Date | May 2021 | Marks available | 1 | Reference code | 21M.1.AHL.TZ2.10 |
Level | Additional Higher Level | Paper | Paper 1 | Time zone | Time zone 2 |
Command term | Find | Question number | 10 | Adapted from | N/A |
Question
A manufacturer of chocolates produces them in individual packets, claiming to have an average of 8585 chocolates per packet.
Talha bought 3030 of these packets in order to check the manufacturer’s claim.
Given that the number of individual chocolates is xx, Talha found that, from his 3030 packets, Σx=2506Σx=2506 and Σx2=209 738Σx2=209738.
Find an unbiased estimate for the mean number (μ)(μ) of chocolates per packet.
Use the formula s2n-1=Σx2-(Σx)2nn-1s2n−1=Σx2−(Σx)2nn−1 to determine an unbiased estimate for the variance of the number of chocolates per packet.
Find a 95%95% confidence interval for μμ. You may assume that all conditions for a confidence interval have been met.
Suggest, with justification, a valid conclusion that Talha could make.
Markscheme
ˉx=Σxn=250630=83.5 (83.5333…)¯x=Σxn=250630=83.5 (83.5333…) A1
[1 mark]
(s2n-1=Σx2-(Σx)2nn-1=) 209738-250623029⎛⎝s2n−1=Σx2−(Σx)2nn−1=⎞⎠ 209738−250623029 (M1)
=13.9 (13.9126…)=13.9 (13.9126…) A1
[2 marks]
(82.1, 84.9) (82.1405…, 84.9261…)(82.1, 84.9) (82.1405…, 84.9261…) A2
[2 marks]
8585 is outside the confidence interval and therefore Talha would suggest that the manufacturer’s claim is incorrect R1
Note: The conclusion must refer back to the original claim.
Allow use of a two sided tt-test giving a pp-value rounding to 0.04<0.050.04<0.05 and therefore Talha would suggest that the manufacturer’s claims in incorrect.
[1 mark]