| Exam Board | CAIE |
|---|---|
| Module | Further Paper 4 (Further Paper 4) |
| Year | 2023 |
| 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 routine calculation of missing expected frequencies using Po(3.5), and part (b) follows the standard test procedure: pooling cells to ensure expected frequencies ≥5, calculating the test statistic, finding degrees of freedom, and comparing to critical value. While it requires careful arithmetic and knowledge of the chi-squared test procedure, it involves no novel problem-solving or conceptual challenges beyond textbook application. |
| Spec | 5.06c Fit other distributions: discrete and continuous |
| Number of breakdowns per day | 0 | 1 | 2 | 3 | 4 | 5 | 6 | 7 | 8 or more |
| Observed frequency | 0 | 5 | 13 | 17 | 21 | 16 | 9 | 5 | 4 |
| Expected frequency | 2.718 | 9.512 | 16.646 | 16.993 | 11.895 | 3.469 | 2.407 |
| Answer | Marks | Guidance |
|---|---|---|
| Answer | Marks | Guidance |
| Frequencies: \(0, 5, 13, 17, 21, 16, 9, 5, 4\) | B1 | Each. |
| Expected: \(2.718, 9.512, 16.646, \mathbf{19.421}, 16.993, 11.895, 6.939, 3.469, 2.407\) | B1 | |
| 2 |
| Answer | Marks | Guidance |
|---|---|---|
| Answer | Marks | Guidance |
| \(H_0\): Po(3.5) is a good fit to the data; \(H_1\): Po(3.5) is not a good fit to the data | B1 | |
| Combine first 2 columns: 5, 12.23 and last 2 columns: 9, 5.876 | M1 | Both. |
| Chi-squared values: \(4.274 + 0.799 + 0.302 + 0.945 + 1.417 + 0.612 + 1.661\) | M1 | Allow if no or incorrect number of columns added. At least two 'correct' values (3 sf) or expressions seen from their grouping (or lack of). \(2.718 + 2.140 + \ldots + 0.6757 + 1.054\) |
| \(10.0\) | A1 | AWRT 10.0. If M0 awarded then SC B1 for 10.0. |
| Tabular value: \(10.64\); \('10.0' < 10.64\), accept \(H_0\)/not significant | M1 | Allow equivalent to 10.64 if columns not combined or only one pair combined (12.02 one pair combined, 13.36 none combined). |
| Insufficient evidence to suggest that Po(3.5) is not a good fit to the data | A1 | Correct conclusion in context, following correct work, level of uncertainty in language. A0 if hypotheses the wrong way round or missing. |
| 6 |
## Question 2(a):
| Answer | Marks | Guidance |
|--------|-------|----------|
| Frequencies: $0, 5, 13, 17, 21, 16, 9, 5, 4$ | B1 | Each. |
| Expected: $2.718, 9.512, 16.646, \mathbf{19.421}, 16.993, 11.895, 6.939, 3.469, 2.407$ | B1 | |
| | **2** | |
## Question 2(b):
| Answer | Marks | Guidance |
|--------|-------|----------|
| $H_0$: Po(3.5) is a good fit **to the data**; $H_1$: Po(3.5) is not a good fit **to the data** | B1 | |
| Combine first 2 columns: 5, 12.23 and last 2 columns: 9, 5.876 | M1 | Both. |
| Chi-squared values: $4.274 + 0.799 + 0.302 + 0.945 + 1.417 + 0.612 + 1.661$ | M1 | Allow if no or incorrect number of columns added. At least two 'correct' values (3 sf) or expressions seen from their grouping (or lack of). $2.718 + 2.140 + \ldots + 0.6757 + 1.054$ |
| $10.0$ | A1 | AWRT 10.0. If M0 awarded then SC B1 for 10.0. |
| Tabular value: $10.64$; $'10.0' < 10.64$, accept $H_0$/not significant | M1 | Allow equivalent to 10.64 if columns not combined or only one pair combined (12.02 one pair combined, 13.36 none combined). |
| Insufficient evidence to suggest that Po(3.5) is not a good fit to the data | A1 | Correct conclusion in context, following correct work, level of uncertainty in language. A0 if hypotheses the wrong way round or missing. |
| | **6** | |2 The number of breakdowns on a particular section of road is recorded each day over a period of 90 days. It is suggested that the number of breakdowns follows a Poisson distribution with mean 3.5. The data is summarised in the table, together with some of the expected frequencies resulting from the suggested Poisson distribution.
\begin{center}
\begin{tabular}{|l|l|l|l|l|l|l|l|l|l|}
\hline
Number of breakdowns per day & 0 & 1 & 2 & 3 & 4 & 5 & 6 & 7 & 8 or more \\
\hline
Observed frequency & 0 & 5 & 13 & 17 & 21 & 16 & 9 & 5 & 4 \\
\hline
Expected frequency & 2.718 & 9.512 & 16.646 & & 16.993 & 11.895 & & 3.469 & 2.407 \\
\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 or not $\operatorname { Po } ( 3.5 )$ is a good fit to the data.
\end{enumerate}
\hfill \mbox{\textit{CAIE Further Paper 4 2023 Q2 [8]}}