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# Standard Error

#### mkparis

##### Consultant
Howdy,
in computing the standard error, in the lecture we have
se(b0) = sqrt(var(b0)).
I am wondering if it is not
se(b0) = sqrt(var(b0))/sqrt(n)

Because in the CLT we have :

Let X1,X2....Xn be n random variables with a sample mean Xmean, a mean mu and a variance of sigma^2
we have.

sqrt(n) ( Xmean - mu) ==> N(0,sigma^2)

mu = Xmean +/- Z(alpha/2) x sigma/sqrt(n)

hence the standard error is sqrt of the variance and divided by the sqrt of n.

Can you please let me know where I am wrong

#### David Harper CFA FRM

##### David Harper CFA FRM
Staff member
Subscriber
Hi mkparis,

I assume b0 refers to estimators of parameters in linear regression (don't i have them as b1, b2 per Gujarati?). If so, your logic is indeed good, but the sqrt(n) is already in the denominator so,

se(bm) = sqrt(var(bm)), where bm = a parameter estimator

because, to your point, the var(bm) already includes a (1/n) which becomes SQRT(1/n) in the standard error, so, i hadn't thought of it this way, but I *think* you are correct that is indeed CLT manifesting but the formula is still the first way

David