\u00d6rne\u011fin, bir ara\u015ft\u0131rmac\u0131n\u0131n \u00fc\u00e7 farkl\u0131 g\u00fcbrenin farkl\u0131 seviyelerde bitki b\u00fcy\u00fcmesi \u00fcretip \u00fcretmedi\u011fini bilmek istedi\u011fini varsayal\u0131m.<\/span><\/p>\n Bunu test etmek i\u00e7in, \u00fc\u00e7 farkl\u0131 teknisyenin her biri d\u00f6rt bitkiye A g\u00fcbresi serpiyor, di\u011fer \u00fc\u00e7 teknisyenin her biri d\u00f6rt bitkiye B g\u00fcbresi serpiyor ve di\u011fer \u00fc\u00e7 teknisyenin her biri d\u00f6rt bitkiye C g\u00fcbresi serpiyor.<\/span><\/p>\n Bu senaryoda yan\u0131t de\u011fi\u015fkeni<\/a> bitki b\u00fcy\u00fcmesidir ve iki fakt\u00f6r teknisyen ve g\u00fcbredir. Teknisyenin g\u00fcbrenin i\u00e7ine yerle\u015fti\u011fi<\/em> ortaya \u00e7\u0131kt\u0131:<\/span> <\/p>\n A\u015fa\u011f\u0131daki ad\u0131m ad\u0131m \u00f6rnek, bu i\u00e7 i\u00e7e ANOVA’n\u0131n R’de nas\u0131l ger\u00e7ekle\u015ftirilece\u011fini g\u00f6sterir.<\/span><\/p>\n \u00d6ncelikle verilerimizi R’de tutacak bir veri \u00e7er\u00e7evesi olu\u015ftural\u0131m:<\/span><\/p>\n R’ye yuvalanm\u0131\u015f bir ANOVA’y\u0131 s\u0131\u011fd\u0131rmak i\u00e7in a\u015fa\u011f\u0131daki s\u00f6zdizimini kullanabiliriz:<\/span><\/p>\n aov(cevap ~ fakt\u00f6r A \/ fakt\u00f6r B)<\/strong><\/span><\/p>\n Alt\u0131n:<\/span><\/p>\n A\u015fa\u011f\u0131daki kod, veri k\u00fcmemiz i\u00e7in i\u00e7 i\u00e7e ANOVA’n\u0131n nas\u0131l s\u0131\u011fd\u0131r\u0131laca\u011f\u0131n\u0131 g\u00f6sterir:<\/span><\/p>\n Her fakt\u00f6r\u00fcn bitki b\u00fcy\u00fcmesi \u00fczerinde istatistiksel olarak anlaml\u0131 bir etkiye sahip olup olmad\u0131\u011f\u0131n\u0131 belirlemek i\u00e7in p-de\u011feri s\u00fctununa bakabiliriz.<\/span><\/p>\n Yukar\u0131daki tablodan, g\u00fcbrenin bitki b\u00fcy\u00fcmesi \u00fczerinde istatistiksel olarak anlaml\u0131 bir etkiye sahip oldu\u011funu (p-de\u011feri < 0,05), ancak teknisyenin olmad\u0131\u011f\u0131n\u0131 (p-de\u011feri = 0,211) g\u00f6rebiliriz.<\/span><\/p>\n Bu bize, bitki b\u00fcy\u00fcmesini art\u0131rmak istiyorsak g\u00fcbreyi uygulayan bireysel teknisyen yerine kullan\u0131lan g\u00fcbreye odaklanmam\u0131z gerekti\u011fini s\u00f6yl\u00fcyor.<\/span><\/p>\n Son olarak, bitki b\u00fcy\u00fcmesinin g\u00fcbre ve teknisyene g\u00f6re da\u011f\u0131l\u0131m\u0131n\u0131 g\u00f6rselle\u015ftirmek i\u00e7in kutu grafiklerini kullanabiliriz:<\/span> <\/p>\n Grafik, \u00fc\u00e7 farkl\u0131 g\u00fcbre aras\u0131nda b\u00fcy\u00fcme a\u00e7\u0131s\u0131ndan \u00f6nemli farkl\u0131l\u0131klar oldu\u011funu, ancak her g\u00fcbre grubundaki teknisyenler aras\u0131nda \u00e7ok fazla farkl\u0131l\u0131k olmad\u0131\u011f\u0131n\u0131 g\u00f6stermektedir.<\/span><\/p>\n Bu, i\u00e7 i\u00e7e ge\u00e7mi\u015f ANOVA’n\u0131n sonu\u00e7lar\u0131yla e\u015fle\u015fiyor gibi g\u00f6r\u00fcn\u00fcyor ve g\u00fcbrelerin bitki b\u00fcy\u00fcmesi \u00fczerinde \u00f6nemli bir etkiye sahip oldu\u011funu ancak bireysel teknisyenlerin bunu yapmad\u0131\u011f\u0131n\u0131 do\u011fruluyor.<\/span><\/p>\n R’de tek y\u00f6nl\u00fc ANOVA nas\u0131l ger\u00e7ekle\u015ftirilir<\/a> Yuvalanm\u0131\u015f ANOVA, en az bir fakt\u00f6r\u00fcn ba\u015fka bir fakt\u00f6r\u00fcn i\u00e7ine yerle\u015ftirildi\u011fi bir ANOVA t\u00fcr\u00fcd\u00fcr (“varyans analizi”). \u00d6rne\u011fin, bir ara\u015ft\u0131rmac\u0131n\u0131n \u00fc\u00e7 farkl\u0131 g\u00fcbrenin farkl\u0131 seviyelerde bitki b\u00fcy\u00fcmesi \u00fcretip \u00fcretmedi\u011fini bilmek istedi\u011fini varsayal\u0131m. Bunu test etmek i\u00e7in, \u00fc\u00e7 farkl\u0131 teknisyenin her biri d\u00f6rt bitkiye A g\u00fcbresi serpiyor, di\u011fer \u00fc\u00e7 teknisyenin her biri d\u00f6rt bitkiye B g\u00fcbresi serpiyor […]<\/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-1578","post","type-post","status-publish","format-standard","hentry","category-rehber"],"yoast_head":"\n![\"\u0130\u00e7](\"https:\/\/statorials.org\/wp-content\/uploads\/2023\/08\/imbrique1-1.png\")
<\/p>\n 1. Ad\u0131m: Verileri olu\u015fturun<\/strong><\/span><\/h3>\n
#create data\n<\/span>df <- data. frame<\/span> (growth=c(13, 16, 16, 12, 15, 16, 19, 16, 15, 15, 12, 15,\n 19, 19, 20, 22, 23, 18, 16, 18, 19, 20, 21, 21,\n 21, 23, 24, 22, 25, 20, 20, 22, 24, 22, 25, 26),\n fertilizer=c(rep(c(' A<\/span> ', ' B<\/span> ', ' C<\/span> '), each= 12<\/span> )),\n tech=c(rep(1:9, each= 4<\/span> )))\n\n#view first six rows of data\n<\/span>head(df)\n\n growth fertilizer tech\n1 13 A 1\n2 16 A 1\n3 16 A 1\n4 12 A 1\n5 15 A 2\n6 16 A 2<\/strong><\/pre>\n Ad\u0131m 2: \u0130\u00e7 \u0130\u00e7e ANOVA’y\u0131 Ayarlay\u0131n<\/strong><\/span><\/h3>\n
\n
#fit nested ANOVA\n<\/span>nest <- aov(df$growth ~ df$fertilizer \/ factor(df$tech))\n\n#view summary of nested ANOVA\n<\/span>summary(nest)\n\n Df Sum Sq Mean Sq F value Pr(>F) \ndf$fertilizer 2 372.7 186.33 53.238 4.27e-10 ***\ndf$fertilizer:factor(df$tech) 6 31.8 5.31 1.516 0.211 \nResiduals 27 94.5 3.50 \n---\nSignificant. codes: 0 '***' 0.001 '**' 0.01 '*' 0.05 '.' 0.1 ' ' 1<\/strong><\/pre>\n 3. Ad\u0131m: Sonucu yorumlay\u0131n<\/strong><\/span><\/h3>\n
4. Ad\u0131m: Sonu\u00e7lar\u0131 g\u00f6rselle\u015ftirin<\/strong><\/span><\/h3>\n
#load ggplot2 data visualization package\n<\/span>library<\/span> (ggplot2)\n\n#create boxplots to visualize plant growth\n<\/span>ggplot(df, aes<\/span> (x=factor(tech), y=growth, fill=fertilizer)) +\n geom_boxplot()\n<\/strong><\/pre>\n![\"\"](\"https:\/\/statorials.org\/wp-content\/uploads\/2023\/08\/imbriquer1.png\")
<\/p>\n Ek kaynaklar<\/strong><\/span><\/h3>\n
R’de iki y\u00f6nl\u00fc ANOVA nas\u0131l ger\u00e7ekle\u015ftirilir<\/a>
R’de tekrarlanan \u00f6l\u00e7\u00fcmler ANOVA nas\u0131l ger\u00e7ekle\u015ftirilir?<\/a><\/p>\n","protected":false},"excerpt":{"rendered":"