| Exam Board | CAIE |
|---|---|
| Module | S2 (Statistics 2) |
| Year | 2021 |
| Session | June |
| Marks | 8 |
| Paper | Download PDF ↗ |
| Mark scheme | Download PDF ↗ |
| Topic | Type I/II errors and power of test |
| Type | State meaning of Type II error |
| Difficulty | Standard +0.3 This is a straightforward hypothesis testing question requiring standard procedures: defining Type II error contextually, performing a one-tailed z-test with given values, and identifying which error type is possible. All steps are routine applications of textbook methods with no novel problem-solving required, making it slightly easier than average. |
| Spec | 2.05a Hypothesis testing language: null, alternative, p-value, significance5.05c Hypothesis test: normal distribution for population mean |
| Answer | Marks | Guidance |
|---|---|---|
| Conclude (mean) (journey) time has not decreased when in fact it has. | B1 | OE in context |
| Answer | Marks | Guidance |
|---|---|---|
| \(H_0\): Pop mean (or \(\mu\)) = 1.4; \(H_1\): Pop mean (or \(\mu\)) < 1.4 | B1 | May be seen in (a) |
| \(\dfrac{1.36-1.4}{\dfrac{0.12}{\sqrt{50}}}\) | M1 | Accept totals method \(\dfrac{68-70}{\sqrt{50} \times 0.12}\); No mixed methods or no standard deviation/variance mixes |
| \(-2.357\) or \(-2.36\) | A1 | Correct \(z\) or correct area if used |
| \(-2.357 < -1.96\) or \(0.0092 < 0.025\) or \(0.9908 > 0.975\); Or CV method \(1.36 < 1.367\) | M1 | Valid comparison |
| There is evidence that (mean) (journey) times have decreased | A1 FT | In context not definite no contradictions; NB use of two tail test scores max B0M1A1M1A0 no ft for two tail test |
| Answer | Marks | Guidance |
|---|---|---|
| \(H_0\) was rejected OE | \*B1 FT | FT \(H_0\) was accepted OE |
| Type I | DB1 FT | FT Type II |
## Question 2:
### Part 2(a):
Conclude (mean) (journey) time has not decreased when in fact it has. | **B1** | OE in context
### Part 2(b):
$H_0$: Pop mean (or $\mu$) = 1.4; $H_1$: Pop mean (or $\mu$) < 1.4 | **B1** | May be seen in (a)
$\dfrac{1.36-1.4}{\dfrac{0.12}{\sqrt{50}}}$ | **M1** | Accept totals method $\dfrac{68-70}{\sqrt{50} \times 0.12}$; No mixed methods or no standard deviation/variance mixes
$-2.357$ or $-2.36$ | **A1** | Correct $z$ or correct area if used
$-2.357 < -1.96$ or $0.0092 < 0.025$ or $0.9908 > 0.975$; Or CV method $1.36 < 1.367$ | **M1** | Valid comparison
There is evidence that (mean) (journey) times have decreased | **A1 FT** | In context not definite no contradictions; NB use of two tail test scores max B0M1A1M1A0 no ft for two tail test
### Part 2(c):
$H_0$ was rejected OE | **\*B1 FT** | FT $H_0$ was accepted OE
Type I | **DB1 FT** | FT Type II
---2 In the past, the time, in hours, for a particular train journey has had mean 1.40 and standard deviation 0.12 . Following the introduction of some new signals, it is required to test whether the mean journey time has decreased.
\begin{enumerate}[label=(\alph*)]
\item State what is meant by a Type II error in this context.
\item The mean time for a random sample of 50 journeys is found to be 1.36 hours.
Assuming that the standard deviation of journey times is still 0.12 hours, test at the $2.5 \%$ significance level whether the population mean journey time has decreased.
\item State, with a reason, which of the errors, Type I or Type II, might have been made in the test in part (b).
\end{enumerate}
\hfill \mbox{\textit{CAIE S2 2021 Q2 [8]}}