Median Test [using chi-square] | Non-Parametric tests | Statistics for All
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- Опубліковано 3 кві 2021
- In this video, we will learn how to Median Test by using chi-square method [Solved Examples] in Non-Parametric tests.
If you like our work, do give it a thumbs up and let us know in the comments below.
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Hello mam
Book me jo formula diya hai or aap jo formula bataye usame different hai kyuki chi-square=
N[|ad-bc|-N/2]²/(a+b)(b+c)(a+c)(b+d)
Ye Formula hai book me
Calculated value = 0.0475 Ho is accepted
You explained very well.. Thank you
Yes correct
Thank you so much
Please check out the whole playlist :
ua-cam.com/play/PLsmamMNLjMzVlML3a49ZAMjxlMTTZYgpT.html
You explain v well ,kruskal Wallis h text b explain kren kia video mn
Mam mne apki sbhi video dekhi h sample wale all topic related like non parametric test and all
And thank u again and again mere sabhi topic phli baar itne ache se clr hue h
Or kl mera exam h
Thank u
Very helpful Thank you
Nice 💯
Your voice is so good ❤ I really liked it
thank you so much
Well explained
hello ma'am
it was amazing, your video cleared all my doubts. ma'am, please make a series on inventory control and queuing theory.
Thank you 💕 please make video Kruskal Wallis
Lovely voice ma'm😍,
And very good explaination✨
Thanks
Again d value is less than 5 and final values of a,b,c,d would also change while substituting their values in final formula right ?
full support to you, for your lovely voice and also more important your best explanation 😎😎😇
PLEASE REPLY
Thanks a lot:)
Nice madam
Keep it up
Thank you. ❤ please make videos about assignment problems
Here it is :
Part 1 :ua-cam.com/video/TRGqgOjybW8/v-deo.html
Part 2:ua-cam.com/video/Aoyg0kFqUNs/v-deo.html
In table u compared abcd with 5. In all the tables we have to compare 5 with abcd or either a small value of u or v?
when you changed the values for a and d , d is still less than a then why did we go ahead?
Kya ap mjy is test ka cirtical value wala table show krwa saktii hain???
mery pss naii hai aur mjy samjh naii aa tha k last sy is test ko solve kaisy krn??
Can you show me this table????
Hlo ma'am meadian test mai sample duble ho to
samajh aa gaya
❤
I have a doubt. Median vlaue is 15.5. then I added 15th and 16th value and divided with 2. Again got an odd no. That is 31.5.
What next to do??
Don't worry median value can be a decimal value, proceed with that value
Hello Mam,
M= 90 in Second sum question....so we have to take Value less than 90...but there is also 90value in 2nd sample....so we have to consider 90 or not?????
90 lesser value u take ....not 90...
In TB they have used different formula and if any value is less then 5 in tabel then they didn't find revised values of it !! They just use one formula for all sums , and answer is still correct ! Can you tell me which formula we have to use
You can use either of them
Both are correct
Good video......but degree of freedom ki value 5% pe to 1.96 hoti hai na???
Ma'am formula are different
For even =N/2
For odd =(N+1)/2
That's why median will be 78
My answer is 0.011 that less than 0.05 so my my hypothesis is null this right or wrong
સરસ
Calculated value is 0.338 ,Ho :M1=M2
Critical value kya hai
ANS = X² C = 0.06
X² T = 3.85
H0 is Accepted .
X² = 0.034. X²c < X²t Ho is accepted
Calculated value of X^2 is 0.0475.
H0 is accepted
Mine is 0.5 is right or wrong tell me as soon as possible dear
hum X²= N[(AD-BC)-(N/2)]²/(A+B)(C+D)(A+C)(B+D)
YETE CORRECTION (.5) ki jarurat nai padhegi ma'am ...
Calculated value is =0.0475
Therefore ho is accepted....
Is it correct mam?
Yes correct
Apne revised a b c d ku value find ki he usme bhi to D ki value less than 5 he.... Is it acceptable ma'am ?
When any value is less than 5, then only we find revised values
Okay... Thanks alot..your videos are really very helpful
Thanks a lot for your support. Please share with your friends also
@@statisticsforall3433 ma'am yhi line smj nhi aye..ki 5 hmne "a" ke According apne app se liya..ya phir ye rule hae ki har questions me 5 he lengy ???
In all questions it is same
mam Vnsgu exam me sirf 1% and 5% level hi pucha jata he na.
Or koi dusra level pucha to nahi jata he na like 99%.
Or d.f. hamesha 1 hi aayega
Please reply
Yes mostly these two level are asked
@@statisticsforall3433 thanks for helping
maam U AND V ki value less then 5 aayi hai toh revised value nikal pdegi ?
maam plz reply
Yes if any value is less than 5 then apply Yates correction method and find revised values
@@statisticsforall3433 then maam mera X^2 ki value m change aarha hai X^2 = 0.0792
Or mene kisi me bhi 0.5 add or minues nahi kiya fir bhi answer aa rahe hai
Calculated value is 0.49
Df=1
Table value of 0.5%=3.84
calculated value is less than table value so , H° is accept?
M i correct or not???
No it's not correct
Calculated value is 0.0475
H0 is accepted
@@statisticsforall3433
A*D= 4.5*3.5 = 15.75
B*C= 5.5*5.5= 30.25
Then
:19(-14.5)²/10*9*10*9
:19*210.25/8100
:3994.75/8100
:0.4931790.....
Where I m wrong plz tell me?🙏🙏
@@Sharmaji-gr8rm original values are A=4, B=6, C=5, D=4. Now solve you will get the right answer.
While counting u and v, count only the values which are less than median and not those which are equal to median.
I think you have included 90 while counting v, remove that and you will get the answer
Isme value jo apne dekhte h new table bnane ke liye, usme 5 se kum hi kyu liya?
In chi square method (2×2 table), if any value is less than 5 , then we have to apply Yate's correction method, that's why..
@@statisticsforall3433 so the value '5' is fixed na
@@bollyhungama7821 yes
I have a confusion tell me null always state that there is no significant difference. U have stated it in positive .way of stating Null ho /wording for Hnot is not correct i guess
Here H0 is already given in the question that's why I have used that.
But if whenever you don't get H0 in the question you can write The difference is not significant.
@@statisticsforall3433 i am pointing on Ho statement not on what is in question simply focus on how to write null hypothesis ? (Framing sentence for null) Then u will get what i am trying to say.
Okay.
H0 is always about the given two variables/attributes are equal. We always assume that there is no difference between them