SOLUTION: I'd have a, maybe somewhat weird, question; I have to draw a scatter plot (which I did), now it says to; "describe the nature of the correlation"... I have started but didnt come

Question 214182: I'd have a, maybe somewhat weird, question; I have to draw a scatter plot (which I did), now it says to; "describe the nature of the correlation"...
I have started but didnt come far;
r=Mxy-MX*My over σx*σy= 88.004-(28)(3.143) over .... I dont know what to set in for σx*σy and I dont understand how the problem is then solved.
Could someone help me out with this please?
values of x; 0, 2, 3, 4, 5, 6, 8 values of y; 1, 2, 2, 4, 3, 4, 6
σ = sigma

Answer by stanbon(75887) (Show Source): You can put this solution on YOUR website!
I have to draw a scatter plot (which I did), now it says to; "describe the nature of the correlation"...
I have started but didnt come far;
r=Mxy-MX*My over σx*σy= 88.004-(28)(3.143) over .... I dont know what to set in for σx*σy and I dont understand how the problem is then solved.
Could someone help me out with this please?
values of x; 0, 2, 3, 4, 5, 6, 8 values of y; 1, 2, 2, 4, 3, 4, 6
σ = sigma
-------------------------
sigma x is the standard deviation of that set of x's: it is 2.4495..
sigma y is the standard deviation of that set of y's: it is 1.5518..
----------------------
When I run a Linear Regression function against the set of your
x/y pairs I get r = 0.93956... which shows that there is a high
level of positive linear correlation.
-----
But you numbers[ 88.004-28*3.143] give a result of zero
Then dividing by (sigma x)(sigma y) you would still get
r = 0 which indicates there is not linear correlation.
Something is wrong with your "r" calculation.
========================================================
Cheers,
Stan H.
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