| Exam Board | OCR |
|---|---|
| Module | S3 (Statistics 3) |
| Year | 2007 |
| Session | June |
| Marks | 9 |
| Paper | Download PDF ↗ |
| Mark scheme | Download PDF ↗ |
| Topic | Chi-squared test of independence |
| Type | Contingency table construction from description |
| Difficulty | Standard +0.3 This is a straightforward chi-squared test of independence with a simple 2×2 contingency table. Part (i) requires basic arithmetic to complete the table, and part (ii) follows a standard procedure (calculate expected frequencies, compute test statistic, compare to critical value). The setup is clear with no conceptual complications, making it slightly easier than average for A-level statistics. |
| Spec | 5.06a Chi-squared: contingency tables |
| Proper | ||||
| \cline { 2 - 4 } \multicolumn{1}{l}{} | Pass | Fail | Total | |
| \cline { 2 - 5 } | Pass | 42 | ||
| \cline { 2 - 5 } Trial | Fail | 13 | ||
| \cline { 2 - 5 } | Total | 36 | 60 | |
| Answer | Marks | Guidance |
|---|---|---|
| P | F | |
| P | 31 | 11 |
| F | 5 | 13 |
| 36 | 24 | 60 |
| B1 | Two correct | |
| B1 | Others correct | |
| 2 marks total |
| Answer | Marks | Guidance |
|---|---|---|
| E-values: 25.2, 16.8, 10.8, 7.2 | M1 | One correct, Fit marginals in (i) |
| A1 | All correct | |
| \(\chi^2 = 5.3^2/(25.2^{-1}+10.8^{-1}+16.8^{-1}+7.2^{-1})\) | M1 | Allow two errors |
| A1 | With Yates' correction art 9.29 | |
| \(= 9.289\) | A1 | |
| Compare correctly with 7.8794 | M1 | Or 7.88 |
| There is evidence that results are not independent. | A1√ | 7 marks total; Ft \(\chi^2_{calc}\) |
(i) Proper contingency table:
| | P | F |
|---|---|---|
| **P** | 31 | 11 | 42 |
| **F** | 5 | 13 | 18 |
| | 36 | 24 | 60 |
| B1 | Two correct
| B1 | Others correct
| | 2 marks total |
(ii) ($H_0$: Trial results and Proper results are independent.)
E-values: 25.2, 16.8, 10.8, 7.2 | M1 | One correct, Fit marginals in (i)
| A1 | All correct
$\chi^2 = 5.3^2/(25.2^{-1}+10.8^{-1}+16.8^{-1}+7.2^{-1})$ | M1 | Allow two errors
| A1 | With Yates' correction art 9.29
$= 9.289$ | A1 |
Compare correctly with 7.8794 | M1 | Or 7.88
There is evidence that results are not independent. | A1√ | 7 marks total; Ft $\chi^2_{calc}$4 The students in a large university department take a trial examination some time before the proper examination. A random sample of 60 students took both examinations during a particular course. 42 students passed the trial examination, 36 passed the proper examination and 13 failed both examinations.\\
(i) Copy and complete the following contingency table.
\begin{center}
\begin{tabular}{ l | l | c | c | c | }
\multicolumn{4}{c}{} & Proper \\
\cline { 2 - 4 }
\multicolumn{1}{l}{} & Pass & Fail & Total & \\
\cline { 2 - 5 }
& Pass & & & 42 \\
\cline { 2 - 5 }
Trial & Fail & & 13 & \\
\cline { 2 - 5 }
& Total & 36 & & 60 \\
\hline
\end{tabular}
\end{center}
(ii) Carry out a test of independence at the $\frac { 1 } { 2 } \%$ level of significance.
\hfill \mbox{\textit{OCR S3 2007 Q4 [9]}}