| Exam Board | CAIE |
|---|---|
| Module | S2 (Statistics 2) |
| Year | 2020 |
| Session | March |
| Marks | 8 |
| Paper | Download PDF ↗ |
| Mark scheme | Download PDF ↗ |
| Topic | Z-tests (known variance) |
| Type | One-tail z-test (lower tail) |
| Difficulty | Standard +0.3 This is a straightforward one-tail hypothesis test with given summary statistics. Part (a) requires standard formulas for unbiased estimates (sample mean and s²), and part (b) is a routine z-test application with clear hypotheses and significance level. The calculations are direct with no conceptual challenges beyond standard S2 content. |
| Spec | 5.05b Unbiased estimates: of population mean and variance5.05c Hypothesis test: normal distribution for population mean |
| Answer | Marks | Guidance |
|---|---|---|
| Answer | Mark | Guidance |
| \(\hat{\mu} = 37.6\) or \(\frac{1504}{40}\) or \(\frac{188}{5}\) | B1 | |
| \(\hat{\sigma^2} = \frac{40}{39}\left[\frac{57760}{40} - 37.6^2\right] = 31.0154 = \frac{2016}{65}\) | M1 | Correct substitution in any correct formula \(\frac{1}{39}\left[57760 - \frac{1504^2}{40}\right]\) |
| \(= 31.(0)\) (3 sf) | A1 | Accept \(\frac{2016}{65}\) or \(31\frac{1}{65}\) |
| Answer | Marks | Guidance |
|---|---|---|
| Answer | Mark | Guidance |
| \(H_0\): Pop mean (or \(\mu\)) \(= 39.2\); \(H_1\): Pop mean (or \(\mu\)) \(< 39.2\) | B1 | Both. Not just 'mean' |
| \(\frac{37.6 - 39.2}{\sqrt{31.0154}/\sqrt{40}}\) | M1 | Allow use of biased variance (30.2), must have \(\sqrt{40}\) |
| \(= -1.817\) | A1 | SC: FT use of biased \(= -1.840\) for A1 |
| \('1.817' > 1.645\) OE | M1 | Valid comparison of *their* 1.817 with 1.645 or valid area comparison \(0.0346 < 0.05\) OE |
| There is evidence that mean time has decreased | A1FT | FT *their* 1.817; in context, not definite, no contradictions. SC: For 2 tail test: \(H_1: \mu \neq 39.2\) and comp 1.96, max B0M1A1M1A0 (no FT for final mark) |
## Question 3(a):
| Answer | Mark | Guidance |
|--------|------|----------|
| $\hat{\mu} = 37.6$ or $\frac{1504}{40}$ or $\frac{188}{5}$ | B1 | |
| $\hat{\sigma^2} = \frac{40}{39}\left[\frac{57760}{40} - 37.6^2\right] = 31.0154 = \frac{2016}{65}$ | M1 | Correct substitution in any correct formula $\frac{1}{39}\left[57760 - \frac{1504^2}{40}\right]$ |
| $= 31.(0)$ (3 sf) | A1 | Accept $\frac{2016}{65}$ or $31\frac{1}{65}$ |
## Question 3(b):
| Answer | Mark | Guidance |
|--------|------|----------|
| $H_0$: Pop mean (or $\mu$) $= 39.2$; $H_1$: Pop mean (or $\mu$) $< 39.2$ | B1 | Both. Not just 'mean' |
| $\frac{37.6 - 39.2}{\sqrt{31.0154}/\sqrt{40}}$ | M1 | Allow use of biased variance (30.2), must have $\sqrt{40}$ |
| $= -1.817$ | A1 | SC: FT use of biased $= -1.840$ for A1 |
| $'1.817' > 1.645$ OE | M1 | Valid comparison of *their* 1.817 with 1.645 or valid area comparison $0.0346 < 0.05$ OE |
| There is evidence that mean time has decreased | A1FT | FT *their* 1.817; in context, not definite, no contradictions. SC: For 2 tail test: $H_1: \mu \neq 39.2$ and comp 1.96, max B0M1A1M1A0 (no FT for final mark) |3 In the past, the mean time taken by Freda for a particular daily journey was 39.2 minutes. Following the introduction of a one-way system, Freda wishes to test whether the mean time for the journey has decreased. She notes the times, $t$ minutes, for 40 randomly chosen journeys and summarises the results as follows.
$$n = 40 \quad \Sigma t = 1504 \quad \Sigma t ^ { 2 } = 57760$$
\begin{enumerate}[label=(\alph*)]
\item Calculate unbiased estimates of the population mean and variance of the new journey time.
\item Test, at the $5 \%$ significance level, whether the population mean time has decreased.
\end{enumerate}
\hfill \mbox{\textit{CAIE S2 2020 Q3 [8]}}