| Exam Board | CAIE |
|---|---|
| Module | S2 (Statistics 2) |
| Year | 2017 |
| Session | June |
| 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-tail hypothesis test with standard calculations. Part (i) requires routine formulas for unbiased estimates (sample mean and variance with n-1 denominator). Part (ii) is a textbook z-test application with given significance level and clear hypotheses. The calculations are mechanical with no conceptual challenges or novel problem-solving required, making it slightly easier than average. |
| Spec | 5.05b Unbiased estimates: of population mean and variance5.05c Hypothesis test: normal distribution for population mean |
| Answer | Marks | Guidance |
|---|---|---|
| Answer | Marks | Guidance |
| \(\bar{x} = 6.7/200\ (= 67/2000 = 0.0335)\) | B1 | |
| \(s^2 = \frac{200}{199}\times\!\left(\frac{0.2312}{200} - \text{"0.0335"}^2\right)\) | M1 | \(s^2 = \frac{0.2312}{200} - 0.0335^2\) M0 |
| \(= 0.0000339(2) = 27/796000\) | A1 | \(= 0.00003375\) A0 |
| Answer | Marks | Guidance |
|---|---|---|
| Answer | Marks | Guidance |
| \(H_0\): Pop mean level \(= 0.034\); \(H_1\): Pop mean level \(\neq 0.034\) | B1 | Not just "mean", but allow just "\(\mu\)" |
| \(\frac{\text{"0.30335"} - 0.034}{\frac{\sqrt{\text{"0.00003392"}}}{\sqrt{200}}}\) | M1 | Must have \(\sqrt{200}\); \(\frac{0.0335-0.034}{\frac{\sqrt{\text{"0.00003375"}}}{\sqrt{200}}}\) M1 |
| \(= -1.21(4)\) (3 sfs) \((-1.22 \leftrightarrow -1.21)\) | A1 | \(= -1.217\) (3 sfs) A1 |
| Compare with \(z = -1.645\) (or \(0.1124 > 0.05\)) | M1 | \(0.112 > 0.05\); valid comparison \(z\) or areas |
| No evidence that (mean) pollutant level has changed, accept \(H_0\) (if correctly defined) | A1FT | Correct conclusion no contradictions. SR: One tail test: B0, M1A1 as normal, M1 (comparison with 1.282 consistent signs) A0 |
## Question 4(i):
| Answer | Marks | Guidance |
|--------|-------|----------|
| $\bar{x} = 6.7/200\ (= 67/2000 = 0.0335)$ | B1 | |
| $s^2 = \frac{200}{199}\times\!\left(\frac{0.2312}{200} - \text{"0.0335"}^2\right)$ | M1 | $s^2 = \frac{0.2312}{200} - 0.0335^2$ **M0** |
| $= 0.0000339(2) = 27/796000$ | A1 | $= 0.00003375$ **A0** |
---
## Question 4(ii):
| Answer | Marks | Guidance |
|--------|-------|----------|
| $H_0$: Pop mean level $= 0.034$; $H_1$: Pop mean level $\neq 0.034$ | B1 | Not just "mean", but allow just "$\mu$" |
| $\frac{\text{"0.30335"} - 0.034}{\frac{\sqrt{\text{"0.00003392"}}}{\sqrt{200}}}$ | M1 | Must have $\sqrt{200}$; $\frac{0.0335-0.034}{\frac{\sqrt{\text{"0.00003375"}}}{\sqrt{200}}}$ **M1** |
| $= -1.21(4)$ (3 sfs) $(-1.22 \leftrightarrow -1.21)$ | A1 | $= -1.217$ (3 sfs) **A1** |
| Compare with $z = -1.645$ (or $0.1124 > 0.05$) | M1 | $0.112 > 0.05$; valid comparison $z$ or areas |
| No evidence that (mean) pollutant level has changed, accept $H_0$ (if correctly defined) | A1FT | Correct conclusion no contradictions. SR: One tail test: **B0**, **M1A1** as normal, **M1** (comparison with 1.282 consistent signs) **A0** |
---4 Last year the mean level of a certain pollutant in a river was found to be 0.034 grams per millilitre. This year the levels of pollutant, $X$ grams per millilitre, were measured at a random sample of 200 locations in the river. The results are summarised below.
$$n = 200 \quad \Sigma x = 6.7 \quad \Sigma x ^ { 2 } = 0.2312$$
(i) Calculate unbiased estimates of the population mean and variance.\\
(ii) Test, at the $10 \%$ significance level, whether the mean level of pollutant has changed.\\
\hfill \mbox{\textit{CAIE S2 2017 Q4 [8]}}