For large n, the sampling distribution of S = 1 n − 1 ∑ i = 1 n ( X i − X ) 2 is sometime approximated with a normal distribution having the mean σ and the variance σ 2 2 n . Show that this approximation leads to the following ( 1 − α ) 100 % large-sample confidence interval for σ: S 1 + Z α / 2 2 n &lt; σ &lt; S 1 − Z α / 2 2 n I'm just unclear what σ represents here. Am looking for a confidence interval where μ = σ? or looking for the confidence interval where σ = σ 2 / n ?? any hints on where to start this?

Question

For large n, the sampling distribution of   S  =            1              n        −        1                    ∑              i        =        1                    n              (          X      i        −    X          )      2       is sometime approximated with a normal distribution having the mean   σ and the variance             σ      2              2      n      . Show that this approximation leads to the following   (  1  −  α  )  100  % large-sample confidence interval for   σ:      S          1      +                        Z                      α                          /                        2                                    2          n                      &amp;lt;  σ  &amp;lt;      S          1      −                        Z                      α                          /                        2                                    2          n                    I&#039;m just unclear what   σ represents here. Am looking for a confidence interval where   μ  =  σ? or looking for the confidence interval where   σ  =            σ      2              /        n  ?? any hints on where to start this?

Grace Moses · Accepted Answer

Step 1  σ is the standard deviation of your initial sample       X    1    ,  …  ,      X    n  , and S is the usual estimator of the standard deviation of a sample where also the mean is unknwon. What you&#039;re told is that  S  ≈      N        (    σ    ,                  σ        2                    2        n              )  Step 2where   ≈ means approximately distributed as. So yes, the mean of S is   σ where   σ is the standard deviation of       X    i  . Note that the above implies that            S      −      σ        σ    =      S    σ    −  1  ≈      N        (    0    ,          1              2        n              )  which enables you to arrive at that confidence interval.

For large n, the sampling distribution of S=sqrt{1/(n=1) sum_{i=1}^{n}(X_i-X)^2} is sometime approximated with a normal distribution having the mean sigma and the variance (sigma^2)/(2n)

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