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Math Help - N(mui,sigma^2)

  1. #1
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    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?
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  2. #2
    MHF Contributor matheagle's Avatar
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    the t converges in distribution to a st normal as v goes to infinity
    since s converges to sigma
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  3. #3
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    Quote Originally Posted by matheagle View Post
    the t converges in distribution to a st normal as v goes to infinity
    since s converges to sigma
    thax alot
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