Standard +0.3 This is a standard chi-squared test of independence with a 2×2 contingency table. Students must construct the table from given information, calculate expected frequencies, compute the test statistic, and compare to critical value. While it requires careful organization of data and multiple computational steps, it follows a completely routine procedure with no novel insight required. The 2×2 case is the simplest contingency table scenario, making this slightly easier than average for A-level Further Maths statistics.
5 Students at a science department of a university are offered the opportunity to study an optional language module, either German or Mandarin, during their second year of study.
From a sample of 50 students who opted to study a language module, 31 were female. Of those who opted to study Mandarin, 8 were female and 12 were male.
Test, using the \(5 \%\) level of significance, whether choice of language is independent of gender.
The sample of students may be regarded as random. [0pt]
[8 marks]
Turn over for the next question
Evaluates \(\chi^2\) model by comparing test statistic and correct critical value, or by comparing \(p\)-value with 0.05
Reject \(H_0\)
E1
Infers that \(H_0\) should be rejected
Significant evidence to suggest that language choice is not independent of gender
E1
Completes test correctly and gives conclusion in context (conclusion should not be definite)
## Question 5:
| Answer/Working | Mark | Guidance |
|---|---|---|
| $H_0$: Language choice is independent of gender; $H_1$: Language choice is not independent of gender | B1 | States both hypotheses using correct notation |
| Contingency table: M: German=7, Mandarin=12, Total=19; F: German=23, Mandarin=8, Total=31; Totals: 30, 20, 50 | M1 | Constructs correct contingency table with frequencies shown (PI) |
| Expected frequencies: M: German=11.4, Mandarin=7.6; F: German=18.6, Mandarin=12.4 | B1 | Finds expected frequencies (PI) |
| $\chi^2 = \sum\dfrac{(O-E)^2}{E}$; $\chi^2 = \sum\dfrac{(\|O-E\|-0.5)^2}{E} = \dfrac{(3.9)^2}{11.4}+\dfrac{(3.9)^2}{7.6}+\dfrac{(3.9)^2}{18.6}+\dfrac{(3.9)^2}{12.4}$ | M1F | Evaluates formula; allow no Yates correction used; FT their observed and expected frequencies |
| $\chi^2 = 5.38$, 1 tail 5% | A1 | Obtains $\chi^2$ test statistic or quotes $p$-value |
| $df=1$, $cv=3.84$, $5.38 > 3.84$; Reject $H_0$ | M1 | Evaluates $\chi^2$ model by comparing test statistic and correct critical value, or by comparing $p$-value with 0.05 |
| Reject $H_0$ | E1 | Infers that $H_0$ should be rejected |
| Significant evidence to suggest that language choice is not independent of gender | E1 | Completes test correctly and gives conclusion in context (conclusion should not be definite) |
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5 Students at a science department of a university are offered the opportunity to study an optional language module, either German or Mandarin, during their second year of study.
From a sample of 50 students who opted to study a language module, 31 were female. Of those who opted to study Mandarin, 8 were female and 12 were male.
Test, using the $5 \%$ level of significance, whether choice of language is independent of gender.
The sample of students may be regarded as random.\\[0pt]
[8 marks]
Turn over for the next question
\hfill \mbox{\textit{AQA Further Paper 3 Statistics Q5 [8]}}