Throw's Pearson's product moment correlation coefficient (r) to establish a measure of linear association between the two variables, where r² is the proportion of the total variance (s²) of Y that can be explained by the linear regression of Y on x, and 1-r² is the proportion that is not explained by the regression. Thus 1-r² = s²xY / s²Y.
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Bet you googled that shizzle.
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Oh my God the apostrophe.
The shame
The shame
Infamy, infamy, they've all got it in for me.
Kills self with Spearman's Rank Rule
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Are you telling us that 1-Rs = linear aggression in your Y-front?
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I had to do a whole year of my degree in the maths faculty to pass a mandatory statistics ancillary course early on and it has left scars but that is the standard definition of Pearson's R
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Student's T test always caused a sreportmei’mracist too (two sugars please m8)
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Next, mutters will solve the Riemann hypothetis. 3, 2, - go!
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It's a hypothesis so cannot be solved but it can be proved.
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I just read the Rieman hypothesis. Twice.
I still don't know what it is.
The Riemann zeta function ζ(s) is a function whose argument s may be any complex number other than 1, and whose values are also complex.
Can you have a complex number other than 1 which doesn't have complex values?
I mean seriously. However much stick lawyers get for jargon, statmaticians should get more. ow much more, I am not sure. There was a formula, but it was written in greek and gibberish. Theta complex argon ringpiece to the power of boron times more.
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I read Du Sautoy's book on it Fool. One of those ones that's absolutely fascinating but then you get to the end and are like "eh?"
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OP seems quite straightforward
What's your point?
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There is nothing odder than the English superstition that you can be innumerate and yet still claim to be well-educated.
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I was not making that claim.
Just to be clear, this was a prolif combined with a moment of arrogant self-regard.
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Germans eh? Just one dig after ANOVA.
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not being able to solve/prove the Riemann hypothesis doesnt make one innumerate. no1 has ever done it.
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Did some fellow in the university of Bristol not do it a couple of years ago?
(Googles)
Yes. some famous maths fellow said in 2017 he had done it, but that he will show everyone his workings 'in a while'.
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Germans? What was I thinking? That's someone else completely.
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