Worked example 2
We will briefly revisit a worked example from Unit 3 where we took two repeated measurements of packed cell volume from each of 10 cows.
Cow No.  PCV  T_{i
} 
1.  2.

1  33  35  68

2  26  23  49

3  29  29  58

4  32  35  67

5  31  28  59

6  31  31  62

7  31  34  65

8  31  29  60

9  35  33  68

10  21  24  45

Sum
   601

By taking the square root of the mean of the variances we estimated:
Withinsubject standard deviation = √(28.5/10) = 1.688
We can obtain the same information by doing a one way analysis of variance:
ANOVA table

Source of variation
 df
 SS
 MS
 F ratio
 P

Cow
 9
 278.45
 30.939
 10.856
 < 0.001

Residual
 10
 28.5
 2.85



Total
 19
 306.95



  
The mean square residual (MS_{W}) is equal to 2.85. The within subject standard deviation is given by √MS_{W} = 1.688, the same as we obtained using the other computational formula.
The added variance component is calculated as below:
s_{A}^{2
}  =
 30.939 − 2.85
 =
 14.0445


2
  
The intraclass correlation coefficient (percentage variation between cows) is then calculated as:
ICC
 = 100 × [
 14.0445
 ]
 =
 83.13%


2.85 + 14.0445
  
This can also be expressed as a percentage measurement error of 16.87%.