| Exam Board | CAIE |
|---|---|
| Module | S2 (Statistics 2) |
| Year | 2019 |
| 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 +0.8 This is a two-part hypothesis testing question requiring (i) a standard one-tailed z-test and (ii) calculation of Type II error probability, which requires finding the power of the test under a specific alternative hypothesis. Part (ii) is conceptually more demanding than typical A-level questions as it requires understanding the sampling distribution under both null and alternative hypotheses and computing tail probabilities accordingly. |
| Spec | 5.05c Hypothesis test: normal distribution for population mean5.05d Confidence intervals: using normal distribution |
| Answer | Marks | Guidance |
|---|---|---|
| Answer | Marks | Guidance |
| \(H_0\): Pop mean \(= 546\); \(H_1\): Pop mean \(> 546\) | B1 | Both. Allow just \(\mu\), but not just 'mean' |
| \(\frac{581 - 546}{\frac{120}{\sqrt{40}}}\) | M1 | Standardising. Need \(\frac{120}{\sqrt{40}}\) |
| \(= 1.845\), allow \(1.844\) | A1 | Allow 1.84 or 1.85 AWRT |
| \(1.845 < 1.96\) | M1 | OE. Or area comparison \(0.0325 > 0.025\) or large probabilities |
| No evidence that mean weekly income has increased | A1FT | No contradictions. If \(H_1: \neq\), and 2.241 used, max B0M1A1M1A0 |
| Total: 5 |
| Answer | Marks | Guidance |
|---|---|---|
| Answer | Marks | Guidance |
| \(\frac{a - 546}{\frac{120}{\sqrt{40}}} = 1.96\) | M1 | Standardise to find \(a\). Need \(\frac{120}{\sqrt{40}}\) and 546 and a value of \(z\) |
| \(a = 583.19\) | A1 | Allow 583 to 3 sf |
| \(\frac{583.19 - 595}{\frac{120}{\sqrt{40}}}\ (= -0.622)\) | M1 | Standardise. Need \(\frac{120}{\sqrt{40}}\) and 595 |
| \(\phi(-0.622) = 1 - \phi(0.622)\) | M1 | Consistent area |
| \(0.267\) | A1 | |
| Total: 5 |
## Question 7(i):
| Answer | Marks | Guidance |
|--------|-------|----------|
| $H_0$: Pop mean $= 546$; $H_1$: Pop mean $> 546$ | B1 | Both. Allow just $\mu$, but not just 'mean' |
| $\frac{581 - 546}{\frac{120}{\sqrt{40}}}$ | M1 | Standardising. Need $\frac{120}{\sqrt{40}}$ |
| $= 1.845$, allow $1.844$ | A1 | Allow 1.84 or 1.85 AWRT |
| $1.845 < 1.96$ | M1 | OE. Or area comparison $0.0325 > 0.025$ or large probabilities |
| No evidence that mean weekly income has increased | A1FT | No contradictions. If $H_1: \neq$, and 2.241 used, max B0M1A1M1A0 |
| **Total: 5** | | |
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## Question 7(ii):
| Answer | Marks | Guidance |
|--------|-------|----------|
| $\frac{a - 546}{\frac{120}{\sqrt{40}}} = 1.96$ | M1 | Standardise to find $a$. Need $\frac{120}{\sqrt{40}}$ and 546 and a value of $z$ |
| $a = 583.19$ | A1 | Allow 583 to 3 sf |
| $\frac{583.19 - 595}{\frac{120}{\sqrt{40}}}\ (= -0.622)$ | M1 | Standardise. Need $\frac{120}{\sqrt{40}}$ and 595 |
| $\phi(-0.622) = 1 - \phi(0.622)$ | M1 | Consistent area |
| $0.267$ | A1 | |
| **Total: 5** | | |7 Bob is a self-employed builder. In the past his weekly income had mean $\$ 546$ and standard deviation $\$ 120$. Following a change in Bob's working pattern, his mean weekly income for 40 randomly chosen weeks was $\$ 581$. You should assume that the standard deviation remains unchanged at $\$ 120$.\\
(i) Test at the $2.5 \%$ significance level whether Bob's mean weekly income has increased.\\
Bob finds his mean weekly income for another random sample of 40 weeks and carries out a similar test at the $2.5 \%$ significance level.\\
(ii) Given that Bob's mean weekly income is now in fact $\$ 595$, 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.\\
\hfill \mbox{\textit{CAIE S2 2019 Q7 [10]}}