Bu e\u015fit varyans varsay\u0131m\u0131n\u0131 yapan en yayg\u0131n istatistiksel testler ve prosed\u00fcrler \u015funlar\u0131 i\u00e7erir:<\/span><\/p>\n 1. ANOVA<\/strong><\/span><\/p>\n 2. t-testleri<\/strong><\/span><\/p>\n 3. Do\u011frusal regresyon<\/strong><\/span><\/p>\n Bu e\u011fitimde her test i\u00e7in yap\u0131lan varsay\u0131m, bu varsay\u0131m\u0131n kar\u015f\u0131lan\u0131p kar\u015f\u0131lanmad\u0131\u011f\u0131n\u0131n nas\u0131l belirlenece\u011fi ve ihlal edilmesi durumunda ne yap\u0131laca\u011f\u0131 a\u00e7\u0131klanmaktad\u0131r.<\/span><\/p>\n \u00dc\u00e7 veya daha fazla ba\u011f\u0131ms\u0131z grubun ortalamalar\u0131 aras\u0131nda anlaml\u0131 bir fark olup olmad\u0131\u011f\u0131n\u0131 belirlemek i\u00e7in ANOVA<\/strong> (\u201cVaryans Analizi\u201d) kullan\u0131l\u0131r.<\/span><\/p>\n ANOVA’y\u0131 ne zaman kullanabilece\u011fimize dair bir \u00f6rnek:<\/span><\/p>\n Diyelim ki bir kilo verme deneyine kat\u0131lmak \u00fczere 90 ki\u015fiyi i\u015fe ald\u0131k. Bir ay boyunca A, B veya C program\u0131n\u0131 kullanmak \u00fczere 30 ki\u015fiyi rastgele se\u00e7iyoruz.<\/span><\/p>\n Program\u0131n kilo kayb\u0131na etkisi olup olmad\u0131\u011f\u0131n\u0131 g\u00f6rmek i\u00e7in tek y\u00f6nl\u00fc ANOVA<\/a> yapabiliriz.<\/span><\/p>\n<\/blockquote>\n ANOVA, gruplar\u0131n her birinin e\u015fit varyansa sahip oldu\u011funu varsayar. Bu hipotezin do\u011fru olup olmad\u0131\u011f\u0131n\u0131 test etmenin iki yolu vard\u0131r:<\/span><\/p>\n 1. Kutu grafikleri olu\u015fturun.<\/strong><\/span> <\/p>\n Kutu grafikleri, varyanslar\u0131n e\u015fitli\u011fi varsay\u0131m\u0131n\u0131 do\u011frulamak i\u00e7in g\u00f6rsel bir yol sa\u011flar.<\/span><\/p>\n Her gruptaki kilo kayb\u0131ndaki farkl\u0131l\u0131k, her kutu grafi\u011finin uzunlu\u011funa g\u00f6re g\u00f6zlemlenebilir. Kutu ne kadar uzun olursa, varyans da o kadar y\u00fcksek olur. \u00d6rne\u011fin Program A ve Program B ile kar\u015f\u0131la\u015ft\u0131r\u0131ld\u0131\u011f\u0131nda Program C kat\u0131l\u0131mc\u0131lar\u0131 i\u00e7in varyans\u0131n biraz daha y\u00fcksek oldu\u011funu g\u00f6rebiliriz.<\/span><\/p>\n 2. Bartlett testini ger\u00e7ekle\u015ftirin.<\/strong><\/span><\/p>\n Bartlett testi,<\/a> \u00f6rneklerin e\u015fit varyansa sahip oldu\u011funu ifade eden bo\u015f hipotezi, \u00f6rneklerin e\u015fit varyansa sahip olmad\u0131\u011f\u0131 alternatif hipotezine kar\u015f\u0131 test eder.<\/span><\/p>\n Testin p de\u011feri belirli bir anlaml\u0131l\u0131k d\u00fczeyinin (0,05 gibi) alt\u0131ndaysa, o zaman \u00f6rneklerin hepsinin e\u015fit varyansa sahip olmad\u0131\u011f\u0131na dair kan\u0131t\u0131m\u0131z olur.<\/span><\/p>\n E\u015fit varyans varsay\u0131m\u0131 kar\u015f\u0131lanmazsa ne olur?<\/strong><\/span><\/p>\n Genel olarak<\/span> ANOVA’lar\u0131n, her grup ayn\u0131 \u00f6rneklem b\u00fcy\u00fckl\u00fc\u011f\u00fcne sahip oldu\u011fu s\u00fcrece e\u015fit varyans varsay\u0131m\u0131n\u0131n ihlallerine kar\u015f\u0131 olduk\u00e7a dayan\u0131kl\u0131 oldu\u011fu kabul edilir.<\/span><\/p>\n Ancak \u00f6rneklem b\u00fcy\u00fckl\u00fckleri ayn\u0131 de\u011filse ve bu varsay\u0131m ciddi \u015fekilde ihlal ediliyorsa bunun yerine tek y\u00f6nl\u00fc ANOVA’n\u0131n parametrik olmayan versiyonu olan Kruskal-Wallis testini<\/a> \u00e7al\u0131\u015ft\u0131rabilirsiniz.<\/span><\/p>\n \u0130ki \u00f6rneklem t testi,<\/a> iki pop\u00fclasyonun ortalamalar\u0131n\u0131n e\u015fit olup olmad\u0131\u011f\u0131n\u0131 test etmek i\u00e7in kullan\u0131l\u0131r.<\/span><\/p>\n Test, varyanslar\u0131n iki grup aras\u0131nda e\u015fit oldu\u011funu varsayar.<\/span> Bu hipotezin do\u011fru olup olmad\u0131\u011f\u0131n\u0131 test etmenin iki yolu vard\u0131r:<\/span><\/p>\n 1. Temel oran kural\u0131n\u0131 kullan\u0131n.<\/strong><\/span><\/p>\n Genel olarak en b\u00fcy\u00fck varyans\u0131n en k\u00fc\u00e7\u00fck varyansa oran\u0131 4’ten k\u00fc\u00e7\u00fckse varyanslar\u0131n yakla\u015f\u0131k olarak e\u015fit oldu\u011funu varsayabilir ve iki \u00f6rnekli t testini kullanabiliriz.<\/span><\/p>\n \u00d6rne\u011fin, 1. numunenin varyans\u0131n\u0131n 24.5 oldu\u011funu ve 2. numunenin varyans\u0131n\u0131n 15.2 oldu\u011funu varsayal\u0131m. En b\u00fcy\u00fck \u00f6rnek varyans\u0131n\u0131n en k\u00fc\u00e7\u00fck \u00f6rnek varyans\u0131na oran\u0131 \u015fu \u015fekilde hesaplanacakt\u0131r: 24,5 \/ 15,2 = 1,61.<\/span><\/p>\n Bu oran\u0131n 4’ten k\u00fc\u00e7\u00fck olmas\u0131 nedeniyle iki grup aras\u0131ndaki farklar\u0131n yakla\u015f\u0131k olarak e\u015fit oldu\u011fu varsay\u0131labilir.<\/span><\/p>\n 2. Bir F testi yap\u0131n.<\/strong><\/span><\/p>\n F testi,<\/strong> \u00f6rneklerin e\u015fit varyansa sahip oldu\u011funu belirten bo\u015f hipotezi, \u00f6rneklerin e\u015fit varyansa sahip olmad\u0131\u011f\u0131 alternatif hipotezine kar\u015f\u0131 test eder.<\/span><\/p>\n Testin p de\u011feri belirli bir anlaml\u0131l\u0131k d\u00fczeyinin (0,05 gibi) alt\u0131ndaysa, o zaman \u00f6rneklerin hepsinin e\u015fit varyansa sahip olmad\u0131\u011f\u0131na dair kan\u0131t\u0131m\u0131z olur.<\/span><\/p>\n E\u015fit varyans varsay\u0131m\u0131 kar\u015f\u0131lanmazsa ne olur?<\/strong><\/span><\/p>\n Bu varsay\u0131m ihlal edilirse, iki \u00f6rnekli t testinin parametrik olmayan bir versiyonu olan ve iki \u00f6rne\u011fin e\u015fit varyansa sahip oldu\u011funu varsaymayan Welch t testini<\/a> uygulayabiliriz.<\/span><\/p>\n Do\u011frusal regresyon,<\/a> bir veya daha fazla \u00f6ng\u00f6r\u00fcc\u00fc de\u011fi\u015fken ile bir yan\u0131t de\u011fi\u015fkeni aras\u0131ndaki ili\u015fkiyi \u00f6l\u00e7mek i\u00e7in kullan\u0131l\u0131r.<\/span><\/p>\n Do\u011frusal regresyon, art\u0131klar\u0131n<\/a> , yorday\u0131c\u0131 de\u011fi\u015fken(ler)in her seviyesinde sabit varyansa sahip oldu\u011funu varsayar. Buna homoskedastisite<\/em> denir. Durum b\u00f6yle olmad\u0131\u011f\u0131nda art\u0131klar de\u011fi\u015fen varyans<\/em> sorunu ya\u015far ve regresyon analizinin sonu\u00e7lar\u0131 g\u00fcvenilmez hale gelir.<\/span><\/p>\n Bu varsay\u0131m\u0131n kar\u015f\u0131lan\u0131p kar\u015f\u0131lanmad\u0131\u011f\u0131n\u0131 belirlemenin en yayg\u0131n yolu, art\u0131klar\u0131n uygun de\u011ferlere kar\u015f\u0131 grafi\u011fini olu\u015fturmakt\u0131r. Bu grafikteki art\u0131klar s\u0131f\u0131r etraf\u0131nda rastgele da\u011f\u0131lm\u0131\u015f gibi g\u00f6r\u00fcn\u00fcyorsa, bu durumda e\u015f varyans varsay\u0131m\u0131 muhtemelen kar\u015f\u0131lanmaktad\u0131r.<\/span><\/p>\n Bununla birlikte, a\u015fa\u011f\u0131daki grafikteki “koni” \u015fekli gibi, art\u0131klarda sistematik bir e\u011filim varsa, o zaman heteroskedastisite bir sorundur:<\/span><\/p>\n E\u015fit varyans varsay\u0131m\u0131 kar\u015f\u0131lanmazsa ne olur?<\/strong><\/span><\/p>\n Bu varsay\u0131m ihlal edilirse sorunu \u00e7\u00f6zmenin en yayg\u0131n yolu, yan\u0131t de\u011fi\u015fkenini \u00fc\u00e7 d\u00f6n\u00fc\u015f\u00fcmden birini kullanarak d\u00f6n\u00fc\u015ft\u00fcrmektir:<\/span><\/p>\n 1. G\u00fcnl\u00fck d\u00f6n\u00fc\u015f\u00fcm\u00fc:<\/strong> yan\u0131t de\u011fi\u015fkenini y’den log(y)’<\/strong> ye d\u00f6n\u00fc\u015ft\u00fcr\u00fcn.<\/span><\/p>\n 2. Karek\u00f6k d\u00f6n\u00fc\u015f\u00fcm\u00fc:<\/strong> Yan\u0131t de\u011fi\u015fkenini y’den \u221ay’ye<\/span><\/strong> d\u00f6n\u00fc\u015ft\u00fcr\u00fcn.<\/span><\/p>\n 3. K\u00fcp k\u00f6k d\u00f6n\u00fc\u015f\u00fcm\u00fc:<\/strong> yan\u0131t de\u011fi\u015fkenini y’den y 1\/3’e<\/sup><\/strong> d\u00f6n\u00fc\u015ft\u00fcr\u00fcn.<\/span><\/p>\n Bu d\u00f6n\u00fc\u015f\u00fcmlerin ger\u00e7ekle\u015ftirilmesiyle de\u011fi\u015fen varyans sorunu genel olarak ortadan kalkar.<\/span><\/p>\n Heteroskedasticity’yi d\u00fczeltmenin ba\u015fka bir yolu da a\u011f\u0131rl\u0131kl\u0131 en k\u00fc\u00e7\u00fck kareler regresyonunu<\/a> kullanmakt\u0131r. Bu regresyon t\u00fcr\u00fc, her veri noktas\u0131na, uydurulan de\u011ferin varyans\u0131na ba\u011fl\u0131 olarak bir a\u011f\u0131rl\u0131k atar.<\/span><\/p>\n Temel olarak bu, daha y\u00fcksek varyansa sahip veri noktalar\u0131na d\u00fc\u015f\u00fck a\u011f\u0131rl\u0131k vererek bunlar\u0131n kalan karelerini azalt\u0131r. Uygun a\u011f\u0131rl\u0131klar kullan\u0131ld\u0131\u011f\u0131nda de\u011fi\u015fen varyans sorunu ortadan kald\u0131r\u0131labilir.<\/span><\/p>\n ANOVA’da form\u00fcle edilen \u00fc\u00e7 hipotez<\/a> Bir\u00e7ok istatistiksel test e\u015fit varyans varsay\u0131m\u0131n\u0131 yapar. Bu varsay\u0131ma uyulmazsa test sonu\u00e7lar\u0131 g\u00fcvenilmez hale gelir. Bu e\u015fit varyans varsay\u0131m\u0131n\u0131 yapan en yayg\u0131n istatistiksel testler ve prosed\u00fcrler \u015funlar\u0131 i\u00e7erir: 1. ANOVA 2. t-testleri 3. Do\u011frusal regresyon Bu e\u011fitimde her test i\u00e7in yap\u0131lan varsay\u0131m, bu varsay\u0131m\u0131n kar\u015f\u0131lan\u0131p kar\u015f\u0131lanmad\u0131\u011f\u0131n\u0131n nas\u0131l belirlenece\u011fi ve ihlal edilmesi durumunda ne yap\u0131laca\u011f\u0131 a\u00e7\u0131klanmaktad\u0131r. […]<\/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-1609","post","type-post","status-publish","format-standard","hentry","category-rehber"],"yoast_head":"\n ANOVA’da varyans e\u015fitli\u011fi varsay\u0131m\u0131<\/strong><\/span><\/h3>\n
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![\"\"](\"https:\/\/statorials.org\/wp-content\/uploads\/2023\/08\/boite-a-moustaches.jpg\")
<\/h3>\n T testlerinde e\u015fit varyans varsay\u0131m\u0131<\/strong><\/span><\/h3>\n
Do\u011frusal Regresyonda E\u015fit Varyans Varsay\u0131m\u0131<\/strong><\/span><\/h3>\n
Ek kaynaklar<\/strong><\/span><\/h3>\n
T testinde form\u00fcle edilen d\u00f6rt hipotez<\/a>
Do\u011frusal regresyonun d\u00f6rt varsay\u0131m\u0131<\/a><\/p>\n","protected":false},"excerpt":{"rendered":"