Thread: N(mui,sigma^2)

The use of t, when x(bar) is estimated is based on the underlying
assumption of a normally distributed variable when sigma is known; if x
is N(mui,sigma^2) then
x - mui/sigma is N(0; 1). It follows that if s^2 is an estimate for
sigma^2, then
x - mui/sigma
is t distributed with some degree of freedom v. What
can you say about s as v tends to infnity? What is the distribution
of
x - mui/sigma
as v tends to infnity?

the t converges in distribution to a st normal as v goes to infinity
since s converges to sigma

Originally Posted by matheagle

the t converges in distribution to a st normal as v goes to infinity
since s converges to sigma

thax alot