The Kolmogorov-Smirnov Goodness-of-fit Test

✔ StudyForce.com
✔ Biology-Forums.com
✔ Ask questions here: Biology-Forums.com/index.php?...
Follow us:
▶ Facebook: / studyforceps
▶ Instagram: / biologyforums
▶ Twitter: / studyforceps
Q. Fasting blood glucose determinations made on 36 non-obese, apparently healthy, adult males are shown below. We wish to know if we may conclude that these data are not from a normally distributed population with a mean of 80 and a standard deviation of 6.
The Kolmogorov-Smirnov test is used when one wishes to know how well the distribution of sample data conforms to some theoretical distribution.
When using the Kolmogrorov-Smirnov goodness-of-fit test, a comparison is made between some theoretical cumulative distribution function, (F_T (x)), and a sample cumulative distribution function, (F_S (x)). The sample is a random sample from a population with unknown cumulative distribution function F(x).
The difference between the theoretical cumulative distribution function and the sample cumulative distribution function is measured by the statistic D, which is the greatest vertical distance between F_S (x) and F_T (x).
"D equals the supreme (greatest), overall x, of the absolute value of the difference F_S (x) minus F_T (x)"

[@sumitchhabra2419]

This is the best explanation I have come across on KS test.
I don't understand why it doesn't appear on the top of youtube algorithms search for KS test.

[@nanhl]

The example makes KS test super easy to understand! This really saves my life

[@marianabfranca]

Thank you so much! I was looking for that since a while!!

[@uglyducklingkpopdata3742]

Your explanation is very clear and so goooooood. Thank you for making it!!!!

[@Tweeteketje]

Thanks, extremely clear and I understand the theory much better now!

[@vl_3171]

Thanks a lot! Very clearly explained!

[@jadeelsa9081]

Thank you for your explanation, very clear and helpful.

[@EW-mb1ih]

There is a minor error in the Ft(x) value. for z=-2, it should be 0.0228 approx 0.023 and not 0.022

[@LuanaSilva-rs5yd]

Thank you!

[@krishnavamsi9326]

So Thankful bro... You made me 2 understand in less than 5 mins

[@jeroenvermunt3372]

In a blog post I saw that you had to compare a certain statistic with the KS-value, not the p-value

[@mellisahgotora9589]

Thank u very much...been struggling to find Ft(x)

[@edgarl.calvadoresii9479]

How is ks test different from chi square goodness of fit?

[@dc33333]

do you have to add a bonferroni correction to the calculated pvalue?

[@NEILEVALAROZA]

kindly check the numbers you are using taken from the Z table. some numbers taken from 0.03 and 0.07....

[@udiibgui2136]

What if we are not given a mean and sigma and still tasked with testing for normality?

[@dmitrii5735]

The video is indeed cool! But, it does not apply to discrete distributions, doest it? In the example shown, we have exactly discrete distribution (the way it is measured).

[@fabioramilli8863]

Dear Sir, thank you for your nice video, but there are some issues that I'm missing. it's not clear why you take the D statistic and consider it as a p-value. Why do you double the p-value for a two-sided test, given that your chose the two-sided critical value of your table? Usually when a statistic exceeds a critical value, then there is a statistically significant difference. Isn't this the case?
Perhaps I misunderstood, but is it possible that the right conclusion for this otherwise excellent video would be that the D statistic is equal to 0.16, and it doesn't exceed the 0.221 critical value (two-sided, alpha=0.05) thus the distribution doesn't differ statistically from the normal distribution (with mu=80 and sigma=6)?

[@johnrecker4352]

same here, i still missing the point about the double D, can someone elaborating?

[@krrsh]

Why is the D value is multiplied by 2 and considered as p-value?

[@sesppsfd3815]

Since it’s 2 tailed test

[@ineshi]

The tables you have been using bliz

[@hosseinpiri5144]

While I like the explanation and the details you provided, I believe that last conclusion is incorrect. You are concluding that since 2D>Critical_level, then our distribution is normal, which is incorrect. Your conclusion implies that for concluding normality, it is better if D is large (i.e., two distributions have a larger difference)!

[@fatyaaaa]

So what is the right conclusion?

[@awge6666]

@@fatyaaaajust compare the D value with the critical value. Reject null hypothesis if D

[@awge6666]

@@fatyaaaabecause D value is essentially the distance (difference) between our obaerved data distribution and theoretical distribution

[@notwavybaby]

This is wrong. You don't multiply the D statistic by 2. Not sure why you did this but it definitely gives the wrong answer.

[@StudyForceOnline]

The solution has been verified by a statistician working at our school

[@anmolmohanty7537]

K-S test is only for one tail test
Because it is a two tailed test (≠)
Not one tailed (

[@Tweeteketje]

@@anmolmohanty7537 But the table already shows at 6:26 that the column for a 97.5% one-tailed test is the same as for a 95% two-tailed test. So I also don't understand why it is doubled.