| Exam Board | OCR MEI |
|---|---|
| Module | S4 (Statistics 4) |
| Year | 2006 |
| Session | June |
| Marks | 24 |
| Paper | Download PDF ↗ |
| Mark scheme | Download PDF ↗ |
| Topic | Chi-squared test of independence |
| Type | Standard 3×3 contingency table |
| Difficulty | Standard +0.3 This is a standard textbook-style ANOVA question requiring routine application of learned procedures: identifying a Latin square design, stating the standard one-way ANOVA model, and performing calculations using given summary statistics. While it involves multiple parts and careful arithmetic, it requires no novel insight or problem-solving beyond direct application of S4 syllabus content, making it slightly easier than average. |
| Spec | 2.01c Sampling techniques: simple random, opportunity, etc2.01d Select/critique sampling: in context5.06a Chi-squared: contingency tables5.06b Fit prescribed distribution: chi-squared test |
| Paint A | Paint B | Paint C | Paint D | Paint E |
| 64 | 66 | 59 | 65 | 64 |
| 58 | 68 | 56 | 78 | 52 |
| 73 | 76 | 69 | 69 | 56 |
| 60 | 70 | 60 | 72 | 61 |
| 67 | 71 | 63 | 71 | 58 |
| Answer | Marks | Guidance |
|---|---|---|
| Answer | Marks | Guidance |
| Latin square | B1 | |
| Layout with correct rows and columns (letters = paints) | B1 | Correct rows and columns |
| A correct arrangement of letters, SC | B1 | For description instead of example allow max 1 out of 2 [3] |
| Answer | Marks | Guidance |
|---|---|---|
| Answer | Marks | Guidance |
| \(X_{ij} = \mu + \alpha_i + e_{ij}\) | B1 | |
| \(\mu =\) population grand mean for whole experiment | B1, B1 | |
| \(\alpha_i =\) population mean amount by which the \(i^{th}\) treatment differs from \(\mu\) | B1, B1 | |
| \(e_{ij}\) are experimental errors \(\sim\) ind \(N(0, \sigma^2)\) | B1, B1, B1, B1 | Allow "uncorrelated"; Mean; Variance [9] |
| Answer | Marks | Guidance |
|---|---|---|
| Answer | Marks | Guidance |
| Totals: 322, 351, 307, 355, 291; Grand total: 1626 | ||
| CF \(= \frac{1626^2}{25} = 105755.04\) | ||
| Total SS \(= 106838 - \text{CF} = 1082.96\) | M1 | |
| Between paints SS \(= \frac{322^2}{5} + \ldots + \frac{291^2}{5} - \text{CF} = 106368 - \text{CF} = 612.96\) | M1 | For correct methods for any two SS |
| Residual SS \(= 1082.96 - 612.96 = 470.00\) | A1 | If each calculated SS is correct |
| ANOVA table: Between paints SS=612.96, df=4, MS=153.24; Residual SS=470.00, df=20, MS=23.5; Total SS=1082.96, df=24 | B1, B1, M1 | df "between paints"; df "residual"; MS column |
| MS ratio \(= \frac{153.24}{23.5} = 6.52\) | M1, A1 | Independent of previous M1; dep only on this M1 |
| Refer to \(F_{4,20}\); upper 5% point is 2.87 | M1, A1, A1 | No ft if wrong; no ft if wrong; ft only c's test statistic and d.o.f.'s |
| Significant; seems performances of paints are not all the same | A1 | ft only c's test statistic and d.o.f.'s [12] |
# Question 4:
## Part (i):
| Answer | Marks | Guidance |
|--------|-------|----------|
| Latin square | B1 | |
| Layout with correct rows and columns (letters = paints) | B1 | Correct rows and columns |
| A correct arrangement of letters, SC | B1 | For description instead of example allow max 1 out of 2 **[3]** |
## Part (ii):
| Answer | Marks | Guidance |
|--------|-------|----------|
| $X_{ij} = \mu + \alpha_i + e_{ij}$ | B1 | |
| $\mu =$ population grand mean for whole experiment | B1, B1 | |
| $\alpha_i =$ population mean amount by which the $i^{th}$ treatment differs from $\mu$ | B1, B1 | |
| $e_{ij}$ are experimental errors $\sim$ ind $N(0, \sigma^2)$ | B1, B1, B1, B1 | Allow "uncorrelated"; Mean; Variance **[9]** |
## Part (iii):
| Answer | Marks | Guidance |
|--------|-------|----------|
| Totals: 322, 351, 307, 355, 291; Grand total: 1626 | | |
| CF $= \frac{1626^2}{25} = 105755.04$ | | |
| Total SS $= 106838 - \text{CF} = 1082.96$ | M1 | |
| Between paints SS $= \frac{322^2}{5} + \ldots + \frac{291^2}{5} - \text{CF} = 106368 - \text{CF} = 612.96$ | M1 | For correct methods for any two SS |
| Residual SS $= 1082.96 - 612.96 = 470.00$ | A1 | If each calculated SS is correct |
| ANOVA table: Between paints SS=612.96, df=4, MS=153.24; Residual SS=470.00, df=20, MS=23.5; Total SS=1082.96, df=24 | B1, B1, M1 | df "between paints"; df "residual"; MS column |
| MS ratio $= \frac{153.24}{23.5} = 6.52$ | M1, A1 | Independent of previous M1; dep only on this M1 |
| Refer to $F_{4,20}$; upper 5% point is 2.87 | M1, A1, A1 | No ft if wrong; no ft if wrong; ft only c's test statistic and d.o.f.'s |
| Significant; seems performances of paints are not all the same | A1 | ft only c's test statistic and d.o.f.'s **[12]** |4 An experiment is carried out to compare five industrial paints, A, B, C, D, E, that are intended to be used to protect exterior surfaces in polluted urban environments. Five different types of surface (I, II, III, IV, V) are to be used in the experiment, and five specimens of each type of surface are available. Five different external locations ( $1,2,3,4,5$ ) are used in the experiment.
The paints are applied to the specimens of the surfaces which are then left in the locations for a period of six months. At the end of this period, a "score" is given to indicate how effective the paint has been in protecting the surface.\\
(i) Name a suitable experimental design for this trial and give an example of an experimental layout.
Initial analysis of the data indicates that any differences between the types of surface are negligible, as also are any differences between the locations. It is therefore decided to analyse the data by one-way analysis of variance.\\
(ii) State the usual model, including the accompanying distributional assumptions, for the one-way analysis of variance. Interpret the terms in the model.\\
(iii) The data for analysis are as follows. Higher scores indicate better performance.
\begin{center}
\begin{tabular}{|l|l|l|l|l|}
\hline
Paint A & Paint B & Paint C & Paint D & Paint E \\
\hline
64 & 66 & 59 & 65 & 64 \\
\hline
58 & 68 & 56 & 78 & 52 \\
\hline
73 & 76 & 69 & 69 & 56 \\
\hline
60 & 70 & 60 & 72 & 61 \\
\hline
67 & 71 & 63 & 71 & 58 \\
\hline
\end{tabular}
\end{center}
[The sum of these data items is 1626 and the sum of their squares is 106838 .]\\
Construct the usual one-way analysis of variance table. Carry out the appropriate test, using a 5\% significance level. Report briefly on your conclusions.\\[0pt]
[12]
\hfill \mbox{\textit{OCR MEI S4 2006 Q4 [24]}}