I have 5 biological samples, each tested at two dilutions (4 technical replicates each). I am using a standard curve to convert the fluorescence units into micrograms of protein and normalizing by weight of samples. I expected the two dilutions of each sample to be close, but they are not and standard deviations are small. I am optimizing a protocol and would like to use a statistical test to evaluate the difference between the two dilutions of a given sample. It seems to me that the t-test is incorrect to use because a normal distribution is not expected. Any help would be much appreciated. Thank you!
1) take the logarithms of your values, since the distribution of protein concentrations is approximately log-normal.
2) average the 4 technical replicates
4) calculate the sample mean difference
3) use a t-test to see if the sample size is sufficient to interpret the sign of the sample mean difference
Hi Shirley,
I think you can find a better answer on "Modern Epidemiology"
book by Kenneth Rothmanthe
The phrase "evaluate the difference between the two dilutions of a given sample" may suggest further consideration.
Is it just the means of each dilution that is of interest ? Is a difference in the spread of values for each dilution meaningful ? Is the shape of the distribution of values important (like maybe one dilution tends to be skewed, and the other bell-shaped) ? And so on.
Maybe comparing the two sets of observations visually and with a variety of summary statistics might bring something to light that is important.
Also, evaluating how "accurate" one dilution is against the other. That is, what if the two dilutions give the same result on average, but the results for a given sample aren't particularly close. Perhaps something like coefficient of variation or root mean square error would make sense.
I would try regression approach:
1. average out your technical replicates as suggested by Jochen Wilhelm
2. plot "dilution-1 outcome" verses the "dilution-2 outcome" (that will be 5 points on the plot)
3. if no difference among 5-means then slope will be zero otherwise test significance of any non-zero slope (using lack-of-fit residual as error)
4. If trend is nonlinear then some simple curve (say) with 2 parameters may be possible (again using lack-of-fit residual as error)
Mewa Singh Dhanoa ,
a) I assume you mean, that if you plot Dilution-1 vs. Dilution-2, the null slope would be 1, not 0.
b) But even in that case, a slope of 1 doesn't mean that the results are same in magnitude. For example, D1 = ( 1, 2, 3, 4, 5); D2 = c(11,12,13,14,15); the slope is 1, but the values for D2 are larger than any in D1.
c) In any case, plotting the data will likely reveal if there is anything interesting between the results of Dilution-1 and Dilution-2.
Thank you to all who responded. I chose to do as Jochen Wilhelm suggested and look at the sample mean difference. I also appreciate Salvatore's suggestions to think about exactly what I need to know. Since the two dilutions of each sample are highly significantly different, this indicates that the sample with the higher concentration is not being fully oxidized under current protocol conditions. Thus, I will do a dilution series to find the optimal dilution for complete oxidation.
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