| Exam Board | OCR |
|---|---|
| Module | S3 (Statistics 3) |
| Year | 2008 |
| Session | January |
| Marks | 11 |
| Paper | Download PDF ↗ |
| Mark scheme | Download PDF ↗ |
| Topic | T-tests (unknown variance) |
| Type | Paired sample t-test |
| Difficulty | Standard +0.3 This is a straightforward application of a paired t-test with clear data and standard procedure. Students must calculate differences, find mean and standard deviation, perform the test, and state the normality assumption. The context makes it obvious that pairing is appropriate. While it requires multiple steps, each is routine for S3 level with no conceptual surprises or novel problem-solving required. |
| Spec | 5.05d Confidence intervals: using normal distribution |
| Gardener | 1 | 2 | 3 | 4 | 5 | 6 | 7 | 8 | 9 |
| Brand \(A\) | 412 | 386 | 389 | 401 | 396 | 394 | 397 | 411 | 391 |
| Brand \(B\) | 422 | 394 | 385 | 408 | 394 | 399 | 397 | 410 | 397 |
| Answer | Marks | Guidance |
|---|---|---|
| Answer | Marks | Guidance |
| Not "independent" Or \(\mu_D = 0, \mu_D > 0\) | B1 | |
| From formula, or B2 from calculator | B1 M1A1 | |
| Accept 1.93. M1A0 if \(t = -1.926\) | M1 A1 B1 M1 | |
| A1 10 |
| Answer | Marks | Guidance |
|---|---|---|
| One valid reason | B1 (11) | Data are clearly paired; Data not independent; Or equivalent |
## Part (i)
Population of differences is normal
$H_0:\mu_A = \mu_B$, $H_1: \mu_A < \mu_B$ where $\mu_A$ and $\mu_B$ denote the population means
$\bar{x}_D = 3.222$
$s_D = 5.019$
$t = \frac{3.222}{(5.019/3)} = 1.926$
$CV = 1.860$
$1.926 > 1.860$
Reject $H_0$: there is evidence that brand A takes less time than brand B
| Answer | Marks | Guidance |
|--------|-------|----------|
| Not "independent" Or $\mu_D = 0, \mu_D > 0$ | B1 | |
| From formula, or B2 from calculator | B1 M1A1 | |
| Accept 1.93. M1A0 if $t = -1.926$ | M1 A1 B1 M1 | |
| | A1 **10** | |
## Part (ii)
One valid reason | B1 **(11)** | Data are clearly paired; Data not independent; Or equivalent |
---5 Of two brands of lawnmower, $A$ and $B$, brand $A$ was claimed to take less time, on average, than brand $B$ to mow similar stretches of lawn. In order to test this claim, 9 randomly selected gardeners were each given the task of mowing two regions of lawn, one with each brand of mower. All the regions had the same size and shape and had grass of the same height. The times taken, in seconds, are given in the table.
\begin{center}
\begin{tabular}{ | l | c | c | c | c | c | c | c | c | c | }
\hline
Gardener & 1 & 2 & 3 & 4 & 5 & 6 & 7 & 8 & 9 \\
\hline
Brand $A$ & 412 & 386 & 389 & 401 & 396 & 394 & 397 & 411 & 391 \\
\hline
Brand $B$ & 422 & 394 & 385 & 408 & 394 & 399 & 397 & 410 & 397 \\
\hline
\end{tabular}
\end{center}
(i) Test the claim using a paired-sample $t$-test at the $5 \%$ significance level. State a distributional assumption required for the test to be valid.\\
(ii) Give a reason why a paired-sample $t$-test should be used, rather than a 2 -sample $t$-test, in this case.
\hfill \mbox{\textit{OCR S3 2008 Q5 [11]}}