MTB > # I'll enter the data. MTB > set c1 DATA> 242 245 244 248 247 248 242 244 246 242 DATA> end MTB > set c2 DATA> 248 246 245 247 248 250 247 246 243 244 DATA> end MTB > set c3 DATA> 246 248 250 252 248 250 246 248 245 250 DATA> end MTB > name c1 '0 mg' c2 '100 mg' c3 '200 mg' MTB > # I'll look at the data and some summary statistics. MTB > dotplot c1-c3; SUBC> same. . : : . . . : ---+---------+---------+---------+---------+---------+---0 mg . . . : : : . ---+---------+---------+---------+---------+---------+---100 mg . . . : : : . ---+---------+---------+---------+---------+---------+---200 mg 242.0 244.0 246.0 248.0 250.0 252.0 MTB > desc c1-c3 N MEAN MEDIAN TRMEAN STDEV SEMEAN 0 mg 10 244.80 244.50 244.75 2.39 0.76 100 mg 10 246.40 246.50 246.37 2.07 0.65 200 mg 10 248.30 248.00 248.25 2.21 0.70 MIN MAX Q1 Q3 0 mg 242.00 248.00 242.00 247.25 100 mg 243.00 250.00 244.75 248.00 200 mg 245.00 252.00 246.00 250.00 MTB > # I'll see that ANOVA F test and K-W test yield. MTB > stack c1-c3 c4; SUBC> subs c5. MTB > name c4 'taps/min' c5 'dose' MTB > oneway c4 c5 ANALYSIS OF VARIANCE ON taps/min SOURCE DF SS MS F p dose 2 61.40 30.70 6.18 0.006 ERROR 27 134.10 4.97 TOTAL 29 195.50 INDIVIDUAL 95 PCT CI'S FOR MEAN BASED ON POOLED STDEV LEVEL N MEAN STDEV ----+---------+---------+---------+-- 1 10 244.80 2.39 (------*------) 2 10 246.40 2.07 (------*------) 3 10 248.30 2.21 (------*-------) ----+---------+---------+---------+-- POOLED STDEV = 2.23 244.0 246.0 248.0 250.0 MTB > krus c4 c5 LEVEL NOBS MEDIAN AVE. RANK Z VALUE 1 10 244.5 10.1 -2.38 2 10 246.5 15.1 -0.20 3 10 248.0 21.4 2.57 OVERALL 30 15.5 H = 8.20 d.f. = 2 p = 0.017 H = 8.40 d.f. = 2 p = 0.015 (adj. for ties) MTB > # I'll do a bunch of M-W tests to get info for J-T test statistic. MTB > mann c2 c1 Mann-Whitney Confidence Interval and Test 100 mg N = 10 Median = 246.50 0 mg N = 10 Median = 244.50 Point estimate for ETA1-ETA2 is 2.00 95.5 pct c.i. for ETA1-ETA2 is (-1.00,4.00) W = 123.5 Test of ETA1 = ETA2 vs. ETA1 n.e. ETA2 is significant at 0.1736 The test is significant at 0.1692 (adjusted for ties) Cannot reject at alpha = 0.05 MTB > mann c3 c1 Mann-Whitney Confidence Interval and Test 200 mg N = 10 Median = 248.00 0 mg N = 10 Median = 244.50 Point estimate for ETA1-ETA2 is 4.00 95.5 pct c.i. for ETA1-ETA2 is (1.00,6.00) W = 140.5 Test of ETA1 = ETA2 vs. ETA1 n.e. ETA2 is significant at 0.0082 The test is significant at 0.0074 (adjusted for ties) MTB > mann c3 c2 Mann-Whitney Confidence Interval and Test 200 mg N = 10 Median = 248.00 100 mg N = 10 Median = 246.50 Point estimate for ETA1-ETA2 is 2.00 95.5 pct c.i. for ETA1-ETA2 is (0.00,4.00) W = 128.0 Test of ETA1 = ETA2 vs. ETA1 n.e. ETA2 is significant at 0.0890 The test is significant at 0.0839 (adjusted for ties) Cannot reject at alpha = 0.05 MTB > # We have u_21 = 68.5, u_31 = 85.5,, and u_32 = 73. The value of the MTB > # J-T test statistic is 227. The null mean and variance are 150 and MTB > # 2075/3. The approximate p-value from an upper-tailed test is MTB > # Phi( (150 - 1/2 -227)/sqrt(2075/3) ). MTB > let k1 = (150 - 1/2 -227)/sqrt(2075/3) MTB > cdf k1 k2; SUBC> norm 0 1. MTB > name k1 'z' k2 'p-value' MTB > print k1 k2 z -2.94682 p-value 0.00160539 *** Note: Exact p-value from StatXact is about 0.0012. MTB > # Now I'll do an approximate version of Page's test. MTB > rank c4 c6 MTB > unstack c6 c11-c13; SUBC> subs c5. MTB > let k11 = mean(c11) MTB > let k12 = mean(c12) MTB > let k13 = mean(c13) MTB > let k14 = k11 + 2*k12 + 3*k13 MTB > # The test statistic value is in k14. MTB > print k11-k14 K11 10.1000 K12 15.0500 K13 21.3500 K14 104.250 MTB > # The null mean and variance are 93 and 15.5, and the approximate MTB > # p-value from an upper-tailed test is MTB > # Phi( (93 - 104.25)/sqrt(15.5) ). MTB > let k1 = (93 - 104.25)/sqrt(15.5) MTB > cdf k1 k2; SUBC> norm 0 1. MTB > print k1 k2 z -2.85750 p-value 0.00213498 *** Note: Exact p-value from StatXact is about 0.0014. MTB > save 'tapping' Saving worksheet in file: tapping.MTW