| Exam Board | CAIE |
|---|---|
| Module | S2 (Statistics 2) |
| Year | 2021 |
| Session | June |
| Marks | 6 |
| Paper | Download PDF ↗ |
| Mark scheme | Download PDF ↗ |
| Topic | Type I/II errors and power of test |
| Type | State meaning of Type I error |
| Difficulty | Standard +0.3 This is a straightforward two-tailed hypothesis test with known variance requiring standard procedure (hypotheses, test statistic, critical value comparison) plus a Type I error definition. The calculations are routine and the context is clearly specified, making it slightly easier than average but still requiring proper statistical reasoning. |
| Spec | 2.05a Hypothesis testing language: null, alternative, p-value, significance5.05c Hypothesis test: normal distribution for population mean |
| Answer | Marks | Guidance |
|---|---|---|
| Answer | Mark | Guidance |
| Conclude that (population) mean time has changed (or is not 42.4) although \(\mu\) has not changed (or is still 42.4) | B1 | OE. In context |
| Answer | Marks | Guidance |
|---|---|---|
| Answer | Mark | Guidance |
| \(H_0\): population mean (or \(\mu\)) \(= 42.4\); \(H_1\): population mean (or \(\mu\)) \(\neq 42.4\) | B1 | Not just 'mean'. (could be seen in (a)) |
| \(\pm\frac{45.6 - 42.4}{\sqrt{38.2 \div 20}}\) | M1 | For standardising (must have \(\sqrt{20}\)) |
| \(\pm\ 2.315\) | A1 | |
| \(2.240 < \text{'2.315'}\) | M1 | For valid comparison (accept 2.241) or \(P(z > 2.315) = 0.0103 < 0.0125\) oe |
| There is evidence that \(\mu\) or mean time has changed | A1 FT | FT *their* \(z\). In context, not definite. No contradictions. Note: Accept correct alternative methods. SC: One tail test no FT. Can score B0 M1 A1 M1 (comparison with 1.96) A0 (maximum 3 out of 5) |
## Question 5(a):
| Answer | Mark | Guidance |
|--------|------|----------|
| Conclude that (population) mean time has changed (or is not 42.4) although $\mu$ has not changed (or is still 42.4) | B1 | OE. In context |
## Question 5(b):
| Answer | Mark | Guidance |
|--------|------|----------|
| $H_0$: population mean (or $\mu$) $= 42.4$; $H_1$: population mean (or $\mu$) $\neq 42.4$ | B1 | Not just 'mean'. (could be seen in **(a)**) |
| $\pm\frac{45.6 - 42.4}{\sqrt{38.2 \div 20}}$ | M1 | For standardising (must have $\sqrt{20}$) |
| $\pm\ 2.315$ | A1 | |
| $2.240 < \text{'2.315'}$ | M1 | For valid comparison (accept 2.241) or $P(z > 2.315) = 0.0103 < 0.0125$ oe |
| There is evidence that $\mu$ or mean time has changed | A1 FT | FT *their* $z$. In context, not definite. No contradictions. **Note:** Accept correct alternative methods. **SC:** One tail test no FT. Can score B0 M1 A1 M1 (comparison with 1.96) A0 (maximum 3 out of 5) |5 The time, in minutes, spent by customers at a particular gym has the distribution $\mathrm { N } ( \mu , 38.2 )$. In the past the value of $\mu$ has been 42.4. Following the installation of some new equipment the management wishes to test whether the value of $\mu$ has changed.
\begin{enumerate}[label=(\alph*)]
\item State what is meant by a Type I error in this context.
\item The mean time for a sample of 20 customers is found to be 45.6 minutes.
Test at the $2.5 \%$ significance level whether the value of $\mu$ has changed.
\end{enumerate}
\hfill \mbox{\textit{CAIE S2 2021 Q5 [6]}}