| Exam Board | CAIE |
|---|---|
| Module | S2 (Statistics 2) |
| Year | 2012 |
| Session | June |
| Marks | 9 |
| Paper | Download PDF ↗ |
| Mark scheme | Download PDF ↗ |
| Topic | Type I/II errors and power of test |
| Type | Identify which error type was made |
| Difficulty | Standard +0.3 This is a straightforward hypothesis testing question requiring standard procedures: identifying test type, calculating a test statistic for a normal distribution with known variance (z-test), and understanding Type I/II errors conceptually. All steps are routine A-level statistics applications with no novel problem-solving required, making it slightly easier than average. |
| Spec | 5.05c Hypothesis test: normal distribution for population mean |
| Answer | Marks | Guidance |
|---|---|---|
| Answer/Working | Marks | Guidance |
| Test is for bias in one direction | B1 | 'Increased' rather than 'changed' or statement that \(\mu > 45.7\) |
| One-tail | B1 (2) | dep 1st B1 |
| Answer | Marks | Guidance |
|---|---|---|
| Answer/Working | Marks | Guidance |
| \(H_0\): pop mean \(= 45.7\) | B1 | Allow \(\mu\), but not 'mean' (follow through their (i)) |
| \(H_1\): pop mean \(> 45.7\) | B1 | |
| \(\bar{x} = 47.375\) or \(47.4\) or \(379/8\) | M1 | Allow without \(\sqrt{}\) |
| \(\frac{47.375 - 45.7}{\frac{3.2}{\sqrt{8}}} = 1.481\) to \(1.503\) | ||
| \(z = 1.645\) | M1 | Explicit comparison with their \(z\) from table; comparison with \(1.645\) or probability (\(0.0664\) to \(0.0693\)) with \(0.05\) |
| \(1.481 < 1.645\); hence no evidence mean time increased (AG) | A1 (5) | Correct conclusion — accept \(H_0\), no errors seen |
| Not rejected \(H_0\) | B1 | dep 1st B1, no contradictions for either mark |
| Type II possible | B1 (2) |
## Question 6:
### Part (i)
| Answer/Working | Marks | Guidance |
|---|---|---|
| Test is for bias in one direction | B1 | 'Increased' rather than 'changed' or statement that $\mu > 45.7$ |
| One-tail | B1 (2) | dep 1st B1 |
### Part (ii)
| Answer/Working | Marks | Guidance |
|---|---|---|
| $H_0$: pop mean $= 45.7$ | B1 | Allow $\mu$, but not 'mean' (follow through their (i)) |
| $H_1$: pop mean $> 45.7$ | B1 | |
| $\bar{x} = 47.375$ or $47.4$ or $379/8$ | M1 | Allow without $\sqrt{}$ |
| $\frac{47.375 - 45.7}{\frac{3.2}{\sqrt{8}}} = 1.481$ to $1.503$ | | |
| $z = 1.645$ | M1 | Explicit comparison with their $z$ from table; comparison with $1.645$ or probability ($0.0664$ to $0.0693$) with $0.05$ |
| $1.481 < 1.645$; hence no evidence mean time increased (AG) | A1 (5) | Correct conclusion — accept $H_0$, no errors seen |
| Not rejected $H_0$ | B1 | dep 1st B1, no contradictions for either mark |
| Type II possible | B1 (2) | |
---6 Last year Samir found that the time for his journey to work had mean 45.7 minutes and standard deviation 3.2 minutes. Samir wishes to test whether his journey times have increased this year. He notes the times, in minutes, for a random sample of 8 journeys this year with the following results.
$$\begin{array} { l l l l l l l l }
46.2 & 41.7 & 49.2 & 47.1 & 47.2 & 48.4 & 53.7 & 45.5
\end{array}$$
It may be assumed that the population of this year's journey times is normally distributed with standard deviation 3.2 minutes.\\
(i) State, with a reason, whether Samir should use a one-tail or a two-tail test.\\
(ii) Show that there is no evidence at the $5 \%$ significance level that Samir's mean journey time has increased.\\
(iii) State, with a reason, which one of the errors, Type I or Type II, might have been made in carrying out the test in part (ii).
\hfill \mbox{\textit{CAIE S2 2012 Q6 [9]}}