| Exam Board | OCR MEI |
|---|---|
| Module | S4 (Statistics 4) |
| Year | 2012 |
| Session | June |
| Marks | 24 |
| Paper | Download PDF ↗ |
| Mark scheme | Download PDF ↗ |
| Topic | Wilcoxon tests |
| Type | Wilcoxon rank-sum test (Mann-Whitney U test) |
| Difficulty | Standard +0.3 This is a straightforward application of standard Wilcoxon rank-sum and t-test procedures with small datasets. The question provides clear context, pre-sorted data, and asks for routine hypothesis testing with standard interpretation. While it requires careful ranking and calculation, it involves no novel insight or complex problem-solving beyond textbook methods for Further Maths Statistics. |
| Spec | 5.05c Hypothesis test: normal distribution for population mean5.07a Non-parametric tests: when to use5.07d Paired vs two-sample: selection |
| Fertiliser A | 9.8 | 10.2 | 10.9 | 11.5 | 12.7 | 13.3 | ||
| Fertiliser B | 10.8 | 11.9 | 12.0 | 12.2 | 12.9 | 13.5 | 13.6 | 13.7 |
| Answer | Marks | Guidance |
|---|---|---|
| \(H_0\): population medians are equal \(H_1\): population median for A \(<\) population median for B Wilcoxon rank sum test (or Mann-Whitney form of test) Ranks are: A: 1 2 4 5 9 11; B: 3 6 7 8 10 12 13 14 \(W = 1 + 2 + 4 + 5 + 9 + 11 = 32\) [or \(0 + 0 + 1 + 1 + 4 + 5 = 11\) if M-W used] Refer to \(W_{6,8}\) [or \(MW_{6,8}\) tables] Lower 5% critical point is 31 [or 10 if M-W used] Result is not significant Seems median yields may be assumed equal | B1, B1, M1, A1, B1, M1, A1, A1 | [Note: "population" must be explicit]. Explicit statement re shapes of distributions (eg that they are the same shape) is not required. More formal statements of hypotheses gain both marks [eg cdfs are \(F(x)\) and \(F(x - \Delta)\), \(H_0\) is \(\Delta = 0\) etc.]. Combined ranking. Correct [allow up to 2 errors; FT provided M1 earned]. No FT if wrong. No FT if wrong. |
| Answer | Marks | Guidance |
|---|---|---|
| \(H_0\): population means are equal \(H_1\): population mean for A \(<\) population mean for B For A: \(\bar{x} = 11.4\), \(s_{n-1}^2 = 1.912\) [\(s_n \cdot = 1.38275\)] For B: \(\bar{y} = 12.575\), \(s_{n-1}^2 = 1.051\) [\(s_{n-1} = 1.025\)] Pooled \(s^2 = \frac{(5 \times 1.912) + (7 \times 1.051)}{12} = \frac{16.915}{12} = 1.4096\) Test statistic \(= \frac{11.4 - 12.575}{\sqrt{1.4096 \sqrt{\frac{1}{5} + \frac{1}{8}}}} = \frac{-1.175}{0.6412} = -1.83(25)\) Refer to \(t_{12}\) Lower single-tailed 5% critical point is \(-1.782\) Significant Seems mean yield for A is less than that for B | B1, B1, M1, A1, M1, A1, A1, A1 | "population" must be explicit, either in words or by notation. For all. Use of \(s_n\) scores B0. For any reasonable attempt at pooling (but not if \(s_n^2\) used). If correct. Ft if incorrect. No FT if wrong. No FT if wrong. Must compare \(-1.83\) with \(-1.782\) unless it is clear and explicit that absolute values are being used. |
| Answer | Marks | Guidance |
|---|---|---|
| \(t\) test is "more sensitive" if Normality is correct. Non-rejection of Normality supports \(t\). But Wilcoxon is more reliable if not Normal – and we do not have proof of Normality. | E1, E1, E1, E1 | [4] |
**Part (i)**
| $H_0$: population medians are equal $H_1$: population median for A $<$ population median for B Wilcoxon rank sum test (or Mann-Whitney form of test) Ranks are: A: 1 2 4 5 9 11; B: 3 6 7 8 10 12 13 14 $W = 1 + 2 + 4 + 5 + 9 + 11 = 32$ [or $0 + 0 + 1 + 1 + 4 + 5 = 11$ if M-W used] Refer to $W_{6,8}$ [or $MW_{6,8}$ tables] Lower 5% critical point is 31 [or 10 if M-W used] Result is not significant Seems median yields may be assumed equal | B1, B1, M1, A1, B1, M1, A1, A1 | [Note: "population" must be explicit]. Explicit statement re shapes of distributions (eg that they are the same shape) is not required. More formal statements of hypotheses gain both marks [eg cdfs are $F(x)$ and $F(x - \Delta)$, $H_0$ is $\Delta = 0$ etc.]. Combined ranking. Correct [allow up to 2 errors; FT provided M1 earned]. No FT if wrong. No FT if wrong. |
**Part (ii)**
| $H_0$: population means are equal $H_1$: population mean for A $<$ population mean for B For A: $\bar{x} = 11.4$, $s_{n-1}^2 = 1.912$ [$s_n \cdot = 1.38275$] For B: $\bar{y} = 12.575$, $s_{n-1}^2 = 1.051$ [$s_{n-1} = 1.025$] Pooled $s^2 = \frac{(5 \times 1.912) + (7 \times 1.051)}{12} = \frac{16.915}{12} = 1.4096$ Test statistic $= \frac{11.4 - 12.575}{\sqrt{1.4096 \sqrt{\frac{1}{5} + \frac{1}{8}}}} = \frac{-1.175}{0.6412} = -1.83(25)$ Refer to $t_{12}$ Lower single-tailed 5% critical point is $-1.782$ Significant Seems mean yield for A is less than that for B | B1, B1, M1, A1, M1, A1, A1, A1 | "population" must be explicit, either in words or by notation. For all. Use of $s_n$ scores B0. For any reasonable attempt at pooling (but not if $s_n^2$ used). If correct. Ft if incorrect. No FT if wrong. No FT if wrong. Must compare $-1.83$ with $-1.782$ unless it is clear and explicit that absolute values are being used. |
**Part (iii)**
| $t$ test is "more sensitive" if Normality is correct. Non-rejection of Normality supports $t$. But Wilcoxon is more reliable if not Normal – and we do not have proof of Normality. | E1, E1, E1, E1 | [4] |
---3 At an agricultural research station, trials are being made of two fertilisers, A and B, to see whether they differ in their effects on the yield of a crop. Preliminary investigations have established that the underlying variances of the distributions of yields using the two fertilisers may be assumed equal. Scientific analysis of the fertilisers has suggested that fertiliser A may be inferior in that it leads, on the whole, to lower yield. A statistical analysis is being carried out to investigate this.
The crop is grown in carefully controlled conditions in 14 experimental plots, 6 with fertiliser A and 8 with fertiliser B. The yields, in kg per plot, are as follows, arranged in ascending order for each fertiliser.
\begin{center}
\begin{tabular}{ l r r r r r r r r }
Fertiliser A & 9.8 & 10.2 & 10.9 & 11.5 & 12.7 & 13.3 & & \\
Fertiliser B & 10.8 & 11.9 & 12.0 & 12.2 & 12.9 & 13.5 & 13.6 & 13.7 \\
\end{tabular}
\end{center}
(i) Carry out a Wilcoxon rank sum test at the $5 \%$ significance level to examine appropriate hypotheses.\\
(ii) Carry out a $t$ test at the $5 \%$ significance level to examine appropriate hypotheses.\\
(iii) Goodness of fit tests based on more extensive data sets from other trials with these fertilisers have failed to reject hypotheses of underlying Normal distributions. Discuss the relative merits of the analyses in parts (i) and (ii).
\hfill \mbox{\textit{OCR MEI S4 2012 Q3 [24]}}