The scores of 10 students on their midterm exam (x) and on their final (y) yielded the following data.
Σx = 638 Σx2 = 43,572
Σy = 690 Σy2 = 49,014
Σxy = 44,636
What is your question?
To find the correlation coefficient between the midterm exam scores (x) and the final exam scores (y), we can use the formula:
r = (Σxy - (Σx)(Σy) / √((Σx^2 - (Σx)^2)(Σy^2 - (Σy)^2))
where Σxy is the sum of the product of x and y, Σx is the sum of x, Σy is the sum of y, Σx^2 is the sum of the squared x values, and Σy^2 is the sum of the squared y values.
Given the following values:
Σx = 638
Σx^2 = 43,572
Σy = 690
Σy^2 = 49,014
Σxy = 44,636
We can substitute these values into the formula:
r = (44,636 - (638)(690)) / √((43,572 - (638)^2)(49,014 - (690)^2))
Now, let's calculate the individual values:
Σx = 638
Σx^2 = 43,572
Σy = 690
Σy^2 = 49,014
Σxy = 44,636
Plugging these values into the formula:
r = (44,636 - (638)(690)) / √((43,572 - (638)^2)(49,014 - (690)^2))
= (44,636 - 439,320) / √((43,572 - 407,044)(49,014 - 476,100))
= (-394,684) / √((-363,472)(-427,086))
Note: We have calculated Σx^2 and Σy^2, but we also need to calculate (Σx)^2 and (Σy)^2:
(Σx)^2 = (638)^2 = 407,044
(Σy)^2 = (690)^2 = 476,100
Now, let's continue substituting values:
r = (-394,684) / √((-363,472)(-427,086))
= (-394,684) / √(155,236,938,592)
Finally, we can simplify the expression:
r ≈ -0.962
Therefore, the correlation coefficient between the midterm exam scores and the final exam scores is approximately -0.962.