Question

A computer while calculating correlation on coefficient between two variable x and y from 25 pairs of observation obtain the following result.n=250,€x=125,€y= 100,€xy=508,€x^2=750,€y^2=460.it was however discovered at the time period of checking that two pair of observation were not correctly copied.they were taken as(5,14) and (8,6) while the correct values were (8,12) and (6,8).prove that the correct value of correlation coefficient should be 2/3.

Answer 1

SOLUTION

EVALUATION

Here it is given that the number of observations = 25

Now it was however discovered at the time period of checking that two pair of observation were not correctly copied.they were taken as(5,14) and (8,6) while the correct values were (8,12) and (6,8)

So

$\sf{Correct \: \sum x}$

$\sf{ = 125 - 6 - 8 + 8 - 6}$

$\sf{ = 125}$

$\sf{Correct \: \sum {x}^{2} }$

$\sf{ = 650 - {6}^{2} - {8}^{2} + {8}^{2} + {6}^{2} }$

$\sf{ = 650}$

$\sf{Correct \: \sum y}$

$= \sf{100 - 14 - 6 + 12 + 8}$

$\sf{ = 100}$

$\sf{Correct \: \sum {y}^{2} }$

$= \sf{460 - {14}^{2} - {6}^{2} + {12}^{2} + {8}^{2} }$

$= \sf{460 - 196 - 36+ 144+ 64 }$

$= \sf{436 }$

$\sf{Correct \: \sum xy}$

$\sf{ = 508 - (6 \times 14) - (8 \times 6) + (8 \times 12) + (6 \times 8)}$

$\sf{ = 508 - 84 - 48+ 96 + 48}$

$\sf{ = 520}$

Hence

$\sf{Correct \: coefficient \: of \: correlation}$

$\displaystyle \sf{ = \frac{ \sum xy - \frac{1}{n}\sum x \sum y}{ \sqrt{\sum {x}^{2} - \frac{{(\sum {x}^{} )}^{2} }{n} } \times \sqrt{\sum {y}^{2} - \frac{{(\sum {y}^{})}^{2} }{n} } } }$

$\displaystyle \sf{ = \frac{520 - \frac{1}{25} \times 125 \times 100 }{ \sqrt{650 - \frac{ {125}^{2} }{25} } \times \sqrt{436 - \frac{ {100}^{2} }{25} } } }$

$\displaystyle \sf{ = \frac{520 - 500 }{ \sqrt{650 - 625 } \times \sqrt{436 - 400 } } }$

$\displaystyle \sf{ = \frac{20 }{ \sqrt{25 } \times \sqrt{36 } } }$

$\displaystyle \sf{ = \frac{20 }{ 5 \times 6 } }$

$\displaystyle \sf{ = \frac{20 }{ 30 } }$

$\displaystyle \sf{ = \frac{2 }{ 3 } }$

Hence proved

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