| Exam Board | OCR |
|---|---|
| Module | S3 (Statistics 3) |
| Session | Specimen |
| Marks | 15 |
| 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 question requiring standard procedures: calculating differences, performing the test, stating assumptions, and computing a confidence interval. All steps are routine S3 material with no novel problem-solving required, making it slightly easier than average. |
| Spec | 5.05c Hypothesis test: normal distribution for population mean |
| Worker | \(A\) | \(B\) | \(C\) | \(D\) | \(E\) | \(F\) | \(G\) | \(H\) | \(I\) | \(J\) | \(K\) | \(L\) |
| Time in seconds for Method 1 | 48 | 38 | 47 | 59 | 62 | 41 | 50 | 52 | 58 | 54 | 49 | 60 |
| Time in seconds for Method 2 | 47 | 40 | 38 | 55 | 57 | 42 | 42 | 40 | 62 | 47 | 47 | 51 |
| Answer | Marks | Guidance |
|---|---|---|
| (i) (a) \(H_0: \mu_d = 0, H_1: \mu_d \neq 0\) | B1 | For both hypotheses stated |
| \(\bar{d} = 4.1667\) | B1 | For correct mean difference (subtraction can be either way round) |
| \(s^2 = \frac{486}{11} - \frac{50^2}{11 \times 12} = 25.2424\) | M1 | For calculation of unbiased variance estimate |
| A1 | For correct value 25.24... | |
| Test statistic is \(\frac{4.1667 - 0}{\sqrt{(25.2424/12)}} = 2.873\) | M1 | For correct standardising process |
| A1 | For correct value of test statistic | |
| This is greater than the critical value 2.718 | M1 | For a relevant comparison using \(t\) tables |
| Hence there is enough evidence to reject \(H_0\) and conclude that there is a difference between the times for the two methods | A1 | 8 |
| (b) Population of differences is normal | B1 | 1 |
| (c) Interval is \(4.1667 \pm 2.201\sqrt{\frac{25.2424}{12}}\) | M1 | For calculation of the form \(\bar{d} \pm t\sqrt{(s^2/n)}\) |
| B1 | For relevant use of \(t = 2.201\) | |
| Hence \(0.97 < \mu_d < 7.36\) | A1 | 3 |
| (ii) (a) Variation in the speed of individual workers is not eliminated, and may be large compared with the difference between the methods that is being tested | B1 | 1 |
| (b) Both samples are from normal populations. The population variances are equal | B1 | For a correct statement about normality |
| B1 | 2 | For a correct statement about the variances |
**(i) (a)** $H_0: \mu_d = 0, H_1: \mu_d \neq 0$ | B1 | For both hypotheses stated
$\bar{d} = 4.1667$ | B1 | For correct mean difference (subtraction can be either way round)
$s^2 = \frac{486}{11} - \frac{50^2}{11 \times 12} = 25.2424$ | M1 | For calculation of unbiased variance estimate
| A1 | For correct value 25.24...
Test statistic is $\frac{4.1667 - 0}{\sqrt{(25.2424/12)}} = 2.873$ | M1 | For correct standardising process
| A1 | For correct value of test statistic
This is greater than the critical value 2.718 | M1 | For a relevant comparison using $t$ tables
Hence there is enough evidence to reject $H_0$ and conclude that there is a difference between the times for the two methods | A1 | 8 | For correctly stated conclusion in context
**(b)** Population of differences is normal | B1 | 1 | For correct statement
**(c)** Interval is $4.1667 \pm 2.201\sqrt{\frac{25.2424}{12}}$ | M1 | For calculation of the form $\bar{d} \pm t\sqrt{(s^2/n)}$
| B1 | For relevant use of $t = 2.201$
Hence $0.97 < \mu_d < 7.36$ | A1 | 3 | For correct interval
**(ii) (a)** Variation in the speed of individual workers is not eliminated, and may be large compared with the difference between the methods that is being tested | B1 | 1 | For any relevant valid statement
**(b)** Both samples are from normal populations. The population variances are equal | B1 | For a correct statement about normality
| B1 | 2 | For a correct statement about the variances7 A factory manager wished to compare two methods of assembling a new component, to determine which method could be carried out more quickly, on average, by the workforce. A random sample of 12 workers was taken, and each worker tried out each of the methods of assembly. The times taken, in seconds, are shown in the table.
\begin{center}
\begin{tabular}{ | l | c c c c c c c c c c c c | }
\hline
Worker & $A$ & $B$ & $C$ & $D$ & $E$ & $F$ & $G$ & $H$ & $I$ & $J$ & $K$ & $L$ \\
\hline
Time in seconds for Method 1 & 48 & 38 & 47 & 59 & 62 & 41 & 50 & 52 & 58 & 54 & 49 & 60 \\
Time in seconds for Method 2 & 47 & 40 & 38 & 55 & 57 & 42 & 42 & 40 & 62 & 47 & 47 & 51 \\
\hline
\end{tabular}
\end{center}
(i) (a) Carry out an appropriate $t$-test, using a $2 \%$ significance level, to test whether there is any difference in the times for the two methods of assembly.\\
(b) State an assumption needed in carrying out this test.\\
(c) Calculate a $95 \%$ confidence interval for the population mean time difference for the two methods of assembly.\\
(ii) Instead of using the same 12 workers to try both methods, the factory manager could have used two independent random samples of workers, allocating Method 1 to the members of one sample and Method 2 to the members of the other sample.\\
(a) State one disadvantage of a procedure based on two independent random samples.\\
(b) State any assumptions that would need to be made to carry out a $t$-test based on two independent random samples.
\hfill \mbox{\textit{OCR S3 Q7 [15]}}