- #1

chwala

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- Homework Statement
- see attached

- Relevant Equations
- stats

Unless there is another alternative method, i would appreciate...ms did not indicate working...thought i should share my working though...

Let Waistline= ##X## and Percentage body fat =##Y## and we know that ##n=11##

##\sum X=992, \sum XY=13,772## and ## \sum Y=150##

Then it follows that,

Correlation coefficient

= ##\dfrac{(11×13,772)-992×150}{\sqrt {(11×89,950)-992^2)(11×2,202)-150^2)}}=\dfrac{151,492-148,800}{3045.4379}=\dfrac{2,692}{3045.4379}=0.8839=0.88## (to two decimal places).

switching ##x## and ##y## would that be appropriate? considered wrong with correct working? ...just asking. By letting ##X## be the Percentage body fat, that is...

...next i would want to determine the equation of least-squares...

Cheers!

Let Waistline= ##X## and Percentage body fat =##Y## and we know that ##n=11##

##\sum X=992, \sum XY=13,772## and ## \sum Y=150##

Then it follows that,

Correlation coefficient

= ##\dfrac{(11×13,772)-992×150}{\sqrt {(11×89,950)-992^2)(11×2,202)-150^2)}}=\dfrac{151,492-148,800}{3045.4379}=\dfrac{2,692}{3045.4379}=0.8839=0.88## (to two decimal places).

switching ##x## and ##y## would that be appropriate? considered wrong with correct working? ...just asking. By letting ##X## be the Percentage body fat, that is...

...next i would want to determine the equation of least-squares...

Cheers!

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