| Exam Board | CAIE |
|---|---|
| Module | FP2 (Further Pure Mathematics 2) |
| Year | 2009 |
| Session | June |
| Marks | 9 |
| Paper | Download PDF ↗ |
| Mark scheme | Download PDF ↗ |
| Topic | Chi-squared goodness of fit |
| Type | Chi-squared goodness of fit: Given ratios |
| Difficulty | Standard +0.3 This is a straightforward chi-squared goodness of fit test with given expected proportions and clear data. The first part requires standard calculation of expected frequencies, test statistic, and comparison with critical value. The second part about minimum sample size requires understanding that expected frequencies must be ≥5, which is a simple calculation (smallest category is 3%, so n ≥ 5/0.03). While this is Further Maths content, it's a textbook application with no conceptual challenges beyond knowing the standard procedure. |
| Spec | 5.06a Chi-squared: contingency tables |
| Blood Type | A | B | AB | O |
| Frequency | 57 | 24 | 9 | 110 |
| Answer | Marks | Guidance |
|---|---|---|
| Working/Answer | Mark | Guidance |
| \(H_0\): Sydney popn. has same freq. \([0.38,\ 0.10,\ 0.03,\ 0.49]\) | B1 | State (at least) null hypothesis (A.E.F.) |
| \(76\ \ 20\ \ 6\ \ 98\) | M1 A1 | Calculate expected values |
| \(\chi^2 \approx 8.5_{[2]}\) | M1 *A1 | Calculate value of \(\chi^2\) (M1 even if cells combined) |
| \(\chi^2_{3,\,0.95} = 7.815\ [\chi^2_2 = 5.991]\) | *B1\(\sqrt{}\) | Compare with consistent tabular value (to 2 dp) |
| Sydney does not conform (AEF) | B1 | Correct conclusion (dep *A1, *B1). Part mark: 7 |
| \(n \geq 5/0.03 = 166.7\) so \(n_{min}\) is \(167\) | M1 A1 | Find smallest size \(n\) with expected values \(\geq 5\). Part mark: 2. Total: [9] |
## Question 9:
| Working/Answer | Mark | Guidance |
|---|---|---|
| $H_0$: Sydney popn. has same freq. $[0.38,\ 0.10,\ 0.03,\ 0.49]$ | B1 | State (at least) null hypothesis (A.E.F.) |
| $76\ \ 20\ \ 6\ \ 98$ | M1 A1 | Calculate expected values |
| $\chi^2 \approx 8.5_{[2]}$ | M1 *A1 | Calculate value of $\chi^2$ (M1 even if cells combined) |
| $\chi^2_{3,\,0.95} = 7.815\ [\chi^2_2 = 5.991]$ | *B1$\sqrt{}$ | Compare with consistent tabular value (to 2 dp) |
| Sydney does **not** conform (AEF) | B1 | Correct conclusion (dep *A1, *B1). Part mark: 7 |
| $n \geq 5/0.03 = 166.7$ so $n_{min}$ is $167$ | M1 A1 | Find smallest size $n$ with expected values $\geq 5$. Part mark: 2. Total: **[9]** |9 The proportions of blood types $\mathrm { A } , \mathrm { B } , \mathrm { AB }$ and O in the Australian population are $38 \% , 10 \% , 3 \%$ and $49 \%$ respectively. In order to test whether the population in Sydney conforms to these figures, a random sample of 200 residents is selected. The table shows the observed frequencies of these types in the sample.
\begin{center}
\begin{tabular}{ | l | c | c | c | c | }
\hline
Blood Type & A & B & AB & O \\
\hline
Frequency & 57 & 24 & 9 & 110 \\
\hline
\end{tabular}
\end{center}
Carry out a suitable test at the 5\% significance level.
Find the smallest sample size that could be used for the test.
\hfill \mbox{\textit{CAIE FP2 2009 Q9 [9]}}