Worked example 1

A die was rolled 180 times with the following results:

Score	1	2	3	4	5	6
Frequency	25	35	28	32	22	38

Test, at the 5% significance level, whether the die is fair.

Show solution outline

Hypotheses $H_0$ : The die is fair. (The data follows a discrete uniform distribution). $H_1$ : The die is not fair.
Expected Frequencies If the die is fair, the probability of each score is $1/6$ . The total number of rolls is 180. Expected frequency for each score, $E_i = 180 \times \frac{1}{6} = 30$ . All expected frequencies are > 5, so the test is valid.

Calculate the Test Statistic We use the formula $\chi^2 = \sum \frac{(O_i - E_i)^2}{E_i}$ .

Score	$O_i$	$E_i$	$O_i - E_i$	$(O_i - E_i)^2$	$(O_i - E_i)^2 / E_i$
1	25	30	-5	25	0.8333
2	35	30	5	25	0.8333
3	28	30	-2	4	0.1333
4	32	30	2	4	0.1333
5	22	30	-8	64	2.1333
6	38	30	8	64	2.1333

$\chi^2_{calc} = 0.8333 + 0.8333 + 0.1333 + 0.1333 + 2.1333 + 2.1333 = 6.2$

Degrees of Freedom and Critical Value Number of categories = 6. Number of restrictions = 1 (since we only used the total frequency). No parameters were estimated. Degrees of freedom, $\nu = 6 - 1 = 5$ . Significance level = 5% = 0.05. From tables, the critical value is $\chi^2_{5}(0.05) = 11.070$ .
Conclusion Since our calculated value $\chi^2_{calc} = 6.2$ is less than the critical value of 11.070, we do not reject $H_0$ . There is insufficient evidence at the 5% significance level to suggest that the die is not fair.

Worked example 2

A survey of 200 sixth form students recorded their year group and preferred subject choice from a selection. The results are shown in the table. Test, at the 5% significance level, whether there is an association between a student's year group and their subject preference.

	Maths	History	Art	Total
Year 12	45	25	20	90
Year 13	55	35	20	110
Total	100	60	40	200

Show solution outline

Hypotheses $H_0$ : There is no association between year group and subject preference. $H_1$ : There is an association between year group and subject preference.
Expected Frequencies We calculate $E_{ij} = \frac{\text{Row Total} \times \text{Column Total}}{\text{Grand Total}}$ . $E_{11} = \frac{90 \times 100}{200} = 45$ $E_{12} = \frac{90 \times 60}{200} = 27$ $E_{13} = \frac{90 \times 40}{200} = 18$ $E_{21} = \frac{110 \times 100}{200} = 55$ $E_{22} = \frac{110 \times 60}{200} = 33$ $E_{23} = \frac{110 \times 40}{200} = 22$ Table of Expected Frequencies:
Maths History Art
Year 12 45 27 18
Year 13 55 33 22
All expected frequencies are > 5.
Calculate the Test Statistic $\chi^2_{calc} = \sum \frac{(O - E)^2}{E}$ $= \frac{(45-45)^2}{45} + \frac{(25-27)^2}{27} + \frac{(20-18)^2}{18} + \frac{(55-55)^2}{55} + \frac{(35-33)^2}{33} + \frac{(20-22)^2}{22}$ $= 0 + \frac{4}{27} + \frac{4}{18} + 0 + \frac{4}{33} + \frac{4}{22}$ $= 0 + 0.1481... + 0.2222... + 0 + 0.1212... + 0.1818...$ $= 0.6733... \approx 0.673$ (3 s.f.)
Degrees of Freedom and Critical Value The table is 2x3. Degrees of freedom, $\nu = (2-1) \times (3-1) = 1 \times 2 = 2$ . Significance level = 5% = 0.05. From tables, the critical value is $\chi^2_{2}(0.05) = 5.991$ .
Conclusion Since $\chi^2_{calc} = 0.673$ is less than the critical value of 5.991, we do not reject $H_0$ . There is insufficient evidence at the 5% significance level to suggest an association between a student's year group and their subject preference.

χ²-tests — practice questions

Worked example 1

Worked example 2