| Exam Board | CAIE |
|---|---|
| Module | S2 (Statistics 2) |
| Year | 2004 |
| Session | June |
| Marks | 8 |
| Paper | Download PDF ↗ |
| Mark scheme | Download PDF ↗ |
| Topic | Type I/II errors and power of test |
| Type | Simultaneous critical region and Type II error |
| Difficulty | Standard +0.3 This is a standard hypothesis testing question requiring calculation of a critical region using normal distribution (part i) and Type II error probability (part ii). While it involves multiple steps and understanding of hypothesis testing concepts, the techniques are routine for S2 level with no novel problem-solving required—students follow standard procedures with given formulas. |
| Spec | 2.05a Hypothesis testing language: null, alternative, p-value, significance5.05c Hypothesis test: normal distribution for population mean |
| Answer | Marks | Guidance |
|---|---|---|
| (i) \(\frac{c-54}{3.1/\sqrt{10}} = -1.282\) | B1 M1 | For \(\pm\) or \(-\) 1.282 seen; For equality/inequality with their \(z\) (\(\pm\)) (must have used tables), no \(\sqrt{10}\) needed (c can be numerical) |
| \(c = 54 - 1.282 \times \frac{3.1}{\sqrt{10}} = 52.74\) | A1 | For correct expression (c can be numerical, but signs must be consistent) |
| 4 | For correct GIVEN answer. No errors seen. | |
| (ii) \(P(x > 52.74) = 1 - \Phi\left(\frac{52.74-51.5}{3.1/\sqrt{10}}\right) = 1 - \Phi(1.265) = 1 - 0.8971 = 0.103\) or \(0.102\) | B1 M1 A1 A1 | For identifying the outcome for a type II error; For standardising, no \(\sqrt{10}\) needed; For \(\pm\) 1.265 (accept 1.26-1.27); For correct answer |
| 4 |
**(i)** $\frac{c-54}{3.1/\sqrt{10}} = -1.282$ | B1 M1 | For $\pm$ or $-$ 1.282 seen; For equality/inequality with their $z$ ($\pm$) (must have used tables), no $\sqrt{10}$ needed (c can be numerical)
$c = 54 - 1.282 \times \frac{3.1}{\sqrt{10}} = 52.74$ | A1 | For correct expression (c can be numerical, but signs must be consistent)
| | **4** | For correct GIVEN answer. No errors seen.
**(ii)** $P(x > 52.74) = 1 - \Phi\left(\frac{52.74-51.5}{3.1/\sqrt{10}}\right) = 1 - \Phi(1.265) = 1 - 0.8971 = 0.103$ or $0.102$ | B1 M1 A1 A1 | For identifying the outcome for a type II error; For standardising, no $\sqrt{10}$ needed; For $\pm$ 1.265 (accept 1.26-1.27); For correct answer
| | **4** | |5 The lectures in a mathematics department are scheduled to last 54 minutes, and the times of individual lectures may be assumed to have a normal distribution with mean $\mu$ minutes and standard deviation 3.1 minutes. One of the students commented that, on average, the lectures seemed too short. To investigate this, the times for a random sample of 10 lectures were used to test the null hypothesis $\mu = 54$ against the alternative hypothesis $\mu < 54$ at the $10 \%$ significance level.\\
(i) Show that the null hypothesis is rejected in favour of the alternative hypothesis if $\bar { x } < 52.74$, where $\bar { x }$ minutes is the sample mean.\\
(ii) Find the probability of a Type II error given that the actual mean length of lectures is 51.5 minutes.
\hfill \mbox{\textit{CAIE S2 2004 Q5 [8]}}