For any k [0 , 1], Y z k n < < Y + z (1 k ) n is a (1 )-confidence interval for

Question

For any k [0 , 1],

Y z k n < < Y + z (1 k ) n

is a (1 )-confidence interval for . Show that the length of a confidence interval of this form is smallest when k = 1 / 2. Hint: the length of a confidence interval is a function of its endpoints.   Indeed, suppose that [ Y + a n , Y + b n ] is a (1 )-confidence interval of .   Then the length of this confidence interval is:

L ( a,b ) = n ( b a ) .

It suffices to show that L ( a,b ) is minimized when a = b = z / 2 .

See picture of full problem below:

(b) For any k e [0,1], rn rL is a (1-a)-confidence interval for Show that the length of a confidence interval of this form is smallest when k 1 /2 Hint: the length of a confidence interval is a function of its endpoints. Indeed, suppose that Y+ a', , Y-bor is a (1-)-confidence interval of . Then the length of this confidence interval is L(a, b) = (b-a) It suffices to show that L(a, b) is minimized when a =-b =-za/2. Hint: To do so, we can formulate the minimum length confidence interval as a nonlinear optimiza- tion problem: min L(a, b) - 2T the constraint ensures that (a, b) give the confidence limits of a (1-)-confidence interval. Next, Lagrange multiplier theory (review this material from Calculus III) implies that the optimal solution (a, b) minimizing L(a, b) must satisfy 00 (a, b) + (a, b) = 0 (5.2.2) Oh Ob (5.2.3) Ob for some Lagrange multiplier , where complete the argument, vou can show (5.2.2) and (5.2.3) are only satisfied when a- -b denotes the partial derivative of f with respect to x. To

Explanation / Answer

length of the interval is / n(Z_(1-k) + Z_k) as k increases  Z_k decreases but  Z_(1-k) increases as (1-k) decreases..... Z_k means that point on real-line above which the probability is k so when k increases k increases (the area under normal curve beyond a point ) then the point must be decreases.....So this is a trade-off between  Z_(1-k) and  Z_k when k increases or decreases ...So we need find such a value of k at which this trade off can be overcome and that value is k=.5 because if we see at k=.5 both  Z_(1-k) is Z _/2 and Z_k is Z _/2 and at this stage the length is 2*Z _/2 and it is the min length brcuase ther is no any trade -off like the previous situation so to judge the min length it is not problematic (both of the components of the sum give us the same value ) as the trade is over-come at k=1/2 and hence it is min at k=1/2

when 1-k increases and k decreases at k<1/2 and at k>1/2 k increases and (1-k) decreases but at k=1/2 they are equal so it is quite natural to get min value of the interval at k=1/2....This is a logical explanation with out rigorous mathematics ....

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