<\/p>\n
Welch’in ANOVA’s\u0131,<\/strong> varyanslar\u0131n e\u015fitli\u011fi varsay\u0131m\u0131<\/a> kar\u015f\u0131lanmad\u0131\u011f\u0131nda tipik tek y\u00f6nl\u00fc ANOVA’ya<\/a> bir alternatiftir.<\/span><\/p>\n A\u015fa\u011f\u0131daki ad\u0131m ad\u0131m \u00f6rnek, Welch’in ANOVA’s\u0131n\u0131n Python’da nas\u0131l ger\u00e7ekle\u015ftirilece\u011fini g\u00f6sterir.<\/span><\/p>\n \u00dc\u00e7 farkl\u0131 \u00e7al\u0131\u015fma tekni\u011finin farkl\u0131 s\u0131nav sonu\u00e7lar\u0131na yol a\u00e7\u0131p a\u00e7mad\u0131\u011f\u0131n\u0131 belirlemek i\u00e7in, bir profes\u00f6r rastgele 10 \u00f6\u011frenciye her tekni\u011fi (Teknik A, B veya C) bir hafta boyunca kullanmalar\u0131 i\u00e7in atar ve ard\u0131ndan her \u00f6\u011frenciye e\u015fit zorlukta bir test verir.<\/span><\/p>\n 30 \u00f6\u011frencinin s\u0131nav sonu\u00e7lar\u0131 a\u015fa\u011f\u0131da sunulmu\u015ftur:<\/span><\/p>\n Daha sonra, her grup aras\u0131ndaki varyanslar\u0131n e\u015fit olup olmad\u0131\u011f\u0131n\u0131 belirlemek i\u00e7in Bartlett testi<\/a> yapabiliriz.<\/span><\/p>\n Test istatisti\u011finin p de\u011feri<\/a> belirli bir anlaml\u0131l\u0131k d\u00fczeyinin alt\u0131ndaysa (\u03b1 = 0,05 gibi), o zaman s\u0131f\u0131r hipotezini reddedebilir ve t\u00fcm gruplar\u0131n ayn\u0131 varyansa sahip olmad\u0131\u011f\u0131 sonucuna varabiliriz.<\/span><\/p>\n Python’da Bartlett testini ger\u00e7ekle\u015ftirmek i\u00e7in a\u015fa\u011f\u0131daki kodu kullanabiliriz:<\/span><\/span><\/p>\n Bartlett testinin p de\u011feri ( .01089<\/strong> ) \u03b1 = .05’ten k\u00fc\u00e7\u00fckt\u00fcr; bu, her grubun ayn\u0131 varyansa sahip oldu\u011fu \u015feklindeki s\u0131f\u0131r hipotezini reddedebilece\u011fimiz anlam\u0131na gelir.<\/span><\/p>\n B\u00f6ylece varyanslar\u0131n e\u015fitli\u011fi varsay\u0131m\u0131 ihlal edilmi\u015f olur ve Welch’in ANOVA’s\u0131na ge\u00e7ebiliriz.<\/span><\/p>\n Python’da Welch’in ANOVA’s\u0131n\u0131 ger\u00e7ekle\u015ftirmek i\u00e7in Penguin paketindeki welch_anova()<\/strong> fonksiyonunu kullanabiliriz.<\/span><\/p>\n \u00d6ncelikle Penguin’i kurmam\u0131z gerekiyor:<\/span><\/p>\n Daha sonra Welch’in ANOVA’s\u0131n\u0131 ger\u00e7ekle\u015ftirmek i\u00e7in a\u015fa\u011f\u0131daki kodu kullanabiliriz:<\/span><\/p>\n ANOVA tablosunun genel p de\u011feri ( 0,001598<\/strong> ) \u03b1 = 0,05’ten k\u00fc\u00e7\u00fckt\u00fcr; bu, s\u0131nav sonu\u00e7lar\u0131n\u0131n \u00fc\u00e7 \u00e7al\u0131\u015fma tekni\u011fi aras\u0131nda e\u015fit oldu\u011funa ili\u015fkin bo\u015f hipotezi reddedebilece\u011fimiz anlam\u0131na gelir.<\/span><\/p>\n Daha sonra tam olarak hangi grup ortalamalar\u0131n\u0131n farkl\u0131 oldu\u011funu belirlemek i\u00e7in Games-Howell post-hoc testini ger\u00e7ekle\u015ftirebiliriz:<\/span><\/p>\n P de\u011ferlerinden a<\/strong> ve b<\/strong> gruplar\u0131 aras\u0131ndaki ortalama fark\u0131n anlaml\u0131 derecede farkl\u0131 oldu\u011funu ve b<\/strong> ve c<\/strong> gruplar\u0131 aras\u0131ndaki ortalama fark\u0131n anlaml\u0131 derecede farkl\u0131 oldu\u011funu g\u00f6rebiliriz.<\/span><\/p>\n Python’da Tek Y\u00f6nl\u00fc ANOVA Nas\u0131l Ger\u00e7ekle\u015ftirilir<\/a> Welch’in ANOVA’s\u0131, varyanslar\u0131n e\u015fitli\u011fi varsay\u0131m\u0131 kar\u015f\u0131lanmad\u0131\u011f\u0131nda tipik tek y\u00f6nl\u00fc ANOVA’ya bir alternatiftir. A\u015fa\u011f\u0131daki ad\u0131m ad\u0131m \u00f6rnek, Welch’in ANOVA’s\u0131n\u0131n Python’da nas\u0131l ger\u00e7ekle\u015ftirilece\u011fini g\u00f6sterir. 1. Ad\u0131m: Verileri olu\u015fturun \u00dc\u00e7 farkl\u0131 \u00e7al\u0131\u015fma tekni\u011finin farkl\u0131 s\u0131nav sonu\u00e7lar\u0131na yol a\u00e7\u0131p a\u00e7mad\u0131\u011f\u0131n\u0131 belirlemek i\u00e7in, bir profes\u00f6r rastgele 10 \u00f6\u011frenciye her tekni\u011fi (Teknik A, B veya C) bir hafta boyunca kullanmalar\u0131 i\u00e7in […]<\/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-1617","post","type-post","status-publish","format-standard","hentry","category-rehber"],"yoast_head":"\n 1. Ad\u0131m: Verileri olu\u015fturun<\/strong><\/span><\/h3>\n
A = [64, 66, 68, 75, 78, 94, 98, 79, 71, 80]\nB = [91, 92, 93, 90, 97, 94, 82, 88, 95, 96]\nC = [79, 78, 88, 94, 92, 85, 83, 85, 82, 81]\n<\/strong><\/pre>\n
Ad\u0131m 2: E\u015fit Farklar Testi<\/strong><\/span><\/h3>\n
import<\/span> scipy. stats<\/span> as<\/span> stats\n\n#perform Bartlett's test \n<\/span>stats. bartlett<\/span> (A, B, C)\n\nBartlettResult(statistic=9.039674395, pvalue=0.010890796567)\n<\/strong><\/span><\/pre>\n Ad\u0131m 3: Welch’in ANOVA’s\u0131n\u0131 ger\u00e7ekle\u015ftirin<\/strong><\/span><\/h3>\n
pip install<\/span> Penguin<\/span>\n<\/strong><\/pre>\n import penguin<\/span> as pg<\/span>\nimport<\/span> pandas as<\/span> pd\nimport<\/span> numpy as<\/span> np\n\n#createDataFrame\n<\/span>df = pd. DataFrame<\/span> ({' score<\/span> ': [64, 66, 68, 75, 78, 94, 98, 79, 71, 80,\n 91, 92, 93, 90, 97, 94, 82, 88, 95, 96,\n 79, 78, 88, 94, 92, 85, 83, 85, 82, 81],\n ' group<\/span> ': np. repeat<\/span> (['a', 'b', 'c'], repeats<\/span> = 10<\/span> )}) \n\n#perform Welch's ANOVA\n<\/span>pg. welch_anova<\/span> (dv=' score<\/span> ', between=' group<\/span> ', data=df)\n\n Source ddof1 ddof2 F p-unc np2\n0 group 2 16.651295 9.717185 0.001598 0.399286\n<\/strong><\/pre>\n pg. pairwise_gameshowell<\/span> (dv=' score<\/span> ', between=' group<\/span> ', data=df)\n\n\n A B mean(A) mean(B) diff se T df pval\t\n0 a b 77.3 91.8 -14.5 3.843754 -3.772354 11.6767 0.0072\n1 a c 77.3 84.7 -7.4 3.952777 -1.872102 12.7528 0.1864\n2 b c 91.8 84.7 7.1 2.179959 3.256942 17.4419 0.0119\n<\/strong><\/pre>\n Ek kaynaklar<\/strong><\/span><\/h3>\n
Python’da \u0130ki Y\u00f6nl\u00fc ANOVA Nas\u0131l Ger\u00e7ekle\u015ftirilir<\/a>
Python’da Tekrarlanan \u00d6l\u00e7\u00fcmler ANOVA Nas\u0131l Ger\u00e7ekle\u015ftirilir<\/a><\/p>\n","protected":false},"excerpt":{"rendered":"