| Exam Board | OCR MEI |
|---|---|
| Module | S4 (Statistics 4) |
| Year | 2009 |
| Session | June |
| Marks | 24 |
| Paper | Download PDF ↗ |
| Mark scheme | Download PDF ↗ |
| Topic | Non-parametric tests |
| Type | Experimental design types |
| Difficulty | Standard +0.3 This is a straightforward Further Maths Statistics question testing standard knowledge of experimental design and ANOVA. Part (i) requires describing a Latin square (textbook recall), part (ii) asks for standard distributional assumptions (e_ij ~ N(0,σ²)), and part (iii) involves routine ANOVA table completion with simple arithmetic (subtraction for SS, counting df, division for MS, F-test). All components are standard bookwork or mechanical calculations with no novel problem-solving required, making it slightly easier than average. |
| Spec | 2.01c Sampling techniques: simple random, opportunity, etc2.01d Select/critique sampling: in context |
| Source of variation | Sums of squares | Degrees of freedom | Mean squares | Mean square ratio |
| Between treatments | ||||
| Residual | 68.76 | |||
| Total | 161.06 |
| Answer | Marks | Guidance |
|---|---|---|
| Answer/Working | Marks | Guidance |
| Description must be in context. If no context given, mark according to scheme and give half-marks, rounded down | ||
| Clear description of "rows" | E1, E1 | |
| And "columns" | E1, E1 | |
| As extraneous factors to be taken account of in the design, with "treatments" to be compared | E1, E1 | |
| Need same numbers of each | E1 | |
| Clear contrast with situations for completely randomised design and randomised trends | E1, E1 |
| Answer | Marks | Guidance |
|---|---|---|
| Answer/Working | Marks | Guidance |
| \(e_{ij} \sim \text{ind } N(0, \sigma^2)\) | 1, 1, 1 | Allow uncorrelated; For 0; For \(\sigma^2\) |
| \(\alpha_i\) is population mean effect by which \(i\)th treatment differs from overall mean | 1, 1 |
| Answer | Marks | Guidance |
|---|---|---|
| Answer/Working | Marks | Guidance |
| ANOVA table: Between Treatments: SS = 92.30, df = 4, MS = 23.075, MS ratio = 5.034 | 1 | |
| Residual: SS = 68.76, df = 15, MS = 4.584 | 1 | |
| Total: SS = 161.06, df = 19 | 1 | |
| Refer to \(F_{4,15}\) | 1 | No FT if wrong |
| Upper 1% point is 4.89; Significant, seems treatments are not all the same | 1, 1 | No FT if wrong |
# Question 4:
## Part (i):
| Answer/Working | Marks | Guidance |
|---|---|---|
| Description must be in context. If no context given, mark according to scheme and give half-marks, rounded down | | |
| Clear description of "rows" | E1, E1 | |
| And "columns" | E1, E1 | |
| As extraneous factors to be taken account of in the design, with "treatments" to be compared | E1, E1 | |
| Need same numbers of each | E1 | |
| Clear contrast with situations for completely randomised design and randomised trends | E1, E1 | | **[9 marks]**
## Part (ii):
| Answer/Working | Marks | Guidance |
|---|---|---|
| $e_{ij} \sim \text{ind } N(0, \sigma^2)$ | 1, 1, 1 | Allow uncorrelated; For 0; For $\sigma^2$ |
| $\alpha_i$ is population mean effect by which $i$th treatment differs from overall mean | 1, 1 | | **[5 marks]**
## Part (iii):
| Answer/Working | Marks | Guidance |
|---|---|---|
| ANOVA table: Between Treatments: SS = 92.30, df = 4, MS = 23.075, MS ratio = 5.034 | 1 | |
| Residual: SS = 68.76, df = 15, MS = 4.584 | 1 | |
| Total: SS = 161.06, df = 19 | 1 | |
| Refer to $F_{4,15}$ | 1 | No FT if wrong |
| Upper 1% point is 4.89; Significant, seems treatments are not all the same | 1, 1 | No FT if wrong | **[10 marks]**
4 (i) Describe, with the aid of a specific example, an experimental situation for which a Latin square design is appropriate, indicating carefully the features which show that a completely randomised or randomised blocks design would be inappropriate.\\
(ii) The model for the one-way analysis of variance may be written, in a customary notation, as
$$x _ { i j } = \mu + \alpha _ { i } + e _ { i j }$$
State the distributional assumptions underlying $e _ { i j }$ in this model. What is the interpretation of the term $\alpha _ { i }$ ?\\
(iii) An experiment for comparing 5 treatments is carried out, with a total of 20 observations. A partial one-way analysis of variance table for the analysis of the results is as follows.
\begin{center}
\begin{tabular}{|l|l|l|l|l|}
\hline
Source of variation & Sums of squares & Degrees of freedom & Mean squares & Mean square ratio \\
\hline
Between treatments & & & & \\
\hline
Residual & 68.76 & & & \\
\hline
Total & 161.06 & & & \\
\hline
\end{tabular}
\end{center}
Copy and complete the table, and carry out the appropriate test using a $1 \%$ significance level.
\hfill \mbox{\textit{OCR MEI S4 2009 Q4 [24]}}