Run a CHi2 test; CHi2=SUM ((of-ei)^2)/ei; where: of=data i from distribution f; ei=data i from distribution i. The suma distributes CHi2 and evaluate its probability or compare with tabulated CHi2 from a CHi2 distribution at the probability level of 0.05. If the probability of the value of CHi2 is less than 0.05, both distributions have different shapes and viceversa. Most important fit a PDF (probabilistic density function for both data sets; evaluate parameters of the pdf for both data sets and conduct a T(Student) test on each parameter of the distribution for both data sets. For example if you fit the 3-parameter Weibull density function, it has 3 parameters (shape, scale and location). To test of the shape parameter is similar between data sets evalauet the shape parameter (its standard error) for both data sets and conduct the T test with these two shape (standard error) parameter values.

This is a complex but most recommended test.

Hope these comments would help You to reach your objectives.

Kind Regards

Sincerely

Jose Navar, Ph.D

Run a CHi2 test; CHi2=SUM ((of-ei)^2)/ei; where: of=data i from distribution f; ei=data i from distribution i. The suma distributes CHi2 and evaluate its probability or compare with tabulated CHi2 from a CHi2 distribution at the probability level of 0.05. If the probability of the value of CHi2 is less than 0.05, both distributions have different shapes and viceversa. Most important fit a PDF (probabilistic density function for both data sets; evaluate parameters of the pdf for both data sets and conduct a T(Student) test on each parameter of the distribution for both data sets. For example if you fit the 3-parameter Weibull density function, it has 3 parameters (shape, scale and location). To test of the shape parameter is similar between data sets evalauet the shape parameter (its standard error) for both data sets and conduct the T test with these two shape (standard error) parameter values.

This is a complex but most recommended test.

Hope these comments would help You to reach your objectives.

Kind Regards

Sincerely

Jose Navar, Ph.D

Use a two sample Kolmogorov-Smirnov test. It has better properties than a Chi square test. Details for doing this in SPSS are in this link:

https://www.google.com/search?client=firefox-b-1&ei=EnFvW7WbAsvOjwSwn7eIAw&q=2+sample+kolmogorov+smirnov+test+SPSS&oq=2+sample+kolmogorov+smirnov+test+SPSS&gs_l=psy-ab.12...6584.10225.0.20184.5.5.0.0.0.0.91.429.5.5.0....0...1.1.64.psy-ab..0.0.0....0.Mg4pbc5fIwo

Best wishes, David Booth

https://stats.stackexchange.com/questions/135737/will-two-distributions-with-identical-5-number-summaries-always-have-the-same-sh

https://statistics.laerd.com/premium-sample/mwut/mann-whitney-test-in-spss-12.php

http://www.aip.de/groups/soe/local/numres/bookcpdf/c14-2.pdf

https://stackoverflow.com/questions/28917917/how-can-i-check-if-two-plots-have-the-same-shape-in-r

https://stackoverflow.com/questions/44929620/histogram-to-decide-whether-two-distributions-have-the-same-shape-in-r?noredirect=1&lq=1

Testing of

(1) the significance of the difference between the skewness of the two curves

and

(2) the significance of the difference between the kurtosis of the two curves

can solve the problem.

I suggest the use of GAMLSS package. There is a function that adjust distributions - fitDist - use it to adjust distributions for each category and compare the results.

I agree with David Eugene Booth, the Kolmogorov-Smirnov test would be more suitable. I'd also include plots of the histograms/density distributions and empirical cumulative distributions (see attached) so that one can visualize differences.

Those visualizations might prompt you to think about using the Anderson-Darling test, similar to the K-S but is more sensitive to differences between curves near the beginning or end of the distributions.

Another thought is you might be able to do permutations of the difference in skewness and kurtosis measures between the samples—though I haven't done this nor do I have a reference handy.

There are many technical answers to this question but start off just thinking & looking at the data. Ask yourself are there reasons why they should be the same or different? Ask if there reasons why a certain type of distribution is to be expected. Ask your self what you mean by different? Mean, Variance, higher moments or general shape. Next look at the distributions using different bin sizes (bin sizes can radically change both the look of the distribution & statistical significance of many tests.) If you just blindly apply the various technical tests that have been developed you will eventually find one that proves your preferred answer is correct.

I do agree with Jack Lothian.

Caleb A. Aldridge K-S test is for testing if two variables arise form the same distribution (thus considering equality of mean, etc.) and cannot test whether two distribution only have the same shape. Actually, to test your hypothesis I created a variable with two categories. The first 100 data were generated using random number (category=0) and the next 100 were just the first 100 data plus 10 (category=1). So we are sure that these two categories have the same SD and shape, and the mean of the second group=10+mean of the first group. So I did a K-S test on this variable and the null hypothesis (Same distribution across two categories) was rejected (p

I'm actually very hopeful of performing the test proposed by Dr.Jose, which is beyond routine medical statistics and my level of knowledge, so I'm waiting for the holiday between two semesters to end so I can refer the calculations to someone from the epidemiology department

Soroush Dianaty,

Indeed the K-S test can detect differences in location, but it can also detect differences in shape. This is because, the K-S statistic quantifies the greatest distance between the empirical distribution functions of two samples and tests if the observed distance is greater than expected. I also mention a similar test, the Anderson-Darling (A-D), that is more sensitive to differences between curves near the beginning or end of the distributions. So hypothetically you can adjust the location of the distributions so they are centered at the same point and then test for differences in shape.

Below, and attached graphs, are some results comparing a few normal distributions that differ in location or scale and a uniform distribution near the same location. Give attention especially to the comparisons of samples "a" (Normal, mean = 0, sd = 1) and "d" (Uniform, min = -1, max = 1). Small P-values suggest that the samples are not from the same distribution—to really interpret how they differ will require use of summary statistics and graphics like Jack Lothian advised.

Again, conceptually, if you wanted to compare shape statistics (e.g., skewness and kurtosis) I think it's possible to do resampling (e.g., permutation) tests. I'll try to put together some reproducible code for this soon.

Cheers,

Caleb

***

> ks.test(a,b);ks.test(a,c);ks.test(a,d)

Two-sample Kolmogorov-Smirnov test

data: a and b

D = 0.76, p-value < 2.2e-16

alternative hypothesis: two-sided

Two-sample Kolmogorov-Smirnov test

data: a and c

D = 0.3, p-value = 0.0002468

alternative hypothesis: two-sided

Two-sample Kolmogorov-Smirnov test

data: a and d

D = 0.27, p-value = 0.001365

alternative hypothesis: two-sided

> kSamples::ad.test(a,b);kSamples::ad.test(c,d);kSamples::ad.test(a,d)

Anderson-Darling k-sample test.

data: a and b

Number of samples: 2

Sample sizes: 100, 100

Number of ties: 0

Mean of Anderson-Darling Criterion: 1

Standard deviation of Anderson-Darling Criterion: 0.75419

T.AD = ( Anderson-Darling Criterion - mean)/sigma

Null Hypothesis: All samples come from a common population.

AD T.AD asympt. P-value

version 1: 55.77 72.62 3.855e-31

version 2: 55.90 72.76 4.184e-31

Anderson-Darling k-sample test.

data: a and c

Number of samples: 2

Sample sizes: 100, 100

Number of ties: 0

Mean of Anderson-Darling Criterion: 1

Standard deviation of Anderson-Darling Criterion: 0.75419

T.AD = ( Anderson-Darling Criterion - mean)/sigma

Null Hypothesis: All samples come from a common population.

AD T.AD asympt. P-value

version 1: 12.33 15.02 4.718e-07

version 2: 12.40 15.12 4.374e-07

Anderson-Darling k-sample test.

data: a and d

Number of samples: 2

Sample sizes: 100, 100

Number of ties: 0

Mean of Anderson-Darling Criterion: 1

Standard deviation of Anderson-Darling Criterion: 0.75419

T.AD = ( Anderson-Darling Criterion - mean)/sigma

Null Hypothesis: All samples come from a common population.

AD T.AD asympt. P-value

version 1: 7.571 8.713 0.0001674

version 2: 7.630 8.785 0.0001582

***

Reproducible R code:

set.seed(1234)

ecdf1