| Exam Board | OCR MEI |
|---|---|
| Module | S4 (Statistics 4) |
| Year | 2006 |
| Session | June |
| Marks | 24 |
| Paper | Download PDF ↗ |
| Mark scheme | Download PDF ↗ |
| Topic | T-tests (unknown variance) |
| Type | Two-sample t-test equal variance |
| Difficulty | Standard +0.3 This is a standard two-part hypothesis testing question requiring (i) a two-sample t-test and (ii) a paired t-test. Both are routine applications of textbook procedures with straightforward calculations. The question clearly signposts which tests to use and provides all necessary data. While it requires careful arithmetic and proper statement of assumptions/hypotheses, it demands no novel insight or problem-solving beyond standard S4 technique. |
| Spec | 5.05c Hypothesis test: normal distribution for population mean5.07b Sign test: and Wilcoxon signed-rank5.07d Paired vs two-sample: selection |
| Employees trained by method A: | 35.2 | 47.8 | 25.8 | 38.0 | 53.6 | 31.0 | 33.9 |
| 35.4 | 21.6 | 42.5 | |||||
| Employees trained by method B: | 43.0 | 57.5 | 68.6 | 20.9 | 31.4 | 44.9 | 62.8 |
| 27.6 | 41.8 | 46.1 | 39.8 | 61.6 |
| Worker | \(W _ { 1 }\) | \(W _ { 2 }\) | \(W _ { 3 }\) | \(W _ { 4 }\) | \(W _ { 5 }\) | \(W _ { 6 }\) | \(W _ { 7 }\) | \(W _ { 8 }\) |
| Time before | 32.6 | 28.5 | 22.9 | 27.6 | 34.9 | 28.8 | 34.2 | 31.3 |
| Time after | 26.2 | 24.1 | 19.0 | 28.6 | 29.3 | 20.0 | 36.0 | 19.2 |
| Answer | Marks | Guidance |
|---|---|---|
| Answer | Marks | Guidance |
| \(\bar{x} = 36.48\), \(s = 9.6307\), \(s^2 = 92.7507\) | B1 | If all correct. No marks for use of \(s_n\) (9.1365 and 14.1823) |
| \(\bar{y} = 45.5\), \(s = 14.8129\), \(s^2 = 219.4218\) | ||
| Assumptions: Normality of both populations; equal variances | B1, B1 | |
| \(H_0: \mu_A = \mu_B \quad H_1: \mu_A \neq \mu_B\) | B1, B1 | Do NOT accept \(\bar{X} = \bar{Y}\) or similar |
| Pooled \(s^2 = \frac{9 \times 92.7507 + 11 \times 219.4218}{20} = \frac{834.756 + 2413664}{20} = 162.4198\) | B1 | \(= (12.7444)^2\) |
| Test statistic \(= \frac{36.48 - 45.5}{\sqrt{162.4198}\sqrt{\frac{1}{10}+\frac{1}{12}}} = \frac{-9.02}{5.4568} = -1.653\) | M1, A1 | |
| Refer to \(t_{20}\); double tailed 5% point is 2.086 | M1, A1 | No ft if wrong |
| Not significant | A1 | ft only c's test statistic |
| No evidence that population mean times differ | A1 | ft only c's test statistic [12] |
| Answer | Marks | Guidance |
|---|---|---|
| Answer | Marks | Guidance |
| Assumption: Normality of underlying population of differences | B1 | |
| \(H_0: \mu_D = 0 \quad H_1: \mu_D > 0\) | B1, B1 | Do NOT accept \(\bar{D}=0\). Direction of \(D\) must be clear |
| Differences: \(6.4, 4.4, 3.9, -1.0, 5.6, 8.8, -1.8, 12.1\) | M1 | |
| \(\bar{x} = 4.8\), \(s = 4.6393\) | A1 can be awarded if NOT awarded in (i). \(s_n = 4.3396\) NOT acceptable | |
| Test statistic \(= \frac{4.8-0}{4.6393/\sqrt{8}} = 2.92(64)\) | M1, A1 | |
| Refer to \(t_7\); single tailed 5% point is 1.895 | M1, A1 | No ft if wrong |
| Significant; seems mean is lowered | A1, A1 | ft only c's test statistic [10] |
| Answer | Marks | Guidance |
|---|---|---|
| Answer | Marks | Guidance |
| The paired comparison in part (ii) eliminates the variability between workers | E2 | (E1, E1) [2] |
# Question 3:
## Part (i):
| Answer | Marks | Guidance |
|--------|-------|----------|
| $\bar{x} = 36.48$, $s = 9.6307$, $s^2 = 92.7507$ | B1 | If all correct. No marks for use of $s_n$ (9.1365 and 14.1823) |
| $\bar{y} = 45.5$, $s = 14.8129$, $s^2 = 219.4218$ | | |
| Assumptions: Normality of both populations; equal variances | B1, B1 | |
| $H_0: \mu_A = \mu_B \quad H_1: \mu_A \neq \mu_B$ | B1, B1 | Do NOT accept $\bar{X} = \bar{Y}$ or similar |
| Pooled $s^2 = \frac{9 \times 92.7507 + 11 \times 219.4218}{20} = \frac{834.756 + 2413664}{20} = 162.4198$ | B1 | $= (12.7444)^2$ |
| Test statistic $= \frac{36.48 - 45.5}{\sqrt{162.4198}\sqrt{\frac{1}{10}+\frac{1}{12}}} = \frac{-9.02}{5.4568} = -1.653$ | M1, A1 | |
| Refer to $t_{20}$; double tailed 5% point is 2.086 | M1, A1 | No ft if wrong |
| Not significant | A1 | ft only c's test statistic |
| No evidence that population mean times differ | A1 | ft only c's test statistic **[12]** |
## Part (ii):
| Answer | Marks | Guidance |
|--------|-------|----------|
| Assumption: Normality of underlying population of differences | B1 | |
| $H_0: \mu_D = 0 \quad H_1: \mu_D > 0$ | B1, B1 | Do NOT accept $\bar{D}=0$. Direction of $D$ must be clear |
| Differences: $6.4, 4.4, 3.9, -1.0, 5.6, 8.8, -1.8, 12.1$ | M1 | |
| $\bar{x} = 4.8$, $s = 4.6393$ | | A1 can be awarded if NOT awarded in (i). $s_n = 4.3396$ NOT acceptable |
| Test statistic $= \frac{4.8-0}{4.6393/\sqrt{8}} = 2.92(64)$ | M1, A1 | |
| Refer to $t_7$; single tailed 5% point is 1.895 | M1, A1 | No ft if wrong |
| Significant; seems mean is lowered | A1, A1 | ft only c's test statistic **[10]** |
## Part (iii):
| Answer | Marks | Guidance |
|--------|-------|----------|
| The paired comparison in part (ii) eliminates the variability between workers | E2 | (E1, E1) **[2]** |
---3 The human resources department of a large company is investigating two methods, A and B, for training employees to carry out a certain complicated and intricate task.\\
(i) Two separate random samples of employees who have not previously performed the task are taken. The first sample is of size 10 ; each of the employees in it is trained by method A. The second sample is of size 12; each of the employees in it is trained by method B. After completing the training, the time for each employee to carry out the task is measured, in controlled conditions. The times are as follows, in minutes.
\begin{center}
\begin{tabular}{ l l l l l l l l }
Employees trained by method A: & 35.2 & 47.8 & 25.8 & 38.0 & 53.6 & 31.0 & 33.9 \\
& 35.4 & 21.6 & 42.5 & & & & \\
Employees trained by method B: & 43.0 & 57.5 & 68.6 & 20.9 & 31.4 & 44.9 & 62.8 \\
& 27.6 & 41.8 & 46.1 & 39.8 & 61.6 & & \\
\end{tabular}
\end{center}
Stating appropriate assumptions concerning the underlying populations, use a $t$ test at the $5 \%$ significance level to examine whether either training method is better in respect of leading, on the whole, to a lower time to carry out the task.\\
(ii) A further trial of method B is carried out to see if the performance of experienced and skilled workers can be improved by re-training them. A random sample of 8 such workers is taken. The times in minutes, under controlled conditions, for each worker to carry out the task before and after re-training are as follows.
\begin{center}
\begin{tabular}{ | l | c | c | c | c | c | c | c | c | }
\hline
Worker & $W _ { 1 }$ & $W _ { 2 }$ & $W _ { 3 }$ & $W _ { 4 }$ & $W _ { 5 }$ & $W _ { 6 }$ & $W _ { 7 }$ & $W _ { 8 }$ \\
\hline
Time before & 32.6 & 28.5 & 22.9 & 27.6 & 34.9 & 28.8 & 34.2 & 31.3 \\
\hline
Time after & 26.2 & 24.1 & 19.0 & 28.6 & 29.3 & 20.0 & 36.0 & 19.2 \\
\hline
\end{tabular}
\end{center}
Stating an appropriate assumption, use a $t$ test at the $5 \%$ significance level to examine whether the re-training appears, on the whole, to lead to a lower time to carry out the task.\\
(iii) Explain how the test procedure in part (ii) is enhanced by designing it as a paired comparison.
\hfill \mbox{\textit{OCR MEI S4 2006 Q3 [24]}}