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Date November 2019 Marks available 1 Reference code 19N.3.AHL.TZ0.Hsp_3
Level Additional Higher Level Paper Paper 3 Time zone Time zone 0
Command term Write down Question number Hsp_3 Adapted from N/A

Question

A random variable XX has a distribution with mean μμ and variance 4. A random sample of size 100 is to be taken from the distribution of XX.

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

State the central limit theorem as applied to a random sample of size nn, taken from a distribution with mean μμ and variance σ2σ2.

[2]
a.

Jack takes a random sample of size 100 and calculates that ˉx=60.2¯x=60.2. Find an approximate 90 % confidence interval for μμ.

[2]
b.

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 μμ, giving your answer correct to three significant figures.

[5]
c.iii.

Markscheme

for nn (sufficiently) large the sample mean ˉX¯X approximately           A1

N(μσ2n)N(μσ2n)           A1

Note: Award the first A1 for nn large and reference to the sample mean (ˉX)(¯X), the second A1 is for normal and the two parameters.

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

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

[2 marks]

a.

[59.9, 60.5]                   A1A1

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

[2 marks]

b.

under H0H0ˉXN(604100)¯XN(604100)                   (A1)

required to find kk such that P(ˉX>k)=0.05P(¯X>k)=0.05                   (M1)

use of any valid method, eg GDC Inv(Normal) or k=60+zσnk=60+zσn                   (M1)

hence critical region is ˉx=60.33¯x=60.33                   A1

[4 marks]

c.i.

0.05                   A1

[1 mark]

c.ii.

PP(Type II error) = PP(H0H0 is accepted / H0H0 is false)       (R1)

Note: Accept Type II error means H0H0 is accepted given H0H0 is false.

P(ˉX<60.33)=0.25P(¯X<60.33)=0.25 when ˉXN(μ4100)¯XN(μ4100)       (M1)

P(ˉXμ210<60.33μ210)=0.25P(¯Xμ210<60.33μ210)=0.25       (M1)

P(Z<60.33μ210)=0.25P(Z<60.33μ210)=0.25  where  ZN(012)ZN(012)

60.33μ210=0.674460.33μ210=0.6744       (A1)

μ=60.33+210×0.6744μ=60.33+210×0.6744

μ=60.5μ=60.5                   A1

[5 marks]

c.iii.

Examiners report

[N/A]
a.
[N/A]
b.
[N/A]
c.i.
[N/A]
c.ii.
[N/A]
c.iii.

Syllabus sections

Topic 4—Statistics and probability » AHL 4.18—T and Z test, type I and II errors
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Topic 4—Statistics and probability

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