| Exam Board | CAIE |
|---|---|
| Module | FP2 (Further Pure Mathematics 2) |
| Year | 2010 |
| Session | June |
| Marks | 8 |
| Paper | Download PDF ↗ |
| Mark scheme | Download PDF ↗ |
| Topic | Chi-squared goodness of fit |
| Type | Chi-squared goodness of fit: Given ratios |
| Difficulty | Challenging +1.2 This is a straightforward chi-squared goodness of fit test with given probabilities. Students must calculate expected frequencies using the logarithmic formula, combine categories appropriately (≥6 requires summing probabilities), compute the test statistic, and compare to critical values. While it involves logarithms and careful arithmetic, it follows a standard procedure with no novel insight required—typical for Further Maths statistics but more involved than basic A-level questions. |
| Spec | 5.06b Fit prescribed distribution: chi-squared test5.06c Fit other distributions: discrete and continuous |
| Leading digit | 1 | 2 | 3 | 4 | 5 | \(\geqslant 6\) |
| Frequency | 22 | 21 | 13 | 11 | 11 | 22 |
| Answer | Marks | Guidance |
|---|---|---|
| Answer/Working | Mark | Guidance |
| State (at least) null hypothesis (A.E.F.): \(H_0\): Data conforms to Law | B1 | |
| Calculate expected values (to 1 dp): \(30.10\quad 17.61\quad 12.49\quad 9.69\quad 7.92\quad 22.18\) | M1 A1 | |
| Calculate value of \(\chi^2\): \(\chi^2 = 4.23 \pm 0.02\) | M1 A1 | |
| Compare with consistent tabular value (to 2 dp): \(\chi^2_{5,\,0.9} = 9.236\) | B1 | |
| Valid method for reaching conclusion: Reject \(H_0\) if \(\chi^2 >\) tabular value | M1 | |
| Correct conclusion (A.E.F., requires correct values): Data conforms to Benford's Law | A1 | Total: 8 |
## Question 7:
| Answer/Working | Mark | Guidance |
|---|---|---|
| State (at least) null hypothesis (A.E.F.): $H_0$: Data conforms to Law | B1 | |
| Calculate expected values (to 1 dp): $30.10\quad 17.61\quad 12.49\quad 9.69\quad 7.92\quad 22.18$ | M1 A1 | |
| Calculate value of $\chi^2$: $\chi^2 = 4.23 \pm 0.02$ | M1 A1 | |
| Compare with consistent tabular value (to 2 dp): $\chi^2_{5,\,0.9} = 9.236$ | B1 | |
| Valid method for reaching conclusion: Reject $H_0$ if $\chi^2 >$ tabular value | M1 | |
| Correct conclusion (A.E.F., requires correct values): Data conforms to Benford's Law | A1 | **Total: 8** |
---7 Benford's Law states that, in many tables containing large numbers of numerical values, the probability distribution of the leading non-zero digit $D$ is given by
$$\mathrm { P } ( D = d ) = \log _ { 10 } \left( \frac { d + 1 } { d } \right) , \quad d = 1,2 , \ldots , 9 .$$
The following table shows a summary of a random sample of 100 non-zero leading digits taken from a table of cumulative probabilities for the Poisson distribution.
\begin{center}
\begin{tabular}{ | l | c | c | c | c | c | c | }
\hline
Leading digit & 1 & 2 & 3 & 4 & 5 & $\geqslant 6$ \\
\hline
Frequency & 22 & 21 & 13 & 11 & 11 & 22 \\
\hline
\end{tabular}
\end{center}
Carry out a suitable goodness of fit test at the 10\% significance level.
\hfill \mbox{\textit{CAIE FP2 2010 Q7 [8]}}