39.5. LINEAR REGRESSION 831

above expression is

1σ2

n

∑k=1

(Xk−

(α̂ + β̂ (tk− t̄)

))2

=1

σ2

n

∑k=1

(Xk−

(X̄ +

(∑

nj=1 X j (t j− t̄)

∑ j (t j− t̄)2

)(tk− t̄)

))2

Thus, to find a confidence interval for the variance, do the following.

PROCEDURE 39.5.4 In the above situation, to find a .95 confidence intervalfor the variance which is unknown do this:

1. If there are n observations, n fairly large, certainly larger than 2, find an interval[a,b] such that if V is a X 2 (n−2) random variable P(V ∈ [a,b])≥ .95.

2. Find X̄ the sample mean and t̄ the average of the t values. Then fill in to find

S≡n

∑k=1

(Xk−

(X̄ +

(∑

nj=1 X j (t j− t̄)

∑ j (t j− t̄)2

)(tk− t̄)

))2

3. Then the .95 confidence interval for σ2 is determined by

a≤ Sσ2 ≤ b

In other words, with probability .95, the variance σ2 satisfies

Sb≤ σ

2 ≤ Sa

I think one is even more interested in β , the slope of the line for the mean. To finda confidence interval for β , recall the T test. The T distribution was the distribution of

W√V/r

where V was X 2 (r) and W was n(0,1). Do we have such random variables above?

Recall it was shown above that β̂ is n(

β , σ2

∑i(ti−t̄)2

). Therefore,

β̂ −β(σ/

√∑i (ti− t̄)2

) is n(0,1)

Therefore,β̂−β(

σ/√

∑i(ti−t̄)2)√

1σ2 ∑

nk=1

(Xk−

(α̂ + β̂ (tk− t̄)

))2/(n−2)