| Exam Board | CAIE |
|---|---|
| Module | Further Paper 4 (Further Paper 4) |
| Year | 2024 |
| Session | November |
| Marks | 8 |
| Paper | Download PDF ↗ |
| Mark scheme | Download PDF ↗ |
| Topic | Chi-squared goodness of fit |
| Type | Chi-squared goodness of fit: Poisson |
| Difficulty | Standard +0.3 This is a standard chi-squared goodness of fit test with a given Poisson distribution. Part (a) requires straightforward calculation of expected frequencies using Po(1.9), and part (b) follows the routine procedure: calculate test statistic, find critical value, and compare. The parameter is given (not estimated), making it slightly easier than average. The only minor complication is combining cells for expected frequencies below 5, which is standard A-level technique. |
| Spec | 5.06b Fit prescribed distribution: chi-squared test |
| Number of calls | 0 | 1 | 2 | 3 | 4 | 5 | 6 or more |
| Observed frequency | 10 | 18 | 35 | 21 | 11 | 4 | 1 |
| Expected frequency | 14.957 | 28.418 | 26.997 | 1.322 |
| Answer | Marks | Guidance |
|---|---|---|
| Answer | Marks | Guidance |
| 17.098, 8.122, 3.086 | B1 | One correct |
| B1 | All correct |
| Answer | Marks | Guidance |
|---|---|---|
| Answer | Marks | Guidance |
| Classes: 0, 1, 2, 3, 4 or more; Observed: 10, 18, 35, 21, 16; Expected: 14.957, 28.418, 26.997, 17.098, 12.53 | M1 | Last two or three columns combined |
| Contributions to test statistic: 1.6428, 3.8192, 2.3724, 0.8905, 0.9609(7) | M1 | May be implied by awrt 9.69 |
| Test statistic is 9.69 | A1 | 9.686 |
| \(H_0\): Po(1.9) is a good fit for the data; \(H_1\): Po(1.9) is not a good fit for the data | B1 | |
| Critical value is 7.779, compare \(9.69 > 7.779\) reject \(H_0\) | M1 | 4 degrees of freedom |
| Sufficient evidence to suggest that Po(1.9) is not a good fit for the data / Sufficient evidence to reject/not support the statistician's claim | A1 | Correct work only, including hypotheses, in context, level of uncertainty in language |
## Question 3(a):
| Answer | Marks | Guidance |
|--------|-------|----------|
| 17.098, 8.122, 3.086 | B1 | One correct |
| | B1 | All correct |
## Question 3(b):
| Answer | Marks | Guidance |
|--------|-------|----------|
| Classes: 0, 1, 2, 3, 4 or more; Observed: 10, 18, 35, 21, 16; Expected: 14.957, 28.418, 26.997, 17.098, 12.53 | M1 | Last two or three columns combined |
| Contributions to test statistic: 1.6428, 3.8192, 2.3724, 0.8905, 0.9609(7) | M1 | May be implied by awrt 9.69 |
| Test statistic is 9.69 | A1 | 9.686 |
| $H_0$: Po(1.9) is a good fit for the data; $H_1$: Po(1.9) is not a good fit for the data | B1 | |
| Critical value is 7.779, compare $9.69 > 7.779$ reject $H_0$ | M1 | 4 degrees of freedom |
| Sufficient evidence to suggest that Po(1.9) is not a good fit for the data / Sufficient evidence to reject/not support the statistician's claim | A1 | Correct work only, including hypotheses, in context, level of uncertainty in language |
---3 A statistician believes that the number of telephone calls received by an advice centre in a 10 -minute interval can be modelled by the Poisson distribution $\mathrm { Po } ( 1.9 )$. The number of calls received in a randomly chosen 10-minute interval was recorded on each of 100 days. The results are summarised in the table, together with some of the expected frequencies corresponding to the distribution $\operatorname { Po } ( 1.9 )$.
\begin{center}
\begin{tabular}{|l|l|l|l|l|l|l|l|}
\hline
Number of calls & 0 & 1 & 2 & 3 & 4 & 5 & 6 or more \\
\hline
Observed frequency & 10 & 18 & 35 & 21 & 11 & 4 & 1 \\
\hline
Expected frequency & 14.957 & 28.418 & 26.997 & & & & 1.322 \\
\hline
\end{tabular}
\end{center}
\begin{enumerate}[label=(\alph*)]
\item Complete the table.
\item Carry out a goodness of fit test, at the $10 \%$ significance level, to determine whether the statistician's belief is reasonable.\\
\includegraphics[max width=\textwidth, alt={}, center]{e2a45d19-7d48-4aa5-93f9-6ef90f99d7c4-07_2726_35_97_20}
\end{enumerate}
\hfill \mbox{\textit{CAIE Further Paper 4 2024 Q3 [8]}}