How To... Perform a Kruskal-Wallis H Test (By Hand)
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- เผยแพร่เมื่อ 15 ก.ค. 2024
- The Kruskal-Wallis H Test is a nonparametric test similar to an ANOVA test. Use it to compare three or more sets of data that could be categorical data or data that are not normally distributed. This video shows how the calculation is done manually.
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Thank you for this great lecture ! I have a question: if the null hypotheses is rejected, are you going to use SPSS to perform pairwise comparison between the samples when a significant difference is found or doesn't matter if the null hypotheses is rejected or not? thank you
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Love these videos but wonder: if the we had fewer than 6 rows and had to use the Kruskal-wallis table how would go about that?
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If nos. 4 and 5 have the same value and you will find the average of the mean the ranking has a decimal like 5.5, will you still put '5.5' in the table? or will you just round it off ?
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How could you tell if you will have to perform an ANOVA ( Parametric) vs Kurstal Wallis ( Non-parametric)? Please provide some example questions. Thank you.
what if there are ties in our data ,can we still use the same formula as that used there when we want to compare means of more than three groups by nonparametric test??
Can we say that when we have observations in points like 2.3, 5.8 etc we will use KW TEST, and when we have whole no we will use ANOVA?
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The content is superb. Would you please tell me how can i retrieve P value from H value of kruskal-wallis test? Is there any chart like chi square to obtain P value for H value? Please let me know. Thanks
He said it follows a chi square distribution, so that is the chart you use
Thank you, illustrated very simply and effectively. I am a student of biology and have very little understanding of statistics, i have a question, on what basis we select p-value, i mean when to use 0.05 and when 0.01? this may be a silly question for someone but i am a beginner so asking
Hi Hilal,
The significance level is decided by you the researcher, depending on what level of risk of making a Type I error you are prepared to take. 95% (p value of 0.05 or less) is generally accepted in statistical tests as an appropriate level in most cases. However, a 99% significance level (a much harder test) would be more appropriate for research in areas such as medicine, drug trials, or where accuracy is vital.
Dr E.
Is this the suitable test to use if group sizes are different?
Thank you very much. You are an excellent teacher. Is there any video after statistically significan to find with one is significant?
I have 6 treatments with equal sample size. Does this method work for it. My data did not meet ANOVA assumptions. None normal distribution. Is it okey to use this method?
Yes - it is OK to use 6 samples.
Sir I have one question, if in place of sample, mark of any subject is given. In that case also I will take rank one to that score which is least among all the marks.
Thank you! I have only one question, is it acceptable to use this formula, as we deal with 3 ties here?
Or the number of the ties is so small that the difference between this simplified model and corrected one is irrelevant?
With only 1 three way tie involved in an 18 item data set, the correction factor is only 1 - (3^3 -3)/(18^3 - 18), or 0.9969. This means that the factor by which we need to increase the test statistic is 1/0.9969, or 1.0031. Since the uncorrected test statistic value of 1.82 will not exceed the critical value of 5.991 even when multiplied by 1.0031, our conclusion will not change. Most of the time, this will be the case unless the number of ties are quite extensive.
Hi Eugene, I have found your video very useful. However my lecturer states that the H formula used is for when no ties are present. Is this correct or will it not make a significant difference? Thanks
Alex Dolan Most of the time, the correction for ties will not make a statistical difference, unless there are quite a number of them. The correction always increases the value of the test statistic, so correction is not needed if the test statistic is greater than the critical value to start with.
In case you were wondering how to do it, you'd find the number of ties in the data and assign each of them a weight. Generally, an N-way tie has a weight of N^3 - N. So a 3-way tie has a weight of 3^3 - 3, or 24. You'd add all the weights for each case of ties and divide that by N^3 - N, where N here represents the total number of data points across all groups. Subtract this fractional result from 1 to get the correction factor, and divide H by this correction factor to get the result you should use. Since you're dividing H by something that's always less than 1, H is always larger after the correction.
Isn’t the H statistic formula in the video, is for when there are no ties in the rankings?
can the ranks for repeated values contain a decimal? or do I round it?
Ex. (30 + 31 + 32)/3 = 30.3
Even though (30 + 31 + 32)/3 = 31 you should not round mean ranks for ties. E.g. for 30th and 31st tied values the mean rank would be (30 + 31)/2 = 30.5
I have three samples, each one of different size. Is it possible and how to perform a Kruskal-Wallis test in this case ?
Thank you in advance
Simple answer - yes!
Dr E.
Totally thought I understood this but I can't get 1.8 on my calculator no matter how I put it in! Keep getting 1.2 or something totally different!
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Ahm Sir, should we use a correction factor since there is a tie between the values of the data?
Kendrick Leyson You can, certainly, but most of the time it won’t make a difference in the conclusion, unless you have quite an extensive number of ties. Did you want to know the procedure for adjusting for ties?
@@woodchuk1 Thank you Maam/Sir, that would be really great! 🤩
Kendrick Leyson Okay, here it is. First, you’d find out how many of each kind of tie (2-way tie, 3-way tie, etc) there are in the entire data set. Next, you need to figure out the weight for each kind of tie. In general, an n-way tie has a weight of (n^3-n), so a 2 way tie has a weight of 6, and a 3-way tie has a weight of 24, and so on. You add up all the weights for each case of ties and divide this total by (x^3-x), where x is the total number of data points across all groups. Subtract this fraction from 1 to get your correction factor. Divide your calculated H by the correction factor to get the true H value you should use. Note that since the correction factor is always less than 1, the corrected H will always be larger after adjusting for ties.
Thank you for this great lecture ! I have a question: As we know that the Kruskal-Wallis test is a nonparametric test, and is used when the assumptions of one-way ANOVA are not met. In 1-way ANOVA, if there are significant difference between 3,4,... groups, we will need to do the tukey test to determine which mean pairs are differ or not. How about the Kruskal Wallis test ? Do I need to do the Tukey test when there are significant difference between ........groups or not ?
Hi Becks Huy,
I use SPSS to perform a pairwise comparison between the samples when a significant difference is found. See ibm.co/2oKHmQi. It is not a Tukey Test.
Thanks for your kind comments!
Dr E.
Thank you for your kind answer ! I have another question related to the formula of Kruskal Wallis Test. I read in the Fundamentals of Biostatistics and compared to the formula test statistics in your great lecture, I found that the formula test statistics in your lecture was applied when there were no ties observations. How about the formula test statistics when there are ties observations ? Thank you in advance
Becks Huy Here's how to correct the H value for ties in the KW test. First find out what kind of ties are involved in the entire data set (2-way ties, 3-way ties, etc.), as well as how many of each kind there are. Next, calculate the weight for each type of tie. An n-way tie has a weight of (n^3-n), so a 2-way tie has a weight of (2^3-2), or 6. A 3-way tie would have a weight of 3^3-3, or 24, and so on. Add up all the weights for all the tied cases and divide this by N^3-N, where N is the total number of data points across all groups. Subtract this fraction from 1 to get the correction factor, then divide the calculated H by the correction factor to obtain the corrected result. It will always increase after the correction, so if the result is significant before adjusting for ties, strictly speaking the adjustment is not necessary, since it makes no difference in the conclusion.
Hi Pham, I am answering very late, but may be it helps you. There are ties in the example solved by @Eugene O'Loughlin. You see score 9.1 appeared thrice.
Why 12 in the first fraction? How do you know its 12, is it always that number?
that is my question too!
tks, I took sometime to go in the formula today and it is applied when the data contain no ties the denominator of the expression.
12 is apart of the formula
Can i ask for help?
what is the example that you show in the test? Is that a survey margin of error or what?
Hi Jops,
The data are fictitious and made up by me for demonstration purposes. The sample sizes are small, and normality is in doubt - hence the use of the Kruskal-Wallis Test.
Hope this helps,
Dr E.
can you do a run through of the calculation step by step, no matter what i do i can't get 1.8
Ben Grief 43^2/6 = 308.167
61^2/6 = 620.167
67^2/6 = 748.167
308.167+620.167+748.167 = 1676.501
(1676.501*12) = 20118.012
(20118.012)/(18*19) = 58.824596
58.824596 - (3)(19) = 1.825
Sir if you don't mind, I have a little question about the formula of your H. Is it small "n" or big "N"?
I think "n" is different from "N". "N" is the total number of observations or the sample rather while "n" is just a part of it.