<\/p>\n
Friedman testi,<\/strong> tekrarlanan \u00f6l\u00e7\u00fcmler ANOVA’ya parametrik olmayan bir alternatiftir. Her grupta ayn\u0131 deneklerin yer ald\u0131\u011f\u0131 \u00fc\u00e7 veya daha fazla grubun ortalamalar\u0131 aras\u0131nda istatistiksel olarak anlaml\u0131 bir fark olup olmad\u0131\u011f\u0131n\u0131 belirlemek i\u00e7in kullan\u0131l\u0131r.<\/span><\/p>\n Bu e\u011fitimde Friedman testinin R’de nas\u0131l ger\u00e7ekle\u015ftirilece\u011fi a\u00e7\u0131klanmaktad\u0131r.<\/span><\/p>\n R’de Friedman testini ger\u00e7ekle\u015ftirmek i\u00e7in a\u015fa\u011f\u0131daki s\u00f6zdizimini kullananriedman.test()<\/strong> fonksiyonunu kullanabiliriz:<\/span><\/p>\n riedman.test (y, gruplar, bloklar)<\/span><\/strong><\/p>\n Alt\u0131n:<\/span><\/p>\n Bu fonksiyon bir Ki-kare test istatisti\u011fi ve buna kar\u015f\u0131l\u0131k gelen bir p de\u011feri \u00fcretir. P de\u011feri belirli bir anlaml\u0131l\u0131k d\u00fczeyinin alt\u0131ndaysa (ortak se\u00e7enekler 0,10, 0,05 ve 0,01’dir), o zaman gruplar\u0131n her biri aras\u0131ndaki ortalamalar\u0131n e\u015fit olmad\u0131\u011f\u0131na dair yeterli kan\u0131t vard\u0131r.<\/span><\/p>\n Bu \u00f6zelli\u011fin nas\u0131l kullan\u0131laca\u011f\u0131n\u0131 g\u00f6stermek i\u00e7in, be\u015f hastan\u0131n d\u00f6rt farkl\u0131 ila\u00e7 tedavisine verdi\u011fi reaksiyon s\u00fcresini g\u00f6steren bir veri seti olu\u015fturaca\u011f\u0131z. Her hasta d\u00f6rt ilac\u0131n her birinde \u00f6l\u00e7\u00fcld\u00fc\u011f\u00fcnden, ortalama reaksiyon s\u00fcresinin ila\u00e7lar aras\u0131nda farkl\u0131l\u0131k g\u00f6sterip g\u00f6stermedi\u011fini belirlemek i\u00e7in Friedman testini kullanaca\u011f\u0131z.<\/span><\/p>\n \u00d6ncelikle veri setimizi olu\u015fturaca\u011f\u0131z:<\/span><\/p>\n Daha sonra, yan\u0131t de\u011fi\u015fkeni olarak puan\u0131<\/strong> , grupland\u0131rma de\u011fi\u015fkeni olarak ilac\u0131<\/strong> ve engelleme de\u011fi\u015fkeni olarak ki\u015fiyi<\/strong> kullanarak Friedman testini ger\u00e7ekle\u015ftirece\u011fiz:<\/span><\/p>\n Ki-kare test istatisti\u011fi 13,56<\/strong> ve kar\u015f\u0131l\u0131k gelen p de\u011feri 0,00357’dir<\/strong> . Bu p de\u011feri 0,05’ten k\u00fc\u00e7\u00fck oldu\u011fundan, ortalama yan\u0131t s\u00fcresinin d\u00f6rt ilac\u0131n t\u00fcm\u00fc i\u00e7in ayn\u0131 oldu\u011fu y\u00f6n\u00fcndeki s\u0131f\u0131r hipotezini reddedebiliriz. Kullan\u0131lan ila\u00e7 t\u00fcr\u00fcn\u00fcn yan\u0131t s\u00fcresinde istatistiksel olarak anlaml\u0131 farkl\u0131l\u0131klara yol a\u00e7t\u0131\u011f\u0131 sonucuna varmak i\u00e7in yeterli kan\u0131t\u0131m\u0131z var.<\/span><\/p>\n Friedman testi bize ila\u00e7lar aras\u0131nda ortalama tepki s\u00fcrelerinde farkl\u0131l\u0131klar olup olmad\u0131\u011f\u0131n\u0131 s\u00f6ylese de, bize spesifik olarak hangi ila\u00e7lar\u0131n farkl\u0131 ortalama tepki s\u00fcrelerine sahip oldu\u011funu s\u00f6ylemez. Bunu anlamak i\u00e7in post-hoc test yapmam\u0131z gerekiyor.<\/span><\/p>\n Bir Friedman testi i\u00e7in uygun post hoc testi, a\u015fa\u011f\u0131daki s\u00f6zdizimi kullan\u0131larak uygulanabilen, Bonferroni d\u00fczeltmeli ikili Wilcoxon s\u0131ra toplam\u0131 testidir:<\/span><\/p>\n pairwise.wilcox.test(veri$puan, veri$ila\u00e7, p.adj = \u201ciyilik\u201d)<\/strong><\/span><\/p>\n Alt\u0131n:<\/span><\/p>\n \u00d6rne\u011fimiz i\u00e7in kullanaca\u011f\u0131m\u0131z s\u00f6zdizimi \u015f\u00f6yledir:<\/span><\/p>\n Bu, her ikili Wilcoxon s\u0131ra toplam\u0131 testi i\u00e7in p de\u011ferini g\u00f6steren bir matris \u00fcretir. \u0130statistiksel olarak 0,10 oran\u0131nda anlaml\u0131 farka sahip olan tek ila\u00e7 grubunun grup 3 ve 4 oldu\u011funu g\u00f6r\u00fcyoruz ( p = 0,072<\/strong> ).<\/span><\/p>\n","protected":false},"excerpt":{"rendered":" Friedman testi, tekrarlanan \u00f6l\u00e7\u00fcmler ANOVA’ya parametrik olmayan bir alternatiftir. Her grupta ayn\u0131 deneklerin yer ald\u0131\u011f\u0131 \u00fc\u00e7 veya daha fazla grubun ortalamalar\u0131 aras\u0131nda istatistiksel olarak anlaml\u0131 bir fark olup olmad\u0131\u011f\u0131n\u0131 belirlemek i\u00e7in kullan\u0131l\u0131r. Bu e\u011fitimde Friedman testinin R’de nas\u0131l ger\u00e7ekle\u015ftirilece\u011fi a\u00e7\u0131klanmaktad\u0131r. \u00d6rnek: R’deki Friedman testi R’de Friedman testini ger\u00e7ekle\u015ftirmek i\u00e7in a\u015fa\u011f\u0131daki s\u00f6zdizimini kullananriedman.test() fonksiyonunu kullanabiliriz: riedman.test […]<\/p>\n","protected":false},"author":1,"featured_media":0,"comment_status":"open","ping_status":"open","sticky":false,"template":"","format":"standard","meta":{"footnotes":""},"categories":[11],"tags":[],"class_list":["post-687","post","type-post","status-publish","format-standard","hentry","category-rehber"],"yoast_head":"\n \u00d6rnek: R’deki Friedman testi<\/strong><\/h3>\n
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#create data<\/span>\ndata <- data.frame(person = rep(1:5, each=4),\n drug = rep(c(1, 2, 3, 4), times=5),\n score = c(30, 28, 16, 34, 14, 18, 10, 22, 24, 20,\n 18, 30, 38, 34, 20, 44, 26, 28, 14, 30))\n\n#view data<\/span>\ndata\n\n person drug score\n1 1 1 30\n2 1 2 28\n3 1 3 16\n4 1 4 34\n5 2 1 14\n6 2 2 18\n7 2 3 10\n8 2 4 22\n9 3 1 24\n10 3 2 20\n11 3 3 18\n12 3 4 30\n13 4 1 38\n14 4 2 34\n15 4 3 20\n16 4 4 44\n17 5 1 26\n18 5 2 28\n19 5 3 14\n20 5 4 30\n<\/strong><\/pre>\n #perform Friedman Test<\/span>\nfriedman.test(y=data$score, groups=data$drug, blocks=data$person)\n\n\tFriedman rank sum test\n\ndata: data$score, data$drug and data$person\nFriedman chi-squared = 13.56, df = 3, p-value = 0.00357\n<\/strong><\/pre>\n\n
<\/span><\/li>\n<\/ul>\n #perform post-hoc tests<\/span>\npairwise.wilcox.test(data$score, data$drug, p.adj = \"bonf\")\n\tPairwise comparisons using Wilcoxon rank sum test \n\ndata: data$score and data$drug \n\n 1 2 3 \n2 1,000 - - \n3 0.449 0.210 - \n4 1,000 1,000 0.072\n\nP value adjustment method: bonferroni \n<\/strong><\/pre>\n