| Exam Board | CAIE |
|---|---|
| Module | FP2 (Further Pure Mathematics 2) |
| Year | 2016 |
| Session | June |
| Marks | 10 |
| Paper | Download PDF ↗ |
| Mark scheme | Download PDF ↗ |
| Topic | Chi-squared goodness of fit |
| Type | Chi-squared goodness of fit: Binomial |
| Difficulty | Standard +0.3 This is a straightforward chi-squared goodness of fit test with a fully specified binomial distribution. Students must calculate expected frequencies using B(6,0.6), combine cells with low expected frequencies, compute the test statistic, and compare to critical values. While it requires careful arithmetic and understanding of the chi-squared procedure, it follows a standard template with no novel problem-solving or conceptual challenges beyond routine application of the test. |
| Spec | 5.06b Fit prescribed distribution: chi-squared test5.06c Fit other distributions: discrete and continuous |
| Number of successful applicants | 0 | 1 | 2 | 3 | 4 | 5 | 6 |
| Number of groups | 1 | 3 | 25 | 51 | 38 | 30 | 2 |
| Answer | Marks | Guidance |
|---|---|---|
| Answer/Working | Marks | Guidance |
| H\(_0\): Given distribution fits data | B1 | State (at least) null hypothesis in full (AEF) |
| Find expected values using \(150\,{}^6C_i\,q^{6-i}\,p^i\) with \(p = 0.6\), \(q = 0.4\) (to 3 s.f.) | M1 A2 | allow A1 if only one error, or if all values correct to 2 s.f. |
| \(0.6144\ \ 5.5296\ \ 20.736\ \ 41.472\ \ 46.656\ \ 27.9936\ \ 6.9984\) | ||
| Combine \(0.6144\) & \(5.5296\) since 1st expected value \(< 5\): \(O_i: 4\ldots\); \(E_i: 6.14[4]\ldots\) | ||
| \(\chi^2 = 0.748 + 0.877 + 2.189 + 1.606 + 0.144 + 3.570 = 9.13\) | B1 | Calculate \(\chi^2\) (result correct to 3 s.f.) |
| 6 cells: \(\chi^2_{5,\,0.95} = 11.07\) | M1 A1 | State or use consistent tabular value (to 3 s.f.); or if no or more cells combined |
| 7 cells: \(\chi^2_{6,\,0.95} = 12.59\) | ||
| 5 cells: \(\chi^2_{4,\,0.95} = 9.488\) | ||
| 4 cells: \(\chi^2_{3,\,0.95} = 7.815\) | ||
| 3 cells: \(\chi^2_{2,\,0.95} = 5.991\) ] | B1\(\checkmark\) | |
| Accept H\(_0\) if \(\chi^2 <\) tabular value | M1 | State or imply valid method for conclusion |
| \(9.13\ [\pm 0.01] < 11.1\) so distribution fits [data] | A1 | Conclusion (AEF, requires both values correct); Total: [10] |
| (S.C.) \(150\,{}^6C_i\,q^i\,p^{6-i}\) can earn B1 M1 B1 M1 B1 M1 (max 6/10) |
## Question 9:
| Answer/Working | Marks | Guidance |
|---|---|---|
| H$_0$: Given distribution fits data | B1 | State (at least) null hypothesis in full (AEF) |
| Find expected values using $150\,{}^6C_i\,q^{6-i}\,p^i$ with $p = 0.6$, $q = 0.4$ (to 3 s.f.) | M1 A2 | allow A1 if only one error, or if all values correct to 2 s.f. |
| $0.6144\ \ 5.5296\ \ 20.736\ \ 41.472\ \ 46.656\ \ 27.9936\ \ 6.9984$ | | |
| Combine $0.6144$ & $5.5296$ since 1st expected value $< 5$: $O_i: 4\ldots$; $E_i: 6.14[4]\ldots$ | | |
| $\chi^2 = 0.748 + 0.877 + 2.189 + 1.606 + 0.144 + 3.570 = 9.13$ | B1 | Calculate $\chi^2$ (result correct to 3 s.f.) |
| 6 cells: $\chi^2_{5,\,0.95} = 11.07$ | M1 A1 | State or use consistent tabular value (to 3 s.f.); or if no or more cells combined |
| 7 cells: $\chi^2_{6,\,0.95} = 12.59$ | | |
| 5 cells: $\chi^2_{4,\,0.95} = 9.488$ | | |
| 4 cells: $\chi^2_{3,\,0.95} = 7.815$ | | |
| 3 cells: $\chi^2_{2,\,0.95} = 5.991$ ] | B1$\checkmark$ | |
| Accept H$_0$ if $\chi^2 <$ tabular value | M1 | State or imply valid method for conclusion |
| $9.13\ [\pm 0.01] < 11.1$ so distribution fits [data] | A1 | Conclusion (AEF, requires both values correct); Total: [10] |
| **(S.C.)** $150\,{}^6C_i\,q^i\,p^{6-i}$ can earn B1 M1 B1 M1 B1 M1 (max 6/10) | | |
---9 Applicants for a national teacher training course are required to pass a mathematics test. Each year, the applicants are tested in groups of 6 and the number of successful applicants in each group is recorded. The overall proportion of successful applicants has remained constant over the years and is equal to $60 \%$ of the applicants. The results from 150 randomly chosen groups are shown in the following table.
\begin{center}
\begin{tabular}{ | l | c | c | c | c | c | c | c | }
\hline
Number of successful applicants & 0 & 1 & 2 & 3 & 4 & 5 & 6 \\
\hline
Number of groups & 1 & 3 & 25 & 51 & 38 & 30 & 2 \\
\hline
\end{tabular}
\end{center}
Test, at the $5 \%$ significance level, the goodness of fit of the distribution $\mathbf { B } ( 6,0.6 )$ for the number of successful applicants in a group.
\hfill \mbox{\textit{CAIE FP2 2016 Q9 [10]}}