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Date	May 2014	Marks available	2	Reference code	14M.1.sl.TZ2.8
Level	SL only	Paper	1	Time zone	TZ2
Command term	Find	Question number	8	Adapted from	N/A

Question

A group of 100 students gave the following responses to the question of how they get to school.

A ${\chi ^2}$ test for independence was conducted at the $5\%$ significance level. The null hypothesis was defined as

${{\text{H}}_0}$ : Method of getting to school is independent of gender.

Find the expected frequency for the females who use public transport to get to school.

[2]

Find the ${\chi ^2}$ statistic.

[2]

The ${\chi ^2}$ critical value is $7.815$ at the $5\%$ significance level.

State whether or not the null hypothesis is accepted. Give a reason for your answer.

[2]

Markscheme

$\frac{{30}}{{100}} \times \frac{{48}}{{100}} \times 100$ OR $\frac{{30 \times 48}}{{100}}$ (M1)

Note: Award (M1) for correct substitution into correct formula.

$= 14.4{\text{ }}\left( {\frac{{72}}{5}} \right)$ (A1) (C2)

[2 marks]

$13.0{\text{ }}(12.9554…)$ (A2) (C2)

Note: Award (A1)(A0) for $12.9$ .

[2 marks]

the null hypothesis is not accepted (A1)(ft)

$\chi _{calc}^2 > \chi _{crit}^2$ OR $13.0 > 7.82$ (R1)

the null hypothesis is not accepted (A1)(ft)

p-value ${\text{(0.0047) (0.00473391}} \ldots {\text{)}} < 0.05$ (R1) (C2)

Notes: Follow through from their answer to part (b).

Do not award (A1)(ft)(R0).

[2 marks]

Examiners report

[N/A]

Syllabus sections

Topic 4 - Statistical applications » 4.4 » The

${\chi ^2}$ test for independence: formulation of null and alternative hypotheses; significance levels; contingency tables; expected frequencies; degrees of freedom;

$p$ -values.

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