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Date May Example question Marks available 11 Reference code EXM.2.AHL.TZ0.24
Level Additional Higher Level Paper Paper 2 Time zone Time zone 0
Command term State and Justify Question number 24 Adapted from N/A

Question

The hens on a farm lay either white or brown eggs. The eggs are put into boxes of six. The farmer claims that the number of brown eggs in a box can be modelled by the binomial distribution, B(6, p ). By inspecting the contents of 150 boxes of eggs she obtains the following data.

Show that this data leads to an estimated value of p = 0.4 .

[1]
a.

Stating null and alternative hypotheses, carry out an appropriate test at the 5 % level to decide whether the farmer’s claim can be justified.

[11]
b.

Markscheme

from the sample, the probability of a brown egg is

0 × 7 + 1 × 32 + 6 × 150 = 360 900 = 0.4        A1

p = 0.4        AG

[1 mark]

a.

if the data can be modelled by a binomial distribution with p = 0.4 , the expected frequencies of boxes are given in the table

          A3

Notes: Deduct one mark for each error or omission.
Accept any rounding to at least one decimal place.

null hypothesis: the distribution is binomial          A1

alternative hypothesis: the distribution is not binomial          A1

for a chi-squared test the last two columns should be combined           R1

χ calc 2 = ( 7 7 ) 2 7 + ( 32 28 ) 2 28 + = 6.05   (Accept 6.06)          (M1)A1

degrees of freedom = 4          A1

critical value = 9.488           A1

Or use of p -value

we conclude that the farmer’s claim can be justified          R1

[11 marks]

b.

Examiners report

[N/A]
a.
[N/A]
b.

Syllabus sections

Topic 4—Statistics and probability » SL 4.11—Expected, observed, hypotheses, chi squared, gof, t-test
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