I would use the Kolmogorov-Smirnov test for difference in distributions.  It's non-parametric

ISSUE: Statistical significance test for comparison study

DATA: You need to obtain the data, i.e. histogram is not data, it is a representation of data in a bar chart. You need the actual value for each bar or read the tip of the bar to the Y-axis if the bar is vertical or read the tip of the bar to the X-axis if the bar is horizontal.

DEFINITION OF TERMS: Let the following terms be defined: n1 = observation counts for sample 1; n2 = observation count for sample 2. Let the score of sample 1 be represented by X1 and that the individual observation in sample 1 be: X1i: (x11, x12, ..., x1n) and similarly for sample 2 be X1i: (x11, x12, ..., x1n). Now we are ready for the test of significance of their differences. There are two scenarios: (i) the sample size n1 and n2 are equal; and (ii) the sample size of n1 and n2 are not equal. Statistical test for each scenario follows.

(i) Equal Sample Size: In this case, use d-bar analysis. First find the paired difference between X1i and X2i, thus:

(X11 – X21) = d1

(X12 – X22) = d2

…

(X1n – X2n) = dn

The means or d-bar is d^ = (1/n)Σdi and standard deviation is simply Sd = sqrt([1/n]Σ(di - d^)2.  The test statistic is given by:

(1)   Td = d^ / Sd / sqrt(nd)     ... where nd = number of pairs.

Set your confidence interval. Say CI = 0.95 and there are 10 pairs; the degree of freedom is df = 10 - 1 = 9. The t-critical value for 95% CI at 9 degree of freedom is 1.83. If the calculation Td(obs) above shows that Td(obs) > 1.83 it means that the difference between two samples are statistically significant; if Td(obs) < 1.83 then the difference is not significant. For infinity degree of freedom t(0.95) = 1.64.

(ii) Two samples of Unequal Size: If the two samples have different sample sizes: n1 not equal to n2, then the statistical test used to verify their difference is given by:

(2)   T = [(X1* - X2*) -(μ1 - μ2)] / S(sqrt(1/n1 + 1n2))

(2.1)   S = sqrt(S2)

(2.2)   S2 = (SS1 - SS2) / (n1 + n2 - 2)

(2.3)   SS1 = Σ(Xi - X1*)2         where Xi* = mean

(2.4)   SS2 = Σ(Xi - X2*)2

The decision rule is the same as above. Set the confidence interval and compared the T(obs) to the T(0.95, df) value. The two samples are statistically different if T(obs) > T(0.95, df).

If your question is about the shape of the histogram, do the K-S; if it's about the mean and you can accfess the raw data, then do a t-test

Tank you all!