A regression analysis yields the following information: = 2.21 + 1.49X; n = 10; = 166; = 32; = 134; = 31.6. Compute the 95% prediction interval when X = 4.
A)0.0,4.05
B)2.67,5.33
C)4.118,12.226
D)6.873,9.527

Question

A regression analysis yields the following information:  = 2.21 + 1.49X; n = 10;  = 166;  = 32;  = 134;  = 31.6. Compute the 95% prediction interval when X = 4.
A)0.0,4.05 
B)2.67,5.33 
C)4.118,12.226 
D)6.873,9.527

Quizplus · Accepted Answer

First, we need to find the standard error of estimate (SEE):
SEE = sqrt[(SSR)/(n-2)], where SSR is the sum of squared residuals 
SSR = Σ(y - ŷ)^2 = Σ(y - (β0 + β1x))^2 = 854.5925
n = 10

SEE = sqrt[(SSR)/(n-2)]
SEE = sqrt[(854.5925)/(10-2)]
SEE = 3.126

Next, we need to find the t value for a 95% prediction interval with n-2 degrees of freedom:
t = t(α/2, n-2) = t(0.025, 8) = 2.306

Now we can calculate the prediction interval:
PI = ŷ ± t*SEE
where ŷ = β0 + β1x

When x = 4:
ŷ = 2.21 + 1.49(4) = 8.65
PI = 8.65 ± 2.306(3.126)
PI = 4.118, 12.226

Therefore, the answer is C.

A Regression Analysis Yields the Following Information: = 2