Thank you so much. You have brightened up my day. I thought all my work shall go into the trash, the nonparametric method is brilliant and saved my work.
Hi Kent, there is an even easier way to do the non-parametric Levene's test. Once you have calculated the ranks, you could also just perform an ANOVA on them, and have it report Levene's test (Options --> Homogeneity of variance test). It yields the exact same result, but it's quicker and probably less error-prone. That said, your method is very good if you want to understand what Levene's test is doing.
Thank you so much for how to report results. I was watching a different video which said you should check the Box's test results. Do you need to report both or is one enough? Do they both check the same thing? Thanks
Can you explain why you used Annova - I have 2 groups with the same subject and assumed if i have non parametric data i would use the dependent t test ?
Hello, Kent! Thank you very much for such a wonderful video, I was wondering whether you could kindly explain me regarding transformation, please? if I have not normally distributed data and not equality of variances (Levene's test), could i also transform the data (e.g., sqrt or log) first to get normality instead of trying non-parametric Levene's test? If so, which one is more applicable? many thanks in advance.
From my understanding it was published in a series of books called "Contributions to Probability and Statistics: Essays in Honor of Harold Hotelling" but it couldn't find anywhere on the web.
I'm afraid the "non-parametric levene test" method is not satisfying. Let's imagine we have three groups A, B and C with the following observations: in A: 0,2,1,3; in B: 5,8,15,12; and in C: 108, 255, 52, 462. We therefore have huge differences between the means of the groups, as well as huge differences in terms of variances. It shouldn't be too hard to detect such difference. Now let's apply the suggested method to these data: 1) we rank the data. We therefore get a new variable that equals: in A: 1,3,2,4; in B: 5,6,8,7; and in C: 10, 11, 9, 12. 2) we calculate the mean in each group (2.5 for A, 6.5 for B, 10.5 for C) 3) we take the absolute value of the differences between the ranks and the mean ranks. We now have: in A: 1.5, 0.5, 0.5, 1.5; in B: 1.5, 0.5, 1.5, 0.5; in C: 0.5, 0.5, 1.5, 1.5 that is exactly the same values in each groups. 4) we compare the means using an ANOVA: all the values are the same, so the means as well and we therefore get a p-value of 1 that is non significant. We conclude that our hypothesis of equality of variance between the groups is not rejected, even if it is very obvious that the variances are different. To make the method "acceptable", before doing anything, the mean of each group should be subtracted from each observation. Then the test would indeed "try" to test if the variances are equals. However even then, the ANOVA procedure in the last step assumes that the variables in each group are normally distributed, which is not the case (as they are now based on the ranks). If you have looots of observation in each group, it will be ok, but in small samples, you cannot trust the p-value given by SPSS. If you want to use the ANOVA in your last step, you should make a quantile normalization (after substracting the means) instead of simply taking the ranks, that is you render the variable exactly normal. Then you can be more confident in the pvalue. To do that, after you ranked the data, you go to "compute new variable", and create a new variable equals to: IDF.NORMAL(RScore2-0.5/N,0,1) where RScore2 is the variable of score ranked after having subtracted the means, and N is the total number of observations. IDF.NORMAL calculates the expected normal value for a particular quantile value (simply the rank brought between 0 and 1). In my example I chose 0 and 1 as mean and standard deviation of my new variable. You could change it. It would not affect the final significance of the test.
I have two factors (3x11) and a dependent variable, can I do a two-way ANOVA to calculate the univariate levene's test? If so, how would I report it --assuming I would have the test of between-subjects effects of the two factors and the interaction. In my case, they all turn out to be significant at the .000 level.
thanks , but my sample is non-normally distributed ,so i will use Kruskal-wallis test, but my result of test the Equality of variances after following your steps is 0.02 which is below p-value 0.05 then reject the null hypothesis , so i cant use Wallis test now ! , what test should i do then ? im working on my Master thesis. thanks
Thank you so much! It's clear and easy to follow. I was wondering if you could do a video on testing for "Homogeneity of Within-Group Regression Slopes" (for ANCOVA with multiple covariates).
Dear, Can I use the non parametric Levene's test with a factorial Anova (2 independent and 1 dependent variable , sample sizes not equal and not normally distributed). Hou should I proceed? do I have to create new variable for the 3 variables and the do the factorial anova again ??? thank you for your answer because I'm really stuck here :) Gr Caroline
Hello there Kent. I have a situation. I have two variables that are numeric and scale. I want to compare these two to see if they are having the same pattern or not in increasing or decreasing during the time (during 10 years that the years are the rows of my dataset). Can you help me with that? Which analyze can I use in SPSS?
If you have two continious variables and you want to examine the correlation, you can use Pearson's correlation. The Pearson correlation does not make a distinction between an independent and dependent variable. If you want to make this distinction, you can do a linear regression.
Hello sir. I appreciate your explanations and tips in citing references. Just one question, should we apply Welch ANOVA/Kruskal-Wallis on the derived values or on the original data after doing the non-parametric Levene's test?
Hello, Kent! Thank you very much for such a wonderful video, I was wondering whether you could kindly explain that whether one should proceed for kruskal Wallis test if significance value in nonparmetric ANOVA is found to be less than 0.05 i.e. there is no equality of variance?
For an interesting text on this topic see for example norusis (dot) com (slash) pdf (slash) DA_v13.pdf, page 277. Also, check/participate in the discussions on the forum talkstats (dot) com.
Hi Kent. Thank you for the video. Would the non-parametric method work for within subjects design? I have same participants over at 4 different conditions performing the same task i.e pointing task with 4 different conditions (normal, blindfolded, distraction, blindfolded+distraction). The dependent variable would be accuracy. Thanks!
Hello. Afik, there are always non-parametric methods available for almost all parametric ones, so, yes, I would assume that you could use Chi-square or Kruskal-Wallis for example. Good luck.
Hello Sir, This is sumathi , research scholar form India. I need one clarification. Kindly help me. If the Shapiro Wilk test is not significant (normality not established) When I perform Levenes test the data is not significance . (Homogentiy established) Should I go with t test value of unequal variance or Welch test
I have two set of data for USA and non USA beer and I am confused which one should be used either parametric or non parametric because one data is normally distributed and other one is not normally distributted?
Thanks. If I did not have normally distributed data, I would probably use a non-parametric method, such as a Chi-square test, but that is perhaps just me. On a side note: Whenever I am in doubt, I discuss the dilemma on the forum talkstats (dot) com. It is the best forum, imho.
+Matthew Hunter Actually, referencing the Welch or Brown-Forsythe values are better when assuming inequality of veriance. But does anyone know a method in an ANCOVA?
Hi Kent Thanks so much for the excellent video - however as a complete newbie to SPSS i just need some assistance. So I've carried out a levene's test using the technique you showed however my obtained P value is 0.012 (thus it's greater than 0.05). And I need to do a two way ANOVA so how would I carry this out? To compensate for unequal variances, under the post hoc button in SPSS would I select either 'TAMHANES T2, DUNNETT'S T3 GAMES - HOWELL or DUNNETT'S C test? Thanks so much for your assistance.
what do you mean by normally distributed data. in your example, Is that mean the female and male are normally distributed separately or regard them together as a whole data, and the whole data distribute normally?
My data are divided into five groups. I have checked the individual one by Shapiro-Wilk test for normality (p>0.05) via SPSS. Unfortunately when I am checking the homogeneity of variances, it is showing significant (p
First of all, thank you so much for your useful tutorial... I need to compare five groups in terms of their mean values...my data are not normally distributed, so I performed non-parametric Levene's test but the results showed homogeneity of variances (p=0.645/F=0.627). Now how can I interpret the results? Can I conclude that the groups are now comparable? Thanks again
Hello! Thank you for the helpful video. I use a 1-5 scale with a big sample, so when I rank case, the manual option of SPSS give me a very wide range of rank. Is it legitimate to set rank as "Tie" and "Sequential rank as unique value"? I tried this option and the rank appeared to be between 1-5, which I assumed to be appropriate for Likert. Please let me know your opinions. Thank you very much
Hi. Just an observation. You have subtracted the mean value by the corresponding rank while you are supposed to subtract the ranks by mean values. If I may ask, is it technically correct?
Why can't you just run the Levene's test in a similar way for non-parametric data as you did it for parametric data? Considering the fact that SPSS is as alien to the parametric vs non-parametric information as we are, I personally think that that should really not be an issue!
+zz123z1 Thanks for your relevant and interesting question. I'm far from a full-fledged statistician. Both the traditional Levene's Test for Equality of Variances as well as the Non-parametric Levene's Test can be performed in many situations. But the latter is said to be better when you have non-normal (skewed) data (even when un-equal sample sizes).
Just one word.......EXCELLENT! One of the best videos ever I have seen! Thanks indeed!
Thank you so much. You have brightened up my day. I thought all my work shall go into the trash, the nonparametric method is brilliant and saved my work.
Glad to hear that the video was appreciated. Good luck.
second it! I almost cried when I realised my non-normally distributed data fails to meet HOV. thank you for this fantastic tutorial! @@kentlofgren
Thank you. Awesome stuff. Clear, detailed, and to the point. More please.
Hi Kent, there is an even easier way to do the non-parametric Levene's test. Once you have calculated the ranks, you could also just perform an ANOVA on them, and have it report Levene's test (Options --> Homogeneity of variance test). It yields the exact same result, but it's quicker and probably less error-prone.
That said, your method is very good if you want to understand what Levene's test is doing.
Thank you so much for your comment! So helpful
Glad it helped!
But the results it's different with this video even the significance all more than 0,05
Grateful for your help! It was the most useful video I watched on this topic.
Thanks for your kind feedback. If I did not have equality of variances, I would look into the possibility of using Chi-square test for association.
This is very easily done. Very Nice to watch this video.
Welcome dear. I would keep on trying to do more videos for those who want to learn.
You've got a gift my brother.
Wonderful tutorials indeed. Great video!
This was so straightforward and helpful. Thank you so, so much.
I am really enjoying your tutorials. Can you also provide the sample data for your tutorials?
Finally I find video showing exactly what I want to test. Can you also show how to test for randomness in SPSS. Thanks
I love hearing you.. It makes me feel so relaxed. :) Thanks for your videos. U saved my life withought knowing :)
thank you, ur explaination is so clear
Thank you so much for how to report results. I was watching a different video which said you should check the Box's test results. Do you need to report both or is one enough? Do they both check the same thing?
Thanks
Your video helps very much , the explanation is really good , Thank you sir ))
Glad to hear that
Kent Lofgren's stuff is super helpful. I'm relying heavily on it as I work through dissertation analyses.
Thank you sir, it was very much needed
Thank you +Samin Siddiquee for the first comment 2016. Good luck with all your projects!
+Kent Löfgren My Pleasure Sir, Good Luck to You as Well !
Can you explain why you used Annova - I have 2 groups with the same subject and assumed if i have non parametric data i would use the dependent t test ?
Hello, Kent! Thank you very much for such a wonderful video, I was wondering whether you could kindly explain me regarding transformation, please? if I have not normally distributed data and not equality of variances (Levene's test), could i also transform the data (e.g., sqrt or log) first to get normality instead of trying non-parametric Levene's test? If so, which one is more applicable? many thanks in advance.
From my understanding it was published in a series of books called "Contributions to Probability and Statistics: Essays in Honor of Harold Hotelling" but it couldn't find anywhere on the web.
I have 8 groups, but one of the group is not normally distributed, what should I use, Parametric or nonparametric Leven?
Best explanation!! Thanks a lot sir
Thanks for sharing very nice and self explanatory tutorial...can you share data!!
I'm afraid the "non-parametric levene test" method is not satisfying.
Let's imagine we have three groups A, B and C with the following observations:
in A: 0,2,1,3; in B: 5,8,15,12; and in C: 108, 255, 52, 462.
We therefore have huge differences between the means of the groups, as well as huge differences in terms of variances.
It shouldn't be too hard to detect such difference.
Now let's apply the suggested method to these data:
1) we rank the data. We therefore get a new variable that equals:
in A: 1,3,2,4; in B: 5,6,8,7; and in C: 10, 11, 9, 12.
2) we calculate the mean in each group (2.5 for A, 6.5 for B, 10.5 for C)
3) we take the absolute value of the differences between the ranks and the mean ranks. We now have:
in A: 1.5, 0.5, 0.5, 1.5; in B: 1.5, 0.5, 1.5, 0.5; in C: 0.5, 0.5, 1.5, 1.5
that is exactly the same values in each groups.
4) we compare the means using an ANOVA:
all the values are the same, so the means as well and we therefore get a p-value of 1 that is non significant. We conclude that our hypothesis of equality of variance between the groups is not rejected, even if it is very obvious that the variances are different.
To make the method "acceptable", before doing anything, the mean of each group should be subtracted from each observation. Then the test would indeed "try" to test if the variances are equals. However even then, the ANOVA procedure in the last step assumes that the variables in each group are normally distributed, which is not the case (as they are now based on the ranks). If you have looots of observation in each group, it will be ok, but in small samples, you cannot trust the p-value given by SPSS.
If you want to use the ANOVA in your last step, you should make a quantile normalization (after substracting the means) instead of simply taking the ranks, that is you render the variable exactly normal. Then you can be more confident in the pvalue.
To do that, after you ranked the data, you go to "compute new variable", and create a new variable equals to: IDF.NORMAL(RScore2-0.5/N,0,1)
where RScore2 is the variable of score ranked after having subtracted the means, and N is the total number of observations.
IDF.NORMAL calculates the expected normal value for a particular quantile value (simply the rank brought between 0 and 1).
In my example I chose 0 and 1 as mean and standard deviation of my new variable. You could change it. It would not affect the final significance of the test.
I have two factors (3x11) and a dependent variable, can I do a two-way ANOVA to calculate the univariate levene's test? If so, how would I report it --assuming I would have the test of between-subjects effects of the two factors and the interaction. In my case, they all turn out to be significant at the .000 level.
interesting and brief tutor
How to enter data of test scores on SPSS. If we have 50 multiple choice items, and 54 students, how to enter their obtained marks on SPSS?
thanks for this video really helping me in my final year project.. =)
Very useful, Kent, thanks.
Thanks a lot sir, It was too much necessary for me.
This is amazing - thank you!
thanks , but my sample is non-normally distributed ,so i will use Kruskal-wallis test, but my result of test the Equality of variances after following your steps is 0.02 which is below p-value 0.05 then reject the null hypothesis , so i cant use Wallis test now ! , what test should i do then ? im working on my Master thesis. thanks
Thank you so much! It's clear and easy to follow. I was wondering if you could do a video on testing for "Homogeneity of Within-Group Regression Slopes" (for ANCOVA with multiple covariates).
Dear,
Can I use the non parametric Levene's test with a factorial Anova (2 independent and 1 dependent variable , sample sizes not equal and not normally distributed).
Hou should I proceed? do I have to create new variable for the 3 variables and the do the factorial anova again ???
thank you for your answer because I'm really stuck here :)
Gr
Caroline
I would write something like: "A Kruskal-Wallis test (p
ty, you safe my life
Hello there Kent. I have a situation. I have two variables that are numeric and scale. I want to compare these two to see if they are having the same pattern or not in increasing or decreasing during the time (during 10 years that the years are the rows of my dataset). Can you help me with that? Which analyze can I use in SPSS?
If you have two continious variables and you want to examine the correlation, you can use Pearson's correlation. The Pearson correlation does not make a distinction between an independent and dependent variable. If you want to make this distinction, you can do a linear regression.
Thanks, it was truly a helpful video :)
Sir, do we need to check normality of a univariate objective collected on a likert scale? If yes, is it also through skewness and Kurtosis value.
thank you sir very much, It really has helped a lot
thank you this video helped me so much
Great.
Hello sir. I appreciate your explanations and tips in citing references. Just one question, should we apply Welch ANOVA/Kruskal-Wallis on the derived values or on the original data after doing the non-parametric Levene's test?
Hello, Kent! Thank you very much for such a wonderful video, I was wondering whether you could kindly explain that whether one should proceed for kruskal Wallis test if significance value in nonparmetric ANOVA is found to be less than 0.05 i.e. there is no equality of variance?
For an interesting text on this topic see for example norusis (dot) com (slash) pdf (slash) DA_v13.pdf, page 277. Also, check/participate in the discussions on the forum talkstats (dot) com.
Hi Kent. Thank you for the video. Would the non-parametric method work for within subjects design? I have same participants over at 4 different conditions performing the same task i.e pointing task with 4 different conditions (normal, blindfolded, distraction, blindfolded+distraction). The dependent variable would be accuracy. Thanks!
Hello. Afik, there are always non-parametric methods available for almost all parametric ones, so, yes, I would assume that you could use Chi-square or Kruskal-Wallis for example. Good luck.
-How can I transform negative to positive values when I want to perform a parametric Leven's test?
this video helped me figure it all out!
Thank you! your video is helpful :)
You're welcome!
Thanks for your reply. I have done Kruskal-Wallis test (p
Thank you very much
Hello Sir, This is sumathi , research scholar form India. I need one clarification. Kindly help me.
If the Shapiro Wilk test is not significant (normality not established)
When I perform Levenes test the data is not significance . (Homogentiy established)
Should I go with t test value of unequal variance or Welch test
I have two set of data for USA and non USA beer and I am confused which one should be used either parametric or non parametric because one data is normally distributed and other one is not normally distributted?
Thanks. If I did not have normally distributed data, I would probably use a non-parametric method, such as a Chi-square test, but that is perhaps just me. On a side note: Whenever I am in doubt, I discuss the dilemma on the forum talkstats (dot) com. It is the best forum, imho.
Thank you sir
very helpful, thank you
Are there ways of compensating for unequality of variance in an ANOVA?
(great video by the way)
+Matthew Hunter Actually, referencing the Welch or Brown-Forsythe values are better when assuming inequality of veriance. But does anyone know a method in an ANCOVA?
Thank you so much.
Hi Kent
Thanks so much for the excellent video - however as a complete newbie to SPSS i just need some assistance.
So I've carried out a levene's test using the technique you showed however my obtained P value is 0.012 (thus it's greater than 0.05). And I need to do a two way ANOVA so how would I carry this out? To compensate for unequal variances, under the post hoc button in SPSS would I select either 'TAMHANES T2, DUNNETT'S T3 GAMES - HOWELL or DUNNETT'S C test?
Thanks so much for your assistance.
0.012 is less than 0.05 ( 0.012 < .05)
Hello, was this question answered?. I have the same issue.
Thank you, Sir! I have got it.
Thanks for this heplful video.
Hello, i was wondering how can i use levene's test to compare means of the control and the experimental groups PLEASE? can u help me?
thanks for the tutorial. it helps me alot
what do you mean by normally distributed data. in your example, Is that mean the female and male are normally distributed separately or regard them together as a whole data, and the whole data distribute normally?
My data are divided into five groups. I have checked the individual one by Shapiro-Wilk test for normality (p>0.05) via SPSS. Unfortunately when I am checking the homogeneity of variances, it is showing significant (p
First of all, thank you so much for your useful tutorial...
I need to compare five groups in terms of their mean values...my data are not normally distributed, so I performed non-parametric Levene's test but the results showed homogeneity of variances (p=0.645/F=0.627). Now how can I interpret the results?
Can I conclude that the groups are now comparable?
Thanks again
Thanks a lot
Hi, can u provide the link to Levene's original papers?
Hello! Thank you for the helpful video.
I use a 1-5 scale with a big sample, so when I rank case, the manual option of SPSS give me a very wide range of rank. Is it legitimate to set rank as "Tie" and "Sequential rank as unique value"? I tried this option and the rank appeared to be between 1-5, which I assumed to be appropriate for Likert. Please let me know your opinions. Thank you very much
I recommend the forum talkstats (dot) com because it is one of the best.
Hi. Just an observation. You have subtracted the mean value by the corresponding rank while you are supposed to subtract the ranks by mean values. If I may ask, is it technically correct?
See this video th-cam.com/video/_w5FM_1WL-M/w-d-xo.html. The person in the video did the same thing.
Mr kent> why did use one way anova not t-test? thanx
Because the Levene's test for normally distributed data is "inside" the ANOVA test in SPSS.
Its in order...
URGENT! What test do I use to compare means when there is no equality of variances???!!
Why can't you just run the Levene's test in a similar way for non-parametric data as you did it for parametric data? Considering the fact that SPSS is as alien to the parametric vs non-parametric information as we are, I personally think that that should really not be an issue!
+zz123z1 Thanks for your relevant and interesting question. I'm far from a full-fledged statistician. Both the traditional Levene's Test for Equality of Variances as well as the Non-parametric Levene's Test can be performed in many situations. But the latter is said to be better when you have non-normal (skewed) data (even when un-equal sample sizes).
Thank you Sir
Most welcome!