| Exam Board | CAIE |
|---|---|
| Module | Further Paper 4 (Further Paper 4) |
| Year | 2020 |
| Session | November |
| Marks | 7 |
| 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 non-standard pdf (quadratic function) to find expected frequencies, followed by a standard hypothesis test. The integration and probability calculations are moderately challenging but systematic, and the test procedure itself is routine. Above average difficulty due to the calculus requirement and Further Maths context, but not exceptionally hard. |
| Spec | 5.06c Fit other distributions: discrete and continuous |
| Interval | \(0 \leqslant x < 0.5\) | \(0.5 \leqslant x < 1\) | \(1 \leqslant x < 1.5\) | \(1.5 \leqslant x < 2\) | \(2 \leqslant x < 2.5\) | \(2.5 \leqslant x < 3\) |
| Observed frequency | 5 | 23 | 40 | 41 | 46 | 45 |
| Interval | \(0 \leqslant x < 0.5\) | \(0.5 \leqslant x < 1\) | \(1 \leqslant x < 1.5\) | \(1.5 \leqslant x < 2\) | \(2 \leqslant x < 2.5\) | \(2.5 \leqslant x < 3\) |
| Expected frequency | \(p\) | \(q\) | 37.96 | 43.52 | 43.52 | 37.96 |
| Answer | Marks | Guidance |
|---|---|---|
| Answer | Marks | Guidance |
| \(\int_0^{0.5}\left(\frac{4}{9}x - \frac{1}{9}x^2\right)dx = \frac{1}{9}\left[2\times0.5^2 - \frac{1}{3}0.5^3\right]\) | M1 | Statement of integration with correct limits |
| \(\left(= 0.0509(26) \text{ or } \frac{11}{216}\right)\) | ||
| \(\text{Freq} = p = 0.0509(26) \times 200 = 10.19\) AG | A1 | Requires sight of \(0.0509\) or \(\frac{11}{216}\) |
| By addition to 200, \(q = 26.85\ \left(\frac{725}{27}\right)\) | B1 | Or by integration |
| Answer | Marks | Guidance |
|---|---|---|
| Answer | Marks | Guidance |
| O: \(5\ \ 23\ \ 40\ \ 41\ \ 46\ \ 45\); E: \(10.19\ \ 26.85\ \ 37.96\ \ 43.52\ \ 43.52\ \ 37.96\) | M1 | At least 4 correct |
| Test statistic \(= 2.6433 + 0.5520 + 0.1096 + 0.1459 + 0.1413 + 1.3056\) | ||
| \(4.89\) or \(4.90\) | A1 | |
| \(4.90 < 11.07\) | M1 | Compare their value with \(11.07\) |
| PDF is a satisfactory model for the data. | A1 FT | Correct conclusion in context, ft only their \(4.90\) |
## Question 3(a):
| Answer | Marks | Guidance |
|--------|-------|----------|
| $\int_0^{0.5}\left(\frac{4}{9}x - \frac{1}{9}x^2\right)dx = \frac{1}{9}\left[2\times0.5^2 - \frac{1}{3}0.5^3\right]$ | M1 | Statement of integration with correct limits |
| $\left(= 0.0509(26) \text{ or } \frac{11}{216}\right)$ | | |
| $\text{Freq} = p = 0.0509(26) \times 200 = 10.19$ AG | A1 | Requires sight of $0.0509$ or $\frac{11}{216}$ |
| By addition to 200, $q = 26.85\ \left(\frac{725}{27}\right)$ | B1 | Or by integration |
---
## Question 3(b):
| Answer | Marks | Guidance |
|--------|-------|----------|
| O: $5\ \ 23\ \ 40\ \ 41\ \ 46\ \ 45$; E: $10.19\ \ 26.85\ \ 37.96\ \ 43.52\ \ 43.52\ \ 37.96$ | M1 | At least 4 correct |
| Test statistic $= 2.6433 + 0.5520 + 0.1096 + 0.1459 + 0.1413 + 1.3056$ | | |
| $4.89$ or $4.90$ | A1 | |
| $4.90 < 11.07$ | M1 | Compare their value with $11.07$ |
| PDF is a satisfactory model for the data. | A1 FT | Correct conclusion in context, ft only their $4.90$ |
---3 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 | }
\hline
Interval & $0 \leqslant x < 0.5$ & $0.5 \leqslant x < 1$ & $1 \leqslant x < 1.5$ & $1.5 \leqslant x < 2$ & $2 \leqslant x < 2.5$ & $2.5 \leqslant x < 3$ \\
\hline
Observed frequency & 5 & 23 & 40 & 41 & 46 & 45 \\
\hline
\end{tabular}
\end{center}
It is required to test the goodness of fit of the distribution with probability density function f given by
$$f ( x ) = \begin{cases} \frac { 1 } { 9 } x ( 4 - x ) & 0 \leqslant x \leqslant 3 \\ 0 & \text { otherwise } \end{cases}$$
Most of 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 | }
\hline
Interval & $0 \leqslant x < 0.5$ & $0.5 \leqslant x < 1$ & $1 \leqslant x < 1.5$ & $1.5 \leqslant x < 2$ & $2 \leqslant x < 2.5$ & $2.5 \leqslant x < 3$ \\
\hline
Expected frequency & $p$ & $q$ & 37.96 & 43.52 & 43.52 & 37.96 \\
\hline
\end{tabular}
\end{center}
\begin{enumerate}[label=(\alph*)]
\item Show that $p = 10.19$ and find the value of $q$.
\item Carry out a goodness of fit test, at the $5 \%$ significance level, to test whether f is a satisfactory model for the data.
\end{enumerate}
\hfill \mbox{\textit{CAIE Further Paper 4 2020 Q3 [7]}}