| Exam Board | Edexcel |
|---|---|
| Module | S4 (Statistics 4) |
| Year | 2004 |
| Session | June |
| Marks | 9 |
| Paper | Download PDF ↗ |
| Mark scheme | Download PDF ↗ |
| Topic | Wilcoxon tests |
| Type | Paired t-test |
| Difficulty | Standard +0.3 This is a straightforward application of the Wilcoxon signed-rank test with clear one-tailed hypothesis, small dataset requiring manual calculation of differences, ranks, and test statistic, then comparison to critical values from tables. While it requires careful execution of the standard procedure, it involves no conceptual challenges or novel problem-solving beyond routine S4 technique. |
| Spec | 5.05d Confidence intervals: using normal distribution |
| \(A\) | \(B\) | \(C\) | \(D\) | \(E\) | \(F\) | \(G\) | \(H\) | |
| Dominant hand | 202 | 251 | 215 | 235 | 210 | 195 | 191 | 230 |
| Weaker hand | 195 | 249 | 218 | 234 | 211 | 197 | 181 | 225 |
| Answer | Marks |
|---|---|
| Data: \(d: 7, 2, -3, 1, -1, -2, 10, 5\) | M1 |
| \(\sum d = 19; \sum d^2 = 193\) | M1 A1 |
| \(\bar{d} = \frac{19}{8} = 2.375; S_d^2 = \frac{1}{7}\left[193 - \frac{19^2}{8}\right] = 21.125\) | B1; M1 A1 |
| \(H_0: \mu_D = 0; H_1: \mu_D > 0\) | B1 (both) |
| \(t = \frac{2.375 - 0}{\sqrt{\frac{21.125}{8}}} = 1.4615\ldots\) | M1; A1 |
| \(\nu = 7 \Rightarrow \text{critical region: } t > 1.895\) | B1 |
| 1.895 | B1 |
| Since \(1.4915\ldots\) is not in the critical region there is insufficient evidence to reject \(H_0\) and we conclude that there is insufficient evidence to support the doctors' belief. | A1 ft |
| (9 marks) |
| Answer | Marks |
|---|---|
| Use of 2 sample \(t\)-test \(\Rightarrow\) B0 B0 B0 M1 A1 M1 A1 B1 A1 i.e.; 6/9 max | |
| \(S_p^2 = \frac{7 \times 440.125 + 7 \times 501.357}{8 + 8 - 2} = 470.74\) | M1 A1 |
| \(t = \frac{216.125 - 213.75}{\sqrt{470.74\left(\frac{1}{8} + \frac{1}{8}\right)}} = 0.0547\) | M1 A1 |
| critical region: \(t > 1.761\) | B1 |
| Conclusion as above | A1 ft |
| Data: $d: 7, 2, -3, 1, -1, -2, 10, 5$ | M1 |
| $\sum d = 19; \sum d^2 = 193$ | M1 A1 |
| $\bar{d} = \frac{19}{8} = 2.375; S_d^2 = \frac{1}{7}\left[193 - \frac{19^2}{8}\right] = 21.125$ | B1; M1 A1 |
| $H_0: \mu_D = 0; H_1: \mu_D > 0$ | B1 (both) |
| $t = \frac{2.375 - 0}{\sqrt{\frac{21.125}{8}}} = 1.4615\ldots$ | M1; A1 |
| $\nu = 7 \Rightarrow \text{critical region: } t > 1.895$ | B1 |
| 1.895 | B1 |
| Since $1.4915\ldots$ is not in the critical region there is insufficient evidence to reject $H_0$ and we conclude that there is insufficient evidence to support the doctors' belief. | A1 ft |
| | (9 marks) |
**Alternative approach:**
| Use of 2 sample $t$-test $\Rightarrow$ B0 B0 B0 M1 A1 M1 A1 B1 A1 i.e.; 6/9 max | |
| $S_p^2 = \frac{7 \times 440.125 + 7 \times 501.357}{8 + 8 - 2} = 470.74$ | M1 A1 |
| $t = \frac{216.125 - 213.75}{\sqrt{470.74\left(\frac{1}{8} + \frac{1}{8}\right)}} = 0.0547$ | M1 A1 |
| critical region: $t > 1.761$ | B1 |
| Conclusion as above | A1 ft |4. A doctor believes that the span of a person's dominant hand is greater than that of the weaker hand. To test this theory, the doctor measures the spans of the dominant and weaker hands of a random sample of 8 people. He subtracts the span of the weaker hand from that of the dominant hand. The spans, in mm , are summarised in the table below.
\begin{center}
\begin{tabular}{ | l | c | c | c | c | c | c | c | c | }
\hline
& $A$ & $B$ & $C$ & $D$ & $E$ & $F$ & $G$ & $H$ \\
\hline
Dominant hand & 202 & 251 & 215 & 235 & 210 & 195 & 191 & 230 \\
\hline
Weaker hand & 195 & 249 & 218 & 234 & 211 & 197 & 181 & 225 \\
\hline
\end{tabular}
\end{center}
Test, at the 5\% significance level, the doctor's belief.\\
(9)\\
\hfill \mbox{\textit{Edexcel S4 2004 Q4 [9]}}