UMBRELLA TESTS for a one-way layout Test whether treatment groups increase to a particular treatment group then decrease afterwards Two versions: Known peak and unknown peak Starting random-number seed: 123456 (Enter a number on the command line to specify a different seed.) FIRST DATA SET: Five treatment groups with 10 observations each (Simulated data) Number of treatments: 5 Total number of observations: 50 #1 (n=10): 308 308 12 206 368 264 88 240 118 264 #2 (n=10): 126 252 262 258 136 204 170 276 314 308 #3 (n=10): 218 364 300 322 46 180 278 454 70 188 #4 (n=10): 358 366 394 252 430 274 216 284 350 290 #5 (n=10): 332 358 356 178 262 158 176 278 240 304 Data means and rank averages for 5 treatment groups: #1 (n=10): DataAv: 217.60 RankAv: 21.35 #2 (n=10): DataAv: 230.60 RankAv: 20.40 #3 (n=10): DataAv: 242.00 RankAv: 24.35 #4 (n=10): DataAv: 321.40 RankAv: 35.40 #5 (n=10): DataAv: 264.20 RankAv: 26.00 THE UMBRELLA TEST for KNOWN p=4 (Mack-Wolfe): Single-umbrella test: Is the data significant for a `peak' in treatment-group medians at p=4? Mann-Whitney differences between the treatment groups: 1. - 51 54 77 59.5 2. 49 - 56 81.5 64.5 3. 46 44 - 69 52.5 4. 23 18.5 31 - 28.5 5. 40.5 35.5 47.5 71.5 - Example of an Ap statistic: Ap(4) = U(1,2)+U(1,3)+U(1,4)+U(2,3)+U(2,4)+U(3,4) + U(5,4) = 51 + 54 + 77 + 56 + 81.5 + 69 + 71.5 = 460 The umbrella statistics Ap with normal approximations Astar: p=1 Ap=406.5 Mean=500.0 Var=3416.67 Astar=-1.600 P=0.94516 (1-sided) p=2 Ap=299.0 Mean=350.0 Var=2391.67 Astar=-1.043 P=0.85149 (1-sided) p=3 Ap=311.0 Mean=300.0 Var=2050.00 Astar= 0.243 P=0.40402 (1-sided) *p=4 Ap=460.0 Mean=350.0 Var=2391.67 Astar= 2.249 P=0.01225 (1-sided) p=5 Ap=593.5 Mean=500.0 Var=3416.67 Astar= 1.600 P=0.05484 (1-sided) Simulating the P-value of Ap=460 (p=4) using 10,000 permutations: P = 112/10,000 = 0.0112 95% CI (0.0091, 0.0112, 0.0133) (1-sided) THE UMBRELLA TEST with UNKNOWN p (Chen-Wolfe form): Note that the one-sided P-values in the table above are NOT multiple-comparison corrected. Multiple-umbrella test: Is there a `peak' in treatment-group medians at SOME p, and if so, at which treatment group? MAX ASTAR=2.249 at p=4 Simulating the P-value of MaxAstar=2.249 using 10,000 permutations: The multiple-comparison-corrected P-value for MaxAstar=2.249 is: P = 516/10,000 = 0.0516 95% CI (0.0473, 0.0516, 0.0559) (1-sided) -------------------------------------------- -------------------------------------------- SECOND DATA SET: Fasting metabolic rates in deer by months Hollander & Wolfe Table 6.8, p215 Number of treatments: 6 Total number of observations: 26 Jan-Feb (n=7): 36.0 33.6 26.9 35.8 30.1 31.2 35.3 Mar-Apr (n=3): 39.9 29.1 43.4 May-Jun (n=5): 44.6 54.4 48.2 55.7 50.0 Jul-Aug (n=4): 53.8 53.9 62.5 46.6 Sep-Oct (n=4): 44.3 34.1 35.7 35.6 Nov-Dec (n=3): 31.7 22.1 30.7 Data means and rank averages for 6 treatment groups: Jan-Feb (n=7): DataAv: 32.70 RankAv: 8.14 Mar-Apr (n=3): DataAv: 37.47 RankAv: 11.33 May-Jun (n=5): DataAv: 50.58 RankAv: 21.60 Jul-Aug (n=4): DataAv: 54.20 RankAv: 22.50 Sep-Oct (n=4): DataAv: 37.43 RankAv: 12.25 Nov-Dec (n=3): DataAv: 28.17 RankAv: 4.33 THE UMBRELLA TEST for KNOWN p=4 (Mack-Wolfe): Single-umbrella test: Is the data significant for a `peak' in treatment-group medians at p=4? Mann-Whitney differences between the treatment groups: 1. - 15 35 28 21 5 2. 6 - 15 12 6 2 3. 0 0 - 12 0 0 4. 0 0 8 - 0 0 5. 7 6 20 16 - 0 6. 16 7 15 12 12 - Example of an Ap statistic: Ap(4) = U(1,2)+U(1,3)+U(1,4)+U(2,3)+U(2,4)+U(3,4) + U(6,5)+U(6,4)+U(5,4) = 15 + 35 + 28 + 15 + 12 + 12 + 12 + 12 + 16 = 157 The umbrella statistics Ap with normal approximations Astar: p=1 Ap=125.0 Mean=138.0 Var=493.17 Astar=-0.585 P=0.72086 (1-sided) p=2 Ap=111.0 Mean= 82.0 Var=269.17 Astar= 1.768 P=0.03856 (1-sided) p=3 Ap=148.0 Mean= 83.0 Var=291.50 Astar= 3.807 P=0.00007 (1-sided) *p=4 Ap=157.0 Mean= 85.5 Var=291.92 Astar= 4.185 P=0.00001 (1-sided) p=5 Ap=156.0 Mean=109.5 Var=383.92 Astar= 2.373 P=0.00882 (1-sided) p=6 Ap=151.0 Mean=138.0 Var=493.17 Astar= 0.585 P=0.27914 (1-sided) Simulating the P-value of Ap=157 (p=4) using 100,000 permutations: P = 0/100,000 = 0.0000 95% CI (0.0000, 0.0000, 0.0000) (1-sided) THE UMBRELLA TEST with UNKNOWN p (Chen-Wolfe form): Note that the one-sided P-values in the table above are NOT multiple-comparison corrected. Multiple-umbrella test: Is there a `peak' in treatment-group medians at SOME p, and if so, at which treatment group? MAX ASTAR=4.185 at p=4 Simulating the P-value of MaxAstar=4.185 using 100,000 permutations: The multiple-comparison-corrected P-value for MaxAstar=4.185 is: P = 2/100,000 = 0.0000 95% CI (-0.0000, 0.0000, 0.0000) (1-sided) Random numbers used: 5,980,000