1) what are the two dimensions that we use to interpret a correlation? How are t

Question

1) what are the two dimensions that we use to interpret a correlation? How are these represented by the correlation coefficient?
2) based on the correlation coefficient, interpret the relationship between the number of slurpees consumed and the number of burpees completed.
3) using an alpha level of .05, interpret the "significance" of your outcome. Was it significant? Accept or reject the null hypothesis?
4) describe the relationship between the coefficient of determination and the coefficient of alienation. What do these coefficients tell us?

3. Using the numbers provided below, calculate the correlation coefficient #of Slurpees(X)-1. # Burpees (Y) X Y Patrick Alan Carson Asher Martin Daniel Sum () 2 9 12 6 15 2 2

Explanation / Answer

Question 1.

calculation procedure for correlation

sum of (x) = x = 12

sum of (y) = y = 51

sum of (x^2)= x^2 = 28

sum of (y^2)= y^2 = 539

sum of (x*y)= x*y = 84

to caluclate value of r( x,y) = covariance ( x,y ) / sd (x) * sd (y)

covariance ( x,y ) = [ x*y - N *(x/N) * (y/N) ]/n-1

= 84 - [ 6 * (12/6) * (51/6) ]/6- 1

= -3

and now to calculate r( x,y) = -3/ (SQRT(1/6*84-(1/6*12)^2) ) * ( SQRT(1/6*84-(1/6*51)^2)

=-3 / (0.816*4.193)

= -0.876

value of correlation is =-0.876

Question 2.

properties of correlation

1. If r = 1 Corrlation is called Perfect Positive Corrlelation

2. If r = -1 Correlation is called Perfect Negative Correlation

3. If r = 0 Correlation is called Zero Correlation

& with above we conclude that correlation ( r ) is = -0.8768< 0, perfect nagative correlation

Question 3.

Given that,

value of r =-0.876

number (n)=6

null, Ho: =0

alternate, H1: !=0

level of significance, = 0.05

from standard normal table, two tailed t /2 =2.776

since our test is two-tailed

reject Ho, if to < -2.776 OR if to > 2.776

we use test statistic (t) = r / sqrt(1-r^2/(n-2))

to=-0.876/(sqrt( ( 1--0.876^2 )/(6-2) )

to =-3.63

|to | =3.63

critical value

the value of |t | at los 0.05% is 2.776

we got |to| =3.63 & | t | =2.776

make decision

hence value of | to | > | t | and here we reject Ho

ANSWERS

---------------

null, Ho: =0

alternate, H1: !=0

test statistic: -3.63

critical value: -2.776 , 2.776

decision: reject Ho

Q4.

coeffcient of determination = ( correlation coeffcient)^2

coeffcient of determination = r^2 = 0.768

( X) ( Y) X^2 Y^2 X*Y 3 7 9 49 21 2 9 4 81 18 1 12 1 144 12 2 6 4 36 12 1 15 1 225 15 3 2 9 4 6

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