Question 3 osevations for 1) yield the following regression r SER = 15.1 }, = 32

Question

Question 3 osevations for 1) yield the following regression r SER = 15.1 }, = 32. I + 66.8.1. R2=0.81. his entered twice researcher is intcerested in the same regression, but he makes an error when he enters the data gram He enters each observation twice, so he has 200 observations (with observati , observ ation 2 entered twice, and so forth). standard error of the regression)? Please show your steps mark if you HTite down the answers only without showing the steps. 200 observations, what results will be produced by his regression program (SER stands f in obtaining the answers. I will not give yo ,SER= -END--

Explanation / Answer

Here the regresssion line will not change. It is because the data values are just twice written. That will not change the regression line.

Here the new regression line is

Let it be y^ = a + bx

where a = [(y) (x2 ) - (x) (xy)]/ [ n (x2 ) - (x)2 ]

here y (200) = 2 * y(100) ; x (200) = 2 * x(100)

(x2 (200) = 4 *  y(100)  

xy (200) = 4 * xy (100)

so putting the value

a (200) = a(100) = 32.1

b =  [ n(xy) - (x)((y)]/ [ n (x2 ) - (x)2 ]

similarly for b also

b (200) = b (100) = 66.8

Y^ = 32.1 + 66.8 X

The value of R2 would also be the same

R2 = [n(xy) - (x)((y)] / sqrt [ (n (x2 ) - (x)2) ( n (y2 ) - (y)2)]

R2( n= 200) = [200 * 4 * xy100 - 4 * (x)100((y)100] / sqrt [(200 * 4 * (x2 )100 - 4 * (x)2100 ) (200 * 4 (y2 )100 - 4 * (y)2100)

R2 = R2(n = 100)

R2  = 0.81

Here standard error of the regression se (100 observations) = 15.1

so MSE = se2 = 15.1 2 = 228.01

here dF = 100 - 2= 98

so, SSR = dF * MSE = 98 * 228.01 = 22344.98

so when there are 100 observation SSR = 22344.98

SO when there are 200 observations, SSR would be

SSR = 22344.98 * (200/100) = 44689.96

dF = 200 - 2 = 198

MSE = 44689.96/198 = 225.70687

so SER (200 observations) = sqrt(MSE) = sqrt (225.70687) = 15.0235

SER (200 observations) = 15.0235

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