistent conclusion on two $\chi^2$ values; A.E.F. |
---8 The numbers of a particular type of laptop computer sold by a store on each of 100 consecutive Saturdays are summarised in the following table.
\begin{center}
\begin{tabular}{ | l | c | c | c | c | c | c | c | c | c | }
\hline
Number sold & 0 & 1 & 2 & 3 & 4 & 5 & 6 & 7 & $\geqslant 8$ \\
\hline
Number of Saturdays & 7 & 20 & 39 & 16 & 14 & 2 & 1 & 1 & 0 \\
\hline
\end{tabular}
\end{center}
Fit a Poisson distribution to the data and carry out a goodness of fit test at the $2.5 \%$ significance level.
\hfill \mbox{\textit{CAIE FP2 2014 Q8 [9]}}