Jarque-Bera statistics guessing Normal or Uniform?
The Classical J.Neyman-E. Pearson NHST scheme to decide if a Null Hypothesis should be rejected can be used in order to choose what the preferential distribution to fit a data-problem is. In fact, allied to systematic simulation concerning the Null, H0, Normal Population, and Alternative Hypotheses, H1, Uniform Population, and the sample response towards a GoF test concerning the Null and an adequate Population parameters we can easily achieve our goal. Here it is intended, to illustrate the progressive ability to discriminate Uniformity from Normality by the Jarque-Bera (ALM) algorithm in conjunction with the Skewness and Excess Kurtosis indexes when sample sizes are let to vary from 50 to 100 (400?000 samples each time). No specific data-sample is used simulated instead does. So, the test will inform how sizes became untenable towards Normality against Uniformity when we compare the observed triple outputs/responses JB, S, and k, codified by  to , 1 indicating significant result, 0 not.
The proposed Method
Firstly of all the 5% critical values: ____________JBcrit(5%)____________1%______ ____50______6.55__6.55_______16.63__16.62__ ____60______6.41__6.41_______16.23__16.19__ ____70______6.41__6.40_______15.68__15.68__ ____80______6.38__6.39_______15.31__15.32__ ____90______6.34__6.33_______14.96__14.92__ ___100______6.36__6.31_______14.71__14.73__
and ________________5%_____________1%_______ ____________Ucrit__Vcrit___ ___50_______3.92___3.44______7.39___11.02__ ___60_______3.91___3.44______7.30___10.81__ ___70_______3.90___3.45______7.25___10.52__ ___80_______3.91___3.47______7.19___10.44__ ___90_______3.90___3.48______7.12___10.15__ __100_______3.90___3.52______7.03____9.99__
Finally the *ouputs* using JBcrit, Ucrit,Vcrit (5% level) After carefully consideering vantages and drawbacks to obtain more precise critical values I opt to choose the following ones, taking into account that semi-quantitative goal of the present account is not extremely demanding.
Supposing that you had found  or , which is, of course, very likely to occur with H1 true, then you decide to reject H0 because p= 0.601+ 0.341 = 94.2% if. The decision has the risk 0.015+ 0.019 = 3.4% to fail. On contrary if you got  the H1 rejection has only 5.5% probability, and to retain H0, p= 91.6%, is advisable.
__n=100____ __0.915__0.015__0.020__0.000__0.001¬__0.019__0.014__0.016_ __0.025__0.436__0.000__0.000__0.001__0.537__0.001__0.000_ __p = 0.436 + 0.537 = 97.3% if H1=true and get  or . Contrasting with p(or | H0 true) = 3.4%.