I really like the way you simplify the process into numerous steps. Now, I know what exactly is the logic behind those calculations when I use STATA. Thank you very much :)
Thank you for your fast reply. I realized my error- I was averaging the wrong column for the mean ranking so I had rank values that were lower than what they should have been. I was trying to compare the responses for the three different groups, but I'm thinking the one-way ANOVA is the best test. Thanks!
what does the h test statistic mean exactly? why do we choose this formula? So let me get this straight my conclusion in this case would tell me me of some difference between my samples and the chi values tell me the probablity of having some number difference between my samples? I understand how to calculate the formula but how is it derived i dont want to just memorize it?
Thank you for the video, very clear explanation, but there is one thing that I do not really understand. Why are we using Chi-Square distribution in this question?
Which test can i use in case i need to use a non-parametric two way ANOVA? From what i understand, Kruskal-Wallis test is the non-parametric alternative for one way ANOVA
Assuming that you have two way ANOVA with replication (where each combination of the two factors has more than one data point), and that all combinations have the same number of elements (that is, the ANOVA is balanced) and that this number is at least 5, you can use a test called the Schierer Ray Hare test as a nonparametric alternative to two way ANOVA. Reply back to me if you want the details.
Hello guys, where can i find the Table with alpha and df online ? Does anyone have a link for me ?:) I need the source from the table for my excercise :) Thanks
hi dear how are you doing? Could Friedman's test be applied to data from a one-way layout in which there are the same ,n, of observations from each of k treatments? Explain. Should Friedman's test be applied to such data? Explain.
P.S. The reason why I was used this test initially was due to the Likert Scale. I read that using a one-way ANOVA isn't a good idea for scores ranging from 1-5, 1=strongly disagree, 5=strongly agree. Thanks again!
Hello. Excellent video! Question for you: In the formula for 'h' there was a numeric value of 3. Is this because there are 3 groups? In other words, if I have two groups (say, males and females) rather than 3, would I use 2 in the formula? Thanks! Mike
I've deleted my previous replies since I screwed up...I was using the F statistic for the KW test when I shouldn't have. My error, sorry. The KW test uses the chi-square distribution, which has only one degree of freedom, equal to the number of groups minus 1. So since we have three groups here, we use 2 degrees of freedom in the chi-square distribution, not the F distribution. At an alpha of 0.05, our critical value is 5.991, and since our chi-square value of 2.854 falls short of the critical value, our test statistic cuts off an area greater than 5% on the right side of the curve. So our p value is greater than 0.05 and our value is not significant. The 18 here just refers to the total number of observations, but it's strictly not needed, since we don't have two separate degrees of freedom like we do in the F distribution. All we need for chi-square is the number of groups minus 1. Sorry for the mixup!
I am analyzing survey scores for three questions using a Likert Scale (1=Strongly disagree to 5=Strongly agree) for three independent groups. Unfortunately, here is the breakdown for the first question: Score # Respones 1 0 2 0 3 1 4 21 5 64 As a result, I have only 3 ranks, low in value, yielding a negative H value. Am I doing the right test?
@statslectures OK nevermind, I missed something you said. At 3:10 you say that you replace the data in your original table with the ranks. That's what I was missing, hence my misinformed comment.
is it possible that the group 1 has 6,6,6,6,6,6,6,8,6,66,8,6 and group 2 8,8,8,8,8,8,6,8,8,8 and group 3 10,10,8,10,10,10,8,10....how can i rank these? oh my g....pls help
First, let's assume that the 66 was meant to be two 6's. Then, I get that we have 12 6's, 13 8's, and 6 10's. If the 6's were all distinct, they would occupy ranks 1-12 in numerical order. Since they are identical, we assign each 6 the average rank of 1-12 (1+2+3+...+12, all divided by 12), which is 6.5. Likewise, the 8's would occupy ranks 13-25 if distinct; averaging as before, I get 19 to be used as the rank for all the 8's. Lastly, the 10's would occupy ranks 26-31, so they would all receive an average rank of 28.5. Run the KW test using these ranks, and divide the result by 0.864919 as a correction factor for the extensive ties. That's your answer. If you want more info, message me back.
+woodchuk1 The correction factor should be 0.861895 instead...sorry for the error! It makes no difference in this case, as the correction factor always increases the value of the test statistic, which was already far over the critical value before the correction in this case. So your conclusion would have been the same either way.
well in my opinion if you have the same score perhaps you can do this: You know in the group one 6 is the smallest score (since the biggest is 8), then it should be put before the 8 score. The problem is you have no idea whats the rank for each 6 score since there are eleven of 6 score, then to know the exact rank all you have to do first is divide the first and second rank so you can get the rank for each 6 score. Example: 6,6,8,7----> supposed to have 4 ranks, but oops we have the same score. But 6 is the smallest, so it must be put in the first and second rank before 7 and 8. But whats the exact rank for them? (1+2)/2: 1,5. So the rank for each of them is 1.5 Since you have more than two of 6 score then you have to do this several times.
I have 24 as my N, and (48sq/8+32sq/8+27sq/8) (the last part is -75). After doing this equation I come up with -64.8575 as my H answer. It does not seem right that it should be negative. Any ideas?
The sum of your ranks must be wrong...if you have 24 scores, the left side of the KW statistic (12/(24*25)) simplifies to 1/50. You would subtract 75 from the end, as you have done. If each group has 8 values, your denominators are correct. The only other part is the numerators...check your ranks again to make sure the sums are correct. They're too low if you're getting a negative result.
You ranked the data but then never used the results of that. Since you are doing a chi-squared test, are you supposed to subtract the actual data from the ordinal rank and square that quantity?
