EXM.2.AHL.TZ0.29

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

A horse breeder records the number of births for each of 100 horses during the past eight years. The results are summarized in the following table:

Stating null and alternative hypotheses carry out an appropriate test at the 5% significance level to decide whether the results can be modelled by B (6, 0.5).

[10]

Without doing any further calculations, explain briefly how you would carry out a test, at the 5% significance level, to decide if the data can be modelled by B(6, $p$ ), where $p$ is unspecified.

[2]

A different horse breeder collected data on the time and outcome of births. The data are summarized in the following table:

Carry out an appropriate test at the 5% significance level to decide whether there is an association between time and outcome.

[8]

Markscheme / solution

METHOD 1

H₀: distribution is B(6, 0.5); H₁: distribution is not B(6, 0.5) A1

$\left( {{E_0} = 100{{\left( {0.5} \right)}^6} = \frac{{25}}{{16}} = 0.015625} \right)$ A3

Combining the first two columns and the last two columns: A1

${\chi ^2} = \sum {\frac{{{O^2}}}{E}} - \sum E$

$= \frac{{{6^2}}}{{\left( {\frac{{175}}{{16}}} \right)}} + \frac{{{{26}^2}}}{{\left( {\frac{{375}}{{16}}} \right)}} + \frac{{{{37}^2}}}{{\left( {\frac{{500}}{{16}}} \right)}} + \frac{{{{18}^2}}}{{\left( {\frac{{375}}{{16}}} \right)}} + \frac{{{{13}^2}}}{{\left( {\frac{{175}}{{16}}} \right)}} - 100$ (M1)

= 5.22 A1

$v$ = 4, so critical value of $\chi _{5{\text{% }}}^2 = 9.488$ A1A1

Since 5.22 < 9.488 the result is not significant and we accept H₀ R1

METHOD 2

H₀: distribution is B(6, 0.5); H₁: distribution is not B(6, 0.5) A1

By GDC, $p = 0.266$ A8

Since 0.266 > 0.05 the result is not significant and we accept H₀ R1

[10 marks]

Estimate $p$ from the data which would entail the loss of one degree of freedom A1A1

[2 marks]

H₀: there is no association H₁: there is an association A1

${\chi ^2} = \frac{{{{68}^2}}}{{64.8}} + \frac{{{{42}^2}}}{{45.2}} + \frac{{{{103}^2}}}{{94.3}} + \ldots + \frac{{{6^2}}}{{10.6}} + \frac{{{{12}^2}}}{{7.4}} - 314$ (M1)

= 15.7

$v = 3{\text{,}}\,\,\chi _{5{\text{% }}}^2\left( 3 \right) = 7.815$ A1A1

Since 15.7 > 7.815 we reject H₀ R1

METHOD 2

H₀: there is no association H₁: there is an association A1

By GDC, $p$ = 0.00129 A6

Since 0.00129 < 0.05 we reject H₀. R1

[8 marks]

Examiners’ report

[N/A]