| Date | May 2014 | Marks available | 2 | Reference code | 14M.1.sl.TZ2.8 |
| Level | SL only | Paper | 1 | Time zone | TZ2 |
| Command term | State | 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]
a.
Find the \({\chi ^2}\) statistic.
[2]
b.
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]
c.
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]
a.
\(13.0{\text{ }}(12.9554…)\) (A2) (C2)
Note: Award (A1)(A0) for \(12.9\).
[2 marks]
b.
the null hypothesis is not accepted (A1)(ft)
\(\chi _{calc}^2 > \chi _{crit}^2\) OR \(13.0 > 7.82\) (R1)
OR
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]
c.
Examiners report
[N/A]
a.
[N/A]
b.
[N/A]
c.
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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