| Exam Board | CAIE |
|---|---|
| Module | S2 (Statistics 2) |
| Year | 2022 |
| Session | March |
| Marks | 6 |
| Paper | Download PDF ↗ |
| Mark scheme | Download PDF ↗ |
| Topic | Type I/II errors and power of test |
| Type | Explain impossibility of one error type |
| Difficulty | Standard +0.3 This is a straightforward one-tailed hypothesis test with known variance followed by a standard Type I error interpretation question. Part (a) requires routine application of the z-test formula and comparison with critical values, while part (b) tests basic understanding that Type I errors occur when we reject Hâ‚€. Both parts are standard textbook exercises with no novel problem-solving required, making it slightly easier than average. |
| 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 | Mark | Guidance |
| \(H_0: \mu = 25.5\), \(H_1: \mu < 25.5\) | B1 | |
| \(\frac{23.7 - 25.5}{5.2 \div \sqrt{40}}\) | M1 | Must have \(\sqrt{40}\) |
| \(= -2.189\) | A1 | |
| \('2.189' < 2.326\) | M1 | For valid comparison. For two-tailed test: allow compare 2.576 if \(H_1: \mu \neq 25.5\) |
| [Accept \(H_0\)] No evidence that mean time has decreased | A1 FT | In context, not definite, no contradictions. FT *their* 2.189 but no FT for two-tailed test. N.B. Use of two-tailed test can score max B0 M1 A1 M1 A0. Condone use of critical value method (23.59 M1 A1 and \(23.7 > 23.59\) M1 A1 correct conclusion or 25.612 M1 A1 and \(25.5 < 25.612\) M1 A1 with correct conclusion) |
| Answer | Marks | Guidance |
|---|---|---|
| Answer | Mark | Guidance |
| No, because \(H_0\) was not rejected | B1 FT | FT *their* conclusion in (a) |
## Question 4:
### Part 4(a):
| Answer | Mark | Guidance |
|--------|------|----------|
| $H_0: \mu = 25.5$, $H_1: \mu < 25.5$ | B1 | |
| $\frac{23.7 - 25.5}{5.2 \div \sqrt{40}}$ | M1 | Must have $\sqrt{40}$ |
| $= -2.189$ | A1 | |
| $'2.189' < 2.326$ | M1 | For valid comparison. For two-tailed test: allow compare 2.576 if $H_1: \mu \neq 25.5$ |
| [Accept $H_0$] No evidence that mean time has decreased | A1 FT | In context, not definite, no contradictions. FT *their* 2.189 but no FT for two-tailed test. N.B. Use of two-tailed test can score max B0 M1 A1 M1 A0. Condone use of critical value method (23.59 M1 A1 and $23.7 > 23.59$ M1 A1 correct conclusion or 25.612 M1 A1 and $25.5 < 25.612$ M1 A1 with correct conclusion) |
### Part 4(b):
| Answer | Mark | Guidance |
|--------|------|----------|
| No, because $H_0$ was not rejected | B1 FT | FT *their* conclusion in (a) |
---4 In the past the time, in minutes, taken by students to complete a certain challenge had mean 25.5 and standard deviation 5.2. A new challenge is devised and it is expected that students will take, on average, less than 25.5 minutes to complete this challenge. A random sample of 40 students is chosen and their mean time for the new challenge is found to be 23.7 minutes.
\begin{enumerate}[label=(\alph*)]
\item Assuming that the standard deviation of the time for the new challenge is 5.2 minutes, test at the $1 \%$ significance level whether the population mean time for the new challenge is less than 25.5 minutes.
\item State, with a reason, whether it is possible that a Type I error was made in the test in part (a).
\end{enumerate}
\hfill \mbox{\textit{CAIE S2 2022 Q4 [6]}}