for an army personnel of strength 25.The regression of weight of kidneys(Y) on weight of heart(X) both measured in ounces is: y-0.399x-6.934=0 and regression of weight of heart on weight of kidney is: x-1.212y+2.461=0.
Find the correlation coefficient between x and y and their mean values. can you find out standard deviation of x and y as well?
Answers
Given : y-0.399x-6.934=0
x-1.212y+2.461=0.
To find : correlation coefficient between x and y
Solution:
y-0.399x-6.934=0
=> y = 0.399x + 6.934
=> byx = 0.399
x-1.212y+2.461=0.
=> x = 1.212y - 2.461
=> bxy = 1.212
r = √bxy . byx
=> r = √(1.212)(0.399)
=> r = 0.6954
Correlation coefficient = 0.6954
bxy = r/variance y
=> 1.212 = 0.6954 /variance y
=> variance y = 0.57376
=> SD y = 0.757
byx = r/variance x
=> 0.399 = 0.6954 /variance x
=> variance x = 1.7428
=> SD x = 1.32
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Question ⤵️
for an army personnel of strength 25.The regression of weight of kidneys(Y) on weight of heart(X) both measured in ounces is: y-0.399x-6.934=0 and regression of weight of heart on weight of kidney is: x-1.212y+2.461=0.
Find the correlation coefficient between x and y and their mean values. can you find out standard deviation of x and y as well?
Given⤵️
y-0.399x-6.934=0
x-1.212y+2.461=0.
To find ⤵️
correlation coefficient between x and y
Solution ⤵️
y-0.399x-6.934=0
=> y = 0.399x + 6.934
=> byx = 0.399
x-1.212y+2.461=0.
=> x = 1.212y - 2.461
=> bxy = 1.212
r = √bxy . byx
=> r = √(1.212)(0.399)
=> r = 0.6954
Correlation coefficient = 0.6954
bxy = r/variance y
=> 1.212 = 0.6954 /variance y
=> variance y = 0.57376
=> SD y = 0.757
byx = r/variance x
=> 0.399 = 0.6954 /variance x
=> variance x = 1.7428
=> SD x = 1.32
Hope it helps.