Find the number of degrees of freedom for the \(\chi^2\) test.

Write down the null hypothesis for the \(\chi^2\) test.

The critical value for the \(\chi^2\) test at the 5% significance level is 12.59. The \(\chi^2\) test statistic is calculated to be 146 with a p-value of 6.62 × 10⁻²⁹ (both numbers given correct to 3 significant figures).

Write down the conclusion reached at the 5 % significance level. Give a clear reason for your answer.

\( (3 - 1)(4 -1) \)     (M1)

= 6     (A1)     (C2)

[2 marks]

The preferred type of drink is independent of age.     (A1)     (C1)

Note: For independent accept “not associated” but do not accept “not related” or “not correlated”

[1 mark]

Reject null hypothesis as critical value \( < \chi_{calc}^2\)     (A1)(R1)(ft)

Note: (ft) from their value in (c).

OR

Reject null hypothesis as p-value \(< 0.05\)     (A1)(R1)     (C2)

Notes: Do not award (A1)(R0).

Award the (R1) for comparison of correct values.

[2 marks]

This question caused significant difficulties for many candidates. It seemed, from the responses, that the purpose of the test is not well understood, even if its procedure on the GDC can be performed.

The test is one of “independence” and it should be stressed to candidates that it is this which is key in stating the hypotheses. Improper terminology, most notably, “not correlated” is not acceptable.

This question caused significant difficulties for many candidates. It seemed, from the responses, that the purpose of the test is not well understood, even if its procedure on the GDC can be performed.

This question caused significant difficulties for many candidates. It seemed, from the responses, that the purpose of the test is not well understood, even if its procedure on the GDC can be performed.

Many candidates did not know the correct figures to compare in order to arrive at the decision. Others gave no reason at all.