1. **Problem statement:** We want to calculate the Chi-square test statistic to determine if there is a significant association between gender and sport. 2. **Formula:** The Chi-square test statistic is given by $$\chi^2 = \sum \frac{(O - E)^2}{E}$$ where $O$ are the observed values and $E$ are the expected values. 3. **Calculation:** Given observed and expected values: $$\chi^2 = \frac{(1590 - 1570.5)^2}{1570.5} + \frac{(1451 - 1470.5)^2}{1470.5} + \frac{(116 - 135.5)^2}{135.5} + \frac{(146 - 126.5)^2}{126.5}$$ Calculate each squared difference: $$19.5^2 = 380.25$$ Substitute: $$\chi^2 = \frac{380.25}{1570.5} + \frac{380.25}{1470.5} + \frac{380.25}{135.5} + \frac{380.25}{126.5}$$ 4. **Simplify each term:** $$\frac{380.25}{1570.5} \approx 0.242$$ $$\frac{380.25}{1470.5} \approx 0.259$$ $$\frac{380.25}{135.5} \approx 2.805$$ $$\frac{380.25}{126.5} \approx 3.005$$ 5. **Sum all terms:** $$\chi^2 \approx 0.242 + 0.259 + 2.805 + 3.005 = 6.311$$ 6. **Degrees of freedom (DOF):** $$DOF = (rækker - 1) \times (kolonner - 1) = (2 - 1) \times (2 - 1) = 1$$ 7. **Critical value:** For $\alpha = 0.05$ and 1 DOF, the critical value is $$\chi^2_{0.05,1} = 3.841$$ 8. **Conclusion:** Since $$6.311 > 3.841$$ we reject the null hypothesis $H_0$. There is statistically significant evidence at the 5% significance level that gender and sport are dependent.