CAIE S2 2023 June — Question 3 6 marks

Exam BoardCAIE
ModuleS2 (Statistics 2)
Year2023
SessionJune
Marks6
PaperDownload PDF ↗
Mark schemeDownload PDF ↗
TopicZ-tests (known variance)
TypeOne-tail z-test (upper tail)
DifficultyStandard +0.3 This is a straightforward one-tail z-test application requiring students to state the normality assumption (or CLT justification), set up hypotheses, calculate a test statistic using given values, and compare to a critical value. All steps are routine and formulaic with no conceptual challenges beyond standard S2 content.
Spec5.05c Hypothesis test: normal distribution for population mean

3 In the past, the annual amount of wheat produced per farm by a large number of similar sized farms in a certain region had mean 24.0 tonnes and standard deviation 5.2 tonnes. Last summer a new fertiliser was used by all the farms, and it was expected that the mean amount of wheat produced per farm would be greater than 24.0 tonnes. In order to test whether this was true, a scientist recorded the amounts of wheat produced by a random sample of 50 farms last summer. He found that the value of the sample mean was 25.8 tonnes. Stating a necessary assumption, carry out the test at the \(1 \%\) significance level.

Question 3:
AnswerMarks Guidance
AnswerMarks Guidance
Assume SD still \(= 5.2\)B1 OE i.e. 'Assume the SD remains unchanged'.
\(H_0: \mu = 24.0 \quad H_1: \mu > 24.0\)B1 Or population mean; not just mean.
\(\frac{25.8 - 24.0}{\frac{5.2}{\sqrt{50}}}\)M1 For standardising (could be implied). Must have \(\sqrt{50}\).
\(= 2.448\)A1 Or \(P(\bar{X} > 25.8) = 0.0071\).
\('2.448' > 2.326\)M1 Or \(0.0071 < 0.01\). For valid comparison.
[Reject \(H_0\)] There is evidence that (mean) amount of wheat is greater.A1FT OE. FT their \(z_\text{calc}\). In context, not definite. CV method: CV \(= 25.71\) M1A1 \(25.71 < 25.8\) M1 A1FT or CV \(= 24.09\) M1 A1 \(24.09 > 24\) M1 A1FT.
Total: 6
## Question 3:

| Answer | Marks | Guidance |
|--------|-------|----------|
| Assume SD still $= 5.2$ | B1 | OE i.e. 'Assume the SD remains unchanged'. |
| $H_0: \mu = 24.0 \quad H_1: \mu > 24.0$ | B1 | Or population mean; not just mean. |
| $\frac{25.8 - 24.0}{\frac{5.2}{\sqrt{50}}}$ | M1 | For standardising (could be implied). Must have $\sqrt{50}$. |
| $= 2.448$ | A1 | Or $P(\bar{X} > 25.8) = 0.0071$. |
| $'2.448' > 2.326$ | M1 | Or $0.0071 < 0.01$. For valid comparison. |
| [Reject $H_0$] There is evidence that (mean) amount of wheat is greater. | A1FT | OE. FT their $z_\text{calc}$. In context, not definite. CV method: CV $= 25.71$ M1A1 $25.71 < 25.8$ M1 A1FT or CV $= 24.09$ M1 A1 $24.09 > 24$ M1 A1FT. |
| **Total: 6** | | |
3 In the past, the annual amount of wheat produced per farm by a large number of similar sized farms in a certain region had mean 24.0 tonnes and standard deviation 5.2 tonnes. Last summer a new fertiliser was used by all the farms, and it was expected that the mean amount of wheat produced per farm would be greater than 24.0 tonnes. In order to test whether this was true, a scientist recorded the amounts of wheat produced by a random sample of 50 farms last summer. He found that the value of the sample mean was 25.8 tonnes.

Stating a necessary assumption, carry out the test at the $1 \%$ significance level.\\

\hfill \mbox{\textit{CAIE S2 2023 Q3 [6]}}