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% confidence interval for the slope of the line of best fit?

Answer 1

Answer:

Given,

Slope $\beta$ 1^ = [ Σxy - 1/n(Σx)(Σy) ] / [Σx^2 - 1/n (Σx)^2]

substitute the given values

= [56089 - 1/10(683*813)] / [47405 - 1/10 (683)^2]

= 0.7421

Syy = Σy^2 - (Σy)^2/n

substitute values

= (66731 - (813^2/10))

= 634.1

Now,

SSE = Syy - $\beta$ 1^ Sxy

substitute values

= 634.1 - 0.7421*561.1

= 217.71

Standard error ($\beta$1^) = sqrt(SSE/(n-2))

substitute values

= sqrt(217.71/8)

= 5.2167

Now consider,

Here 99% CI for beta$\beta$1 = $\beta$ 1^ +/- t*SE($\beta$1^)

substitute values

= 0.7421 +/- 3.2498*5.2167

= 0.7421 +/- 16.9532

= (- 16.2111 , 17.6953)