| Exam Board | CAIE |
|---|---|
| Module | S2 (Statistics 2) |
| Year | 2013 |
| Session | November |
| Marks | 8 |
| Paper | Download PDF ↗ |
| Mark scheme | Download PDF ↗ |
| Topic | Z-tests (known variance) |
| Type | Two-tail z-test |
| Difficulty | Moderate -0.3 This is a straightforward two-tailed hypothesis test with clear data summaries provided. Part (i) requires standard formulas for unbiased estimates (sample mean and variance with n-1 denominator), and part (ii) is a routine z-test application with given significance level. The question involves no conceptual challenges—just direct application of learned procedures with arithmetic calculation, making it slightly easier than average for A-level statistics. |
| Spec | 5.05b Unbiased estimates: of population mean and variance5.05c Hypothesis test: normal distribution for population mean |
| Answer | Marks | Guidance |
|---|---|---|
| est \((\mu) = 2866\) or \(2870\) (3 s.f.) | B1 | Accept 143300/50 o.e. |
| est \((\sigma^2) = \frac{1}{49}(410900000 - \frac{143300^2}{50})\) | M1 | Correct subst in correct formula |
| \((= 4126.53)\) | ||
| \(= 4130\) (3 sf) | A1 [3] |
| Answer | Marks | Guidance |
|---|---|---|
| \(H_0\): Pop mean (or \(\mu\)) \(= 2850\) | B1 | Both. Not just 'mean' |
| \(H_1\): Pop mean (or \(\mu\)) \(\neq 2850\) | ||
| \(\frac{143300}{50} - 2850\) | M1 | Allow '4126.53' without \(\sqrt{\phantom{x}}\), but must have all \(\sqrt{50}\) |
| \(\frac{\sqrt{4126.53'}}{\sqrt{50}}\) | ||
| \(= 1.761\) | A1 | Or correct c.v. (2867.81) for all method |
| '1.761' \(< 1.96\) | M1 | For valid comparison of \(z\) values, areas or c.v. |
| No evidence mean distance changed | A1If | Dep 1.96; If their 1.761 If \(H_1\): \(\mu > 2850\) and c.f. 1.645, max B0M1A1M1A0 |
| [5] | (c.v. for 1 tail test 2864.94) |
**(i)**
est $(\mu) = 2866$ or $2870$ (3 s.f.) | B1 | Accept 143300/50 o.e.
est $(\sigma^2) = \frac{1}{49}(410900000 - \frac{143300^2}{50})$ | M1 | Correct subst in correct formula
$(= 4126.53)$ | |
$= 4130$ (3 sf) | A1 [3] |
**(ii)**
$H_0$: Pop mean (or $\mu$) $= 2850$ | B1 | Both. Not just 'mean'
$H_1$: Pop mean (or $\mu$) $\neq 2850$ | |
$\frac{143300}{50} - 2850$ | M1 | Allow '4126.53' without $\sqrt{\phantom{x}}$, but must have all $\sqrt{50}$
$\frac{\sqrt{4126.53'}}{\sqrt{50}}$ | |
$= 1.761$ | A1 | Or correct c.v. (2867.81) for all method
'1.761' $< 1.96$ | M1 | For valid comparison of $z$ values, areas or c.v.
No evidence mean distance changed | A1If | Dep 1.96; If their 1.761 If $H_1$: $\mu > 2850$ and c.f. 1.645, max B0M1A1M1A0
| | [5] | (c.v. for 1 tail test 2864.94)3 Following a change in flight schedules, an airline pilot wished to test whether the mean distance that he flies in a week has changed. He noted the distances, $x \mathrm {~km}$, that he flew in 50 randomly chosen weeks and summarised the results as follows.
$$n = 50 \quad \Sigma x = 143300 \quad \Sigma x ^ { 2 } = 410900000$$
(i) Calculate unbiased estimates of the population mean and variance.\\
(ii) In the past, the mean distance that he flew in a week was 2850 km . Test, at the $5 \%$ significance level, whether the mean distance has changed.
\hfill \mbox{\textit{CAIE S2 2013 Q3 [8]}}