# How does one calculate a percentage measurement error and variance using mean square error?

I'm trying to calculate a measurement error for some snail body length measurements I made over the summer. I was measuring the body length of 40 snails and because I was using calipers and sometimes measurement markers weren't always that obvious, I want to determine the proportion of variance my error contributed to the measurement overall.

I measured each animal three separate times and am using an equation by Yezerinac (1992) that is as follows:

$ME\text{%} = s_{within}^2/(s_{within}^2+ s_{among}^2 )$

where $s_{among}^2 = (MS_{among}-MS_{within})/m$

and m is the repeated number of measurements. In the paper, they state that "Mean squared deviations of scores within individuals ($MS_{within}$) estimated the within-individual component of variance ($s_{within}^2$)" which makes sense to me because $MS$ is an estimate of variance.

I measured each animal three times and then ran an anova on the measurements (Length) sorted by the 1st, 2nd or 3rd time measured (Group). An example output would be:

Response: Length

                   Df         Sum Sq   Mean Sq     F value   Pr(>F)
Group          2   0.00126   0.0006308   0.1936    0.8243
Residuals    117   0.38133   0.0032592


The way I understand it, the $MS$ is equal to the variance. Residual $MS$ would be '$s_{within}^2$' and group MS would be "$s_{among}^2$". Do you know why they use the equation (second equation) to calculate $s_{among}^2$? When I use that equation, assuming MS for group = MSamong, I end up with a very high $MS\text{%}$ which doesn't make sense.

Hopefully my description is clear!

The problem I was having involved my ANOVA set up. To partition variance to within individual animals and across all individuals (among), individuals should be set as a Group which would mean the MS of the residuals approximates the MSwithin. Here's sample of my data, code and output:

    Groups Meas
1      Bg1 0.15
2      Bg1 0.15
3      Bg1 0.19
4      Bg2 0.29
5      Bg2 0.28
6      Bg2 0.29
7      Bg3 0.32
8      Bg3 0.28


I ran an ANOVA:

boxplot(bg$Meas~bg$Groups)
bgaov=aov(bg$Meas~bg$Groups)
summary(bgaov)

Df Sum Sq  Mean Sq F value Pr(>F)
bg\$Groups   39 0.3739 0.009586   87.81 <2e-16 ***
Residuals   80 0.0087 0.000109


Using the MS for residuals as MSwithin and calculating s2among using the MS of groups as MSamong the measurement error for this example is 3.335%.

• This was calculated like this: s2among = (0.009586-0.000109)/3; 0.000109/(0.000109+s2among)*100 Commented Jun 17, 2019 at 13:25