| Exam Board | OCR MEI |
|---|---|
| Module | S2 (Statistics 2) |
| Year | 2006 |
| Session | June |
| Marks | 18 |
| Paper | Download PDF ↗ |
| Mark scheme | Download PDF ↗ |
| Topic | Chi-squared test of independence |
| Type | Interpret association after test |
| Difficulty | Standard +0.3 This is a standard chi-squared test of independence with clearly presented data in a 3×3 contingency table. Students must calculate expected frequencies, compute chi-squared contributions, compare to critical value, and interpret results. While it requires careful arithmetic and understanding of the test procedure, it follows a routine template with no conceptual surprises, making it slightly easier than average for an A-level statistics question. |
| Spec | 5.06a Chi-squared: contingency tables |
| \multirow{2}{*}{} | Musical preference | \multirow{2}{*}{Row totals} | |||
| Pop | Classical | Jazz | |||
| \multirow{3}{*}{Age group} | Under 25 | 57 | 15 | 12 | 84 |
| 25-50 | 43 | 21 | 21 | 85 | |
| Over 50 | 22 | 32 | 27 | 81 | |
| Column totals | 122 | 68 | 60 | 250 | |
| Answer | Marks |
|---|---|
| \(H_0\): no association between musical preference and age group | B1 |
| \(H_1\): there is association | B1 |
| Expected values calculated: e.g. \(E(\text{Under 25, Pop}) = \frac{84 \times 122}{250} = 41.136\) | M1 A1 |
| Full table of expected values and \((O-E)^2/E\) contributions | M1 A1 |
| \(X^2 = \sum\frac{(O-E)^2}{E}\) calculated | M1 |
| \(X^2 \approx 12.75\) | A1 |
| Degrees of freedom \(= (3-1)(3-1) = 4\) | B1 |
| Critical value \(\chi^2_4\) at 5% \(= 9.488\) | B1 |
| \(12.75 > 9.488\), reject \(H_0\) | M1 |
| Significant evidence of association between musical preference and age group | A1 |
| Answer | Marks |
|---|---|
| Under 25s prefer pop, least classical/jazz | B1 |
| Over 50s prefer classical, less pop | B1 |
| 25–50 roughly average across preferences | B1 |
| Reference to specific large contributions | B1 B1 B1 |
# Question 4:
## Part (i)
$H_0$: no association between musical preference and age group | B1 |
$H_1$: there is association | B1 |
Expected values calculated: e.g. $E(\text{Under 25, Pop}) = \frac{84 \times 122}{250} = 41.136$ | M1 A1 |
Full table of expected values and $(O-E)^2/E$ contributions | M1 A1 |
$X^2 = \sum\frac{(O-E)^2}{E}$ calculated | M1 |
$X^2 \approx 12.75$ | A1 |
Degrees of freedom $= (3-1)(3-1) = 4$ | B1 |
Critical value $\chi^2_4$ at 5% $= 9.488$ | B1 |
$12.75 > 9.488$, reject $H_0$ | M1 |
Significant evidence of association between musical preference and age group | A1 |
## Part (ii)
Under 25s prefer pop, least classical/jazz | B1 |
Over 50s prefer classical, less pop | B1 |
25–50 roughly average across preferences | B1 |
Reference to specific large contributions | B1 B1 B1 |4 A survey of a random sample of 250 people is carried out. Their musical preferences are categorized as pop, classical or jazz. Their ages are categorized as under 25, 25 to 50, or over 50. The results are as follows.
\begin{center}
\begin{tabular}{|l|l|l|l|l|l|}
\hline
\multicolumn{2}{|c|}{\multirow{2}{*}{}} & \multicolumn{3}{|c|}{Musical preference} & \multirow{2}{*}{Row totals} \\
\hline
& & Pop & Classical & Jazz & \\
\hline
\multirow{3}{*}{Age group} & Under 25 & 57 & 15 & 12 & 84 \\
\hline
& 25-50 & 43 & 21 & 21 & 85 \\
\hline
& Over 50 & 22 & 32 & 27 & 81 \\
\hline
\multicolumn{2}{|c|}{Column totals} & 122 & 68 & 60 & 250 \\
\hline
\end{tabular}
\end{center}
(i) Carry out a test at the $5 \%$ significance level to examine whether there is any association between musical preference and age group. State carefully your null and alternative hypotheses. Your working should include a table showing the contributions of each cell to the test statistic.\\
(ii) Discuss briefly how musical preferences vary between the age groups, as shown by the contributions to the test statistic.
\hfill \mbox{\textit{OCR MEI S2 2006 Q4 [18]}}