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,

$W = \frac{S}{\sqrt{n}}\left [ Z_y + Z_{\alpha-y} \right ]$
To find, the value of y, for which W is minimum.

Thus we have,
$W = \frac{S}{\sqrt{n}}\left [ Z_y + Z_{\alpha-y} \right ]$
       $= \frac{S}{\sqrt{n}}\left [ \Phi^{-1} (1-y) + \Phi^{-1} (1-\alpha+y) \right ]$ where, $\Phi^{-1}$ is the inverse CDF of standard normal variable.

Now,

$\frac{\partial W}{\partial y} = \frac{S}{\sqrt{n}}\left [ -\frac{1}{\phi(1-y)} + \frac{1}{\phi(1-\alpha +y)}\right ]$ where, $\phi(x) = \frac{\partial \Phi(x) }{\partial x}$ is the PDF of standard normal variable.

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

$\frac{\partial W}{\partial y} = 0$
$or, \ \frac{S}{\sqrt{n}}\left [ -\frac{1}{\phi(1-y)} + \frac{1}{\phi(1-\alpha +y)}\right ]= 0$
$or, \ -\phi(1-y) + \phi(1-\alpha +y)= 0$
$or, \ \phi(1-y) = \phi(1-\alpha +y)$

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

$or, \ 1-y=1-\alpha+y$
$or, \ y=\alpha/2$

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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