📅  最后修改于: 2021-01-23 06:51:03             🧑  作者: Mango
通过两次测量同一个人并计算两组测量值之间的相关性来获得的测试或测量仪器的准确性的测量值。
可靠性系数由以下函数定义和给出:
$ {Reliability \ Coefficient,\ RC =(\ frac {N} {(N-1)})\ times(\ frac {(Total \ Variance \-Sum \ of \ Variance)} {Total Variance}}} $
哪里-
$ {N} $ =任务数
问题陈述:
经历过三个人(P)的工作,并且为他们分配了三个不同的任务(T)。发现可靠性系数?
P0-T0 = 10
P1-T0 = 20
P0-T1 = 30
P1-T1 = 40
P0-T2 = 50
P1-T2 = 60
解:
给定的学生数(P)= 3任务数(N)=3。要查找可靠性系数,请按照以下步骤操作:
给我们机会首先计算人员及其任务的平均分数
The average score of Task (T0) = 10 + 20/2 = 15
The average score of Task (T1) = 30 + 40/2 = 35
The average score of Task (T2) = 50 + 60/2 = 55
接下来,计算以下变量的方差:
Variance of P0-T0 and P1-T0:
Variance = square (10-15) + square (20-15)/2 = 25
Variance of P0-T1 and P1-T1:
Variance = square (30-35) + square (40-35)/2 = 25
Variance of P0-T2 and P1-T2:
Variance = square (50-55) + square (50-55)/2 = 25
现在,计算P 0 -T 0和P 1 -T 0 ,P 0 -T 1和P 1 -T 1 ,P 0 -T 2和P 1 -T 2的个体方差。为了确定个体差异值,我们应该包括所有以上计算的变化值。
Total of Individual Variance = 25+25+25=75
计算总变化
Variance= square ((P0-T0)
- normal score of Person 0)
= square (10-15) = 25
Variance= square ((P1-T0)
- normal score of Person 0)
= square (20-15) = 25
Variance= square ((P0-T1)
- normal score of Person 1)
= square (30-35) = 25
Variance= square ((P1-T1)
- normal score of Person 1)
= square (40-35) = 25
Variance= square ((P0-T2)
- normal score of Person 2)
= square (50-55) = 25
Variance= square ((P1-T2)
- normal score of Person 2)
= square (60-55) = 25
现在,包括每一种特质并计算出总体变化
Total Variance= 25+25+25+25+25+25 = 150
最后,用下面提供的等式中的质量来发现