Home > Statistical Methods calculators > Pearson's Correlation Coefficient for bivariate grouped data calculator

Solve any problem
(step by step solutions)
Input table (Matrix, Statistics)
Mode :
SolutionHelp
Solution will be displayed step by step (In 2 parts)
Solution
Problem: Correlation Coefficient {{50-55,55-60,60-65,65-70},{90-100,100-110,110-120,120-130},{{4,7,5,2},{6,10,7,4},{6,12,10,7},{3,8,6,3}}} [ Calculator, Method and examples ]

Solution:
Your problem `->` Correlation Coefficient {{50-55,55-60,60-65,65-70},{90-100,100-110,110-120,120-130},{{4,7,5,2},{6,10,7,4},{6,12,10,7},{3,8,6,3}}}


C.I.`(y)`90 - 100100 - 110110 - 120120 - 130
M.V.`(y)`95105115125
C.I.`(x)`M.V.`(x)`
`dy`
`dx`
-1012`f_x``fdx``fdx^2``fdxdy`
50 - 5552.5-1[4]4[0]7[-5]5[-4]218-1818-5
55 - 6057.50[0]6[0]10[0]7[0]427000
60 - 6562.51[-6]6[0]12[10]10[14]735353518
65 - 7067.52[-6]3[0]8[12]6[12]320408018
`f_y`193728161005713331
`fdy`-190283241
`fdy^2`1902864111
`fdxdy`-80172231


Correlation Coefficient r :
`r = (n * sum fdxdy - sum fdx * sum fdy)/(sqrt(n * sum f dx^2 - (sum f dx)^2) * sqrt(n * sum f dy^2 - (sum f dy)^2))`

`=(100 * 31 - 57 * 41 )/(sqrt(100 * 133 - (57)^2) * sqrt(100 * 111 - (41)^2)`

`=(3100 - 2337)/(sqrt(13300 - 3249) * sqrt(11100 - 1681))`

`=763/( sqrt(10051) * sqrt(9419))`

`=763/( 100.2547 * 97.0515)`

`=763/9729.8699`

`=0.0784`



Correlation Coefficient r with Population Cov(x,y) :
Population `Cov(x,y) = (sum fdxdy - (sum fdx * sum fdy)/n)/(n)`

`=(31 - (57 xx 41)/100)/100`

`=(31 - (2337)/100)/100`

`=(31 - 23.37)/100`

`=(7.63)/100`

`=0.0763`


Population Standard deviation `sigma_x = sqrt((sum fdx^2 - (sum fdx)^2/n)/(n))`

`=sqrt((133 - (57)^2/100)/100)`

`=sqrt((133 - 32.49)/100)`

`=sqrt(100.51/100)`

`=sqrt(1.0051)`

`=1.0025`

Population Standard deviation `sigma_y = sqrt((sum fdy^2 - (sum fdy)^2/n)/(n))`

`=sqrt((111 - (41)^2/100)/100)`

`=sqrt((111 - 16.81)/100)`

`=sqrt(94.19/100)`






Solution provided by AtoZmath.com
Any wrong solution, solution improvement, feedback then Submit Here
Want to know about AtoZmath.com and me
  
 

 
Copyright © 2019. All rights reserved. Terms, Privacy





We use cookies to improve your experience on our site and to show you relevant advertising. By browsing this website, you agree to our use of cookies. Learn more