| Exam Board | CAIE |
|---|---|
| Module | S2 (Statistics 2) |
| Year | 2023 |
| Session | March |
| Marks | 9 |
| Paper | Download PDF ↗ |
| Mark scheme | Download PDF ↗ |
| Topic | Type I/II errors and power of test |
| Type | Carry out hypothesis test |
| Difficulty | Standard +0.3 This is a straightforward hypothesis testing question covering standard procedures: identifying test type, interpreting z-values against critical values, finding significance levels from tables, and reverse-calculating a population mean. All parts use routine techniques with no novel problem-solving required, making it slightly easier than average for A-level statistics. |
| Spec | 5.05a Sample mean distribution: central limit theorem5.05c Hypothesis test: normal distribution for population mean |
| Answer | Marks | Guidance |
|---|---|---|
| Answer | Mark | Guidance |
| He is expecting a decrease (in \(\mu\)) | B1 | OE |
| Answer | Marks | Guidance |
|---|---|---|
| Answer | Mark | Guidance |
| \(-2.02 < -1.96\) | M1 | For valid comparison. Allow \(2.02 > 1.96\) or \(0.0217 < 0.025\) or \(0.9783 > 0.975\). |
| (Reject \(H_0\)) There is evidence to suggest that this year's (mean) time is less than 25 | A1 | OE (such as evidence to support Akash's belief), in context, not definite. No contradictions. |
| Answer | Marks | Guidance |
|---|---|---|
| Answer | Mark | Guidance |
| \(1 - \phi(2.14) = 0.0162\) | M1 | |
| \(1.62\) | A1 | Allow 1.62% or 1.6 or 1.6%. |
| \(\alpha \geq 1.62\) (3 sf) | A1ft | FT their 1.62. Allow \(\alpha \geq 1.62\%\) or 1.6 or 1.6%. Condone \(>\). |
| Answer | Marks | Guidance |
|---|---|---|
| Answer | Mark | Guidance |
| \(\frac{24.8 - m}{3.9 \div 10}\) | M1 | For standardising. |
| \(\frac{24.8 - m}{3.9 \div 10} = -1.645\) | M1 | Equate their standardised value to \(-1.645\) (signs must be consistent). |
| \(m = 25.4\) (3 sf) | A1 |
## Question 6(a):
| Answer | Mark | Guidance |
|--------|------|----------|
| He is expecting a **decrease** (in $\mu$) | B1 | OE |
## Question 6(b):
| Answer | Mark | Guidance |
|--------|------|----------|
| $-2.02 < -1.96$ | M1 | For valid comparison. Allow $2.02 > 1.96$ or $0.0217 < 0.025$ or $0.9783 > 0.975$. |
| (Reject $H_0$) There is evidence to suggest that this year's (mean) time is less than 25 | A1 | OE (such as evidence to support Akash's belief), in context, not definite. No contradictions. |
## Question 6(c):
| Answer | Mark | Guidance |
|--------|------|----------|
| $1 - \phi(2.14) = 0.0162$ | M1 | |
| $1.62$ | A1 | Allow 1.62% or 1.6 or 1.6%. |
| $\alpha \geq 1.62$ (3 sf) | A1ft | FT their 1.62. Allow $\alpha \geq 1.62\%$ or 1.6 or 1.6%. Condone $>$. |
## Question 6(d):
| Answer | Mark | Guidance |
|--------|------|----------|
| $\frac{24.8 - m}{3.9 \div 10}$ | M1 | For standardising. |
| $\frac{24.8 - m}{3.9 \div 10} = -1.645$ | M1 | Equate their standardised value to $-1.645$ (signs must be consistent). |
| $m = 25.4$ (3 sf) | A1 | |6 Last year, the mean time taken by students at a school to complete a certain test was 25 minutes. Akash believes that the mean time taken by this year's students was less than 25 minutes. In order to test this belief, he takes a large random sample of this year's students and he notes the time taken by each student. He carries out a test, at the $2.5 \%$ significance level, for the population mean time, $\mu$ minutes. Akash uses the null hypothesis $\mathrm { H } _ { 0 } : \mu = 25$.
\begin{enumerate}[label=(\alph*)]
\item Give a reason why Akash should use a one-tailed test.\\
Akash finds that the value of the test statistic is $z = - 2.02$.
\item Explain what conclusion he should draw.\\
In a different one-tailed hypothesis test the $z$-value was found to be 2.14 .
\item Given that this value would lead to a rejection of the null hypothesis at the $\alpha \%$ significance level, find the set of possible values of $\alpha$.\\
The population mean time taken by students at another school to complete a test last year was $m$ minutes. Sorin carries out a one-tailed test to determine whether the population mean this year is less than $m$, using a random sample of 100 students. He assumes that the population standard deviation of the times is 3.9 minutes. The sample mean is 24.8 minutes, and this result just leads to the rejection of the null hypothesis at the 5\% significance level.
\item Find the value of $m$.\\
If you use the following lined page to complete the answer(s) to any question(s), the question number(s) must be clearly shown.
\end{enumerate}
\hfill \mbox{\textit{CAIE S2 2023 Q6 [9]}}