| Exam Board | OCR MEI |
|---|---|
| Module | S4 (Statistics 4) |
| Year | 2010 |
| 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 straightforward application of one-way ANOVA with all calculations scaffolded by given summaries. Parts (i) and (ii) test standard recall of experimental design and model interpretation. Part (iii) requires routine computation of SS_between and SS_within using provided summaries, then a standard F-test lookup. No conceptual challenges or novel problem-solving required—slightly easier than average A-level due to the computational scaffolding. |
| Spec | 5.06c Fit other distributions: discrete and continuous5.06d Goodness of fit: chi-squared test |
| Variety | |||
| A | B | C | D |
| 12.3 | 14.2 | 14.1 | 13.6 |
| 11.9 | 13.1 | 13.2 | 12.8 |
| 12.8 | 13.1 | 14.6 | 13.3 |
| 12.2 | 12.5 | 13.7 | 14.3 |
| 13.5 | 12.7 | 13.4 | 13.8 |
| Answer | Marks |
|---|---|
| Randomised blocks | B1 |
| Plots in strips (blocks) | M1 |
| Correctly aligned w.r.t. fertility trend | E1 |
| Each letter occurs at least once in each block | M1 |
| In a random arrangement | E1 |
| Answer | Marks |
|---|---|
| \(\mu =\) population grand mean for whole experiment | B1 |
| \(\alpha_i =\) population mean amount by which the \(i\)th treatment differs from \(\mu\) | B1 |
| \(e_{ij} \sim\) ind \(N\) [accept "uncorrelated"] \((0, \sigma^2)\) | B1, B1, B1 |
| Answer | Marks | Guidance |
|---|---|---|
| Totals: \(62.7,\ 65.6,\ 69.0,\ 67.8\), all from samples of size 5 | ||
| Grand total \(265.1\); CF \(= 265.1^2/20 = 3513.9005\) | ||
| Total SS \(= 3524.31 - \text{CF} = 10.4095\) | M1 | Attempt to form three sums of squares |
| Between varieties SS \(= \frac{62.7^2}{5} + \frac{65.6^2}{5} + \frac{69.0^2}{5} + \frac{67.8^2}{5} - \text{CF} = 4.5975\) | M1, A1 | Correct method; each calculated SS correct |
| Residual SS \(= 10.4095 - 4.5975 = 5.8120\) | A1 | |
| Between varieties: SS \(= 4.5975\), df \(= 3\), MS \(= 1.5325\), MS ratio \(= 4.22\) | B1, M1, A1 cao | |
| Residual: SS \(= 5.8120\), df \(= 16\), MS \(= 0.36325\) | B1 | |
| Refer MS ratio to \(F_{3,16}\) | M1 | No FT if wrong |
| Upper 5% point is \(3.24\); Significant | A1 | No FT if wrong |
| Seems the mean yields of the varieties are not all the same | E1, E1 |
# Question 4:
## Part (i)
| Randomised blocks | B1 | |
|---|---|---|
| Plots in strips (blocks) | M1 | |
| Correctly aligned w.r.t. fertility trend | E1 | |
| Each letter occurs at least once in each block | M1 | |
| In a random arrangement | E1 | |
## Part (ii) - Model
| $\mu =$ population grand mean for whole experiment | B1 | |
|---|---|---|
| $\alpha_i =$ population mean amount by which the $i$th treatment differs from $\mu$ | B1 | |
| $e_{ij} \sim$ ind $N$ [accept "uncorrelated"] $(0, \sigma^2)$ | B1, B1, B1 | |
## Part (ii) - ANOVA
| Totals: $62.7,\ 65.6,\ 69.0,\ 67.8$, all from samples of size 5 | | |
|---|---|---|
| Grand total $265.1$; CF $= 265.1^2/20 = 3513.9005$ | | |
| Total SS $= 3524.31 - \text{CF} = 10.4095$ | M1 | Attempt to form three sums of squares |
| Between varieties SS $= \frac{62.7^2}{5} + \frac{65.6^2}{5} + \frac{69.0^2}{5} + \frac{67.8^2}{5} - \text{CF} = 4.5975$ | M1, A1 | Correct method; each calculated SS correct |
| Residual SS $= 10.4095 - 4.5975 = 5.8120$ | A1 | |
| Between varieties: SS $= 4.5975$, df $= 3$, MS $= 1.5325$, MS ratio $= 4.22$ | B1, M1, A1 cao | |
| Residual: SS $= 5.8120$, df $= 16$, MS $= 0.36325$ | B1 | |
| Refer MS ratio to $F_{3,16}$ | M1 | No FT if wrong |
| Upper 5% point is $3.24$; Significant | A1 | No FT if wrong |
| Seems the mean yields of the varieties are not all the same | E1, E1 | |4 At an agricultural research station, a trial is made of four varieties (A, B, C, D) of a certain crop in an experimental field. The varieties are grown on plots in the field and their yields are measured in a standard unit.\\
(i) It is at first thought that there may be a consistent trend in the natural fertility of the soil in the field from the west side to the east, though no other trends are known. Name an experimental design that should be used in these circumstances and give an example of an experimental layout.
Initial analysis suggests that any natural fertility trend may in fact be ignored, so the data from the trial are analysed by one-way analysis of variance.\\
(ii) The usual model for one-way analysis of variance of the yields $y _ { i j }$ may be written as
$$y _ { i j } = \mu + \alpha _ { i } + e _ { i j }$$
where the $e _ { i j }$ represent the experimental errors. Interpret the other terms in the model. State the usual distributional assumptions for the $e _ { i j }$.\\
(iii) The data for the yields are as follows, each variety having been used on 5 plots.
\begin{center}
\begin{tabular}{ | c | c | c | c | }
\hline
\multicolumn{4}{|c|}{Variety} \\
\hline
A & B & C & D \\
\hline
12.3 & 14.2 & 14.1 & 13.6 \\
11.9 & 13.1 & 13.2 & 12.8 \\
12.8 & 13.1 & 14.6 & 13.3 \\
12.2 & 12.5 & 13.7 & 14.3 \\
13.5 & 12.7 & 13.4 & 13.8 \\
\hline
\end{tabular}
\end{center}
$$\left[ \Sigma \Sigma y _ { i j } = 265.1 , \quad \Sigma \Sigma y _ { i j } ^ { 2 } = 3524.31 . \right]$$
Construct the usual one-way analysis of variance table and carry out the usual test, at the 5\% significance level. Report briefly on your conclusions.
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\hfill \mbox{\textit{OCR MEI S4 2010 Q4 [24]}}