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What is the p value of the Kolmogorov Smirnov test?
From the output we can see that the test statistic is 0.97725 and the corresponding p-value is 2.2e-16. Since the p-value is less than .05, we reject the null hypothesis. We have sufficient evidence to say that the sample data does not come from a normal distribution.
Are there exact p values for the KS function?
Exact p-values are not available for the one-sided two-sample case, or in the case of ties if y is continuous.
Do you have to give the arguments of pnorm to ks.test?
As pointed out in the ks.test help, you have to give to the ks.test function the arguments of pnorm. If you do not precise mean and standard variation, the test is done on a standard gaussian distribution. Thanks for contributing an answer to Stack Overflow!
How to interpret p-value of K-S test?
The p-value returned by the k-s test has the same interpretation as other p-values. You reject the null hypothesis that the two samples were drawn from the same distribution if the p-value is less than your significance level. You can find tables online for the conversion of the D statistic into a p-value if you are interested in the procedure.
What is the d statistic of the K-S test?
The D statistic is the absolute max distance (supremum) between the CDFs of the two samples. The closer this number is to 0 the more likely it is that the two samples were drawn from the same distribution. Check out the Wikipedia page for the k-s test.
Why does Student’s t-test give a high p-value?
Because the sample mean and standard deviation are highly similar the Student’s T-Test gives a very high p-value. KS Test can detect the variance. In this case the red distribution has a slightly binomial distribution which KS detect. In other words: Student’s T-Test says that there is 79.3% chances the two samples come from the same distribution.
Is the p value of ks.test true?
I’m confused about the appropriate interpretation of p-values returned by the two-sample Kolmogorov-Smirnov test (ks.test) in R. In slide 23 of this presentation about non-parametric two-sample tests, the author states that when analyzing the ks.test results: needs to be multiplied by 2 for a 2-tail test. Thus, P = 0.05194 Is that true?
How to calculate the KINV of the Kolmogorov distribution?
where c(α) = the inverse of the Kolmogorov distribution at α, which can be calculated in Excel as Dm,n,α = KINV (α)*SQRT ((m+n)/ (m*n)) where KINV is defined in Kolmogorov Distribution. The values of c(α) are also the numerators of the last entries in the Kolmogorov-Smirnov Table.
When to use a low p-value test?
Low p-values can occur with low sample sizes (as your example provides) or the presence of interesting but small differences, e.g. superimposed oscillatory disturbances. If you are working with situations with large numbers of ties it suggests you may need to use a test that more closely fits your data situation.