Question

Let 0 <  γ  <  α . Then a 100(1 −  α )% CI for μ when n is large is Xbar+/-z_γ*(s/sqrt(n))The choice γ = α /2 yields the usual interval derived in Section 8.2; if γ ≠ α /2, this confidence interval is not symmetric about . The width of the interval is W=s(z_γ+ z_α-γ)/sqrt(n). Show that w is minimized for the choice γ = α /2, so that the symmetric interval is the shortest. [ Hints : (a) By definition of z α , Φ( z_α )  =  1 −  α , so that z_α =  Φ − 1 (1 −  α ); (b) the relationship between the derivative of a function y = f ( x ) and the inverse function x = f − 1 ( y ) is ( d / dy ) f − 1 ( y )  =  1/ f′ ( x ).]

Answer 1

Given,

To find, the value of y, for which W is minimum.

Thus we have,

       where, is the inverse CDF of standard normal variable.

Now,

where, is the PDF of standard normal variable.

To minimize 'W' with respect to 'y',

Now since, is a one to one function, thus the above statement is true only under the following condition,

Hence, it is proved that W is minimized for the choice γ = α /2, so that the symmetric interval is the shortest.

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