Bir testteki farkl\u0131 maddeler taraf\u0131ndan \u00fcretilen yan\u0131tlar\u0131n g\u00fcvenilirli\u011fini de\u011ferlendirmek i\u00e7in kullan\u0131l\u0131r. Bir testin de\u011ferlendiriciler aras\u0131 g\u00fcvenilirli\u011fi d\u00fc\u015f\u00fckse, bu durum test maddelerinin kafa kar\u0131\u015ft\u0131r\u0131c\u0131, belirsiz ve hatta yarars\u0131z oldu\u011funu g\u00f6sterebilir.<\/span><\/p>\n De\u011ferlendiriciler aras\u0131 g\u00fcvenilirli\u011fi \u00f6l\u00e7menin iki yayg\u0131n yolu vard\u0131r:<\/span><\/p>\n 1. Anla\u015fma y\u00fczdesi<\/strong><\/span><\/p>\n De\u011ferlendiriciler aras\u0131 g\u00fcvenilirli\u011fi \u00f6l\u00e7menin basit yolu, yarg\u0131\u00e7lar\u0131n \u00fczerinde anla\u015ft\u0131\u011f\u0131 maddelerin y\u00fczdesini hesaplamakt\u0131r.<\/span><\/p>\n Buna y\u00fczde anla\u015fma<\/strong> ad\u0131 verilir ve bu de\u011fer her zaman 0 ile 1 aras\u0131ndad\u0131r; 0, de\u011ferlendiriciler aras\u0131nda hi\u00e7bir anla\u015fma olmad\u0131\u011f\u0131n\u0131, 1 ise de\u011ferlendiriciler aras\u0131nda m\u00fckemmel anla\u015fmay\u0131 belirtir.<\/span><\/p>\n \u00d6rne\u011fin, iki yarg\u0131c\u0131n bir testteki 10 maddenin zorlu\u011funu 1’den 3’e kadar bir \u00f6l\u00e7ekte derecelendirmesinin istendi\u011fini varsayal\u0131m. Sonu\u00e7lar a\u015fa\u011f\u0131da g\u00f6sterilmektedir:<\/span> <\/p>\n Her soru i\u00e7in her iki yarg\u0131\u00e7 da kat\u0131l\u0131yorsa \u201c1\u201d, kat\u0131lm\u0131yorsa \u201c0\u201d yazabiliriz.<\/span> <\/p>\n J\u00fcrinin \u00fczerinde mutabakata vard\u0131\u011f\u0131 sorular\u0131n y\u00fczdesi 7\/10 = %70<\/strong> idi.<\/span><\/p>\n 2. Cohen’in Kappa’s\u0131<\/strong><\/span><\/p>\n De\u011ferlendiriciler aras\u0131 g\u00fcvenilirli\u011fi \u00f6l\u00e7menin en zor (ve en kat\u0131) yolu, De\u011ferlendiricilerin yaln\u0131zca belirli \u00f6\u011feler \u00fczerinde hemfikir olabilece\u011fini hesaba katarak, de\u011ferlendiricilerin \u00fczerinde hemfikir oldu\u011fu maddelerin y\u00fczdesini hesaplayan Cohen’in Kappa’s\u0131n\u0131<\/a> kullanmakt\u0131r. Neyse ki.<\/span><\/p>\n Cohen’in kappa form\u00fcl\u00fc \u015fu \u015fekilde hesaplan\u0131r:<\/span><\/p>\n k = (p o<\/sub> \u2013 p e<\/sub> ) \/ (1 \u2013 p e<\/sub> )<\/strong><\/span><\/p>\n Alt\u0131n:<\/span><\/p>\n Cohen’in Kappa’s\u0131 her zaman 0 ile 1 aras\u0131nda de\u011fi\u015fir; 0, de\u011ferlendiriciler aras\u0131nda hi\u00e7bir anla\u015fma olmad\u0131\u011f\u0131n\u0131, 1 ise de\u011ferlendiriciler aras\u0131nda m\u00fckemmel anla\u015fmay\u0131 g\u00f6sterir.<\/span><\/p>\n Cohen’in Kappa’s\u0131n\u0131n nas\u0131l hesaplanaca\u011f\u0131na ili\u015fkin ad\u0131m ad\u0131m bir \u00f6rnek i\u00e7in bu e\u011fitime<\/a> bak\u0131n.<\/span><\/p>\n De\u011ferlendiriciler aras\u0131 g\u00fcvenilirlik ne kadar y\u00fcksek olursa, birden fazla yarg\u0131\u00e7, benzer puanlara sahip bir testteki \u00f6\u011feleri veya sorular\u0131 o kadar tutarl\u0131 bir \u015fekilde derecelendirir.<\/span><\/p>\n Genel olarak, bir testin g\u00fcvenilir say\u0131lmas\u0131 i\u00e7in \u00e7o\u011fu alanda de\u011ferlendiriciler aras\u0131 en az %75’lik bir anla\u015fma gerekir.<\/span> Ancak belirli alanlarda de\u011ferlendiriciler aras\u0131 g\u00fcvenilirli\u011fin daha y\u00fcksek olmas\u0131 gerekebilir.<\/span><\/p>\n \u00d6rne\u011fin, %75’lik de\u011ferlendiriciler aras\u0131 g\u00fcvenilirlik, bir televizyon program\u0131n\u0131n ne kadar iyi kar\u015f\u0131lanaca\u011f\u0131n\u0131 belirlemeye y\u00f6nelik bir test i\u00e7in kabul edilebilir.<\/span><\/p>\n \u00d6te yandan, birden fazla doktorun belirli bir hastada belirli bir tedavinin kullan\u0131l\u0131p kullan\u0131lmamas\u0131 gerekti\u011fine karar verdi\u011fi t\u0131bbi ortamlarda de\u011ferlendiriciler aras\u0131 g\u00fcvenilirli\u011fin %95 olmas\u0131 gerekebilir.<\/span><\/p>\n \u00c7o\u011fu akademik ortamda ve titiz ara\u015ft\u0131rma alanlar\u0131nda, de\u011ferlendiriciler aras\u0131 g\u00fcvenilirli\u011fi hesaplamak i\u00e7in Cohen’in Kappa’s\u0131n\u0131n kullan\u0131ld\u0131\u011f\u0131n\u0131 unutmay\u0131n.<\/span><\/p>\n G\u00fcvenilirlik Analizine H\u0131zl\u0131 Bir Giri\u015f<\/a> \u0130statistikte de\u011ferlendiriciler aras\u0131 g\u00fcvenilirlik , birden fazla de\u011ferlendirici veya yarg\u0131\u00e7 aras\u0131ndaki anla\u015fman\u0131n d\u00fczeyini \u00f6l\u00e7menin bir yoludur. Bir testteki farkl\u0131 maddeler taraf\u0131ndan \u00fcretilen yan\u0131tlar\u0131n g\u00fcvenilirli\u011fini de\u011ferlendirmek i\u00e7in kullan\u0131l\u0131r. Bir testin de\u011ferlendiriciler aras\u0131 g\u00fcvenilirli\u011fi d\u00fc\u015f\u00fckse, bu durum test maddelerinin kafa kar\u0131\u015ft\u0131r\u0131c\u0131, belirsiz ve hatta yarars\u0131z oldu\u011funu g\u00f6sterebilir. De\u011ferlendiriciler aras\u0131 g\u00fcvenilirli\u011fi \u00f6l\u00e7menin iki yayg\u0131n yolu vard\u0131r: 1. Anla\u015fma […]<\/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-1487","post","type-post","status-publish","format-standard","hentry","category-rehber"],"yoast_head":"\n![\"\"](\"https:\/\/statorials.org\/wp-content\/uploads\/2023\/08\/evaluateur1.png\")
<\/p>\n![\"\"](\"https:\/\/statorials.org\/wp-content\/uploads\/2023\/08\/evaluateur2.png\")
<\/p>\n\n
De\u011ferlendiriciler aras\u0131 g\u00fcvenilirlik nas\u0131l yorumlan\u0131r?<\/strong><\/span><\/h3>\n
Ek kaynaklar<\/strong><\/span><\/h3>\n
G\u00fcvenilirlik ikiye b\u00f6l\u00fcnm\u00fc\u015f nedir?<\/a>
Test-tekrar test g\u00fcvenilirli\u011fi nedir?<\/a>
Paralel formlar\u0131n g\u00fcvenilirli\u011fi nedir?<\/a>
Standart \u00f6l\u00e7\u00fcm hatas\u0131 nedir?<\/a>
Cohen’in Kappa Hesaplay\u0131c\u0131s\u0131<\/p>\n","protected":false},"excerpt":{"rendered":"