| Exam Board | CAIE |
|---|---|
| Module | S2 (Statistics 2) |
| Year | 2017 |
| 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 a Type II error calculation. Part (i) requires routine application of a z-test for a mean with known variance (estimated from sample), while part (ii) involves calculating the probability of Type II error given the true mean—a mechanical calculation once the critical region is established. The multi-step nature and the Type II error component elevate it slightly above average difficulty, but it follows a standard template without requiring novel insight. |
| Spec | 2.05a Hypothesis testing language: null, alternative, p-value, significance5.05c Hypothesis test: normal distribution for population mean |
| Answer | Marks | Guidance |
|---|---|---|
| Answer | Mark | Guidance |
| \(\bar{x} = 27/150 (= 0.18)\) | B1 | |
| \(s = \sqrt{\frac{150}{149}} \times \sqrt{\frac{5.01}{150} - 0.18^2}\) or variance \((= 0.031729)\) \((\text{var} = 3/2980 = 0.0010067)\) | M1 | or \(\text{var} = \frac{1}{149}(5.01 - 27.0^2/150)\) |
| \(H_0\): Pop mean \(= 0.185\) \(H_1\): Pop mean \(< 0.185\) | B1 | allow just '\(\mu\)' |
| \(\frac{0.18 - 0.185}{\frac{0.031729}{\sqrt{150}}}\) | M1 | standardising, need \(\sqrt{150}\) |
| \(= (-)\, 1.930\) (3 sfs) or \(1.93\) | A1 | |
| Compare with \(z = (-)\, 2.326\) | M1 | consistent signs or using probs \(0.0268 > 0.01\) or \(0.9732 < 0.99\) or using \(x_{\text{crit}}\ 0.18 > 0.17897\) |
| There is no evidence (at \(1\%\) level) that concentration with drug is less than without drug | A1 FT | conclusion FT, no contradictions |
| Total: 7 |
| Answer | Marks | Guidance |
|---|---|---|
| Answer | Mark | Guidance |
| \(\frac{cv - 0.185}{\frac{0.031729}{\sqrt{150}}} = -2.326\) | M1 | must use \(0.185\) and \(\sqrt{150}\) |
| \(= 0.17897\) or \(0.179\) | A1 | acceptance region (for \(H_0\)) is \(> 0.179\) |
| \(\frac{0.17897 - 0.175}{\frac{0.031729}{\sqrt{150}}}\) \((= 1.534)\) | M1 | must use \(0.175\) and \(\sqrt{150}\) |
| \(1 - \phi(\text{``}1.534\text{''})\) | M1 | indep mark |
| \(= 0.0625\) (3 sf) | A1 | Accept \(0.0610\) to \(0.0628\) |
| Total: 5 |
# Question 8(i):
| Answer | Mark | Guidance |
|--------|------|----------|
| $\bar{x} = 27/150 (= 0.18)$ | B1 | |
| $s = \sqrt{\frac{150}{149}} \times \sqrt{\frac{5.01}{150} - 0.18^2}$ or variance $(= 0.031729)$ $(\text{var} = 3/2980 = 0.0010067)$ | M1 | or $\text{var} = \frac{1}{149}(5.01 - 27.0^2/150)$ |
| $H_0$: Pop mean $= 0.185$ $H_1$: Pop mean $< 0.185$ | B1 | allow just '$\mu$' |
| $\frac{0.18 - 0.185}{\frac{0.031729}{\sqrt{150}}}$ | M1 | standardising, need $\sqrt{150}$ |
| $= (-)\, 1.930$ (3 sfs) or $1.93$ | A1 | |
| Compare with $z = (-)\, 2.326$ | M1 | consistent signs or using probs $0.0268 > 0.01$ or $0.9732 < 0.99$ or using $x_{\text{crit}}\ 0.18 > 0.17897$ |
| There is no evidence (at $1\%$ level) that concentration with drug is less than without drug | A1 FT | conclusion FT, no contradictions |
| **Total: 7** | | |
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# Question 8(ii):
| Answer | Mark | Guidance |
|--------|------|----------|
| $\frac{cv - 0.185}{\frac{0.031729}{\sqrt{150}}} = -2.326$ | M1 | must use $0.185$ and $\sqrt{150}$ |
| $= 0.17897$ or $0.179$ | A1 | acceptance region (for $H_0$) is $> 0.179$ |
| $\frac{0.17897 - 0.175}{\frac{0.031729}{\sqrt{150}}}$ $(= 1.534)$ | M1 | must use $0.175$ and $\sqrt{150}$ |
| $1 - \phi(\text{``}1.534\text{''})$ | M1 | indep mark |
| $= 0.0625$ (3 sf) | A1 | Accept $0.0610$ to $0.0628$ |
| **Total: 5** | | |8 In order to test the effect of a drug, a researcher monitors the concentration, $X$, of a certain protein in the blood stream of patients. For patients who are not taking the drug the mean value of $X$ is 0.185 . A random sample of 150 patients taking the drug was selected and the values of $X$ were found. The results are summarised below.
$$n = 150 \quad \Sigma x = 27.0 \quad \Sigma x ^ { 2 } = 5.01$$
The researcher wishes to test at the $1 \%$ significance level whether the mean concentration of the protein in the blood stream of patients taking the drug is less than 0.185 .\\
(i) Carry out the test.\\
(ii) Given that, in fact, the mean concentration for patients taking the drug is 0.175 , find the probability of a Type II error occurring in the test.\\
\hfill \mbox{\textit{CAIE S2 2017 Q8 [12]}}