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Date	November 2019	Marks available	2	Reference code	19N.3.AHL.TZ0.Hsp_3
Level	Additional Higher Level	Paper	Paper 3	Time zone	Time zone 0
Command term	Find	Question number	Hsp_3	Adapted from	N/A

Question

A random variable $X$ has a distribution with mean $\mu$ and variance 4. A random sample of size 100 is to be taken from the distribution of $X$ .

Josie takes a different random sample of size 100 to test the null hypothesis that $\mu = 60$ against the alternative hypothesis that $\mu > 60$ at the 5 % level.

State the central limit theorem as applied to a random sample of size $n$ , taken from a distribution with mean $\mu$ and variance ${\sigma ^2}$ .

[2]

Jack takes a random sample of size 100 and calculates that $\bar x = 60.2$ . Find an approximate 90 % confidence interval for $\mu$ .

[2]

Find the critical region for Josie’s test, giving your answer correct to two decimal places.

[4]

c.i.

Write down the probability that Josie makes a Type I error.

[1]

c.ii.

Given that the probability that Josie makes a Type II error is 0.25, find the value of $\mu$ , giving your answer correct to three significant figures.

[5]

c.iii.

Markscheme

for $n$ (sufficiently) large the sample mean ${\bar X}$ approximately A1

$\sim {\text{N}}\left( {\mu {\text{, }}\frac{{{\sigma ^2}}}{n}} \right)$ A1

Note: Award the first A1 for $n$ large and reference to the sample mean $\left( {\bar X} \right)$ , the second A1 is for normal and the two parameters.

Note: Award the second A1 only if the first A1 is awarded.

Note: Allow ‘ $n$ tends to infinity’ or ‘ $n$ ≥ 30’ in place of ‘large’.

[2 marks]

[59.9, 60.5] A1A1

Note: Accept answers which round to the correct 3sf answers.

[2 marks]

under ${H_0}$ , $\bar X \sim {\text{N}}\left( {60{\text{, }}\frac{4}{{100}}} \right)$ (A1)

required to find $k$ such that $P\left( {\bar X > k} \right) = 0.05$ (M1)

use of any valid method, eg GDC Inv(Normal) or $k = 60 + z\frac{\sigma }{{\sqrt n }}$ (M1)

hence critical region is $\bar x = 60.33$ A1

[4 marks]

c.i.

0.05 A1

[1 mark]

c.ii.

${\text{P}}$ (Type II error) = ${\text{P}}$ ( ${H_0}$ is accepted / ${H_0}$ is false) (R1)

Note: Accept Type II error means ${H_0}$ is accepted given ${H_0}$ is false.

$\Rightarrow {\text{P}}\left( {\bar X < 60.33} \right) = 0.25$ when $\bar X \sim {\text{N}}\left( {\mu {\text{, }}\frac{4}{{100}}} \right)$ (M1)

$\Rightarrow {\text{P}}\left( {\frac{{\bar X - \mu }}{{\frac{2}{{10}}}} < \frac{{60.33 - \mu }}{{\frac{2}{{10}}}}} \right) = 0.25$ (M1)

$\Rightarrow {\text{P}}\left( {Z < \frac{{60.33 - \mu }}{{\frac{2}{{10}}}}} \right) = 0.25$ where $Z \sim {\text{N}}\left( {0{\text{, }}{1^2}} \right)$