| Exam Board | CAIE |
|---|---|
| Module | S2 (Statistics 2) |
| Year | 2020 |
| Session | June |
| Marks | 7 |
| Paper | Download PDF ↗ |
| Mark scheme | Download PDF ↗ |
| Topic | Type I/II errors and power of test |
| Type | Carry out hypothesis test |
| Difficulty | Moderate -0.3 This is a straightforward hypothesis testing question requiring standard assumptions (normality and independence) and a routine one-sample z-test calculation. While it involves multiple steps, all procedures are textbook-standard with no novel problem-solving required, making it slightly easier than average. |
| Spec | 5.05a Sample mean distribution: central limit theorem5.05c Hypothesis test: normal distribution for population mean |
| Answer | Marks | Guidance |
|---|---|---|
| Answer | Mark | Guidance |
| Assume standard deviation unchanged or standard deviation \(= 0.08\) | B1 | |
| Assume yields normally distributed | B1 |
| Answer | Marks | Guidance |
|---|---|---|
| Answer | Mark | Guidance |
| \(H_0\): Population mean yield (or \(\mu\)) \(= 0.56\); \(H_1\): Population mean yield (or \(\mu\)) \(> 0.56\) | B1 | |
| \(\dfrac{0.61 - 0.56}{\dfrac{0.08}{\sqrt{10}}}\) | M1 | |
| \(1.976\) | A1 | |
| Compare \(1.96\) | M1 | |
| There is evidence that mean yield has increased | A1 |
## Question 2:
**Part 2(a):**
| Answer | Mark | Guidance |
|--------|------|----------|
| Assume standard deviation unchanged or standard deviation $= 0.08$ | B1 | |
| Assume yields normally distributed | B1 | |
**Part 2(b):**
| Answer | Mark | Guidance |
|--------|------|----------|
| $H_0$: Population mean yield (or $\mu$) $= 0.56$; $H_1$: Population mean yield (or $\mu$) $> 0.56$ | B1 | |
| $\dfrac{0.61 - 0.56}{\dfrac{0.08}{\sqrt{10}}}$ | M1 | |
| $1.976$ | A1 | |
| Compare $1.96$ | M1 | |
| There is evidence that mean yield has increased | A1 | |
---2 In the past the yield of a certain crop, in tonnes per hectare, had mean 0.56 and standard deviation 0.08 Following the introduction of a new fertilizer, the farmer intends to test at the $2.5 \%$ significance level whether the mean yield has increased. He finds that the mean yield over 10 years is 0.61 tonnes per hectare.
\begin{enumerate}[label=(\alph*)]
\item State two assumptions that are necessary for the test.
\item Carry out the test.
\end{enumerate}
\hfill \mbox{\textit{CAIE S2 2020 Q2 [7]}}