Standard errors for 3 averages - Some clarification (Sep/26/2005 )

Hi Statistics gurus,

I have some very basic questions (which is being highly debated in the lab)...

I have 3 replicates of the same experiments. I need to have the average and standard error. Hence, I take the averages of the 3 replicates and calculate their average and their standard deviation. This I denote as the Average of the 3 replicates and Standard Error.....Am I right??

2nd: I have to convert averages into percentiles...therfore I designate the control group average as 100 and then compute the averages in percent for the other treatment groups. Some of my labmates use something called as a Standard error percent (?). What they do is take the SE for the 3 samples (in one experiment) and divide it by the average of that group and multiply it by 100.

1. For Eg: (What my labmates do)
Control 20, 25 and 50 Average is 30 and SE is 0.1 (The av and SE is just an example not actual values)
Average % : 100
SE % : 0.1/30*100

2. For Eg: (What I do)
Control 20, 25 and 50 Average is 30 and SE is 0.1
Average % : 100
SE : 0.1 ( I dont take the percntile)

Which is the right way to do things?? 1. or 2.?

Help me out...There are some Donuts at stake here...

hi
i've putted in the list 20, 25, 50
mean is 31,6.
Variance (S) is : somme(i-mean)²/number of sample
S = [(20-31,6)²+(31-31,6)²+(50-31,6)²]/3
S = [134.6+2.56+338.56]/3
S = 475.68/3
S = 158.56

Std dev is the square root of variance
it's the same as variance^(1/2)
"sigma" = sq root of S
"sigma" = sq root of 158,56
"sigma" = 12,59

statc calc is unfortunately a shareware bu may help. just type statcalc in google
but excel can perform such analysis.

hope that helps.
fred

I am sorry I had a nice long reply for you with math worked out and all, but somehow it just erased. suffice it to say that your collegues are both right and wrong..

you are ignoring the variability in your control construct, when you divide by control to normalize to 100% you are supposed to add the std. deviations together from the control and the construct you are dividing by control... equation looks like this:

normalized standard deviation = normalized average * (sqrt of (st.dev1/ave1)^2 +(stdevcontrol/ave control)^2)

for the numbers listed this statistically correct sd = 71.78%

If you decide biologically to ignore this variability then your collegues do it right in example 1.

For the numbers above the sd = 50.756%

make sure you are using the correct method for error bars... standard error is usually smaller because it = sd/(sqrt of N) but that does not mean it is necessarily the correct way to show your data...

HTH