Date | May 2019 | Marks available | 1 | Reference code | 19M.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
In a large population of hens, the weight of a hen is normally distributed with mean μ kg and standard deviation σ kg. A random sample of 100 hens is taken from the population.
The mean weight for the sample is denoted by ˉX.
The sample values are summarized by ∑x=199.8 and ∑x2=407.8 where x kg is the weight of a hen.
It is found that σ = 0.27 . It is decided to test, at the 1 % level of significance, the null hypothesis μ = 1.95 against the alternative hypothesis μ > 1.95.
State the distribution of ˉX giving its mean and variance.
Find an unbiased estimate for μ.
Find an unbiased estimate for σ2.
Find a 90 % confidence interval for μ.
Find the p-value for the test.
Write down the conclusion reached.
Markscheme
* This question is from an exam for a previous syllabus, and may contain minor differences in marking or structure.
ˉX∼N(μ,σ2100) A1
Note: Accept n in place of 100.
[1 mark]
ˆμ=∑xn=199.8100=1.998 A1
Note: Accept 2.00, 2.0 and 2.
[1 mark]
sn−12=nn−1(∑x2n−ˉx2)=10099(407.8100−1.9982) (M1)
= 0.086864
unbiased estimate for σ2 is 0.0869 A1
Note: Accept any answer which rounds to 0.087.
[2 marks]
90 % confidence interval is 1.998±1.660√0.0869100 (M1)
= (1.95, 2.05) A1A1
Note: FT their σ from (c).
Note: Condone the use of the z-value 1.645 since n is large.
Note: Accept any values that round to 1.95 and 2.05.
[3 marks]
p-value is 0.0377 A2
Note: Award A1 for the 2-tail value 0.0754.
Note: Award A2 for 0.0377 and A1 for any other value that rounds to 0.038.
Note: FT their estimated mean from (b), note that 2 gives p = 0.032(0).
[2 marks]
accept the null hypothesis A1
Note: FT their p-value.
[1 mark]