Question 4 - A-Level Maths

CAIE S2 2014 June — Question 4 6 marks

Exam Board	CAIE
Module	S2 (Statistics 2)
Year	2014
Session	June
Marks	6
Paper	Download PDF ↗
Mark scheme	Download PDF ↗
Topic	Z-tests (known variance)
Type	Two-tail z-test
Difficulty	Moderate -0.8 This is a straightforward two-tailed z-test with all necessary information provided. Students need to calculate sample mean (1.65) and variance from summations, then apply the standard z-test formula—all routine procedures for S2 with no conceptual challenges or novel problem-solving required.
Spec	5.05b Unbiased estimates: of population mean and variance 5.05c Hypothesis test: normal distribution for population mean

4 The weights, $X$ kilograms, of rabbits in a certain area have population mean $\mu \mathrm { kg }$. A random sample of 100 rabbits from this area was taken and the weights are summarised by $$\Sigma x = 165 , \quad \Sigma x ^ { 2 } = 276.25 .$$ Test at the $5 \%$ significance level the null hypothesis $\mathrm { H } _ { 0 } : \mu = 1.6$ against the alternative hypothesis $\mathrm { H } _ { 1 } : \mu \neq 1.6$.

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Question 4:

Answer	Marks	Guidance
Answer/Working	Mark	Guidance
$\bar{x} = 1.65$	B1
$\text{est}(\sigma^2) = \frac{100}{99}\left(\frac{276.25}{100} - 1.65^2\right)$	B1
$= 0.040404\ldots = 4/99$
$(\pm)\frac{1.65 - 1.6}{\sqrt{\frac{0.040404}{100}}}$	M1	Without $\frac{100}{99}$: $\frac{1.65-1.6}{\sqrt{\frac{0.04}{100}}}$ B1 B0 M1
$= (\pm)\ 2.487/2.488$, accept 2.49; or 0.0065/0.0064 if area comparison done	A1	$= 2.50$ A1; CV Method M1 must use 1.96 A1 for 1.639 or 1.6106
Compare with 1.96	M1	For valid comparison ($z/z$ signs consistent or area/area cv)
There is evidence that $\mu$ is not 1.6	A1$\checkmark$ [6]	Accept/Reject $H_0$; no contradictions

## Question 4:

| Answer/Working | Mark | Guidance |
|---|---|---|
| $\bar{x} = 1.65$ | B1 | |
| $\text{est}(\sigma^2) = \frac{100}{99}\left(\frac{276.25}{100} - 1.65^2\right)$ | B1 | |
| $= 0.040404\ldots = 4/99$ | | |
| $(\pm)\frac{1.65 - 1.6}{\sqrt{\frac{0.040404}{100}}}$ | M1 | Without $\frac{100}{99}$: $\frac{1.65-1.6}{\sqrt{\frac{0.04}{100}}}$ B1 B0 M1 |
| $= (\pm)\ 2.487/2.488$, accept 2.49; or 0.0065/0.0064 if area comparison done | A1 | $= 2.50$ A1; CV Method M1 must use 1.96 A1 for 1.639 or 1.6106 |
| Compare with 1.96 | M1 | For valid comparison ($z/z$ signs consistent or area/area cv) |
| There is evidence that $\mu$ is not 1.6 | A1$\checkmark$ [6] | Accept/Reject $H_0$; no contradictions |

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4 The weights, $X$ kilograms, of rabbits in a certain area have population mean $\mu \mathrm { kg }$. A random sample of 100 rabbits from this area was taken and the weights are summarised by

$$\Sigma x = 165 , \quad \Sigma x ^ { 2 } = 276.25 .$$

Test at the $5 \%$ significance level the null hypothesis $\mathrm { H } _ { 0 } : \mu = 1.6$ against the alternative hypothesis $\mathrm { H } _ { 1 } : \mu \neq 1.6$.

\hfill \mbox{\textit{CAIE S2 2014 Q4 [6]}}

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