| Exam Board | CAIE |
|---|---|
| Module | S2 (Statistics 2) |
| Year | 2018 |
| Session | November |
| Marks | 12 |
| 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 | Challenging +1.2 This is a standard hypothesis testing question with Type II error calculation. Part (i) involves routine one-sample z-test mechanics with given summary statistics. Part (ii) requires understanding of Type II errors and power, but follows a standard formula once the critical region is established. The calculations are straightforward with no conceptual surprises, making it moderately above average difficulty for A-level statistics. |
| Spec | 5.05b Unbiased estimates: of population mean and variance5.05c Hypothesis test: normal distribution for population mean |
| Answer | Marks | Guidance |
|---|---|---|
| Answer | Mark | Guidance |
| \(H_0: \mu = 51 \quad H_1: \mu < 51\) | B1 | Or popn mean… |
| \(\bar{x} = \frac{7480}{150} = 49.8667 = 49.9\) | B1 | |
| \(s^2 = \frac{150}{149}\left(\frac{380000}{150} - \left(\frac{748}{15}\right)^2\right) = 46.9620 = 47.0\) or \(s = 6.85\) | M1 | Correct subst in \(s^2\) or \(\sqrt{s^2}\) formula. Biased var scores M0 |
| \(\frac{49.8667 - 51}{\sqrt{\frac{46.962}{150}}}\) allow \(\frac{49.9 - 51}{\sqrt{\frac{47}{150}}}\) | M1 | Allow 49.8667 to 49.9 in numerator. Need sqrt 150 |
| \(= (-)2.025 = (-)1.965\) | A1 | Accept 2.02 or 2.03; Accept \(-2.0264\) to \(-1.9651\) provided correct working |
| comp \(z = 1.96\) | M1 | or comp \(1 - \Phi(2.025)\) with 0.025 |
| There is evidence that \(\mu < 51\) | A1ft | No contradictions; biased var B1B1M0M1A0M1A1ft (max 5/7); accept cv method: \(x_{\text{crit}} = 49.9028\) M1A1; \(49867 < 49.9\ldots\) M1A1 |
| Total: 7 |
| Answer | Marks | Guidance |
|---|---|---|
| Answer | Mark | Guidance |
| \(\frac{\bar{x}-51}{\frac{6.856}{\sqrt{150}}} = -1.96\) | M1 | Need 51 and sqrt 150 and correct form |
| \(\bar{x} = 51 - 1.097 = 49.9\); Rejection region is \(\bar{x} < 49.9\) | A1 | This may have been found in part (i) |
| \(\frac{49.9 - 49}{\frac{6.856}{\sqrt{150}}} \quad (= 1.608 \text{ to } 1.614)\) | M1 | Need 49 and sqrt 150 and correct form |
| \(P(\bar{x} > 49.9 \mid \mu = 49) = 1 - \Phi(\text{'1.608'})\) | M1 | |
| \(P(\text{Type II error}) = 0.0539\) | A1 | Allow 0.0533 to 0.0539 |
| Total: 5 |
# Question 7(i):
| Answer | Mark | Guidance |
|--------|------|----------|
| $H_0: \mu = 51 \quad H_1: \mu < 51$ | **B1** | Or popn mean… |
| $\bar{x} = \frac{7480}{150} = 49.8667 = 49.9$ | **B1** | |
| $s^2 = \frac{150}{149}\left(\frac{380000}{150} - \left(\frac{748}{15}\right)^2\right) = 46.9620 = 47.0$ or $s = 6.85$ | **M1** | Correct subst in $s^2$ or $\sqrt{s^2}$ formula. Biased var scores M0 |
| $\frac{49.8667 - 51}{\sqrt{\frac{46.962}{150}}}$ allow $\frac{49.9 - 51}{\sqrt{\frac{47}{150}}}$ | **M1** | Allow 49.8667 to 49.9 in numerator. Need sqrt 150 |
| $= (-)2.025 = (-)1.965$ | **A1** | Accept 2.02 or 2.03; Accept $-2.0264$ to $-1.9651$ provided correct working |
| comp $z = 1.96$ | **M1** | or comp $1 - \Phi(2.025)$ with 0.025 |
| There is evidence that $\mu < 51$ | **A1ft** | No contradictions; biased var B1B1M0M1A0M1A1ft (max 5/7); accept cv method: $x_{\text{crit}} = 49.9028$ M1A1; $49867 < 49.9\ldots$ M1A1 |
| **Total: 7** | | |
# Question 7(ii):
| Answer | Mark | Guidance |
|--------|------|----------|
| $\frac{\bar{x}-51}{\frac{6.856}{\sqrt{150}}} = -1.96$ | **M1** | Need 51 and sqrt 150 and correct form |
| $\bar{x} = 51 - 1.097 = 49.9$; Rejection region is $\bar{x} < 49.9$ | **A1** | This may have been found in part (i) |
| $\frac{49.9 - 49}{\frac{6.856}{\sqrt{150}}} \quad (= 1.608 \text{ to } 1.614)$ | **M1** | Need 49 and sqrt 150 and correct form |
| $P(\bar{x} > 49.9 \mid \mu = 49) = 1 - \Phi(\text{'1.608'})$ | **M1** | |
| $P(\text{Type II error}) = 0.0539$ | **A1** | Allow 0.0533 to 0.0539 |
| **Total: 5** | | |7 A mill owner claims that the mean mass of sacks of flour produced at his mill is 51 kg . A quality control officer suspects that the mean mass is actually less than 51 kg . In order to test the owner's claim she finds the mass, $x \mathrm {~kg}$, of each of a random sample of 150 sacks and her results are summarised as follows.
$$n = 150 \quad \Sigma x = 7480 \quad \Sigma x ^ { 2 } = 380000$$
(i) Carry out the test at the $2.5 \%$ significance level.\\
You may now assume that the population standard deviation of the masses of sacks of flour is 6.856 kg . The quality control officer weighs another random sample of 150 sacks and carries out another test at the 2.5\% significance level.\\
(ii) Given that the population mean mass is 49 kg , find the probability of a Type II error.\\
If you use the following lined page to complete the answer(s) to any question(s), the question number(s) must be clearly shown.\\
\hfill \mbox{\textit{CAIE S2 2018 Q7 [12]}}