Hallo, Mr. McGukian. This video is a very good learning resource. By the way your good name is very interesting too. It seems like the perfect way to phrase an invitation to dinner to someone, because "Dane McGukian" sounds like "Dine, my cooking!"
Thanks for the amazing video. I was a little confused at the end though. I thought the test statistic was 104 not 106. Could anyone explain why in the end we checked if 106 fit between TL and TU instead of 104?
He explains that usually you use the set of data with fewer ranks/observations for the test statistic. In this case they both have 10 so it doesn't matter. The test is to see whether or not two samples have different medians. In this case the two rank sums are very similar, but if it had been for example 85 and 125, then the 85 would correspond with the 79 critical value and the 125 with the 130. (Still not rejecting H0) One is the lower limit(TL) and one is the upper limit(TU).
Hello sir plz what this question The urinary fluoride concentration (parts per million) was measured for a sample of livestock grazing in an area previously exposed to fluoride pollution and the measurements are provided: 25.8 36.6 26.3 21.8 27.2 Does the data indicate strongly that the true average fluoride concentration for livestock grazing in the polluted region exceeds 25? Test at 5% level of significance using i. the t-test. ii. the Wilcoxon signed-rank test. Specify the assumptions required for each test.
When i use normal method of U1n U2 I got same result that is Ho is accepted .can we use the normal method for the independent one or the methods differ 🤔Iam confused
Hello, thanks for the clear video. I just had a question; you used the sum of the ranks as the test statistic, in my lectures I have to use U as the test statistic, whereby U1 = n1*n2 + n1*(n1+1)/2 - sum of rank1 and U2 = n1*n2 + n2*(n2+1)/2 - sum of rank2. I'm just a little confused by this, since you did it differently.
Hey, lets say T1 and T2 were 104 and 78 and had the same sample size. Then you use the table to find the critical values and find that if you use T1 you do not reject Ho and if you use T2 as test stat you do reject Ho. What happens then?
It’s not possible for T1 and T2 to be 104 and 78. If your sample sizes are both 10. If you have 10 + 10 = 20 values, the ranks will total to 20(21)/2 = 210. If one of the rank totals is 104, the other must be 106.
@@dmcguckian understood. Thank you. If you have unequal sample sizes, you use the larger rank sum as the W stat, correct? I am just learning this stuff. Thanks for the video
You have to handle ties by giving them preliminary ranks then adjusting those ranks by giving each one the average of the ranks they were given. This video explains how to handle ties: www.statsprofessor.com/video.php?chapterId=7&id=424
@@dmcguckian Ah ok thanks, now I understand why you averaged them in the video! I initially thought this is pointless because all ties were in the same sample in your example
Excusme sir please the normal test for Wilcoxon ranked test the n that we gonna use it is it the original n that given in the problem or the n= number of positive signs + number of negative signs? That we gonna substitute it in the formula which is Z= (w-n(n+1)÷4)/sqrt(n(n+1)(2n+1)÷24??
I am not certain that I am answering your question, but if you threw out a data value in a nonparametric procedure, you normally would change n to reflect the new sample size.
One of the best , most simplified videos for Wilcox. Thank you!
And thank you for sharing your knowledge with the world 🌍
thank you for sharing you knowledge. it did help me. I was a bit confused on the rangnumbers because my teacher explained it a bit vague.
You’re welcome
Hallo, Mr. McGukian. This video is a very good learning resource. By the way your good name is very interesting too. It seems like the perfect way to phrase an invitation to dinner to someone, because "Dane McGukian" sounds like "Dine, my cooking!"
:)
Thank you for your "like", Mr. Dane. I too "like"...oh, no...I "love" your teaching 😇
You saved my life, thank you!!
Mine too
Thank u very much kind Sir , u rescued me !
Glad it was helpful!
Thanks...that was really helpful
Great!
Great video! Thanks :)
You’re welcome
very well explained.. thank you sir
Thank you so much !!!
You’re welcome!
Thank you
I often mishear "Wilcoxon Rank Sum Test" as "Wisconsin, raise some glass!" 🥂🍻
OMG You are an angel!!!! Thank you!!1!
You’re welcome!
how do you calculate p value when you are given T(+),T(-) and n?
thank you sir.
Thanks for the amazing video. I was a little confused at the end though. I thought the test statistic was 104 not 106. Could anyone explain why in the end we checked if 106 fit between TL and TU instead of 104?
He explains that usually you use the set of data with fewer ranks/observations for the test statistic. In this case they both have 10 so it doesn't matter.
The test is to see whether or not two samples have different medians. In this case the two rank sums are very similar, but if it had been for example 85 and 125, then the 85 would correspond with the 79 critical value and the 125 with the 130. (Still not rejecting H0) One is the lower limit(TL) and one is the upper limit(TU).
You're welcome!
thanks
You’re welcome
Hello sir plz what this question The urinary fluoride concentration (parts per million) was measured for a sample
of livestock grazing in an area previously exposed to fluoride pollution and the
measurements are provided:
25.8 36.6 26.3 21.8 27.2
Does the data indicate strongly that the true average fluoride concentration for
livestock grazing in the polluted region exceeds 25? Test at 5% level of
significance using
i. the t-test.
ii. the Wilcoxon signed-rank test.
Specify the assumptions required for each test.
When i use normal method of U1n U2 I got same result that is Ho is accepted .can we use the normal method for the independent one or the methods differ 🤔Iam confused
Hello, thanks for the clear video.
I just had a question; you used the sum of the ranks as the test statistic, in my lectures I have to use U as the test statistic, whereby U1 = n1*n2 + n1*(n1+1)/2 - sum of rank1 and U2 = n1*n2 + n2*(n2+1)/2 - sum of rank2.
I'm just a little confused by this, since you did it differently.
u1 = sum of r1 - n1*(n1+n2+1)/2 etc is specially Mann-Whitney. maybe there's a slight diff between wilcoxon sum and mann whitney
Hey, lets say T1 and T2 were 104 and 78 and had the same sample size. Then you use the table to find the critical values and find that if you use T1 you do not reject Ho and if you use T2 as test stat you do reject Ho. What happens then?
could you use an average of T1 and T2?
It’s not possible for T1 and T2 to be 104 and 78. If your sample sizes are both 10. If you have 10 + 10 = 20 values, the ranks will total to 20(21)/2 = 210. If one of the rank totals is 104, the other must be 106.
@@dmcguckian understood. Thank you. If you have unequal sample sizes, you use the larger rank sum as the W stat, correct? I am just learning this stuff. Thanks for the video
Nagu watch the concept video at my website. It’s under section 14.5. www.statsprofessor.com/chapters.php?id=7
What about df is it 18
What happens if the same number appears in the first and the second sample? Can I number them alternately?
You have to handle ties by giving them preliminary ranks then adjusting those ranks by giving each one the average of the ranks they were given. This video explains how to handle ties: www.statsprofessor.com/video.php?chapterId=7&id=424
@@dmcguckian Ah ok thanks, now I understand why you averaged them in the video! I initially thought this is pointless because all ties were in the same sample in your example
Excusme sir please the normal test for Wilcoxon ranked test the n that we gonna use it is it the original n that given in the problem or the n= number of positive signs + number of negative signs? That we gonna substitute it in the formula which is Z= (w-n(n+1)÷4)/sqrt(n(n+1)(2n+1)÷24??
I am not certain that I am answering your question, but if you threw out a data value in a nonparametric procedure, you normally would change n to reflect the new sample size.
@@dmcguckian I mean that when I get the p-value should I change n?
Can we get the table you used please 12:01
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Good luck with your studies!