Question

We have a dataset with n= 10pairs of observations(xi,yi), and n∑i=1xi= 683,n∑i=1yi= 813,n∑i=1x2i= 47,405,n∑i=1xiyi= 56,089,n∑i=1y2i= 66,731.What is an approximate 99% prediction interval for the responsey0atx0= 60?

Answer 1

Ʃx =	683
Ʃy =	813
Ʃxy =	56089
Ʃx² =	47405
Ʃy² =	66731
Sample size, n =	10
x̅ = Ʃx/n = 683/10 =	68.3
y̅ = Ʃy/n = 813/10 =	81.3
SSxx = Ʃx² - (Ʃx)²/n = 47405 - (683)²/10 =	756.1
SSyy = Ʃy² - (Ʃy)²/n = 66731 - (813)²/10 =	634.1
SSxy = Ʃxy - (Ʃx)(Ʃy)/n = 56089 - (683)(813)/10 =	561.1

Sum of Square error, SSE = SSyy -SSxy²/SSxx = 634.1 - (561.1)²/756.1 = 217.70903

Standard error, se = √(SSE/(n-2)) = √(217.70903/(10-2)) = 5.2167

Slope, b = SSxy/SSxx = 561.1/756.1 = 0.7420976

y-intercept, a = y̅ -b* x̅ = 81.3 - (0.7421)*68.3 = 30.614734

Regression equation :

ŷ = 30.6147 + (0.7421) x

Predicted value of y at x = 60

ŷ = 30.6147 + (0.7421) * 60 = 75.1406

Critical value, t_c = T.INV.2T(0.01, 8) = 3.3554

99% Prediction interval :

Lower limit = ŷ - tc*se*√(1 + (1/n) + ((x-x̅)²/(SSxx)))

= 75.1406 - 3.3554*5.2167*√(1 + (1/10) + ((60 - 68.3)²/(756.1))) = 56.037

Upper limit = ŷ + tc*se*√(1 + (1/n) + ((x-x̅)²/(SSxx)))

= 75.1406 + 3.3554*5.2167*√(1 + (1/10) + ((60 - 68.3)²/(756.1))) = 94.244