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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=360900=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

χ2calc=(77)27+(3228)228+=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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