χ²-tests — practice questions

1. Hypotheses $H_0$: The die is fair. (The data follows a discrete uniform distribution). $H_1$: The die is not fair.
2. 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.
3. 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 |

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

4. 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 $\chi^2_{5}(0.05) = 11.070$. 5. 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.

1. 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.
2. 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. 3. 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.) 4. 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 $\chi^2_{2}(0.05) = 5.991$. 5. 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.