Comparing the means of two independent populations

#1
I need assistance please in determining how best to compare the means of two independent populations for significance. I am trying to create an Excel template to allow school districts to compare outcomes between Group A, a non-participant group, and Group B, the participant group. Group B is a subpopulation of Group A, but the data for Group A exclude Group B.

Districts have access to the summary data for each group (sample size and mean), but not the entire data set of individual observations. The template would include multiple variables (e.g. comparison of groups by ethnicity and gender).

I *think* I can use an independent t-test for two populations for each variable (row in the workbook). I realize that some may suggest using ANOVA since there are multiple variables, but in this instance it is not necessary to compare between variables, just each variable to determine if there is a significant difference between Group B and Group A.

My questions are:
  1. Is it possible to test for significance only knowing the means and sample size of the two populations?
  2. If so, what is the best statistic to use for:
    • Cases where the sample size will be large and we can assume normal distribution and equal variance; and
    • Cases where the sample size will be small (<20) with no expectation of equal variance or normal distribution?
  3. Is it possible to do this in Excel? If so, could you please share with me how to set it up?
Thank you so much in advance for your help. I appreciate it!
 

katxt

Active Member
#2
It sounds like you have (non overlapping) groups A and B and the only data available is four numbers, the mean and size of A and the mean and size of B.
If that's the case, the answer is No - it's not possible to test for significance. You need some information about the spread of numbers within each group - the SD or the mean of each participant.
 

Dason

Ambassador to the humans
#3
Yeah it doesn't matter if we can assume equal variance if we have no information on what the variance might actually be.