Suppose you fit a least squares line to 28 data points and the calculated value

Question

Suppose you fit a least squares line to 28 data points and the calculated value of SSE is

8.91.

a. Find s2, the estimator of sigma squared2 (the variance of the random error term ).

b. What is the largest deviation that you might expect between any one of the

28 points and the least squares line?

a. Find s2, the estimator of 2 (the variance of the random error term ).

s2=

(Round to four decimal places as needed.)

b. What is the largest deviation that you might expect between any one of the

28 points and the least squares line?

(Round to three decimal places as needed.)

Explanation / Answer

ans=

Best Answer:  
s² = Sum of Squared Errors
.........................................................
Degrees of Freedom

so

s² =SSE/n-2
= 8.91/26 =0.3426..
Degrees of Freedom = 26 because
the twenty six data points are used to
find the estimates of TWO PARAMETERS when
calculating the least squares line

y = + ßx

the estimators are a and b .

Ironically ß assumes the role of m the slope of the line
and assumes the role of algebra class b{not our estimator b}
in algebra class's linear model
y=mx+b the slope intercept form for a line's equation.

Most importantly DF=26 because 28 data points are used to
estimate two parameters.

The largest deviation you would expect would be
based on the assumption that the errors are
normal and identically distributed . 99.7% of
observations are expected to be within three
standard deviations of the mean so we use

3s as the representative for 3

and that value is 3*0.3426 = 1.756

I just recalled it is conventional, however, to use the
ninety five percent confidence interval so one would
replace 3 with 1.96, or the easier to use number of 2

{2 gives a confidence of 95.44%}

Using 1.96 to make it exactly 95%, your largest expected
deviation is 1.96 x 0.3426 = 1.147

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