| Exam Board | CAIE |
|---|---|
| Module | Further Paper 4 (Further Paper 4) |
| Year | 2023 |
| Session | June |
| Marks | 9 |
| 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 with a non-standard continuous pdf requiring integration of a rational function to find expected frequencies, followed by a standard hypothesis test. The integration and probability calculations are moderately challenging but follow established techniques, making this above average difficulty but not exceptionally hard for Further Maths students. |
| Spec | 5.06b Fit prescribed distribution: chi-squared test |
| Interval | \(1 \leqslant x < 1.5\) | \(1.5 \leqslant x < 2\) | \(2 \leqslant x < 2.5\) | \(2.5 \leqslant x < 3\) | \(3 \leqslant x < 3.5\) | \(3.5 \leqslant x \leqslant 4\) |
| Observed frequency | 3 | 3 | 8 | 11 | 13 | 12 |
| Interval | \(1 \leqslant x < 1.5\) | \(1.5 \leqslant x < 2\) | \(2 \leqslant x < 2.5\) | \(2.5 \leqslant x < 3\) | \(3 \leqslant x < 3.5\) | \(3.5 \leqslant x \leqslant 4\) |
| Expected frequency | 4.4271 | \(a\) | 6.1285 | 8.4549 | \(b\) | 14.9678 |
| Answer | Marks | Guidance |
|---|---|---|
| \(\int_{1.5}^{2} \frac{1}{24}\left(\frac{4}{x^2} + x^2\right)dx = \frac{1}{24}\left[-\frac{4}{x} + \frac{x^3}{3}\right]\) | M1 | Integration with correct powers and correct limits seen |
| \(a = 50 \times 0.092014 = 4.6007\) | A1 | AG; \(\frac{53}{576}\) or \(\frac{1325}{288}\) or 0.092014 or 4.60069 seen |
| \(b = 11.4211\) | B1 | Or 11.421(0) |
| Answer | Marks | Guidance |
|---|---|---|
| Combine first two columns: 6, 9.0278 | M1 | Must be seen, or implied by 1.0155 |
| Calculate value of chi-squared: \(1.0155 + 0.5715 + 0.7661 + 0.2183 + 0.5885\) | M1 | At least 2 correct values (at least 3sf) or expressions seen. Allow columns not combined or three columns combined |
| \(3.16\) | A1 | CWO. Dependent on M1M1 scored. SCB1 for 3.16 with no working |
| \(H_0\): f is a good fit for the data; \(H_1\): f is not a good fit for the data | B1 | |
| Tabular value of chi-squared: 7.78; \(3.16 < 7.78\) and accept \(H_0\) | M1 | Correct tabular value: allow correct FT value if columns not combined (9.236) or three columns combined (6.251) |
| Insufficient evidence to suggest that f is not a good fit for the data. Condone: sufficient evidence to suggest that f is a good fit for the data | A1 | Correct conclusion in context, following correct work, level of uncertainty in language (not 'prove'), not 'there is no evidence'), no contradictions. A0 if hypotheses reversed |
## Question 3(a):
| $\int_{1.5}^{2} \frac{1}{24}\left(\frac{4}{x^2} + x^2\right)dx = \frac{1}{24}\left[-\frac{4}{x} + \frac{x^3}{3}\right]$ | M1 | Integration with correct powers and correct limits seen |
|---|---|---|
| $a = 50 \times 0.092014 = 4.6007$ | A1 | AG; $\frac{53}{576}$ or $\frac{1325}{288}$ or 0.092014 or 4.60069 seen |
| $b = 11.4211$ | B1 | Or 11.421(0) |
---
## Question 3(b):
| Combine first two columns: 6, 9.0278 | M1 | Must be seen, or implied by 1.0155 |
|---|---|---|
| Calculate value of chi-squared: $1.0155 + 0.5715 + 0.7661 + 0.2183 + 0.5885$ | M1 | At least 2 correct values (at least 3sf) or expressions seen. Allow columns not combined or three columns combined |
| $3.16$ | A1 | CWO. Dependent on M1M1 scored. SCB1 for 3.16 with no working |
| $H_0$: f is a good fit for the data; $H_1$: f is not a good fit for the data | B1 | |
| Tabular value of chi-squared: 7.78; $3.16 < 7.78$ and accept $H_0$ | M1 | Correct tabular value: allow correct FT value if columns not combined (9.236) or three columns combined (6.251) |
| Insufficient evidence to suggest that f is not a good fit for the data. Condone: sufficient evidence to suggest that f is a good fit for the data | A1 | Correct conclusion in context, following correct work, level of uncertainty in language (not 'prove'), not 'there is no evidence'), no contradictions. A0 if hypotheses reversed |
---3 A random sample of 50 values of the continuous random variable $X$ was taken. These values are summarised in the following table.
\begin{center}
\begin{tabular}{ | l | c | c | c | c | c | c | }
\hline
Interval & $1 \leqslant x < 1.5$ & $1.5 \leqslant x < 2$ & $2 \leqslant x < 2.5$ & $2.5 \leqslant x < 3$ & $3 \leqslant x < 3.5$ & $3.5 \leqslant x \leqslant 4$ \\
\hline
Observed frequency & 3 & 3 & 8 & 11 & 13 & 12 \\
\hline
\end{tabular}
\end{center}
It is required to test the goodness of fit of the distribution with probability density function $f$ given by
$$f ( x ) = \begin{cases} \frac { 1 } { 24 } \left( \frac { 4 } { x ^ { 2 } } + x ^ { 2 } \right) & 1 \leqslant x \leqslant 4 \\ 0 & \text { otherwise } \end{cases}$$
The expected frequencies, correct to 4 decimal places, are given in the following table.
\begin{center}
\begin{tabular}{ | l | c | c | c | c | c | c | }
\hline
Interval & $1 \leqslant x < 1.5$ & $1.5 \leqslant x < 2$ & $2 \leqslant x < 2.5$ & $2.5 \leqslant x < 3$ & $3 \leqslant x < 3.5$ & $3.5 \leqslant x \leqslant 4$ \\
\hline
Expected frequency & 4.4271 & $a$ & 6.1285 & 8.4549 & $b$ & 14.9678 \\
\hline
\end{tabular}
\end{center}
\begin{enumerate}[label=(\alph*)]
\item Show that $a = 4.6007$ and find the value of $b$.
\item Carry out a goodness of fit test, at the $10 \%$ significance level, to test whether f is a satisfactory model for the data.
\end{enumerate}
\hfill \mbox{\textit{CAIE Further Paper 4 2023 Q3 [9]}}