Mohammad Amin Moeini, I think you might not understand something that is very basic in statistics. The two sets of figures you have provided in your question have the same amount of variance (essentially, spread of scores) - and, sorry, I don't know what you mean by convergence.

Might you provide some details about why you came to your conclusions, please? If you do so, it might be possible to help you.

Mohammad Amin Moeini, I think you might not understand something that is very basic in statistics. The two sets of figures you have provided in your question have the same amount of variance (essentially, spread of scores) - and, sorry, I don't know what you mean by convergence.

Might you provide some details about why you came to your conclusions, please? If you do so, it might be possible to help you.

Mohammad Amin Moeini , are you meaning to transform the variables, so if you two variables are x1 and x2, and if you transform these, say x1/mean(x1) and x2/mean(x2), and then take the variance, these will be different. You could log them, calculate the variance, then exponentiate to get the geometric variance. I think when most people read variance is is assumed arithmetic variance so is unaffected by added a constant. That said, I am not sure that I understand your question. Can you say why you are doing this, what your research question(s) is(are), and something about your study/data?

Mohammad Amin Moeini Variances can be added but standard deviations cannot. However, you can use coefficient of variation (= Relative Standard Deviation) to show the differences between your 2 example sets. As Robert Trevethan correctly points out, the magnitude of the variance is identical in both sets.

I am agree with all of the above answers by Rober, Danel and Alan.

Hi Mohammad Amin Moeini ,

preamble of your question is a bit confusing.

I will answer your direct question: " Is there a way to show how the results are close or far from the average without dependence on the data values? "

Two link, in which is explained "accurancy" and "precision".

Accuracy measures how close your values are from a reference value.

Precision measures how close your values are each other.

https://chem.libretexts.org/Bookshelves/Introductory_Chemistry/Book%3A_Introductory_Chemistry_(CK-12)/03%3A_Measurements/3.12%3A_Accuracy_and_Precision

https://chem.libretexts.org/Courses/University_of_Arkansas_Little_Rock/Chem_1402%3A_General_Chemistry_1_(Belford)/Text/1.B%3A_Review_of_the_Tools_of_Quantitative_Chemistry/1B.2%3A__Making_Measurements%3A__Experimental_Error%2C_Accuracy%2C_Precision%2C_Standard_Deviation_and_Significant_Figures

The datasets should be normalized by dividing by the covariance. See Mahalanobis distance for details.

Yes it, the greater the variance or data dependency among themselves leads to errors.

Dear Daniel Wright

Many thanks for your answer, to explain my objective more clearly: I am measuring a rheological parameter lets call it P for 4 mixtures. I am measuring this parameter of the 4 mixtures using various experimental methods lets call them M1, M2, and M3.

To make comparison between the measurement methods I would like to point out the capacity of each method (M1-M3) to differentiate the rheological parameter (P) between the mixtures. For example it is favorable if using a certain testing method I am able to observe a higher difference of P between my mixtures.

So I was thinking to report the variance of P of my mixtures for each testing method and compare the variances of the methods and say the greater variance means better testing method. However, since the average of the P for different methods is different, I am seeing that as the numbers get higher the variance gets higher as well even though the results are in better approximation.

Using M1 the value of P is measured as follows for 4 mixtures: 26, 16, 20, 15

Using M2 the value of P is measured as follows for 4 mixtures: 504, 100, 250, 150

variance of P for M1 is about 1000 times lower than that of M2 which is affected by the values. Is there anyway to remove the effect of the values ?

Thanks Mohammad Amin Moeini for the details. For many processes, the standard deviation with each group is roughly proportional to the mean. Because this form of heterogeniety (i.e., differences in variances/standard deviations) can still be modeled with least squares people have been assuming this particular relationship for a while for convenience. I'm in the mountains, so away from my books, but some commentator should be able to provide a reference. Other ways of directly modeling the heterogeniety are more flexible, and again, covered in regression books and implemented in much statistics software. Alternatively, transforming the data may help (the first try is often the Box-Cox transformation or ln(x+a), sometimes with a=0 for simplicity if there are no zeroes in the data). THese would allow you to compare the means modeling how the variances also differ. I think this is what you are after.

Many thanks Daniel Wright ! Enjoy your time in nature :)

Yes, but variance is a measure of the square of the deviation. The measure of the deviation is the standard deviation - the square root of the variance. When interpreting the standard deviation as a measure of scatter, the same constraints apply as when interpreting the mathematical deviation as a representative value. If abnormally large or small values ​​are encountered, then the standard deviation is greater than the deviation of most values ​​from the mean.

@ Amin Moeini

You write

· 1,2,3,4,5

· 101,102,103,104,105

The variance will be higher for the second set of data however the second data show better convergence.

Is there a way to show how the results are close or far from the average without dependence on the data values?

• BEFORE asking a question one MUST study!

LATER you write

• So I was thinking to report the variance of P of my mixtures for each testing method and compare the variances of the methods and say the greater variance means better testing method.

THIS makes little Sense!!!

Massimo Sivo , on the variance being higher, can you tell us how much higher? One set is just a constant plus the other.

Massimo Sivo, Daniel Wright asks a possibly revealing question and makes a very important, and correct point.

One way to test whether the variance is different in the two data sets is simply to calculate the variance (and SD if the SD is more familiar to you). There are sites on the web where that can easily be done - and they demonstrate that the variance and SD are identical for both sets of numbers - which should be obvious, anyway, without resorting to calculations.

That aside, the original questioner seems to have revealed the issue to be different / more complex than indicated in the original question - and, forgive me, but I haven't taken the trouble to think through what the new issue is.

Many times, reading Questions and Answers, I wander if People use their rationality.

I think that one should study before making question and before answering.

IF one writes

• · 1,2,3,4,5 and 101,102,103,104,105 The variance will be higher for the second set of data however the second data show better convergence.
• · Is there a way to show how the results are close or far from the average without dependence on the data values?

it DOES MEAN that he does not know what variance is!!!

SO I do not understand why Daniel Wright added an answer Massimo Sivo , on the variance being higher, can you tell us how much higher? One set is just a constant plus the other. Recommend Share 2 Recommendations

AND Robert Trevethan added an answer “omissis … One way to test whether the variance is different in the two data sets is simply to calculate the variance omissis …” Recommend Share 2 Recommendations

TO TEST if two (computed) variances are significantly different ONE MUST make a Test of Hypothesis with the APPROPRIATE distribution….!!!!

• IF the data 1,2,3,4,5 and 101,102,103,104,105 were from an EXPONENTIAL distribution the VARIANCES (NOT the “computed variances”!!!) would be QUITE different!!!!

@ Oh it's Massimo again with his well tempered and absolute calm answers. As always a productive contribution to a scienitific discussion (irony off).

I may be making a silly point here, but wouldn't the coefficient of variation (SD/Mean) be a simple solution to the OP's needs? That way, he would simply compare the CV of each of his testing methods.

Please correct me if I'm wrong.

Rainer Duesing , yea you can see a trend in his responding across questions.