Stata 是康格'；s卡帕在斯塔塔有售吗？
康格（1980）提出的方案在斯塔塔可用吗？我不知道Stata是否有Conger's kappa for multiple raters，但是可以通过Stata 是康格'；s卡帕在斯塔塔有售吗？,stata,Stata,康格（1980）提出的方案在斯塔塔可用吗？我不知道Stata是否有Conger's kappa for multiple raters，但是可以通过irr软件包，使用kappam.fleiss函数并指定exact选项，在R中获得它。有关R中的irr包的信息，请参阅 在R中安装并加载irr包后，您可以使用以下代码查看演示数据集和Conger的kappa计算 data(diagnoses) print(diagnoses) kappam.fleiss(diagnoses, exact=TRUE)
irrkappam.fleissexactirrirrdata(diagnoses)
print(diagnoses)
kappam.fleiss(diagnoses, exact=TRUE)
我希望这里的其他人能够按照您的要求提供Stata解决方案，但如果您在Stata中找不到，这至少可以提供一个解决方案。对于Dimitry下面的评论，我相信Stata的本机
kappa原海报可能还想考虑StATA中的<代码> ICC < /代码>命令，它允许多个唯一的估计器。
< P>由于没有人用STATA解决方案，我开发了一些代码来计算康格的Kappa，使用Gwet提供的公式，K. L.（2012）。马里兰州盖瑟斯堡：高级分析有限责任公司《评级机构间可靠性手册》（第三版），特别见第34-35页 毫无疑问，我的代码没有其他人编写的那么高效，我欢迎其他人对代码或程序格式进行任何改进cap prog drop congerkappa
prog def congerkappa
* This program has only been tested with Stata 11.2, 12.1, and 13.0.
preserve
* Number of judges
scalar judgesnum = _N
* Subject IDs
quietly ds
local vlist `r(varlist)'
local removeit = word("`vlist'",1)
local targets: list vlist - removeit
* Sums of ratings by each judge
egen judgesum = rowtotal(`targets')
* Sum of each target's ratings
foreach i in `targets' {
quietly summarize `i', meanonly
scalar mean`i' = r(mean)
}
* % each target rating of all target ratings
foreach i in `targets' {
gen `i'2 = `i'/judgesum
}
* Variance of each target's % ratings
foreach i in `targets' {
quietly summarize `i'2
scalar s2`i'2 = r(Var)
}
* Mean variance of each target's % ratings
foreach i in `targets' {
quietly summarize `i'2, meanonly
scalar mean`i'2 = r(mean)
}
* Square of mean of each target's % ratings
foreach i in `targets' {
scalar mean`i'2sq = mean`i'2^2
}
* Sum of variances of each target's % ratings
scalar sumvar = 0
foreach i in `targets' {
scalar sumvar = sumvar + s2`i'2
}
* Sum of means of each target's % ratings
scalar summeans = 0
foreach i in `targets' {
scalar summeans = summeans + mean`i'2
}
* Sum of meansquares of each target's % ratings
scalar summeansqs = 0
foreach i in `targets' {
scalar summeansqs = summeansqs + mean`i'2sq
}
* Conger's kappa
scalar conkappa = summeansqs -(sumvar/judgesnum)
di _n "Conger's kappa = " conkappa
restore
end
Judge S1 S2 S3 S4 S5 S6
Rater1 2 4 2 1 1 4
Rater2 2 3 2 2 2 3
Rater3 2 5 3 3 3 5
Rater4 3 3 2 3 2 3
use http://www.stata-press.com/data/r12/judges.dta, clear
sort judge
list, sepby(judge)
reshape wide rating, i(judge) j(target)
rename rating* S*
list, noobs
* Run congerkappa program on demo data set in memory
congerkappa
我仅针对Gwet（第35页）表2.16中的数据对该代码进行了一次验证测试，并复制了Gwet在第35页计算的Conger's kappa=.23343。34在依赖之前，请使用已知的Conger's kappas在其他数据上测试此代码。这是一个老问题，但如果有人仍在寻找——SSC包
kappaetckappa