| Exam Board | OCR |
|---|---|
| Module | S3 (Statistics 3) |
| Year | 2012 |
| 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 paired t-test application with clear data, requiring standard procedure (calculate differences, check assumptions, perform test) and interpretation at a given significance level. While it requires understanding of paired tests and assumptions, it's a routine S3 question with no conceptual complications or novel problem-solving—slightly easier than average due to small sample size and clear structure. |
| Spec | 5.05c Hypothesis test: normal distribution for population mean |
| Twin pair | 1 | 2 | 3 | 4 |
| Time for first-born | 46 | 38 | 44 | 49 |
| Time for second-born | 40 | 41 | 37 | 46 |
| Answer | Marks | Guidance |
|---|---|---|
| Answer | Marks | Guidance |
| Assumes population of time differences is normal | B1 | Need population. If 2 sample test used: Popns N, equal var B1 |
| \(H_0: \mu_F = \mu_S\), \(H_1: \mu_F \neq \mu_S\) | B1 | Allow \(\mu_d=0\) etc. Hypotheses B1 |
| Differences: \(6, -3, 7, 3\) | M1 | Means 44.25, 41 B1 |
| Sample mean \(= 3.25\) | A1 | (vars 21.583, 14) |
| Sample variance \(= 4.5^2 = 20.25\) | A1 | \((Var=(3\times21.583+3\times14)/6\), \(M0 = 17.792\) A0) |
| \(TS = 3.25/("4.5"/2)\) | M1 | One tail, \(TS=("44.25"\text{-}"41")/\sqrt{"17.792"(\frac{1}{4}+\frac{1}{4})}\) M1 |
| \(= 1.44(4)\) | A1 | 1.09, A0 |
| Not \(\geq 2.353\), do not reject \(H_0\) | M1 | Comparison with correct \(t\) and correct first conclusion. "\(1.09\)"\(<1.943\) do not reject NH. M1 |
| There is insufficient evidence of a difference in the mean times of the two populations | A1ft | Ft TS. Not over-assertive. Conclusion A1\(\sqrt{}\). Max 6/9 |
| [9] |
# Question 2:
| Answer | Marks | Guidance |
|--------|-------|----------|
| Assumes population of time differences is normal | B1 | Need population. If 2 sample test used: Popns N, equal var B1 |
| $H_0: \mu_F = \mu_S$, $H_1: \mu_F \neq \mu_S$ | B1 | Allow $\mu_d=0$ etc. Hypotheses B1 |
| Differences: $6, -3, 7, 3$ | M1 | Means 44.25, 41 B1 |
| Sample mean $= 3.25$ | A1 | (vars 21.583, 14) |
| Sample variance $= 4.5^2 = 20.25$ | A1 | $(Var=(3\times21.583+3\times14)/6$, $M0 = 17.792$ A0) |
| $TS = 3.25/("4.5"/2)$ | M1 | One tail, $TS=("44.25"\text{-}"41")/\sqrt{"17.792"(\frac{1}{4}+\frac{1}{4})}$ M1 |
| $= 1.44(4)$ | A1 | 1.09, A0 |
| Not $\geq 2.353$, do not reject $H_0$ | M1 | Comparison with correct $t$ and correct first conclusion. "$1.09$"$<1.943$ do not reject NH. M1 |
| There is insufficient evidence of a difference in the mean times of the two populations | A1ft | Ft TS. Not over-assertive. Conclusion A1$\sqrt{}$. Max 6/9 |
| | **[9]** | |
---2 Four pairs of randomly chosen twins were each given identical puzzles to solve. The times taken (in seconds) are shown in the following table.
\begin{center}
\begin{tabular}{ | l | l | l | l | l | }
\hline
Twin pair & 1 & 2 & 3 & 4 \\
\hline
Time for first-born & 46 & 38 & 44 & 49 \\
\hline
Time for second-born & 40 & 41 & 37 & 46 \\
\hline
\end{tabular}
\end{center}
Stating any necessary assumption, test at the $10 \%$ significance level whether there is a difference between the population mean times of first-born and second-born twins.
\hfill \mbox{\textit{OCR S3 2012 Q2 [9]}}