Hypothesis testing


by MaxManus
Tags: hypothesis, testing
MaxManus
MaxManus is offline
#1
Feb5-09, 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((1-a)/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?
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Mastermind_14
Mastermind_14 is offline
#2
Feb7-09, 05:00 AM
P: 16
Quote Quote by MaxManus View Post
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((1-a)/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?

se==? and does "b0 = 6,85" means "b0=6.85"?
MaxManus
MaxManus is offline
#3
Feb7-09, 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 point-estimated to be 6,85 or 6.85 if the comma was what you asked about.

Mastermind_14
Mastermind_14 is offline
#4
Feb7-09, 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= n-1 =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((1-a)/2,18) = 2.09 ..................(At least in the table i posses)

I Hope you are satisfied with my reply.
MaxManus
MaxManus is offline
#5
Feb7-09, 09:33 AM
P: 297
Thanks:)


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