chi‐square testempirical distribution functionKolmogorov‐SmirnovpowerThe normal (Gaussian) distribution is of central importance in statistical theory, and numerous techniques for judging departures from normality have been developed. The article describes moment tests, normal probability plots, the chi-...
DistributionsUnderstandingtheNormalDistributionPropertiesoftheNormalDistributionBox-CoxTransformation TypesofDataandDistributions DiscreteData(Attribute)➢Binomial➢Poisson DiscreteDistributions ContinuousData(Variable)➢Normal ➢Exponential ➢Weibull ➢Lognormal ContinuousDistributions ➢t ➢Chi-square ➢F Sh...
NormalityTest(double[] x) Constructor for NormalityTest.Method Summary double ChiSquaredTest(int n) Performs the chi-squared goodness-of-fit test. double getChiSquared() Returns the chi-square statistic for the chi-squared goodness-of-fit test. double getDegreesOfFreedom() Returns the deg...
The simulation results were remarkably consistent, with the Anderson-Darling (AD) test almost always rejecting normality and the Ryan-Joiner (RJ) test almost always failing to reject normality. The Kolmogorov-Smirnov (KS) and Chi-square (CS) tests were included in the simulation too. The CS te...
The omnibus chi-square test is updated from the original method to adjust for the fact that the test statistic is not quite distributed as chi-square with 2 degrees of freedom (Royston 1991b).Significant P values for any of the tests indicate non-normality. There is no perfect test for ...
This tool complements the "Distribution fitting" tool, which allows you to determine the value of the parameters of the normal distribution and to test the goodness of fit using a Chi-square or a Kolmogorov Smirnov test. View all tutorials ...
It has the advantage over the chi-square test in that it can be used for small samples and does not require that data frequencies be larger than 5. For additional information and some examplesclick here. Lilliefors Test The Lilliefors Test is a version of the Kolmogorov-Smirnov test that is...
Test Statistic: JB = 62 * (1.05 ^ 2 / 6 + 1.82 ^ 2 / 24) = 19.89 p = 0.0000 (using Chi-Square distribution) Reject the null hypothesis at the 0.05 significance level. The data are not sampled from a normal distribution. SHAPIRO-WILK W TEST ...
If a variable fails a normality test, it is critical to look at the histogram and the normal probability plot to see if an outlier or a small subset of outliers has caused the non-normality. If there are no outliers, you might try a transformation (such as, the log or square root) ...
Many statistical tests and procedures assume thatdatafollows a normal (bell-shaped)distribution. For example, all of the following statistical tests, statistics, or methods assume that data is normally distributed: Hypothesis tests such as t-tests, Chi-Square tests, F tests ...