| Exam Board | CAIE |
|---|---|
| Module | FP2 (Further Pure Mathematics 2) |
| Year | 2014 |
| 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 question requires integration of a non-standard pdf (1/(x ln 8)) to find expected frequencies, calculation of chi-squared test statistic, and hypothesis testing. While the integration is straightforward (ln x), students must correctly apply probability theory, handle the given pdf, and execute a complete hypothesis test. It's more demanding than routine chi-squared tests due to the continuous distribution with a less familiar pdf, but follows standard A-level procedures once the setup is understood. |
| Spec | 5.06b Fit prescribed distribution: chi-squared test5.06c Fit other distributions: discrete and continuous |
| Interval | \(1 \leqslant x < 2\) | \(2 \leqslant x < 3\) | \(3 \leqslant x < 4\) | \(4 \leqslant x < 5\) | \(5 \leqslant x < 6\) | \(6 \leqslant x < 7\) | \(7 \leqslant x < 8\) |
| Observed frequency | 63 | 45 | 32 | 25 | 22 | 7 | 6 |
| Interval | \(1 \leqslant x < 2\) | \(2 \leqslant x < 3\) | \(3 \leqslant x < 4\) | \(4 \leqslant x < 5\) | \(5 \leqslant x < 6\) | \(6 \leqslant x < 7\) | \(7 \leqslant x < 8\) |
| Expected frequency | 66.67 | \(p\) | 27.67 | \(q\) | 17.54 | 14.83 | 12.84 |
| Answer | Marks | Guidance |
|---|---|---|
| Answer/Working | Marks | Guidance |
| \(p = 200 \int_2^3 (1/x \ln 8) \, dx\) | ||
| \(= (200/\ln 8)[\ln x]_2^3\) | ||
| \(= 200 \times 0.1950 = 39.00\) | M1A1 | A.G. |
| \(q = 21.46\) or \(21.45\) | M1A1 | |
| 4 marks | ||
| \(H_0\): \(f(x)\) fits data | B1 | A.E.F. |
| \(\chi^2 = 0.202 + 0.923 + 0.678 + 0.584 + 1.134 + 4.134 + 3.644 = 11.3\) | M1A1 | |
| \(\chi_{6, 0.95}^2 = 12.59\) | B1 | |
| Accept \(H_0\) if \(\chi^2 \leqslant\) tabular value | M1 | |
| Distribution fits observations | A1 | A.E.F. |
| 6 marks, Total: 10 |
# Question 9:
| Answer/Working | Marks | Guidance |
|---|---|---|
| $p = 200 \int_2^3 (1/x \ln 8) \, dx$ | | |
| $= (200/\ln 8)[\ln x]_2^3$ | | |
| $= 200 \times 0.1950 = 39.00$ | M1A1 | A.G. |
| $q = 21.46$ or $21.45$ | M1A1 | |
| | | 4 marks |
| $H_0$: $f(x)$ fits data | B1 | A.E.F. |
| $\chi^2 = 0.202 + 0.923 + 0.678 + 0.584 + 1.134 + 4.134 + 3.644 = 11.3$ | M1A1 | |
| $\chi_{6, 0.95}^2 = 12.59$ | B1 | |
| Accept $H_0$ if $\chi^2 \leqslant$ tabular value | M1 | |
| Distribution fits observations | A1 | A.E.F. |
| | | 6 marks, Total: 10 |
---9 A random sample of 200 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 | c | c | }
\hline
Interval & $1 \leqslant x < 2$ & $2 \leqslant x < 3$ & $3 \leqslant x < 4$ & $4 \leqslant x < 5$ & $5 \leqslant x < 6$ & $6 \leqslant x < 7$ & $7 \leqslant x < 8$ \\
\hline
Observed frequency & 63 & 45 & 32 & 25 & 22 & 7 & 6 \\
\hline
\end{tabular}
\end{center}
It is required to test the goodness of fit of the distribution with probability density function $f$ given by
$$f ( x ) = \begin{cases} \frac { 1 } { x \ln 8 } & 1 \leqslant x < 8 \\ 0 & \text { otherwise } \end{cases}$$
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 | c | c | }
\hline
Interval & $1 \leqslant x < 2$ & $2 \leqslant x < 3$ & $3 \leqslant x < 4$ & $4 \leqslant x < 5$ & $5 \leqslant x < 6$ & $6 \leqslant x < 7$ & $7 \leqslant x < 8$ \\
\hline
Expected frequency & 66.67 & $p$ & 27.67 & $q$ & 17.54 & 14.83 & 12.84 \\
\hline
\end{tabular}
\end{center}
Show that $p = 39.00$, correct to 2 decimal places, and find the value of $q$.
Carry out a goodness of fit test at the 5\% significance level.
\hfill \mbox{\textit{CAIE FP2 2014 Q9 [10]}}