| Exam Board | Edexcel |
|---|---|
| Module | S3 (Statistics 3) |
| Year | 2005 |
| Session | June |
| Paper | Download PDF ↗ |
| Mark scheme | Download PDF ↗ |
| Topic | Chi-squared goodness of fit |
| Type | Chi-squared test of independence |
| Difficulty | Standard +0.3 This is a standard chi-squared test of independence with a 3×3 contingency table. Students must calculate expected frequencies, compute the test statistic, find degrees of freedom (4), and compare to critical value. While it requires multiple computational steps and careful organization, it follows a completely routine procedure taught explicitly in S3 with no novel problem-solving or conceptual insight required. The 11 marks reflect computational length rather than difficulty. |
| Spec | 5.06a Chi-squared: contingency tables |
| No action | Remove diseased branches | Spray with chemicals | |
| Tree died within 1 year | 10 | 5 | 6 |
| Tree survived for 1–4 years | 5 | 9 | 7 |
| Tree survived beyond 4 years | 5 | 6 | 7 |
| Answer | Marks | Guidance |
|---|---|---|
| Content | Marks | Guidance |
| Hypotheses: \(H_0\): Treatment and Survival are independent (not associated) \(H_1\): Treatment and Survival are not independent (associated) | M1, B1 | |
| Degrees of freedom: \(\nu = (3-1) \times (3-1) = 4\) | B1 | |
| Critical region: \(\chi^2 > 9.488\) | B1 | At \(\alpha = 0.05\) |
| Chi-squared test statistic: \(\sum \frac{(O-E)^2}{E} = \frac{3}{7} + \frac{25}{7} + \frac{1}{7} + \frac{25}{7} + \frac{2}{6} + 0 + \frac{1}{6} + 0 + \frac{1}{6}\) Using \(\sum \frac{(o-e)^2}{E}\) Any 2 values \(\sum \frac{(o-e)^2}{E} = 3.47619...\) | M1, A1, A1 | |
| Conclusion: Since \(3.47619...\) is NOT in the critical region (i.e. \(29.4 ff\)), there is insufficient evidence to reject \(H_0\). There is no evidence of association between treatment and length of survival. | M1, Confirmation, Co-dURAn | A1√ (ii) |
| Content | Marks | Guidance |
|---------|-------|----------|
| **Hypotheses:** $H_0$: Treatment and Survival are independent (not associated) $H_1$: Treatment and Survival are not independent (associated) | M1, B1 | |
| **Degrees of freedom:** $\nu = (3-1) \times (3-1) = 4$ | B1 | |
| **Critical region:** $\chi^2 > 9.488$ | B1 | At $\alpha = 0.05$ |
| **Chi-squared test statistic:** $\sum \frac{(O-E)^2}{E} = \frac{3}{7} + \frac{25}{7} + \frac{1}{7} + \frac{25}{7} + \frac{2}{6} + 0 + \frac{1}{6} + 0 + \frac{1}{6}$ Using $\sum \frac{(o-e)^2}{E}$ Any 2 values $\sum \frac{(o-e)^2}{E} = 3.47619...$ | M1, A1, A1 | |
| **Conclusion:** Since $3.47619...$ is NOT in the critical region (i.e. $29.4 ff$), there is insufficient evidence to reject $H_0$. There is no evidence of association between treatment and length of survival. | M1, Confirmation, Co-dURAn | A1√ (ii) |
---A researcher carried out a survey of three treatments for a fruit tree disease. The contingency table below shows the results of a survey of a random sample of 60 diseased trees.
\begin{center}
\begin{tabular}{|l|c|c|c|}
\hline
& No action & Remove diseased branches & Spray with chemicals \\
\hline
Tree died within 1 year & 10 & 5 & 6 \\
\hline
Tree survived for 1–4 years & 5 & 9 & 7 \\
\hline
Tree survived beyond 4 years & 5 & 6 & 7 \\
\hline
\end{tabular}
\end{center}
Test, at the 5\% level of significance, whether or not there is any association between the treatment of the trees and their survival. State your hypotheses and conclusion clearly.
(Total 11 marks)
\hfill \mbox{\textit{Edexcel S3 2005 Q3}}