| Exam Board | CAIE |
|---|---|
| Module | S2 (Statistics 2) |
| Year | 2019 |
| Session | June |
| 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 z-test with standard steps: calculate sample statistics, set up hypotheses, compute test statistic, and compare to critical value. The calculations are routine (finding mean and variance from summations, then applying the z-test formula). Slightly above average difficulty due to the two-part structure and requiring knowledge of hypothesis testing procedure, but all steps are standard A-level statistics techniques with no novel problem-solving required. |
| Spec | 5.05b Unbiased estimates: of population mean and variance5.05c Hypothesis test: normal distribution for population mean |
| Answer | Marks | Guidance |
|---|---|---|
| Answer | Marks | Guidance |
| \(\text{Est}(\mu) = 1.85\) | B1 | |
| \(\text{Est}(\sigma^2) = \frac{50}{49}\left(\frac{175.25}{50} - 1.85^2\right)\) | M1 | Allow \(\sqrt{\frac{50}{49}\left(\frac{175.25}{150} - 1.85^2\right)}\) or 0.0290 for M1 |
| \(= 0.0842\) (3 sf) or \(\frac{33}{392}\) | A1 | Cao. If \(\frac{50}{49}\) omitted (giving var = 0.0825 or sd = 0.287) M0A0 |
| Answer | Marks | Guidance |
|---|---|---|
| Answer | Marks | Guidance |
| \(H_0\): Pop mean time \(= 1.9\) (h); \(H_1\): Pop mean time \(< 1.9\) (h) | B1 | Allow '\(\mu\)' but not just 'mean' |
| \(\pm\frac{1.85 - 1.9}{\sqrt{\frac{0.0842}{50}}}\) | M1 | \(\pm\frac{1.85-1.9}{\frac{0.290}{\sqrt{50}}}\) Accept totals method \((92.5-95)/\sqrt{4.21}\) |
| \(= -1.22\) | A1 | \(= -1.22\) |
| comp \(z = -1.645\) | M1 | Or other valid comparison 0.888 or \(0.889 < 0.95\) OR 0.111 or \(0.112 > 0.05\) |
| No evidence that mean time \(< 1.9\) h | A1 | FT their z. Correct conclusion. No contradictions. If \(\frac{50}{49}\) not used in (i): var = 0.8225, sd = 0.907, cr = 1.17 can score all marks in (ii). Note: 2 tail test can score B0 M1 A1 M1 (comparison with 1.96) A0 (no ft) max3/5 |
## Question 3(i):
| Answer | Marks | Guidance |
|--------|-------|----------|
| $\text{Est}(\mu) = 1.85$ | B1 | |
| $\text{Est}(\sigma^2) = \frac{50}{49}\left(\frac{175.25}{50} - 1.85^2\right)$ | M1 | Allow $\sqrt{\frac{50}{49}\left(\frac{175.25}{150} - 1.85^2\right)}$ or 0.0290 for M1 |
| $= 0.0842$ (3 sf) or $\frac{33}{392}$ | A1 | Cao. If $\frac{50}{49}$ omitted (giving var = 0.0825 or sd = 0.287) M0A0 |
## Question 3(ii):
| Answer | Marks | Guidance |
|--------|-------|----------|
| $H_0$: Pop mean time $= 1.9$ (h); $H_1$: Pop mean time $< 1.9$ (h) | B1 | Allow '$\mu$' but not just 'mean' |
| $\pm\frac{1.85 - 1.9}{\sqrt{\frac{0.0842}{50}}}$ | M1 | $\pm\frac{1.85-1.9}{\frac{0.290}{\sqrt{50}}}$ Accept totals method $(92.5-95)/\sqrt{4.21}$ |
| $= -1.22$ | A1 | $= -1.22$ |
| comp $z = -1.645$ | M1 | Or other valid comparison 0.888 or $0.889 < 0.95$ OR 0.111 or $0.112 > 0.05$ |
| No evidence that mean time $< 1.9$ h | A1 | FT their z. Correct conclusion. No contradictions. If $\frac{50}{49}$ not used in (i): var = 0.8225, sd = 0.907, cr = 1.17 can score all marks in (ii). Note: 2 tail test can score B0 M1 A1 M1 (comparison with 1.96) A0 (no ft) max3/5 |3 It is claimed that, on average, a particular train journey takes less than 1.9 hours. The times, $t$ hours, taken for this journey on a random sample of 50 days were recorded. The results are summarised below.
$$n = 50 \quad \Sigma t = 92.5 \quad \Sigma t ^ { 2 } = 175.25$$
(i) Calculate unbiased estimates of the population mean and variance.\\
(ii) Test the claim at the $5 \%$ significance level.\\
\hfill \mbox{\textit{CAIE S2 2019 Q3 [8]}}