1. **Problem:** Test if there is a difference in the age of onset of Alzheimer's symptoms between men and women at a 5% significance level. 2. **Hypotheses:** - Null hypothesis $H_0$: $\mu_{men} = \mu_{women}$ (no difference in mean ages) - Alternative hypothesis $H_a$: $\mu_{men} \neq \mu_{women}$ (difference in mean ages) 3. **Data:** Men: 67, 73, 70, 62, 65, 59, 80, 66 Women: 70, 68, 57, 66, 74, 67, 61, 72, 64 4. **Method:** Use a two-sample t-test for means with unequal sample sizes. 5. **Calculate sample means:** $$\bar{x}_{men} = \frac{67+73+70+62+65+59+80+66}{8} = \frac{542}{8} = 67.75$$ $$\bar{x}_{women} = \frac{70+68+57+66+74+67+61+72+64}{9} = \frac{599}{9} \approx 66.56$$ 6. **Calculate sample variances:** Men variance $s_{men}^2$: $$s_{men}^2 = \frac{\sum (x_i - \bar{x}_{men})^2}{n-1} = \frac{(67-67.75)^2 + \cdots + (66-67.75)^2}{7} = 34.93$$ Women variance $s_{women}^2$: $$s_{women}^2 = \frac{\sum (x_i - \bar{x}_{women})^2}{n-1} = \frac{(70-66.56)^2 + \cdots + (64-66.56)^2}{8} = 33.98$$ 7. **Calculate standard error (SE):** $$SE = \sqrt{\frac{s_{men}^2}{n_{men}} + \frac{s_{women}^2}{n_{women}}} = \sqrt{\frac{34.93}{8} + \frac{33.98}{9}} = \sqrt{4.37 + 3.78} = \sqrt{8.15} = 2.86$$ 8. **Calculate t-statistic:** $$t = \frac{\bar{x}_{men} - \bar{x}_{women}}{SE} = \frac{67.75 - 66.56}{2.86} = \frac{1.19}{2.86} = 0.42$$ 9. **Degrees of freedom (Welch's approximation):** $$df = \frac{(s_{men}^2/n_{men} + s_{women}^2/n_{women})^2}{\frac{(s_{men}^2/n_{men})^2}{n_{men}-1} + \frac{(s_{women}^2/n_{women})^2}{n_{women}-1}} \approx 14$$ 10. **Find p-value:** For $t=0.42$ and $df=14$, two-tailed p-value $\approx 0.68$ (using t-distribution table or calculator). 11. **Decision:** Since $p=0.68 > 0.05$, we fail to reject $H_0$. 12. **Conclusion:** There is no statistically significant difference in the age of onset of Alzheimer's symptoms between men and women at the 5% significance level.