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The direct of admissions of a small college administered a newly designed entrance test to 20 students selected at random from the new freshman class in a study to determine whether a student’s grade point average (GPA) at the end of the freshman year (y) can be predicted from the entrance test score (x). The results are summarized as follows.

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Answer #1

e)
y^ = 1.7 + 0.84 x
y^ = 1.7 + 0.84 * 6
= 6.74

standard error of prediction interval = t * sigma * sqrt( 1 + 1/n + (Xbar - xi)^2/Sxx)
t = t.inv.2t(0.01,18)
= 2.8784
sigma = 0.4967   {from g) part}

hence
= 2.8784* 0.4967*sqrt(1 + 1/20 + (5 -6 )^2/509.12)
= 1.4664

(6.74 - 1.4664 , 6.74 + 1.4664 )
= (5.2736,8.2064)

f)
standard error of slope = sigma/Sqrt(Sxx)
=   0.4967/sqrt(509.12)
= 0.022

g)
SS Residual = Syy - Sxy^2 / Sxx
= 134.84 - 257.66^2/509.12
= 4.4411242929

sigma^2 = (SS residual/(n-2))
= ( 4.4411242929/18)
= 0.2467

sigma
= 0.4967

h)
R^2 = Sxy^2/(Sxx * Syy)
= 257.66^2/(509.12*134.84)
= 0.9671

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Answer 1

e)
y^ = 1.7 + 0.84 x
y^ = 1.7 + 0.84 * 6
= 6.74

standard error of prediction interval = t * sigma * sqrt( 1 + 1/n + (Xbar - xi)^2/Sxx)
t = t.inv.2t(0.01,18)
= 2.8784
sigma = 0.4967   {from g) part}

hence
= 2.8784* 0.4967*sqrt(1 + 1/20 + (5 -6 )^2/509.12)
= 1.4664

(6.74 - 1.4664 , 6.74 + 1.4664 )
= (5.2736,8.2064)

f)
standard error of slope = sigma/Sqrt(Sxx)
=   0.4967/sqrt(509.12)
= 0.022

g)
SS Residual = Syy - Sxy^2 / Sxx
= 134.84 - 257.66^2/509.12
= 4.4411242929

sigma^2 = (SS residual/(n-2))
= ( 4.4411242929/18)
= 0.2467

sigma
= 0.4967

h)
R^2 = Sxy^2/(Sxx * Syy)
= 257.66^2/(509.12*134.84)
= 0.9671

Please don’t forget to give thumbs up if the solution helped.

If you think answer is wrong or if there is any doubt, feel free to comment. You will be helped