in following example, fourteen raters (
n
{\displaystyle n}
) assign ten subjects (
n
{\displaystyle n}
) total of 5 categories (
k
{\displaystyle k}
). categories presented in columns, while subjects presented in rows. each cell lists number of raters assigned indicated (row) subject indicated (column) category.
data
see table right.
n
{\displaystyle n}
= 10,
n
{\displaystyle n}
= 14,
k
{\displaystyle k}
= 5
sum of cells = 140
sum of
p
i
{\displaystyle p_{i}\,}
= 3.780
calculations
the value
p
j
{\displaystyle p_{j}}
proportion of assignments (
n
×
n
{\displaystyle n\times n}
, here
10
×
14
=
140
{\displaystyle 10\times 14=140}
) made
j
{\displaystyle j}
th category. example, taking first column,
p
1
=
0
+
0
+
0
+
0
+
2
+
7
+
3
+
2
+
6
+
0
140
=
0.143
{\displaystyle p_{1}={\frac {0+0+0+0+2+7+3+2+6+0}{140}}=0.143}
and taking second row,
p
2
=
1
14
(
14
−
1
)
(
0
2
+
2
2
+
6
2
+
4
2
+
2
2
−
14
)
=
0.253
{\displaystyle p_{2}={\frac {1}{14(14-1)}}\left(0^{2}+2^{2}+6^{2}+4^{2}+2^{2}-14\right)=0.253}
in order calculate
p
¯
{\displaystyle {\bar {p}}}
, need know sum of
p
i
{\displaystyle p_{i}}
,
∑
i
=
1
n
p
i
=
1.000
+
0.253
+
⋯
+
0.286
+
0.286
=
3.780
{\displaystyle \sum _{i=1}^{n}p_{i}=1.000+0.253+\cdots +0.286+0.286=3.780}
over whole sheet,
p
¯
=
1
(
10
)
(
3.780
)
=
0.378
{\displaystyle {\bar {p}}={\frac {1}{(10)}}(3.780)=0.378}
p
¯
e
=
0.143
2
+
0.200
2
+
0.279
2
+
0.150
2
+
0.229
2
=
0.213
{\displaystyle {\bar {p}}_{e}=0.143^{2}+0.200^{2}+0.279^{2}+0.150^{2}+0.229^{2}=0.213}
κ
=
0.378
−
0.213
1
−
0.213
=
0.210
{\displaystyle \kappa ={\frac {0.378-0.213}{1-0.213}}=0.210}
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