Question 2 - A-Level Maths

CAIE S2 2017 June — Question 2 6 marks

Exam Board	CAIE
Module	S2 (Statistics 2)
Year	2017
Session	June
Marks	6
Paper	Download PDF ↗
Mark scheme	Download PDF ↗
Topic	Z-tests (known variance)
Type	One-tail z-test (lower tail)
Difficulty	Moderate -0.3 This is a straightforward one-tail z-test application with all values provided directly. Students must state the normality assumption (or CLT justification), set up hypotheses, calculate a test statistic using the given formula, and compare to a critical value. It requires standard procedure execution rather than problem-solving insight, making it slightly easier than average.
Spec	5.05a Sample mean distribution: central limit theorem 5.05c Hypothesis test: normal distribution for population mean

2 Past experience has shown that the heights of a certain variety of plant have mean 64.0 cm and standard deviation 3.8 cm . During a particularly hot summer, it was expected that the heights of plants of this variety would be less than usual. In order to test whether this was the case, a botanist recorded the heights of a random sample of 100 plants and found that the value of the sample mean was 63.3 cm . Stating a necessary assumption, carry out the test at the \(2.5 \%\) significance level.

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Question 2:

Answer	Marks	Guidance
Answer	Marks	Guidance
Assume sd still \(= 3.8\)	B1	or sd unchanged
\(H_0: \mu = 64.0 \quad H_1: \mu < 64.0\)	B1
\(\frac{63.3 - 64.0}{\frac{3.8}{\sqrt{100}}}\)	M1	Standardising with their values (no sd/var mixes). Must have \(\sqrt{100}\)
\(= -1.842\)	A1
comp \("1.842"\) with \(z\)-value: \("1.842" < 1.96\)	M1	comp \(+\)ve with \(+\)ve or \(-\)ve with \(-\)ve, or comp \(\Phi("1.842")\) with \(0.975\), \(0.9672 < 0.975\) OE
No evidence that heights are shorter	A1FT	OE FT their \(z_\text{calc}\)

## Question 2:

| Answer | Marks | Guidance |
|--------|-------|----------|
| Assume sd still $= 3.8$ | B1 | or sd unchanged |
| $H_0: \mu = 64.0 \quad H_1: \mu < 64.0$ | B1 | |
| $\frac{63.3 - 64.0}{\frac{3.8}{\sqrt{100}}}$ | M1 | Standardising with their values (no sd/var mixes). Must have $\sqrt{100}$ |
| $= -1.842$ | A1 | |
| comp $"1.842"$ with $z$-value: $"1.842" < 1.96$ | M1 | comp $+$ve with $+$ve or $-$ve with $-$ve, or comp $\Phi("1.842")$ with $0.975$, $0.9672 < 0.975$ OE |
| No evidence that heights are shorter | A1FT | OE FT their $z_\text{calc}$ |

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2 Past experience has shown that the heights of a certain variety of plant have mean 64.0 cm and standard deviation 3.8 cm . During a particularly hot summer, it was expected that the heights of plants of this variety would be less than usual. In order to test whether this was the case, a botanist recorded the heights of a random sample of 100 plants and found that the value of the sample mean was 63.3 cm . Stating a necessary assumption, carry out the test at the $2.5 \%$ significance level.\\

\hfill \mbox{\textit{CAIE S2 2017 Q2 [6]}}

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