| Exam Board | CAIE |
|---|---|
| Module | S2 (Statistics 2) |
| Year | 2017 |
| Session | November |
| 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 summary statistics provided. Students must calculate sample mean and standard deviation, perform a standard hypothesis test procedure, and demonstrate understanding of when CLT is needed. The calculations are routine and the conceptual demand is modest—slightly easier than average since the test structure is standard and no unusual insight is 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 |
| \(\bar{m} = \dfrac{98.2}{100} = 0.982\) | B1 | Accept either |
| \(s = \sqrt{\dfrac{100}{99}} \times \sqrt{\dfrac{104.52}{100} - 0.982^2}\) \((= 0.28582)\) or \(\text{var} = 0.08169\) | M1 | |
| \(H_0\): Pop mean mass \(= 1.01\); \(H_1\): Pop mean mass \(< 1.01\) | B1 | not just 'mean', but allow just '\(\mu\)' |
| \(\pm\dfrac{0.982 - 1.01}{\frac{0.28582}{\sqrt{100}}}\) | M1 | \(\pm\dfrac{0.982 - 1.01}{\frac{0.284387}{\sqrt{100}}}\) |
| \(= -0.980\) (3 sf) accept \(\pm\) | A1 | \(= -0.985\) (3 sfs) accept \(\pm\) |
| Comp with \(z = -1.645\) (or areas \(0.1635 > 0.05\)) | M1 | Valid comparison of \(z\)'s or area's |
| No evidence that (mean) mass is less than 1.01 | A1 FT | Correct conclusion FT their \(z\) |
| Total: 7 |
| Answer | Marks | Guidance |
|---|---|---|
| Answer | Marks | Guidance |
| Distr of \(X\) normal (so distr of \(\bar{X}\) normal); Must state or imply No | B1 | \(X\)/parent population |
| Total: 1 |
## Question 3:
**Part 3(i):**
| Answer | Marks | Guidance |
|--------|-------|----------|
| $\bar{m} = \dfrac{98.2}{100} = 0.982$ | B1 | Accept either |
| $s = \sqrt{\dfrac{100}{99}} \times \sqrt{\dfrac{104.52}{100} - 0.982^2}$ $(= 0.28582)$ or $\text{var} = 0.08169$ | M1 | |
| $H_0$: Pop mean mass $= 1.01$; $H_1$: Pop mean mass $< 1.01$ | B1 | not just 'mean', but allow just '$\mu$' |
| $\pm\dfrac{0.982 - 1.01}{\frac{0.28582}{\sqrt{100}}}$ | M1 | $\pm\dfrac{0.982 - 1.01}{\frac{0.284387}{\sqrt{100}}}$ |
| $= -0.980$ (3 sf) accept $\pm$ | A1 | $= -0.985$ (3 sfs) accept $\pm$ |
| Comp with $z = -1.645$ (or areas $0.1635 > 0.05$) | M1 | Valid comparison of $z$'s or area's |
| No evidence that (mean) mass is less than 1.01 | A1 FT | Correct conclusion FT their $z$ |
| | **Total: 7** | |
**Part 3(ii):**
| Answer | Marks | Guidance |
|--------|-------|----------|
| Distr of $X$ normal (so distr of $\bar{X}$ normal); Must state or imply No | B1 | $X$/parent population |
| | **Total: 1** | |
---3 The masses, $m \mathrm {~kg}$, of packets of flour are normally distributed. The mean mass is supposed to be 1.01 kg . A quality control officer measures the masses of a random sample of 100 packets. The results are summarised below.
$$n = 100 \quad \Sigma m = 98.2 \quad \Sigma m ^ { 2 } = 104.52$$
(i) Test at the $5 \%$ significance level whether the population mean mass is less than 1.01 kg .\\
(ii) Explain whether it was necessary to use the Central Limit theorem in your answer to part (i).\\
\hfill \mbox{\textit{CAIE S2 2017 Q3 [8]}}