| Exam Board | Edexcel |
|---|---|
| Module | S3 (Statistics 3) |
| Year | 2012 |
| Session | June |
| Marks | 10 |
| Paper | Download PDF ↗ |
| Mark scheme | Download PDF ↗ |
| Topic | Chi-squared test of independence |
| Type | Standard 2×3 contingency table |
| Difficulty | Standard +0.3 This is a standard chi-squared test of independence with a 2×3 contingency table. Students must state hypotheses, calculate expected frequencies, compute the test statistic, find critical value, and conclude. While it requires multiple steps and careful arithmetic, it follows a completely routine procedure taught directly in S3 with no novel problem-solving or insight required. Slightly easier than average due to small table size and straightforward context. |
| Spec | 5.06a Chi-squared: contingency tables |
| \backslashbox{Breed}{Egg yield} | Low | Medium | High |
| Leghorn | 22 | 52 | 26 |
| Cornish | 14 | 32 | 4 |
| Answer | Marks | Guidance |
|---|---|---|
| Answer/Working | Marks | Guidance |
| \(H_0\): Egg yield and breed of chicken are independent (not associated) | B1 | Must mention "yield" and "breed" in both hypotheses; "relationship" or "correlation" or "connection" is B0 |
| \(H_1\): Egg yield and breed of chicken are dependent (associated) | ||
| Expected frequencies: Leghorn Low \(= \frac{100 \times 36}{150} = 24\), Medium \(= \frac{100 \times 84}{150} = 56\), High \(= \frac{100 \times 30}{150} = 20\); Cornish Low \(= \frac{50 \times 36}{150} = 12\), Medium \(= \frac{50 \times 84}{150} = 28\), High \(= \frac{50 \times 30}{150} = 10\) | M1A1 | M1 for use of \(\frac{\text{Row Total} \times \text{Col. Total}}{\text{Grand Total}}\); A1 for all expected frequencies correct |
| \(\sum \frac{(O-E)^2}{E}\): \(0.1667 + 0.2857 + 1.8 + 0.3333 + 0.5714 + 3.6\) | M1A1 | M1 for at least two correct terms; A1 for all correct (2 sf or better) |
| \(\sum \frac{(O-E)^2}{E} = 6.757...\) | A1 | |
| \(\nu = 2\), \(\chi^2_2(5\%) = 5.991\) | B1B1ft | |
| \(6.757 > 5.991\) so sufficient evidence to reject \(H_0\) | M1 | Must be \(\chi^2\) not normal |
| Egg yield and breed of chicken are dependent (associated) | A1 | Must mention "egg yield" and "breed of chicken"; condone "relationship"/"connection" but not "correlation" |
# Question 4:
| Answer/Working | Marks | Guidance |
|---|---|---|
| $H_0$: Egg yield and breed of chicken are independent (not associated) | B1 | Must mention "yield" and "breed" in both hypotheses; "relationship" or "correlation" or "connection" is B0 |
| $H_1$: Egg yield and breed of chicken are dependent (associated) | | |
| Expected frequencies: Leghorn Low $= \frac{100 \times 36}{150} = 24$, Medium $= \frac{100 \times 84}{150} = 56$, High $= \frac{100 \times 30}{150} = 20$; Cornish Low $= \frac{50 \times 36}{150} = 12$, Medium $= \frac{50 \times 84}{150} = 28$, High $= \frac{50 \times 30}{150} = 10$ | M1A1 | M1 for use of $\frac{\text{Row Total} \times \text{Col. Total}}{\text{Grand Total}}$; A1 for all expected frequencies correct |
| $\sum \frac{(O-E)^2}{E}$: $0.1667 + 0.2857 + 1.8 + 0.3333 + 0.5714 + 3.6$ | M1A1 | M1 for at least two correct terms; A1 for all correct (2 sf or better) |
| $\sum \frac{(O-E)^2}{E} = 6.757...$ | A1 | |
| $\nu = 2$, $\chi^2_2(5\%) = 5.991$ | B1B1ft | |
| $6.757 > 5.991$ so sufficient evidence to reject $H_0$ | M1 | Must be $\chi^2$ not normal |
| Egg yield and breed of chicken are dependent (associated) | A1 | Must mention "egg yield" and "breed of chicken"; condone "relationship"/"connection" but not "correlation" |
---\begin{enumerate}
\item Two breeds of chicken are surveyed to measure their egg yield. The results are shown in the table below.
\end{enumerate}
\begin{center}
\begin{tabular}{|l|l|l|l|}
\hline
\backslashbox{Breed}{Egg yield} & Low & Medium & High \\
\hline
Leghorn & 22 & 52 & 26 \\
\hline
Cornish & 14 & 32 & 4 \\
\hline
\end{tabular}
\end{center}
Showing each stage of your working clearly, test, at the $5 \%$ significance level, whether or not there is an association between egg yield and breed of chicken. State your hypotheses clearly.
\hfill \mbox{\textit{Edexcel S3 2012 Q4 [10]}}