Help! Thanks!
@amistre64
this seems to ahve worked
I guess. soooo... could you help out?
using the scatter plot tool .. where is your tool?
I dont have one, and neither does the website, sooo i've just plotted it on an actual graph. :P
correlation coeff is a bugger to calculate by hand and my ti83 gave up the ghost a while back
you got excel?
No. I only have word and powerpoint. :/
ive got mine loading, openoffice is a msoffice clone
Oh okay. Was it easy to download?
Is it still loading @amistre64
it was simple to download yes but im used to excel commands, i cant find the corresponding commands in this one tho. it has a scatter plot but i havent messed with it enough to know how to work it properly :)
we can work it by hand ... line of best fit, but then ill ahve to recall the pearson stuff
Oh. Ok. If you can, i would like to work it by hand!
but i just found the help file :) correl(a;b)
got it we would need to define the slope of each best fit line: x with respect to y, and y with respect to x
Understood.
\[b=\frac{n\sum xy-\sum x\sum y}{n\sum xx-\sum x\sum x}\] \[b'=\frac{n\sum xy-\sum x\sum y}{n\sum yy-\sum y\sum y}\] \[r^2=bb'\]
Ok, then?
well, r = sqrt(r^2)
Hmm. ok.
So, do I have to plug this formula for every problem?
if you insist on doing it by hand, sure. otherwise use ... the .. scatter ... plot ... tool n=8 xx x y yy xy 460865 1909 1051 139529 253318
r = sqrt((8(253318)-1909(1051))^2/((8(460865)-1909^2)(8(139529)-1051^2)))
hmmm. so the correlation is ......?
correlation coefficient*
ive given you the process, and even filled in your values .. its up to you to simplify it
ok.
1.00?
Is 1. correct? Thats what i seemed to get.
@amistre64
Oh. Okay. so 0.9 and the second answer is July. Thanks!
yep
Join our real-time social learning platform and learn together with your friends!