1. **State the problem:** We have two samples and want to test if they come from the same population at a 5% significance level. Sample 1: 15, 25, 16, 20, 22, 24, 21, 17, 19, 23 (n1=10) Sample 2: 35, 31, 25, 38, 26, 29, 32, 34, 33, 27, 29, 31 (n2=12) 2. **Choose the test:** Since we want to compare two independent samples without assuming normality, we use the Mann-Whitney U test (a non-parametric test). 3. **Combine and rank all values:** Combined data sorted: 15,16,17,19,20,21,22,23,24,25,25,26,27,29,29,31,31,32,33,34,35,38 Assign ranks, averaging ties: - 15(1),16(2),17(3),19(4),20(5),21(6),22(7),23(8),24(9),25(10.5),25(10.5),26(12),27(13),29(14.5),29(14.5),31(16.5),31(16.5),32(18),33(19),34(20),35(21),38(22) 4. **Sum ranks for each sample:** Sample 1 ranks: 1+10.5+2+5+7+9+6+3+4+8 = 55.5 Sample 2 ranks: sum of remaining ranks = total ranks sum - 55.5 Total ranks sum = (22*23)/2 = 253 Sample 2 ranks sum = 253 - 55.5 = 197.5 5. **Calculate U statistics:** $$U_1 = n_1 n_2 + \frac{n_1(n_1+1)}{2} - R_1 = 10 \times 12 + \frac{10 \times 11}{2} - 55.5 = 120 + 55 - 55.5 = 119.5$$ $$U_2 = n_1 n_2 - U_1 = 120 - 119.5 = 0.5$$ 6. **Find critical value or p-value:** For $n_1=10$, $n_2=12$, at 5% significance level (two-tailed), critical U is approximately 37. 7. **Decision:** Since $U_2=0.5 < 37$, we reject the null hypothesis. **Conclusion:** There is sufficient evidence at the 5% level to conclude the two samples are from different populations.