Q-test TEST FOR OUTLIERS (non-parametric) ug/l sorted Q 1 14.27 2 13.43 3 14.25 4 14.83 5 14.64 6 14.09 7 15.19 8 12.93 9 13.94 10 11.20 alpha= range= Qcrit= H0….. ##### Sheet/List 2 ##### Confidence interval https://www.youtube.com/watch?v=cxUWQCwxQgk for MEDIAN (non-parametric distributions) Find 95%-confidence interval of medians for both samples A and B: species A species B 16 34 32 36 37 38 39 45 40 50 41 54 42 56 50 59 82 69 91 ##### Sheet/List 3 ##### Wilcoxon signed-rank test FOR PAIRS (DEPENDENT SAMPLES) A new drug for blood pressure correction was tested on 9 patients. The pressure was measured before and after application. "Decide, if there is an effect of the drug!" after 84 75 88 91 85 65 71 90 75 before 97 72 93 110 95 78 69 115 75 differences abs rank ##### Sheet/List 4 ##### Mann-Whitney U-test "There were selected 11 fields of similar quality. On 5 randomly selected fields a new fertilizer was tested, the rest was kept without the fertilizer." "Yields of wheat obtained in August were: 5.1, 6.7, 5.6, 6.3, 5.9 t/ha and 4.5, 5.4, 4.8, 4.4, 5.3, 5.0 t/ha, resp. " "Find, if the fertilizer has an effect." fertilizer no fertil. 5.1 4.5 6.7 5.4 5.6 4.8 6.3 4.4 5.9 5.3 5.0 ##### Sheet/List 5 ##### Kruskal-Wallis test "In 3 samples of crude oils, Ni was determined by AAS. Decide (applying Kruskal-Wallis test), " if there is a significant difference among the samples. Sample 1 14.2 16.8 19.1 15.5 16 15.9 2 14.5 20 18 15.4 16.1 17.7 3 18.3 20.1 17.7 17.9 19.3 16.9 alpha=0.05 ##### Sheet/List 6 ##### NON-PARAMETRIC PARAMETRIC data sorted Q data sorted G 1 32.56 30.10 0.35 "< 0,526" 1 32.56 30.10 1.8745 "< 2,1266" 2 35.14 32.56 2 35.14 32.56 3 34.10 33.33 3 34.10 33.33 4 33.33 34.10 4 33.33 34.10 5 37.10 34.45 5 37.10 34.45 6 34.45 35.12 6 34.45 35.12 7 35.12 35.14 7 35.12 35.14 8 30.10 37.10 0.28 "< 0,526" 8 30.10 37.10 1.5008 "< 2,1266" Q-test Grubbs test alpha: 0.05 mean: 33.9875 range: 7.00 stand. dev. (sample): 2.0739 Qcritical: 0.526 OK Gcritical: 2.1266 OK H0: Accepted. There is no outlier present H0: Accepted. There is no outlier present Confidence interval estimation Student's confident interval data sorted data sorted 1 32.56 30.10 1 32.56 30.10 2 35.14 32.56 2 35.14 32.56 3 34.10 33.33 3 34.10 33.33 4 33.33 34.10 4 33.33 34.10 5 37.10 34.45 5 37.10 34.45 6 34.45 35.12 6 34.45 35.12 7 35.12 35.14 7 35.12 35.14 8 30.10 37.10 8 30.10 37.10 count = 8 https://www.youtube.com/watch?v=MUD390jtgQs count: 8 mean = 33.99 95% CI = "<30,10;35,14>" stand. dev. (sample) = 2.0739 v = n - 1 v = 8 - 1 v = 7 alpha/2 = 0.025 tv;alpha/2 = 2.3646 95% CI = mean +/- tv;alpha/2 * (stand. dev. (sample) / SQUARE(count)) 95% CI = "33,99 +/-" 1.7338 32.254 35.721 95% CI = "<32,254;35,721>" Conclusion: In my opinion in this case is better Student's CI because it is more acuracy than CI estimation. Non-parametric is much more suitable when we don't know distribution data.