EXM.2.AHL.TZ0.24

pestleMathematicsAIHLPaper 2EXM· sl-4-11-expected-observed-hypotheses-chi-squared-gof-t-testsource ↗

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]

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]

Markscheme / solution

from the sample, the probability of a brown egg is

$\frac{{0 \times 7 + 1 \times 32 + \ldots }}{{6 \times 150}} = \frac{{360}}{{900}} = 0.4$ A1

$p = 0.4$ AG

[1 mark]

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

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

$\chi _{{\text{calc}}}^2 = \frac{{{{\left( {7 - 7} \right)}^2}}}{7} + \frac{{{{\left( {32 - 28} \right)}^2}}}{{28}} + \ldots = 6.05$ (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]

Examiners’ report

[N/A]