1. **Problem Statement:** Test the hypothesis that the two laboratories have the same samples at 5% significance level using: i. Median test ii. Mann-Whitney U test iii. Wald run test 2. **Data:** Lab A: 9, 3, 1, 3, 0, 7, 2, 11 Lab B: 12, 6, 6, 4, 8, 5, 4, 0 --- ### i. Median Test 3. **Step 1: Find the combined median** Combine all data: 0,0,1,2,3,3,4,4,5,6,6,7,8,9,11,12 Sorted combined data has 16 values, median is average of 8th and 9th values: $$\text{Median} = \frac{4 + 5}{2} = 4.5$$ 4. **Step 2: Create a 2x2 contingency table** Count values above and below median for each lab: - Lab A above median (>4.5): 7,9,11 (3 values) - Lab A below median (<=4.5): 0,1,2,3,3 (5 values) - Lab B above median: 5,6,6,8,12 (5 values) - Lab B below median: 0,4,4 (3 values) | | > Median | <= Median | |--------|----------|-----------| | Lab A | 3 | 5 | | Lab B | 5 | 3 | 5. **Step 3: Perform Chi-square test** Expected counts under null hypothesis: $$E_{ij} = \frac{(\text{row total})(\text{column total})}{\text{grand total}}$$ Row totals: Lab A = 8, Lab B = 8 Column totals: >Median = 8, <=Median = 8 Grand total = 16 Expected counts: $$E_{LabA,>Median} = \frac{8 \times 8}{16} = 4$$ $$E_{LabA,<=Median} = 4$$ $$E_{LabB,>Median} = 4$$ $$E_{LabB,<=Median} = 4$$ 6. **Step 4: Calculate Chi-square statistic** $$\chi^2 = \sum \frac{(O - E)^2}{E} = \frac{(3-4)^2}{4} + \frac{(5-4)^2}{4} + \frac{(5-4)^2}{4} + \frac{(3-4)^2}{4} = 1$$ 7. **Step 5: Decision** Degrees of freedom = (2-1)(2-1) = 1 Critical value at 5% significance = 3.841 Since $1 < 3.841$, fail to reject null hypothesis. --- ### ii. Mann-Whitney U Test 8. **Step 1: Rank all observations combined** Data combined and sorted: 0,0,1,2,3,3,4,4,5,6,6,7,8,9,11,12 Ranks assigned (average ranks for ties): 0(1.5),0(1.5),1(3),2(4),3(5.5),3(5.5),4(7.5),4(7.5),5(9),6(10.5),6(10.5),7(12),8(13),9(14),11(15),12(16) 9. **Step 2: Sum ranks for each lab** Lab A values and ranks: 9(14),3(5.5),1(3),3(5.5),0(1.5),7(12),2(4),11(15) Sum Lab A ranks: $$14 + 5.5 + 3 + 5.5 + 1.5 + 12 + 4 + 15 = 60.5$$ Lab B values and ranks: 12(16),6(10.5),6(10.5),4(7.5),8(13),5(9),4(7.5),0(1.5) Sum Lab B ranks: $$16 + 10.5 + 10.5 + 7.5 + 13 + 9 + 7.5 + 1.5 = 75.5$$ 10. **Step 3: Calculate U statistics** $$U_A = n_A n_B + \frac{n_A(n_A+1)}{2} - R_A = 8 \times 8 + \frac{8 \times 9}{2} - 60.5 = 64 + 36 - 60.5 = 39.5$$ $$U_B = n_A n_B - U_A = 64 - 39.5 = 24.5$$ 11. **Step 4: Determine critical value** For $n_A = n_B = 8$ and $\alpha = 0.05$, two-tailed critical value is 13. Since $U_B = 24.5 > 13$, fail to reject null hypothesis. --- ### iii. Wald Run Test 12. **Step 1: Combine and order data by value, label by lab** Sorted combined data with labels: 0(B),0(A),1(A),2(A),3(A),3(A),4(B),4(B),5(B),6(B),6(B),7(A),8(B),9(A),11(A),12(B) 13. **Step 2: Count runs** A run is a sequence of identical labels: Sequence: B, A, A, A, A, B, B, B, B, B, A, B, A, A, B Runs: 1) B 2) A (0(A),1(A),2(A),3(A)) 3) B (4(B),4(B),5(B),6(B),6(B)) 4) A (7(A)) 5) B (8(B)) 6) A (9(A),11(A)) 7) B (12(B)) Total runs = 7 14. **Step 3: Calculate expected runs and variance** $$n_1 = 8, n_2 = 8, n = 16$$ Expected runs: $$E(R) = \frac{2 n_1 n_2}{n} + 1 = \frac{2 \times 8 \times 8}{16} + 1 = 9$$ Variance: $$Var(R) = \frac{2 n_1 n_2 (2 n_1 n_2 - n)}{n^2 (n-1)} = \frac{2 \times 8 \times 8 (2 \times 8 \times 8 - 16)}{16^2 \times 15} = \frac{128 (128 - 16)}{256 \times 15} = \frac{128 \times 112}{3840} = 3.733$$ 15. **Step 4: Calculate Z statistic** $$Z = \frac{R - E(R)}{\sqrt{Var(R)}} = \frac{7 - 9}{\sqrt{3.733}} = \frac{-2}{1.932} = -1.036$$ 16. **Step 5: Decision** At $\alpha = 0.05$, critical Z values are $\pm 1.96$ Since $-1.036$ is within $-1.96$ and $1.96$, fail to reject null hypothesis. --- **Final conclusion:** All three tests (Median, Mann-Whitney U, Wald run) fail to reject the null hypothesis at 5% significance level, indicating no significant difference between the two laboratories' samples.