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.
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.
Find the critical region for Josie’s test, giving your answer correct to two decimal places.
Write down the probability that Josie makes a Type I error.
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.
Markscheme
for nn (sufficiently) large the sample mean ˉX¯X approximately A1
∼N(μ, σ2n)∼N(μ, σ2n) 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]
[59.9, 60.5] A1A1
Note: Accept answers which round to the correct 3sf answers.
[2 marks]
under H0H0, ˉX∼N(60, 4100)¯X∼N(60, 4100) (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+zσ√nk=60+zσ√n (M1)
hence critical region is ˉx=60.33¯x=60.33 A1
[4 marks]
0.05 A1
[1 mark]
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)=0.25⇒P(¯X<60.33)=0.25 when ˉX∼N(μ, 4100)¯X∼N(μ, 4100) (M1)
⇒P(ˉX−μ210<60.33−μ210)=0.25⇒P(¯X−μ210<60.33−μ210)=0.25 (M1)
⇒P(Z<60.33−μ210)=0.25⇒P(Z<60.33−μ210)=0.25 where Z∼N(0, 12)Z∼N(0, 12)
60.33−μ210=−0.6744…60.33−μ210=−0.6744… (A1)
μ=60.33+210×0.6744…μ=60.33+210×0.6744…
μ=60.5μ=60.5 A1
[5 marks]