
#1
Feb509, 06:03 AM

P: 297

1. The problem statement, all variables and given/known data
I have used OLS and found that: b0 = 6,85 b1 = 3,88 with se= 0,1121 n = 20 Person x claims that b1 = 5 Choose an alternative hypothesis. Does your estimated relationship support this claim? Use a 5 % significance level 3. The attempt at a solution H0 = b1 = 5 H1 b1!= 5 t((1a)/2,18) = 2,101 t = ([tex]\overline{x}[/tex]  h0)/se(b1) = (3,88  5)/(0.1122) = 9.99 t lies not in the interval + 2,101 so I reject H0 Is this correct? 



#2
Feb709, 05:00 AM

P: 16

se==? and does "b0 = 6,85" means "b0=6.85"? 



#3
Feb709, 05:12 AM

P: 297

Thanks for replying
Se = standard error I'm not sure about what you mean with the last question, but b0 was pointestimated to be 6,85 or 6.85 if the comma was what you asked about. 



#4
Feb709, 09:24 AM

P: 16

Hypothesis testing
From what I know of Hypothesis test and student T distribution (from Statistics course which I am still undertaking), your answer is correct ie "t" lies not in the acceptance region but in the left rejection region thus "H0" is discarded and "Ha" is accepted instead.
A very small error that i have noticed is that you were supposed to look for the range of acceptance region corresponding to "significance level =5% " and "degrees of freedom= n1 =19" where as you seem to have looked it up for degrees of freedom = 20 which is not correct and might result in loss of a few marks in exams even though your answer is still correct. So t((1a)/2,18) = 2.09 ..................(At least in the table i posses) I Hope you are satisfied with my reply. 



#5
Feb709, 09:33 AM

P: 297

Thanks:)



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