The question is: Are the intercept and slope coefficient significantly different from zero at the 5%
level of significance?
The answer is: The t-statistic to test the statistical significance of the intercept and slope coefficient is the parameter estimate divided by its standard error. We reject the null hypothesis and conclude the coefficients are statistically significant if the absolute value of the t-statistic is greater than the two-tail 5% critical t-value with 43 degrees of freedom, which is 2.02.
What I need to clarify is:
Why is this a t-test?
Why is the t-statistic calculated as parameter estimate divided by its standard error?
Why is it a two-tail test?
Why is it 43 degrees of freedom?
The t-distribution table only has 40 and 60 degrees of freedom; it does not have 43. Why do we take the critical value at 40 degree of freedom, instead of 60 degree of freedom?
With all due respect you need to go back and study sampling and hypotheis testing from L1 other wise you are going to find thsi topic extremely tough,
All the points below were covered at level 1
quote=“glsy, post:1, topic:153006”]
Why is this a t-test?
We are testing if parameter is statistically different from zero. For samples we use a t-test using the t-distribution. Similar to normla distribution but adapted for small samples.
Why is the t-statistic calculated as parameter estimate divided by its standard error?
(observation - hypothesis)/s.error. Hypothesis = 0
Why is it a two-tail test?
It can be higher or lower than zero
Why is it 43 degrees of freedom?
sample size - k - 1, Here is have one slope variable k = 1
The t-distribution table only has 40 and 60 degrees of freedom; it does not have 43. Why do we take the critical value at 40 degree of freedom, instead of 60 degree of freedom?
40 is closer
[/quote]
Thank you very very much. Yes, I did review my handwritten notes before, but I guess I was not thorough enough. I spent quite a few hours going through all the L1 content again, and you are right that most are covered at L1.
Just to clarify on 43 degrees of freedom. L1 content states that dof for a t-test is n-1. So why do we need to subtract k in this instance?
Thanks very much once again for your guidance!
It has been many years since I passed L1 and statistics has always been my weak topic. Please bear with me
pulling values from an online table
40 2.0211
43 2.0167
60 2.0003
The critical value decreases as the number of degrees of freedom increases.
If the t-statistic is greater than the critical value with 40 degrees of freedom (2.0211) it will also be greater than the critical value with 43 degrees of freedom (2.0167) because that is less than the critical value with 40 degrees of freedom
The critical value with 60 degrees of freedom (2.0003) is less than the critical value with 43 degrees of freedom (2.0167).
We want
t-statistic > critical value with 43 degrees of freedom (2.0167)
If we use the critical value with 40 degrees of freedom (2.0211), then we have
t-statistic > critical value with 40 degrees of freedom (2.0211) > critical value with 43 degrees of freedom (2.0167)
which gives us what we want
When we tested dor the mean we were predict “1” item the mean from the sample.
So dof = n - 1
When doing regressiion we have “1 interecept” and “k” slope variables
so dof = n - k - 1
I.e
y = b0 + b1X1 +E
k = 1 dof = n - k - 1 = n -1 -1 = n- 2
]
y= bo + b1X1 + b2X2 + b3X3 + e
k = 3 dof = n - k - 1 = n - 3 - 1 = n-4
broadly with dof I think.
How man things I am predict? say 4 (intercept and co-efficeints for 3 independant variables )
dof = n - 4
