| Exam Board | CAIE |
|---|---|
| Module | FP2 (Further Pure Mathematics 2) |
| Year | 2008 |
| Session | November |
| 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 +0.3 This is a straightforward chi-squared goodness of fit test with expected frequencies already provided. Students must verify one expected frequency calculation (integrating a simple pdf) and perform a standard hypothesis test. The pdf integration is routine for Further Maths students, and the chi-squared test procedure is mechanical once expected values are known. Slightly easier than average due to the provided expected frequencies and standard test structure. |
| Spec | 5.06b Fit prescribed distribution: chi-squared test |
| Interval | \(1 \leqslant t < 1.5\) | \(1.5 \leqslant t < 2\) | \(2 \leqslant t < 2.5\) | \(2.5 \leqslant t < 3\) |
| Frequency | 64 | 17 | 16 | 3 |
| Interval | \(1 \leqslant t < 1.5\) | \(1.5 \leqslant t < 2\) | \(2 \leqslant t < 2.5\) | \(2.5 \leqslant t < 3\) |
| Expected frequency | 62.5 | 21.875 | 10.125 | 5.5 |
| Answer | Marks | Guidance |
|---|---|---|
| Answer/Working | Mark | Guidance |
| \(F(t) = -9/8t^2\) | M1 | Integrate \(f(t)\) to give \(F(t)\) |
| \(F(2.5) - F(2) = -(9/8)(2.5^{-2} - 2^{-2})\) | A1 | Apply limits |
| \(10.125\) A.G. | A1 | Evaluate and multiply by 100; Part total: 3 |
| \(H_0\): \(f(t)\) fits data, \(H_1\): doesn't fit | B1 | State hypotheses (A.E.F.) |
| \(\chi^2 = 1.5^2/62.5 + 4.875^2/21.875 + 5.875^2/10.125 + 2.5^2/5.5\) | M1 A1 | Find \(\chi^2\) (A1 if at least 3 terms correct) |
| \(= 0.036 + 1.086 + 3.409 + 1.136\) | ||
| \(= 5.67\) \([\pm 0.01]\) | *A1 | Evaluate \(\chi^2\) |
| \(\chi_{3,0.9}^2 = 6.251\) | *B1\(\sqrt{}\) | Compare with consistent tabular value (to 2 dp) |
| (\(\chi_{2,0.9}^2 = 4.605\), \(\chi_{1,0.9}^2 = 2.706\)) | ||
| Distribution fits data (A.E.F.) | M1\(\sqrt{}\) A1 | Consistent conclusion (A1 dep *A1, *B1); Total: 10 |
## Question 9:
| Answer/Working | Mark | Guidance |
|---|---|---|
| $F(t) = -9/8t^2$ | M1 | Integrate $f(t)$ to give $F(t)$ |
| $F(2.5) - F(2) = -(9/8)(2.5^{-2} - 2^{-2})$ | A1 | Apply limits |
| $10.125$ **A.G.** | A1 | Evaluate and multiply by 100; **Part total: 3** |
| $H_0$: $f(t)$ fits data, $H_1$: doesn't fit | B1 | State hypotheses (A.E.F.) |
| $\chi^2 = 1.5^2/62.5 + 4.875^2/21.875 + 5.875^2/10.125 + 2.5^2/5.5$ | M1 A1 | Find $\chi^2$ (A1 if at least 3 terms correct) |
| $= 0.036 + 1.086 + 3.409 + 1.136$ | | |
| $= 5.67$ $[\pm 0.01]$ | *A1 | Evaluate $\chi^2$ |
| $\chi_{3,0.9}^2 = 6.251$ | *B1$\sqrt{}$ | Compare with consistent tabular value (to 2 dp) |
| ($\chi_{2,0.9}^2 = 4.605$, $\chi_{1,0.9}^2 = 2.706$) | | |
| Distribution fits data (A.E.F.) | M1$\sqrt{}$ A1 | Consistent conclusion (A1 dep *A1, *B1); **Total: 10** |
---9 A sample of 100 observations of the continuous random variable $T$ was obtained and the values are summarised in the following table.
\begin{center}
\begin{tabular}{ | l | c | c | c | c | }
\hline
Interval & $1 \leqslant t < 1.5$ & $1.5 \leqslant t < 2$ & $2 \leqslant t < 2.5$ & $2.5 \leqslant t < 3$ \\
\hline
Frequency & 64 & 17 & 16 & 3 \\
\hline
\end{tabular}
\end{center}
It is required to test the goodness of fit of the distribution with probability density function given by
$$f ( t ) = \begin{cases} \frac { 9 } { 4 t ^ { 3 } } & 1 \leqslant t < 3 \\ 0 & \text { otherwise } \end{cases}$$
The relevant expected values are as follows.
\begin{center}
\begin{tabular}{ | l | c | c | c | c | }
\hline
Interval & $1 \leqslant t < 1.5$ & $1.5 \leqslant t < 2$ & $2 \leqslant t < 2.5$ & $2.5 \leqslant t < 3$ \\
\hline
Expected frequency & 62.5 & 21.875 & 10.125 & 5.5 \\
\hline
\end{tabular}
\end{center}
Show how the expected value 10.125 is obtained.
Carry out the test, at the $10 \%$ significance level.
\hfill \mbox{\textit{CAIE FP2 2008 Q9 [10]}}