Regarding the specific calculation from a sample, I can't tell you, but this is from the standpoint of the theoretical math involved here:

Standard deviation (in combinations) is dependent upon the underlying distribution of whatever phenomenon you are looking at.  For instance, if you had two normally distributed variables, you can add their averages and then add their variances to get the mean and variance for the sum.  Depending what distribution you have, that rule of calculating combination changes.  That is probably why you have a hard time finding the way to calculate this specifically because it would have a dependency upon the samples themselves.

That being said, what I would suggest for you to do is to figure out what kind of distribution your individual samples have (or should have).  That should hold the answer to calculating those kinds of standard deviation calculations.

Thank you for your answer.

I would presume that the distribution of the measurements on one sample is a gaussian distribution which has a lower standard deviation than when comparing data of two samples at the same time. So I'll stick to my current approach (step 1-3) and conclude that the widest spread is resulting from variation between experiments and not between measurements of a single sample.

Nils,

Matthew is correct in stating that "Standard deviation (in combinations) is dependent upon the underlying distribution of whatever phenomenon you are looking at".  The mean and standard deviation are dependent on the presence of a near normal distribution of the data.  To test that, it is possible to use the 'skewness' function in Excel.  If the value is below 1, the data are close to normal distribution.  If above 1, the data are less normally distributed.  If more than 3, then probably best to use the median and quartile method, but much depends on the overall number of data items being tested.

Regards

George

Presumably the samples same composition, i.e. T1 11 and T1 12 are supposed to have the same concentration. It is reasonable that there will be differences in the obtained results. The experimental design is somewhat unclear to me but it seems that you perform the same reaction twice and measure the quantity at various times. You will end up with two time series consisting of eight samples each.with tree replicates. A standard deviation calculated from only three observations is not worth very much. A complicating fact is the possibility of a bias between the time series but if that can be neglected the best you can do is probably to perform a ANOVA with two groups each with three observations. I wold not take the results too seriously, though.

Hey,

Since I posed this question I saw it has been viewed many times so let me clarify what I did and how I solved it.

What did I do?

1) I made a solution of chemicals, lets say 50ml

2) I took 2 times 25ml of this solution (Solution A and B) and let it react

3) At various times (T1, T2 etc.) I took a sample of both solutions (Sample 1, 2 ,3 )

4) I measured the concentration of a certain species of each sample 3 times

This gave me the following table

Time...... Samples of Solution A.......Sample of Solution B

(Min)....... (/)....................................... (/)

10.............Sample 1A........................ Sample 1B

20.............Sample 2A........................ Sample 2B

30............ Sample 3A........................Sample 3B

...

After taken the samples I measured the concentrations of each sample 3 times/

Time.........Concentration of Solution A...Concentration of Solution B

(Min)........(mol/L)..................................... (mol/L)

10............. 0.1; 0.3; 0.2 .............................0.3; 0.2; 0.2

20............. 1; 1.1 ;1.3 .................................1.3; 1.2; 1.2

30............. 2; 1.9 ;2.2 .................................2.1; 1.9; 1.3

...

Now how to report the standard deviation? (My original question).

Wel there are two sources of variation:

1) caused by running the experiment twice (solution A and B)

2) caused by measuring one sample multiple times.

How did I solve it?

If you did everything perfectly both solutions are identical and thus the samples are too. In this case all values at a certain time are coming from the same 'population' . We can then report an average and standard deviation at each time using all number and thus:

using 0.1; 0.3; 0.2; 0.3; 0.2; 0.2 at 10 mins gives 0.2 (average) and 0.1 (standard deviation).

If you did not do everything perfectly (which happens) we need to check if the results of concentration A and B differ significantly. This can be done using for example a paired T-test to check if the average concentration at each time differs significantly. If the test indicates a significant difference at least one of your experiments did not go as planned. If you see a significant difference you can run a third experiment and compare it with the previous two or if that is not possible pick the results that make the most sense and report poor repeatability was observed.

Several answers bring up that the 'underlying distribution' of the measurements needs to be known in order to validate any tests or assumptions but with the little amount of repeated measurements (3 per solution) there is not much that can be proven. Typically in hard to repeat measurements a normal distribution is assumed.

Where does this problem arise?

I see in many fields that 'error bars' are not calculated or indicated correctly. Specifically the repetition of experiments is necessary to show any conclusive evidence where typically you would run a minimum of 2 or 3 experiments.