Statistical Signifance Test for discrete data I'm fairly new to statistics and want to perform a statistical analysis (calculation of p value)
for data I collected, but I'm unsure if I'm doing it correctly and would like some feedback/help.
I have 180 ratings from 1 to 10 for two categories, so 90 ratings per category.
This is the distribution of answers for the categories:

My goal is to prove that these datapoints result from the same underlying distribution, i.e.,
there is no statistical significant difference.
But I'm not sure what statistical test I should use, since there seem to be a lot of conditions and caveats.
The datapoints don't follow a normal distribution, so I should use a non parametric test.
My data is numerical and discrete, which also excludes a lot of options.
I tried the chisquare test but it results in NaN for the scipy.stats.chisquare method, since the values 1 and 10 have a frequency of 0. I could just exclude 1 and 10, but I don't know if thats valid. The output I get is:
Power_divergenceResult(statistic=9.400411679823446, pvalue=0.22517169474250875)

Another test I tried is the Mann Whitney U test, which gave me the following result:
MannwhitneyuResult(statistic=4061.5, pvalue=0.9742929619325416)

Can I use the latter approach? Did I make any mistake so far?
 A: As an opening observation, you should never begin your statistical analysis with a goal of getting a particular result in your test --- rather than aiming to prove that the two sets of data come from the same distribution, your goal should merely be to inquire to see whether or not this is the case, with an a priori openness to the possibility that the underlying distributions may be different.
As to the method of testing, assuming your data values are not paired, it would be reasonable to use any well-performing two-sample test for sameness of distribution.  There are a number of these tests, including the two-sample Kolmogorov-Smirnov test, the two-sample Kuiper test, the two-sample Anderson Darling test, or the two-sample Pearson chi-squared test.$^\dagger$

$^\dagger$ Note: The former tests are created for continuous distributions so you would need to determine the appropriate null distribution via simulation (e.g., using bootstrapping methods).  The latter test treats the variables as categorical, which loses some of the ordering structure, but it is appropriate for discrete distributions under broad circumstances.
A: Assuming that you are treating the responses for Confidence as ordinal categories, your bar plot of counts for Confidence responses for each Group suggests that there is no systematic difference in the central tendency of the responses for the groups.
To test this, you might use a Wilcoxon-Mann-Whitney test, which assesses the probability that a response from Group A is greater than a response from Group B.
You could also compare median responses from the Groups with Mood's median test or another test of medians.
If you wanted to take a bit of a leap of faith and compare the mean responses --- you could assume that the responses are interval in nature and that the marginal distribution of the responses are normal in distribution --- and use a t test.
You might be a little careful with your conclusions. You can't prove that these samples come from the same distribution.  But it looks like you don't have good evidence that they come from different distributions (based on the tests described).
Below is the approximate data extracted from the plot, and a Wilcoxon-Mann-Whitney test, in R.
The vda() function then calculates the probability that an observation from A will be greater than an observation from B.  The result, close to 0.5, suggests that neither group is very likely to have observations greater than the other group.
if(!require(tidyr)){install.packages("tidyr")}

Data = read.table(header=TRUE, stringsAsFactors=TRUE, text="

Group Confidence Count
A 1   0
A 2   2
A 3   8
A 4   8
A 5   8
A 6  20
A 7  24
A 8  17
A 9   3
A 10  0
B 1   0
B 2   2
B 3   7
B 4   9
B 5   9
B 6  14
B 7  34
B 8  11
B 9   4
B 10  0
")

library(tidyr)

Long = uncount(Data, Count)

rownames(Long) = seq(1:nrow(Long))

Long

library(FSA)

Summarize(Confidence ~ Group, data=Long)

wilcox.test(Confidence ~ Group, data=Long)

    ### W = 4038.5, p-value = 0.9743


library(rcompanion)

vda(Confidence ~ Group, data=Long, verbose=TRUE)

   ###            Statistic Value
   ### 1 Proportion Ya > Yb 0.406
   ### 2 Proportion Ya < Yb 0.409
   ### 3    Proportion ties 0.185
   ###  
   ### VDA 
   ### 0.499

