| Exam Board | CAIE |
|---|---|
| Module | S2 (Statistics 2) |
| Year | 2022 |
| Session | November |
| Marks | 10 |
| Paper | Download PDF ↗ |
| Mark scheme | Download PDF ↗ |
| Topic | Type I/II errors and power of test |
| Type | Hypothesis test then Type II error probability |
| Difficulty | Standard +0.3 This is a straightforward hypothesis testing question requiring standard normal distribution techniques. Part (a) is routine application of a one-tailed z-test with known variance. Part (b) requires understanding of Type II error and calculating a probability using the alternative hypothesis mean, which is a standard S2 topic but slightly elevates difficulty above pure routine. |
| Spec | 2.05a Hypothesis testing language: null, alternative, p-value, significance5.05c Hypothesis test: normal distribution for population mean |
| Answer | Marks | Guidance |
|---|---|---|
| Answer | Marks | Guidance |
| \(H_0\): Population mean time (or \(\mu\)) \(= 32.5\) \(\quad H_1\): Population mean time (or \(\mu\)) \(< 32.5\) | B1 | Not just "mean". |
| \(\pm\dfrac{31.8 - 32.5}{3.1 \div \sqrt{50}}\) | M1 | Must have \(\sqrt{50}\). Could be implied. |
| \(= \pm 1.597\) | A1 | |
| \(`-1.597` < -1.406\) [or \(`1.597` > 1.406\)] | M1 | Valid comparison of their \(z_\text{calc}\) with \(\pm 1.406\). or \(0.0551 < 0.08\) (or \(0.0552 < 0.08\)). |
| [reject \(H_0\)] There is evidence that [population] [mean] time has decreased | A1 FT | In context, not definite, no contradictions. Note: Accept critical value method \(31.88\) \((31.9)\) M1 A1 and \(31.8 < 31.88\) M1 conclusion A1. |
| Answer | Marks | Guidance |
|---|---|---|
| Answer | Marks | Guidance |
| \(\dfrac{a - 32.5}{3.1 \div \sqrt{50}} = -1.406\) | M1 | Standardise with \(32.5\) and \(\sqrt{50}\) and \(z\) value on RHS. |
| \(a = 31.88\) or \(31.9\) | A1 | May be seen in part (a). Can score M1A1 here as well using a similar approach to (a). |
| \(\dfrac{\textit{their}\;`31.88` - 31.5}{3.1 \div \sqrt{50}} \quad [= 0.8668 \text{ to } 0.8760]\) | M1 | Standardise with *their* cv and mean \(= 31.5\). Must have \(\sqrt{50}\). |
| \(1 - \Phi(`0.8668`)\) | M1 | For area consistent with their working. |
| \(= 0.190\) to \(0.193\) (3 sf) | A1 |
## Question 7(a):
| Answer | Marks | Guidance |
|--------|-------|----------|
| $H_0$: Population mean time (or $\mu$) $= 32.5$ $\quad H_1$: Population mean time (or $\mu$) $< 32.5$ | B1 | Not just "mean". |
| $\pm\dfrac{31.8 - 32.5}{3.1 \div \sqrt{50}}$ | M1 | Must have $\sqrt{50}$. Could be implied. |
| $= \pm 1.597$ | A1 | |
| $`-1.597` < -1.406$ [or $`1.597` > 1.406$] | M1 | Valid comparison of their $z_\text{calc}$ with $\pm 1.406$. or $0.0551 < 0.08$ (or $0.0552 < 0.08$). |
| [reject $H_0$] There is evidence that [population] [mean] **time** has **decreased** | A1 FT | In context, not definite, no contradictions. Note: Accept critical value method $31.88$ $(31.9)$ **M1 A1** and $31.8 < 31.88$ **M1** conclusion **A1**. |
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## Question 7(b):
| Answer | Marks | Guidance |
|--------|-------|----------|
| $\dfrac{a - 32.5}{3.1 \div \sqrt{50}} = -1.406$ | M1 | Standardise with $32.5$ and $\sqrt{50}$ and $z$ value on RHS. |
| $a = 31.88$ or $31.9$ | A1 | May be seen in part **(a)**. Can score M1A1 here as well using a similar approach to **(a)**. |
| $\dfrac{\textit{their}\;`31.88` - 31.5}{3.1 \div \sqrt{50}} \quad [= 0.8668 \text{ to } 0.8760]$ | M1 | Standardise with *their* cv and mean $= 31.5$. Must have $\sqrt{50}$. |
| $1 - \Phi(`0.8668`)$ | M1 | For area consistent with their working. |
| $= 0.190$ to $0.193$ (3 sf) | A1 | |7 In the past Laxmi's time, in minutes, for her journey to college had mean 32.5 and standard deviation 3.1. After a change in her route, Laxmi wishes to test whether the mean time has decreased. She notes her journey times for a random sample of 50 journeys and she finds that the sample mean is 31.8 minutes. You should assume that the standard deviation is unchanged.
\begin{enumerate}[label=(\alph*)]
\item Carry out a hypothesis test, at the $8 \%$ significance level, of whether Laxmi's mean journey time has decreased.\\
Later Laxmi carries out a similar test with the same hypotheses, at the $8 \%$ significance level, using another random sample of size 50 .
\item Given that the population mean is now 31.5, find the probability of a Type II error.\\
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 2022 Q7 [10]}}