#4 answer all parts to question Suppose you fit a least squares line to 24 data

Question

#4 answer all parts to question

Suppose you fit a least squares line to 24 data points and the calculated value of SSE is 8.19. a. Find s2, the estimator of o he variance of the random error term e). b. What is the largest deviation that you might expect between any one of the 24 points and the least squares line? a. Find s the estimator of o he variance of the random error term e s2- (Round to four decimal places as needed.) b. What is the largest deviation that you might expect between any one of the 24 points and the least squares line? (Round to three decimal places as needed.) Enter your answer in each of the answer boxes.

Explanation / Answer

No. of observations (n) = 24

Sum of squares of error (SSE) = ei2 =8.19

a) The estimator of 2 is given by

s2 = ei2/(n-2) = SSE/(n-2) = 8.19/(24-2) = 8.19/22 = 0.3723 (rounded upto 4 decimal places)

(where n-2 are degrees of freedom. We estimate two parameters in least square line. i.e. intercept & slope. That's why minus 2 is there)

b) In least square fit, we assume that the error term is iid normal.

error means difference between actual point and fitted point.

For normal data, the highest standard deviation is 3, which include 99.7% data.

We can use sample standard deviation(s) as estimate for .

s = (0.3723) = 0.6101

confidence level        largest deviation

   99%                         2.576*s = 2.576*0.6101 = 1.572

• s2= 0.3723

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