| Exam Board | CAIE |
|---|---|
| Module | S2 (Statistics 2) |
| Year | 2015 |
| Session | November |
| Marks | 9 |
| Paper | Download PDF ↗ |
| Mark scheme | Download PDF ↗ |
| Topic | Type I/II errors and power of test |
| Type | Hypothesis test then Type II error probability |
| Difficulty | Standard +0.8 This is a two-part hypothesis testing question requiring (i) a standard one-tailed z-test with known population SD, and (ii) calculation of Type II error probability (beta) which requires finding P(accept H₀|H₁ true). Part (i) is routine A-level statistics, but part (ii) requires understanding of power functions and working backwards from the critical region, which is more conceptually demanding than typical S2 questions. |
| Spec | 2.05a Hypothesis testing language: null, alternative, p-value, significance2.05e Hypothesis test for normal mean: known variance |
| Answer | Marks | Guidance |
|---|---|---|
| Answer | Mark | Guidance |
| \(H_0: \mu = 2.60\); \(H_1: \mu > 2.60\) | B1 | Allow pop mean, not just 'mean' |
| \(\pm\frac{2.64 - 2.6}{0.2 \div \sqrt{75}}\) | M1 | |
| \(= \pm 1.732\) | A1 | Accept \(\pm 1.73\) (3 sf) |
| \(\text{'1.732'} > 1.645\); Reject \(H_0\). There is evidence that \(\mu\) has increased | B1\(\checkmark\) [4] | Valid comparison with 1.645 (or \(0.0416 < 0.05\)) and correct conclusion \(\checkmark\) their 1.732, no contradictions; (or CV method \(x_{crit} = 2.638\) M1A1, comp \(2.64 > 2.638\) and conclusion B1\(\checkmark\)); SR two tail test, using 1.96 (or 0.025) can score B0M1A1B1ft max 3/4 |
| Answer | Marks | Guidance |
|---|---|---|
| Answer/Working | Mark | Guidance |
| \(\frac{x - 2.6}{0.2 \div \sqrt{75}} = 1.645 \quad (x = 2.638)\) | M1 | |
| \(\pm \frac{2.638' - 2.68}{0.2 \div \sqrt{75}}\) | M1 | for standardising with their "2.638" using 2.68 |
| \(= \pm 1.819\) | A1 | accept 1.82 (3 sf) |
| \(\Phi(-1.819') = 1 - \Phi(1.819')\) | M1 | indep M mark, calculate correct area/prob consistent with their working |
| \(= 0.0345\) or \(0.0344\) | A1 | [5] |
# Question 6:
## Part (i)
| Answer | Mark | Guidance |
|--------|------|----------|
| $H_0: \mu = 2.60$; $H_1: \mu > 2.60$ | B1 | Allow pop mean, not just 'mean' |
| $\pm\frac{2.64 - 2.6}{0.2 \div \sqrt{75}}$ | M1 | |
| $= \pm 1.732$ | A1 | Accept $\pm 1.73$ (3 sf) |
| $\text{'1.732'} > 1.645$; Reject $H_0$. There is evidence that $\mu$ has increased | B1$\checkmark$ [4] | Valid comparison with 1.645 (or $0.0416 < 0.05$) and correct conclusion $\checkmark$ their 1.732, no contradictions; (or CV method $x_{crit} = 2.638$ M1A1, comp $2.64 > 2.638$ and conclusion B1$\checkmark$); SR two tail test, using 1.96 (or 0.025) can score B0M1A1B1ft max 3/4 |
# Question 6(ii):
| Answer/Working | Mark | Guidance |
|---|---|---|
| $\frac{x - 2.6}{0.2 \div \sqrt{75}} = 1.645 \quad (x = 2.638)$ | M1 | |
| $\pm \frac{2.638' - 2.68}{0.2 \div \sqrt{75}}$ | M1 | for standardising with their "2.638" using 2.68 |
| $= \pm 1.819$ | A1 | accept 1.82 (3 sf) |
| $\Phi(-1.819') = 1 - \Phi(1.819')$ | M1 | indep M mark, calculate correct area/prob consistent with their working |
| $= 0.0345$ or $0.0344$ | A1 | [5] |
**Total: [9]**
---6 Parcels arriving at a certain office have weights $W \mathrm {~kg}$, where the random variable $W$ has mean $\mu$ and standard deviation 0.2 . The value of $\mu$ used to be 2.60 , but there is a suspicion that this may no longer be true. In order to test at the 5\% significance level whether the value of $\mu$ has increased, a random sample of 75 parcels is chosen. You may assume that the standard deviation of $W$ is unchanged.\\
(i) The mean weight of the 75 parcels is found to be 2.64 kg . Carry out the test.\\
(ii) Later another test of the same hypotheses at the $5 \%$ significance level, with another random sample of 75 parcels, is carried out. Given that the value of $\mu$ is now 2.68 , calculate the probability of a Type II error.
\hfill \mbox{\textit{CAIE S2 2015 Q6 [9]}}