| Exam Board | CAIE |
|---|---|
| Module | FP2 (Further Pure Mathematics 2) |
| Year | 2019 |
| Session | June |
| Marks | 10 |
| Paper | Download PDF ↗ |
| Mark scheme | Download PDF ↗ |
| Topic | Chi-squared goodness of fit |
| Type | Chi-squared goodness of fit: Other continuous |
| Difficulty | Standard +0.8 This is a Further Maths chi-squared test requiring integration of a square root function to find probabilities, then standard hypothesis testing. The integration (substitution with u=4-x) is moderately challenging but routine for FM students, and the test procedure is standard. Above average difficulty due to the non-trivial pdf integration, but well within expected FM scope. |
| Spec | 5.06b Fit prescribed distribution: chi-squared test |
| Interval | \(0 \leqslant x < 0.8\) | \(0.8 \leqslant x < 1.6\) | \(1.6 \leqslant x < 2.4\) | \(2.4 \leqslant x < 3.2\) | \(3.2 \leqslant x < 4\) |
| Observed frequency | 18 | 16 | 8 | 6 | 2 |
| Interval | \(0 \leqslant x < 0.8\) | \(0.8 \leqslant x < 1.6\) | \(1.6 \leqslant x < 2.4\) | \(2.4 \leqslant x < 3.2\) | \(3.2 \leqslant x < 4\) |
| Expected frequency | 14.22 | 12.54 | 10.59 | 8.18 | 4.47 |
| Answer | Marks | Guidance |
|---|---|---|
| Answer | Marks | Guidance |
| \(E_3 = \frac{3}{16}\int_{1.6}^{2.4}(4-x)^{1/2}\,dx = \frac{3}{16}\left[-\frac{2}{3}(4-x)^{3/2}\right]_{1.6}^{2.4}\) | M1 | State or imply expression for required expected value \(E_3\) of \(X\) |
| \(= (2\cdot4^{3/2} - 1\cdot6^{3/2})/8 = 1.694/8\) or \(0.2118\) | A1 | Find expected value \(E_3\) (may be implied in finding \(50E_3\)) (M1 A1 requires adequate explicit working) |
| \(50\,E_3 = 10.59\) AG | A1 | Hence verify corresponding expected frequency |
| 3 |
| Answer | Marks | Guidance |
|---|---|---|
| Answer | Marks | Guidance |
| \(H_0\): Distribution fits data (AEF) | B1 | State (at least) null hypothesis in full |
| \(O_i\): 18, 16, 8, 8; \(E_i\): 14.22, 12.54, 10.59, 12.65 | M1 | Combine values consistent with all exp. values \(\geq 5\) |
| \(X^2 = 1.005 + 0.955 + 0.633 + 1.709 = 4.30\) | M1 A1 | Find value of \(X^2\) from \(\sum(E_i - O_i)^2/E_i\) [or \(\sum O_i^2/E_i - n\)] |
| No. \(n\) of cells: 5, 4, 3; \(\chi^2_{n-1,\,0.95}\): 9.488, 7.815, 5.991 (to 3 s.f.) | B1 | FT on number \(n\) of cells used to find \(X^2\). State or use consistent tabular value \(\chi^2_{n-1,\,0.95}\) |
| Accept \(H_0\) if \(X^2 <\) tabular value (AEF) | M1 | State or imply valid method for conclusion |
| \(4.30\ [\pm 0.01] < 7.815\) so distribution fits [data] or distribution is a suitable model (AEF) | A1 | Conclusion (requires both values approx. correct) |
| 7 |
## Question 9(i):
| Answer | Marks | Guidance |
|--------|-------|----------|
| $E_3 = \frac{3}{16}\int_{1.6}^{2.4}(4-x)^{1/2}\,dx = \frac{3}{16}\left[-\frac{2}{3}(4-x)^{3/2}\right]_{1.6}^{2.4}$ | M1 | State or imply expression for required expected value $E_3$ of $X$ |
| $= (2\cdot4^{3/2} - 1\cdot6^{3/2})/8 = 1.694/8$ or $0.2118$ | A1 | Find expected value $E_3$ (may be implied in finding $50E_3$) (M1 A1 requires adequate explicit working) |
| $50\,E_3 = 10.59$ AG | A1 | Hence verify corresponding expected frequency |
| | **3** | |
## Question 9(ii):
| Answer | Marks | Guidance |
|--------|-------|----------|
| $H_0$: Distribution fits data (AEF) | B1 | State (at least) null hypothesis in full |
| $O_i$: 18, 16, 8, 8; $E_i$: 14.22, 12.54, 10.59, 12.65 | M1 | Combine values consistent with all exp. values $\geq 5$ |
| $X^2 = 1.005 + 0.955 + 0.633 + 1.709 = 4.30$ | M1 A1 | Find value of $X^2$ from $\sum(E_i - O_i)^2/E_i$ [or $\sum O_i^2/E_i - n$] |
| No. $n$ of cells: 5, 4, 3; $\chi^2_{n-1,\,0.95}$: 9.488, **7.815**, 5.991 (to 3 s.f.) | B1 | **FT** on number $n$ of cells used to find $X^2$. State or use consistent tabular value $\chi^2_{n-1,\,0.95}$ |
| Accept $H_0$ if $X^2 <$ tabular value (AEF) | M1 | State or imply valid method for conclusion |
| $4.30\ [\pm 0.01] < 7.815$ so distribution fits [data] or distribution is a suitable model (AEF) | A1 | Conclusion (requires both values approx. correct) |
| | **7** | |9 A random sample of 50 observations of the continuous random variable $X$ was taken and the values are summarised in the following table.
\begin{center}
\begin{tabular}{ | l | c | c | c | c | c | }
\hline
Interval & $0 \leqslant x < 0.8$ & $0.8 \leqslant x < 1.6$ & $1.6 \leqslant x < 2.4$ & $2.4 \leqslant x < 3.2$ & $3.2 \leqslant x < 4$ \\
\hline
Observed frequency & 18 & 16 & 8 & 6 & 2 \\
\hline
\end{tabular}
\end{center}
It is required to test the goodness of fit of the distribution with probability density function $f$ given by
$$f ( x ) = \begin{cases} \frac { 3 } { 16 } ( 4 - x ) ^ { \frac { 1 } { 2 } } & 0 \leqslant x < 4 \\ 0 & \text { otherwise. } \end{cases}$$
The relevant expected frequencies, correct to 2 decimal places, are given in the following table.
\begin{center}
\begin{tabular}{ | l | c | c | c | c | c | }
\hline
Interval & $0 \leqslant x < 0.8$ & $0.8 \leqslant x < 1.6$ & $1.6 \leqslant x < 2.4$ & $2.4 \leqslant x < 3.2$ & $3.2 \leqslant x < 4$ \\
\hline
Expected frequency & 14.22 & 12.54 & 10.59 & 8.18 & 4.47 \\
\hline
\end{tabular}
\end{center}
(i) Show how the expected frequency for $1.6 \leqslant x < 2.4$ is obtained.\\
(ii) Carry out a goodness of fit test at the $5 \%$ significance level.\\
\hfill \mbox{\textit{CAIE FP2 2019 Q9 [10]}}