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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 μ 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 μ = 60  against the alternative hypothesis that μ > 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 μ and variance σ 2 .

[2]
a.

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

[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 μ , giving your answer correct to three significant figures.

[5]
c.iii.

Markscheme

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

N ( μ σ 2 n )            A1

Note: Award the first A1 for n large and reference to the sample mean ( 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 ‘ n tends to infinity’ or ‘ n ≥ 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  H 0 X ¯ N ( 60 4 100 )                    (A1)

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

use of any valid method, eg GDC Inv(Normal) or  k = 60 + z σ n                    (M1)

hence critical region is  x ¯ = 60.33                    A1

[4 marks]

c.i.

0.05                   A1

[1 mark]

c.ii.

P (Type II error) = 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.

P ( X ¯ < 60.33 ) = 0.25 when  X ¯ N ( μ 4 100 )        (M1)

P ( X ¯ μ 2 10 < 60.33 μ 2 10 ) = 0.25        (M1)

P ( Z < 60.33 μ 2 10 ) = 0.25   where   Z N ( 0 1 2 )

60.33 μ 2 10 = 0.6744        (A1)

μ = 60.33 + 2 10 × 0.6744

μ = 60.5                    A1

[5 marks]

c.iii.

Examiners report

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

Syllabus sections

Topic 4—Statistics and probability » AHL 4.16—Confidence intervals
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Topic 4—Statistics and probability

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