| Exam Board | CAIE |
|---|---|
| Module | S2 (Statistics 2) |
| Year | 2011 |
| Session | November |
| Marks | 7 |
| 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 +0.8 This question requires understanding of hypothesis testing mechanics including critical region calculation using normal distribution, and Type II error probability computation. While the individual steps are standard S2 content, part (ii) requires careful manipulation of the sampling distribution under the alternative hypothesis, which goes beyond routine application and demands solid conceptual understanding of both error types. |
| Spec | 2.05a Hypothesis testing language: null, alternative, p-value, significance2.05c Significance levels: one-tail and two-tail5.05c Hypothesis test: normal distribution for population mean |
| Answer | Marks | Guidance |
|---|---|---|
| Answer/Working | Mark | Guidance |
| \(\pm 1.645\) used | B1 | |
| \(\dfrac{\bar{x} - 22}{\dfrac{3.5}{\sqrt{12}}} > 1.645\) | M1 | |
| \(\bar{x} > 23.66(20)\); \(\bar{x} > 23.7\) AG | A1 [3] | Accept \(=\); standardising using \(23.7\) scores M1A0 or \(\bar{x} = 23.66(20)\) |
| Answer | Marks | Guidance |
|---|---|---|
| Answer/Working | Mark | Guidance |
| \(P(\bar{x} < 23.7 \mid \mu = 25.8)\) | M1 | For attempt type II error and standardising |
| \(\dfrac{23.662 - 25.8}{\dfrac{3.5}{\sqrt{12}}} = -2.116\) | A1 | \(\dfrac{23.7 - 25.8}{\dfrac{3.5}{\sqrt{12}}} = -2.078\) |
| \(\Phi(-2.116) = 1 - \Phi(2.116)\) \((= 1 - 0.9828)\) | M1 | \(\Phi(-2.078) = 1 - \Phi(-2.078)\) \((= 1 - 0.9812)\) |
| \(= 0.0172\) (3 sf) | A1 [4] | \(= 0.0188\) |
## Question 5:
### Part (i):
| Answer/Working | Mark | Guidance |
|---|---|---|
| $\pm 1.645$ used | B1 | |
| $\dfrac{\bar{x} - 22}{\dfrac{3.5}{\sqrt{12}}} > 1.645$ | M1 | |
| $\bar{x} > 23.66(20)$; $\bar{x} > 23.7$ **AG** | A1 [3] | Accept $=$; standardising using $23.7$ scores M1A0 or $\bar{x} = 23.66(20)$ |
### Part (ii):
| Answer/Working | Mark | Guidance |
|---|---|---|
| $P(\bar{x} < 23.7 \mid \mu = 25.8)$ | M1 | For attempt type II error and standardising |
| $\dfrac{23.662 - 25.8}{\dfrac{3.5}{\sqrt{12}}} = -2.116$ | A1 | $\dfrac{23.7 - 25.8}{\dfrac{3.5}{\sqrt{12}}} = -2.078$ |
| $\Phi(-2.116) = 1 - \Phi(2.116)$ $(= 1 - 0.9828)$ | M1 | $\Phi(-2.078) = 1 - \Phi(-2.078)$ $(= 1 - 0.9812)$ |
| $= 0.0172$ (3 sf) | A1 [4] | $= 0.0188$ |
---5 The management of a factory thinks that the mean time required to complete a particular task is 22 minutes. The times, in minutes, taken by employees to complete this task have a normal distribution with mean $\mu$ and standard deviation 3.5. An employee claims that 22 minutes is not long enough for the task. In order to investigate this claim, the times for a random sample of 12 employees are used to test the null hypothesis $\mu = 22$ against the alternative hypothesis $\mu > 22$ at the $5 \%$ significance level.\\
(i) Show that the null hypothesis is rejected in favour of the alternative hypothesis if $\bar { x } > 23.7$ (correct to 3 significant figures), where $\bar { x }$ is the sample mean.\\
(ii) Find the probability of a Type II error given that the actual mean time is 25.8 minutes.
\hfill \mbox{\textit{CAIE S2 2011 Q5 [7]}}