{"id":3429,"date":"2023-07-17T13:07:49","date_gmt":"2023-07-17T13:07:49","guid":{"rendered":"https:\/\/statorials.org\/tr\/rde-iki-ornek-t-testi\/"},"modified":"2023-07-17T13:07:49","modified_gmt":"2023-07-17T13:07:49","slug":"rde-iki-ornek-t-testi","status":"publish","type":"post","link":"https:\/\/statorials.org\/tr\/rde-iki-ornek-t-testi\/","title":{"rendered":"R'de iki \u00f6rnekli t testi nas\u0131l yap\u0131l\u0131r"},"content":{"rendered":"

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\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

R’de iki \u00f6rnekli bir t testi ger\u00e7ekle\u015ftirmek i\u00e7in a\u015fa\u011f\u0131daki temel s\u00f6zdizimini kullanabilirsiniz:<\/span><\/p>\n

 t. test<\/span> (group1, group2, var. equal<\/span> = TRUE<\/span> ) \n<\/strong><\/pre>\n
 Not<\/strong> : var.equal=TRUE<\/strong> belirterek, R’ye iki \u00f6rnek aras\u0131ndaki varyanslar\u0131n e\u015fit oldu\u011funu varsaymas\u0131n\u0131 s\u00f6yl\u00fcyoruz.<\/span><\/p>\n
 A\u015fa\u011f\u0131daki \u00f6rnek, pratikte R’de iki \u00f6rnekli bir t testinin nas\u0131l ger\u00e7ekle\u015ftirilece\u011fini g\u00f6sterir.<\/span><\/p>\n
 \u00d6rnek: R’de \u0130ki \u00d6rnekli T Testi<\/strong><\/span><\/h2>\n
 \u0130ki farkl\u0131 bitki t\u00fcr\u00fcn\u00fcn ayn\u0131 ortalama y\u00fcksekli\u011fe sahip olup olmad\u0131\u011f\u0131n\u0131 bilmek istedi\u011fimizi varsayal\u0131m.<\/span><\/p>\n
 Bunu test etmek i\u00e7in her t\u00fcrden 12 bitkiden olu\u015fanbasit rastgele bir \u00f6rnek<\/a> topluyoruz.<\/span><\/p>\n
 A\u015fa\u011f\u0131daki kod, iki t\u00fcr aras\u0131ndaki ortalama y\u00fcksekli\u011fin e\u015fit olup olmad\u0131\u011f\u0131n\u0131 belirlemek i\u00e7in R’de iki \u00f6rnekli bir t testinin nas\u0131l ger\u00e7ekle\u015ftirilece\u011fini g\u00f6sterir:<\/span><\/p>\n
 #create vectors to hold plant heights from each sample\n<\/span>group1 <- c(8, 8, 9, 9, 9, 11, 12, 13, 13, 14, 15, 19)\ngroup2 <- c(11, 12, 13, 13, 14, 14, 14, 15, 16, 18, 18, 19)\n\n<\/strong>#perform two sample t-tests\n<\/span>t. test<\/span> (group1, group2, var. equal<\/span> = TRUE<\/span> )\n\n\tTwo Sample t-test\n\ndata: group1 and group2\nt = -2.5505, df = 22, p-value = 0.01823\nalternative hypothesis: true difference in means is not equal to 0\n95 percent confidence interval:\n -5.5904820 -0.5761847\nsample estimates:\nmean of x mean of y \n 11.66667 14.75000 \n<\/strong><\/pre>\n
 Test sonu\u00e7lar\u0131n\u0131 nas\u0131l yorumlayaca\u011f\u0131n\u0131z a\u015fa\u011f\u0131da a\u00e7\u0131klanm\u0131\u015ft\u0131r:<\/span><\/p>\n
 veriler:<\/strong> \u00f6rnek verileri i\u00e7eren vekt\u00f6rlerin adlar\u0131.<\/span><\/p>\n
 t:<\/strong> t testi istatisti\u011fi. Bu durumda -2,5505’tir<\/strong> .<\/span><\/p>\n
 df<\/strong> : n 1<\/sub> + n 2<\/sub> \u2013 2 = 12 + 12 \u2013 2 = 22<\/strong> olarak hesaplanan serbestlik derecesi.<\/span><\/p>\n
 p de\u011feri:<\/strong> -2,5505 ve df = 22 test istatisti\u011fine kar\u015f\u0131l\u0131k gelen p de\u011feri. P de\u011feri .01823<\/strong> olarak ortaya \u00e7\u0131k\u0131yor. Bu de\u011feri T Puan\u0131 – P De\u011feri hesaplay\u0131c\u0131s\u0131n\u0131<\/a> kullanarak do\u011frulayabiliriz.<\/span><\/p>\n
 %95 G\u00fcven Aral\u0131\u011f\u0131:<\/strong> \u0130ki grup aras\u0131ndaki ortalamalar aras\u0131ndaki ger\u00e7ek fark i\u00e7in %95 g\u00fcven aral\u0131\u011f\u0131. [-5,59, -,576]<\/strong> oldu\u011fu ortaya \u00e7\u0131kt\u0131.<\/span><\/p>\n
 \u00d6rnek tahminleri:<\/strong> her grubun \u00f6rnek ortalamas\u0131<\/a> . Bu durumda Grup 1’in \u00f6rnek ortalamas\u0131 11.667<\/strong> , Grup 2’nin \u00f6rnek ortalamas\u0131 14.75<\/strong> oldu.<\/span><\/p>\n
 Bu \u00f6zel iki \u00f6rnekli t testi i\u00e7in bo\u015f ve alternatif hipotezler a\u015fa\u011f\u0131daki gibidir:<\/span><\/p>\n
 H 0<\/sub> :<\/strong> \u00b5 1<\/sub> = \u00b5 2<\/sub> (iki pop\u00fclasyon ortalamas\u0131 e\u015fittir)<\/span><\/p>\n
 HA<\/sub> :<\/strong> \u00b5 1<\/sub> \u2260 \u00b5 2<\/sub> (iki pop\u00fclasyon ortalamas\u0131 e\u015fit de\u011fildir<\/em> )<\/span><\/p>\n
 Testimizin p de\u011feri (0,01823)<\/strong> 0,05’ten k\u00fc\u00e7\u00fck oldu\u011fundan s\u0131f\u0131r hipotezini reddediyoruz.<\/span><\/p>\n
 Bu, iki t\u00fcr aras\u0131ndaki ortalama bitki boyunun e\u015fit olmad\u0131\u011f\u0131 sonucuna varmak i\u00e7in yeterli kan\u0131t\u0131m\u0131z oldu\u011fu anlam\u0131na geliyor.<\/span><\/p>\n
 Teknik<\/span> notlar<\/span><\/strong><\/h2>\n
 R’deki t.test()<\/strong> i\u015flevi a\u015fa\u011f\u0131daki s\u00f6zdizimini kullan\u0131r:<\/span><\/p>\n
 t. test<\/span> (x, y, alternative=\"two.sided\", mu=0, paired=FALSE, var.equal=FALSE, conf.level=0.95)<\/span><\/strong><\/pre>\n
 Alt\u0131n:<\/span><\/p>\n
 Bu varsay\u0131m\u0131 yapmak istemiyorsan\u0131z, bu arg\u00fcman\u0131 bir kenara b\u0131rak\u0131n; R bunun yerine Welch’in t-testini<\/a> ger\u00e7ekle\u015ftirecektir; bu, \u00f6rnekler aras\u0131nda varyanslar\u0131n e\u015fit oldu\u011funu varsaymaz.<\/span><\/p>\n