Can someone help to solve this question? thanks :) Let pi denote the true propot

Question

Can someone help to solve this question? thanks :)

Let pi denote the true propotion of 1s in a Let pi denote the true propotion of 1s in a population of 0s and 18: ve a CI for it. But someo population, i.e. 1-pi. pi barely touches the ci for 1-pi. Bint: For simplicity suppose p is answer wil1 depend on z and n, and will involve soiving a quadratie egn simple formula for the CI. know how to build ne else may choose to build a cI for the proportion of Os in the lation, 1.e. 1-pi. Pind t the value of p, i.e. sample proportion, such that the c1 for Bint: For simp licity suppose p is less than 0.5: the on z* and nr and will involve solving a quadratic egn. Aiso, use the

Explanation / Answer

Solution:

For simplicity we will assume p is less than and confidence interval is alpha for which critical z - value is z* and sample size is n.

so as p is less than 0.5 then upper confidence value must be equal to lower confidence value of (1-p)

so upper confidence value for p = p + z* sqrt [p*(1-p)/n]

and lower confidence value for (1-p) = (1-p) -  z* sqrt [p*(1-p)/n]

as both are qual

p + z* sqrt [p*(1-p)/n] = (1-p) -  z* sqrt [p*(1-p)/n]

2z* sqrt [p*(1-p)/n] = 1 - 2p

4 z*2 [p(1-p)/n) = (1 -2p)2 = 1 + 4p2 - 4p

4 z*2 p(1-p) = n (1 + 4p2 - 4p )

4 z*2 p - 4 z*2 p2 = n + 4np2 - 4np

p2 (4n + 4z*2 ) - p (4n + 4z*2) + n = 0

this is the quardetic quation and will given the value of p.

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