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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=209738Σx2=209738.

Find an unbiased estimate for the mean number (μ)(μ) of chocolates per packet.

[1]
a.

Use the formula s2n-1=Σx2-(Σx)2nn-1s2n1=Σx2(Σx)2nn1 to determine an unbiased estimate for the variance of the number of chocolates per packet.

[2]
b.

Find a 95%95% confidence interval for μμ. You may assume that all conditions for a confidence interval have been met.

[2]
c.

Suggest, with justification, a valid conclusion that Talha could make.

[1]
d.

Markscheme

ˉx=Σxn=250630=83.5  (83.5333)¯x=Σxn=250630=83.5  (83.5333)        A1


[1 mark]

a.

(s2n-1=Σx2-(Σx)2nn-1=) 209738-250623029s2n1=Σx2(Σx)2nn1= 209738250623029           (M1)

=13.9  (13.9126)=13.9  (13.9126)           A1


[2 marks]

b.

(82.1, 84.9)  (82.1405, 84.9261)(82.1, 84.9)  (82.1405, 84.9261)       A2


[2 marks]

c.

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]

d.

Examiners report

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Syllabus sections

Topic 4—Statistics and probability » AHL 4.14—Linear transformation of a single RV, E(X) and VAR(X), unbiased estimators
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Topic 4—Statistics and probability » AHL 4.16—Confidence intervals
Topic 4—Statistics and probability

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