A t - distribution's convergence to a Normal distributionI'm currently messing around with confidence intervals and I can't really understand how a t distribution converges to a normal distribution for large n.For example, suppose we want to construct a 95% confidence interval when we have a sample mean X―=74.8 and sample variance S=1.23 with n=143.I would construct the confidence interval using,(X―−z1−α2Sn,X―+z1−α2Sn)since n&gt;30. If n&lt;30 I would have used a t-distribution.My question is why does the t - distribution approach a normal distribution for relatively large n?

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A t - distribution&#039;s convergence to a Normal distributionI&#039;m currently messing around with confidence intervals and I can&#039;t really understand how a t distribution converges to a normal distribution for large n.For example, suppose we want to construct a 95% confidence interval when we have a sample mean X―=74.8 and sample variance S=1.23 with n=143.I would construct the confidence interval using,(X―−z1−α2Sn,X―+z1−α2Sn)since n&amp;gt;30. If n&amp;lt;30 I would have used a t-distribution.My question is why does the t - distribution approach a normal distribution for relatively large n?

Luis Melton · Accepted Answer

The t distribution arises because you estimate the population standard deviation σ by the sample standard deviation S. For smaller n, there is some significant chance that S is quite a bit smaller than σ; thus for fixed c and n, there is significant probability that X― is within cσ of μ and not within cS of μ. (The reverse is possible too, but less likely.) But for large n, S is essentially guaranteed to be very close to σ, because it is an asymptotically consistent estimator for σ. And of course, if S is close to σ,X― being within cσ and being within cS of μ are nearly equivalent.Note that strictly speaking you should always use the t distribution for confidence intervals from a normally distributed population with unknown standard deviation. It is just negligibly different from the same interval constructed with the normal distribution if n is large enough. How large n needs to be really depends on how small a difference can be treated as negligible.

fodlonwyrgxc · Accepted Answer

Step 1Suppose          X      1        ,    …    ,          X      n        ∼          i      .      i      .      d      .      N        ⁡    (    μ    ,          σ      2        )    .  Let&amp;nbsp;X―=dX1+⋯+Xnn.Let&amp;nbsp;S2=d1n−1((X1−X―)2+⋯+(Xn−X―)2).Step 2Then&amp;nbsp;dX―−μσn∼N(0,1)&amp;nbsp;and&amp;nbsp;dX―−μSn∼tn−1. The second one has S where the first has&amp;nbsp;σ. If n is large then the probability that S is close to&amp;nbsp;σ&amp;nbsp;is large, so these two random variables are nearly the same.

A t - distribution's convergence to a Normal

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