# Average standard deviation?

#### peptideguy

##### New Member
Hi Everyone,

Sorry if a similar question has been asked before. I didn't see a thread focused on my particular query.

I have a large dataset of peptide fragments associated with mean and standard deviation values from area-under-the-curve integrals (experiments were run 3x to obtain the mean + SD for each peptide). Parts of the amino acid sequence overlap between fragments, giving some redundancy (e.g., Y92 appears in 10 peptide fragments, as highlighted in the attached screenshot).

What I would like to do is re-average each signal containing the same amino acid residue, such as Y92, but also calculate a new standard deviation for that re-average. I’m not exactly sure how to go about the latter… Should I compute an average standard deviation from the existing SD values? If so, which formula would be appropriate?

Best,
Chris

#### GretaGarbo

##### Human
I did not understand this one.

Maybe the OP is asking for a "congeneric measurement model" for the latent variable Y92 which is observed in the marked in the attached image.

I wonder why such a model is not used more?

#### peptideguy

##### New Member
Hi GretaGarbo,

My apologies---I couldn't fully grasp the link you provided. My data are from mass spectrometry (not SEM) measurements of peptides, though I'm not sure if the congeneric measurement model still applies. Nevertheless, I will try to better explain my situation.

Each peptide (e.g. RMVYQHENGN) was measured 3 times and the resulting intensities were averaged to give a mean (e.g. D18) and standard deviation (e.g. E18) as shown in the attached image from Excel. Y92 (underlined in RMVYQHENGN) is just one component of all the highlighted peptides. I want to obtain one new average and standard deviation for that group of highlighted peptides containing Y92. The mean is easy enough to calculate, simply by averaging D18-D27. However, I'm not sure how to compute the new standard deviation. Averaging them seems somehow incorrect, but I might be mistaken.

Thanks!

Best,
Chris