| Exam Board | CAIE |
|---|---|
| Module | FP2 (Further Pure Mathematics 2) |
| Year | 2011 |
| Session | November |
| Marks | 11 |
| 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 question requires finding expected frequencies from a given pdf (involving integration and probability calculations), then performing a chi-squared goodness of fit test. While the integration of kx² is straightforward, students must correctly handle the continuous distribution setup, combine cells appropriately due to low expected frequencies, and determine degrees of freedom correctly (accounting for no estimated parameters). The multi-step nature and requirement to apply chi-squared methodology properly places this above average difficulty for A-level, though it's a fairly standard application once the technique is known. |
| Spec | 5.06b Fit prescribed distribution: chi-squared test |
| Interval | \(0 \leqslant x < 1\) | \(1 \leqslant x < 2\) | \(2 \leqslant x < 3\) | \(3 \leqslant x < 4\) | \(4 \leqslant x < 5\) | \(5 \leqslant x < 6\) |
| Observed frequency | 1 | 3 | 15 | 31 | 59 | 107 |
| Interval | \(0 \leqslant x < 1\) | \(1 \leqslant x < 2\) | \(2 \leqslant x < 3\) | \(3 \leqslant x < 4\) | \(4 \leqslant x < 5\) | \(5 \leqslant x < 6\) |
| Expected frequency | 1 | 7 | \(a\) | \(b\) | \(c\) | 91 |
| Answer | Marks | Guidance |
|---|---|---|
| Answer/Working | Marks | Guidance |
| Find \(k\) by integrating \(f(x)\): \([\frac{1}{3}kx^3]_0^6 = 1\), \(k = 3/6^3 = 1/72\) | B1 | |
| State and evaluate \(a\): \(a = 216[\frac{1}{3}kx^3]_2^3 = 3^3 - 2^3 = 19\) | B1 | A.G. |
| Find \(b\): \(b = 216[\frac{1}{3}kx^3]_3^4 = 37\) | ||
| Find \(c\): \(c = 216[\frac{1}{3}kx^3]_4^5\) or \(216 - 155 = 61\) | M1 A1 | MR: \(f(x)\) as distn. of table max \(3/4\) |
| Answer | Marks | Guidance |
|---|---|---|
| Answer/Working | Marks | Guidance |
| State null hypothesis: \(H_0\): \(f(x)\) fits data | B1 | A.E.F. |
| Combine first 2 cells (exp. value \(< 5\)): \(O\): \(4\ldots\), \(E\): \(8\ldots\) | B1 | |
| Calculate \(\chi^2\) (to 2 dp): \(\chi^2 = 6.69[4]\) | M1 *A1 | |
| Compare tabular value: \(\chi_{4,\,0.9}^2 = 7.779\), \([\chi_{3,\,0.9}^2 = 6.251,\;\chi_{5,\,0.9}^2 = 9.236]\) | *B1 | or if 3 or 0 cells combined |
| Valid method for conclusion: Accept \(H_0\) if \(\chi^2 <\) tabular value | M1 | |
| Conclusion: \(6.69 < 7.78\) so \(f(x)\) does fit | A1 | A.E.F., dep *A1, *B1 |
| Subtotal: 7 marks | Total: 11 marks |
## Question 8(i):
| Answer/Working | Marks | Guidance |
|---|---|---|
| Find $k$ by integrating $f(x)$: $[\frac{1}{3}kx^3]_0^6 = 1$, $k = 3/6^3 = 1/72$ | B1 | |
| State and evaluate $a$: $a = 216[\frac{1}{3}kx^3]_2^3 = 3^3 - 2^3 = 19$ | B1 | A.G. |
| Find $b$: $b = 216[\frac{1}{3}kx^3]_3^4 = 37$ | | |
| Find $c$: $c = 216[\frac{1}{3}kx^3]_4^5$ or $216 - 155 = 61$ | M1 A1 | MR: $f(x)$ as distn. of table max $3/4$ |
**Subtotal: 4 marks**
## Question 8(ii):
| Answer/Working | Marks | Guidance |
|---|---|---|
| State null hypothesis: $H_0$: $f(x)$ fits data | B1 | A.E.F. |
| Combine first 2 cells (exp. value $< 5$): $O$: $4\ldots$, $E$: $8\ldots$ | B1 | |
| Calculate $\chi^2$ (to 2 dp): $\chi^2 = 6.69[4]$ | M1 *A1 | |
| Compare tabular value: $\chi_{4,\,0.9}^2 = 7.779$, $[\chi_{3,\,0.9}^2 = 6.251,\;\chi_{5,\,0.9}^2 = 9.236]$ | *B1 | or if 3 or 0 cells combined |
| Valid method for conclusion: Accept $H_0$ if $\chi^2 <$ tabular value | M1 | |
| Conclusion: $6.69 < 7.78$ so $f(x)$ does fit | A1 | A.E.F., dep *A1, *B1 |
**Subtotal: 7 marks | Total: 11 marks**
---8 A sample of 216 observations of the continuous random variable $X$ was obtained and the results are summarised in the following table.
\begin{center}
\begin{tabular}{ | l | c | c | c | c | c | c | }
\hline
Interval & $0 \leqslant x < 1$ & $1 \leqslant x < 2$ & $2 \leqslant x < 3$ & $3 \leqslant x < 4$ & $4 \leqslant x < 5$ & $5 \leqslant x < 6$ \\
\hline
Observed frequency & 1 & 3 & 15 & 31 & 59 & 107 \\
\hline
\end{tabular}
\end{center}
It is suggested that these results are consistent with a distribution having probability density function f given by
$$f ( x ) = \begin{cases} k x ^ { 2 } & 0 \leqslant x < 6 \\ 0 & \text { otherwise } \end{cases}$$
where $k$ is a positive constant. The relevant expected frequencies are given in the following table.
\begin{center}
\begin{tabular}{ | l | c | c | c | c | c | c | }
\hline
Interval & $0 \leqslant x < 1$ & $1 \leqslant x < 2$ & $2 \leqslant x < 3$ & $3 \leqslant x < 4$ & $4 \leqslant x < 5$ & $5 \leqslant x < 6$ \\
\hline
Expected frequency & 1 & 7 & $a$ & $b$ & $c$ & 91 \\
\hline
\end{tabular}
\end{center}
(i) Show that $a = 19$ and find the values of $b$ and $c$.\\
(ii) Carry out a goodness of fit test at the $10 \%$ significance level.
\hfill \mbox{\textit{CAIE FP2 2011 Q8 [11]}}