Tek y\u00f6nl\u00fc ANOVA’da kullan\u0131lan varsay\u0131mlar \u015funlard\u0131r:<\/span><\/p>\n ANOVA’n\u0131n p de\u011feri belirli bir anlaml\u0131l\u0131k d\u00fczeyinin alt\u0131ndaysa (\u03b1 = 0,05 gibi), s\u0131f\u0131r hipotezini reddedebilir ve grup ortalamalar\u0131ndan en az birinin di\u011ferlerinden farkl\u0131 oldu\u011fu sonucuna varabiliriz.<\/span><\/p>\n Ancak hangi gruplar\u0131n birbirinden farkl\u0131 oldu\u011funu tam olarak bilmek i\u00e7in post-hoc bir test yapmam\u0131z gerekiyor.<\/span><\/p>\n Yayg\u0131n olarak kullan\u0131lan bir post-hoc testi , Fisher’in en az anlaml\u0131 fark (LSD) testidir<\/strong> .<\/span><\/p>\n Bu testi R’de ger\u00e7ekle\u015ftirmek i\u00e7in agricolae<\/strong> paketindeki LSD.test()<\/strong> i\u015flevini kullanabilirsiniz.<\/span><\/p>\n A\u015fa\u011f\u0131daki \u00f6rnekte bu fonksiyonun pratikte nas\u0131l kullan\u0131laca\u011f\u0131 g\u00f6sterilmektedir.<\/span><\/p>\n Bir profes\u00f6r\u00fcn, \u00fc\u00e7 farkl\u0131 \u00e7al\u0131\u015fma tekni\u011finin \u00f6\u011frenciler aras\u0131nda farkl\u0131 test puanlar\u0131na yol a\u00e7\u0131p a\u00e7mad\u0131\u011f\u0131n\u0131 bilmek istedi\u011fini varsayal\u0131m.<\/span><\/p>\n Bunu test etmek i\u00e7in rastgele 10 \u00f6\u011frenciyi her \u00e7al\u0131\u015fma tekni\u011fini kullanmalar\u0131 i\u00e7in g\u00f6revlendiriyor ve s\u0131nav sonu\u00e7lar\u0131n\u0131 kaydediyor.<\/span><\/p>\n A\u015fa\u011f\u0131daki tablo,<\/span> kullan\u0131lan \u00e7al\u0131\u015fma tekni\u011fine g\u00f6re her \u00f6\u011frencinin s\u0131nav sonu\u00e7lar\u0131n\u0131 g\u00f6sterir:<\/span> <\/p>\n R’de bu veri k\u00fcmesini olu\u015fturmak ve \u00fczerinde tek y\u00f6nl\u00fc ANOVA ger\u00e7ekle\u015ftirmek i\u00e7in a\u015fa\u011f\u0131daki kodu kullanabiliriz:<\/span><\/p>\n ANOVA tablosundaki p de\u011feri (0,0188) 0,05’ten k\u00fc\u00e7\u00fck oldu\u011fundan, \u00fc\u00e7 grup aras\u0131ndaki ortalama s\u0131nav puanlar\u0131n\u0131n hepsinin e\u015fit olmad\u0131\u011f\u0131 sonucuna varabiliriz.<\/span><\/p>\n B\u00f6ylece hangi grup ortalamalar\u0131n\u0131n farkl\u0131 oldu\u011funu belirlemek i\u00e7in Fisher’in LSD testini yapabiliriz.<\/span><\/p>\n A\u015fa\u011f\u0131daki kod bunun nas\u0131l yap\u0131laca\u011f\u0131n\u0131 g\u00f6sterir:<\/span><\/p>\n Sonucun bizi en \u00e7ok ilgilendiren k\u0131sm\u0131 $groups<\/strong> ad\u0131 verilen b\u00f6l\u00fcmd\u00fcr. Gruplar<\/strong> s\u00fctununda farkl\u0131 karakterlere sahip olan teknikler \u00e7ok farkl\u0131d\u0131r.<\/span><\/p>\n Sonu\u00e7tan \u015funu g\u00f6rebiliriz:<\/span><\/p>\n A\u015fa\u011f\u0131daki e\u011fitimlerde R’de di\u011fer ortak g\u00f6revlerin nas\u0131l ger\u00e7ekle\u015ftirilece\u011fi a\u00e7\u0131klanmaktad\u0131r:<\/span><\/p>\n R’de tek y\u00f6nl\u00fc ANOVA nas\u0131l ger\u00e7ekle\u015ftirilir<\/a> \u00dc\u00e7 veya daha fazla ba\u011f\u0131ms\u0131z grubun ortalamalar\u0131 aras\u0131nda istatistiksel olarak anlaml\u0131 bir fark olup olmad\u0131\u011f\u0131n\u0131 belirlemek i\u00e7in tek y\u00f6nl\u00fc ANOVA kullan\u0131l\u0131r. Tek y\u00f6nl\u00fc ANOVA’da kullan\u0131lan varsay\u0131mlar \u015funlard\u0131r: H 0 : Ortalamalar her grup i\u00e7in e\u015fittir. H A : En az\u0131ndan yollardan biri di\u011ferlerinden farkl\u0131. ANOVA’n\u0131n p de\u011feri belirli bir anlaml\u0131l\u0131k d\u00fczeyinin alt\u0131ndaysa (\u03b1 = 0,05 […]<\/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-3063","post","type-post","status-publish","format-standard","hentry","category-rehber"],"yoast_head":"\n\n
\u00d6rnek: R’de Fisher’in LSD testi<\/strong><\/span><\/h3>\n
![\"\"](\"https:\/\/statorials.org\/wp-content\/uploads\/2023\/08\/moinssigdiff1.png\")
<\/p>\n #create data frame\ndf <- data. frame<\/span> (technique = rep(c(\"tech1\", \"tech2\", \"tech3\"), each = 10<\/span> ),\n score = c(72, 73, 73, 77, 82, 82, 83, 84, 85, 89,\n 81, 82, 83, 83, 83, 84, 87, 90, 92, 93,\n 77, 78, 79, 88, 89, 90, 91, 95, 95, 98))\n\n#view first six rows of data frame\n<\/span>head(df)\n\n technical score\n1 tech1 72\n2 tech1 73\n3 tech1 73\n4 tech1 77\n5 tech1 82\n6 tech1 82\n\n#fit one-way ANOVA\n<\/span>model <- aov(score ~ technique, data = df)\n\n#view summary of one-way ANOVA\n<\/span>summary(model)\n\n Df Sum Sq Mean Sq F value Pr(>F) \ntechnical 2 341.6 170.80 4.623 0.0188 *\nResiduals 27,997.6 36.95 \n---\nSignificant. codes: 0 '***' 0.001 '**' 0.01 '*' 0.05 '.' 0.1 ' ' 1<\/span><\/span><\/strong><\/pre>\n library<\/span> (agricolae)<\/span>\n\n#perform Fisher's LSD\nprint( LSD.test<\/span> (model,\" technic<\/span> \"))\n \n$statistics\n MSerror Df Mean CV t.value LSD\n 36.94815 27 84.6 7.184987 2.051831 5.57767\n\n$parameters\n test p.adjusted name.t ntr alpha\n Fisher-LSD none technical 3 0.05\n\n$means\n std score r LCL UCL Min Max Q25 Q50 Q75\ntech1 80.0 5.868939 10 76.05599 83.94401 72 89 74.00 82.0 83.75\ntech2 85.8 4.391912 10 81.85599 89.74401 81 93 83.00 83.5 89.25\ntech3 88.0 7.557189 10 84.05599 91.94401 77 98 81.25 89.5 94.00\n\n$comparison\nNULL\n\n$groups\n score groups\ntech3 88.0 a\ntech2 85.8a\ntech1 80.0 b\n\nattr(,\"class\")\n[1] \u201cgroup\u201d\n<\/span><\/span><\/strong><\/pre>\n\n
Ek kaynaklar<\/strong><\/span><\/h3>\n
R’de Bonferroni post hoc testi nas\u0131l yap\u0131l\u0131r<\/a>
R’de Scheffe post-hoc testi nas\u0131l ger\u00e7ekle\u015ftirilir?<\/a><\/p>\n","protected":false},"excerpt":{"rendered":"