Question

show me all work for the problem i,ii,iii

Exercise 1 (Sample size for estimating the mean). Let X1,...,x, be i.i.d. samples from some un- known distribution of mean u. Let X and S denote the sample mean and sample variance. Fix a E (0,1) and € >0. (i) Suppose the population distribution is N(uo?) for known op > 0. Recall that we have the following 100(1 - a)% confidence interval for : (1) Deduce that plue (x-Zalze in 2+ zarze )=1-a. n2 (22120/e)? zarote Plue [X-,x+])=1-a. (ii) Suppose the population distribution is not necessarily normal but has known variance o2 > 0. Recall that CLT gives us the following approximate 100(1-a)% confidence interval for : Deduce that Plue (x-zare come X + 2avne ul) <1-a. nz (22a/201es? – zavezanec pſue[- ,X+) =1-a. (iii) Suppose the population distribution is normal but unknown variance o?. Recall that we have the following 100(1-a)% confidence interval for : plue [x-tarabu v . 8 + tarln – 1. a)=1-a. (5) nz (tar2(n = 1510j? tarzin 1) P(ue(8- ,x+])=1-a. (6) Deduce that (iv) In each cases of (1)-(ii), suppose o2 = 5. If we would like to get a 95% confidence interval for u of length <0.01, what is the smallest sample size n that guarantees this? (v) In the cases of (iii), suppose s = 5 and n = 30. If we would like to get a confidence interval of length = 0.01, what is the largest confidence we can guarantee?

Answer 1

ANSWER:

Solution: Let x,.... In be Samples from some unknown distribution of mean M. and X and st denote the sample mean Sample variance. Fix a E (0) and aeso i .P (Me [x-Za 12 = x + Z012 I D=1-a. - P[ue [a-zaj v x + 2)2* - P[/x-ul < Zarzt u on = 1-a...o and again if p {ue [x - e x + J - = p (1x-ul = ) - nz (2 za 12 E) a Zala in se lo [7.- 2012 x 2 + 204 x ] = { x - x x + ] => 1-2=P [[-M1 < Zalz X ] 5P (X-M159)

· P (x-M15 ) =1-6 To the population distribution is not necessarily xl-a aso. 'P (ME/X-2012 T x+29/ 2 Pf nz (zza12 Ote)". arguments is then using we CLT and the can prove this part D=1-a ③ P (Me [2-tap, (n-1) = x + tape (0-1) if nz (2tal2 (n-1) sye)" then using the we can arguments in ① prove this part Q=0.05 afere A=30 3-5 225 0.01 Zal = 20.005 1.96 e=0.01 n~ (220.025 - 12) => 768320.

>> 1768320 o suppose 525 n=30 like get to a interval if we would confidence to length so.ol. n = 30 87-5 e=0.01 tal2 (29) * g = cle - tal2 (29) < 0.0122 jane taiz(29) ♡ 0.0122 l-a/2 = 0.5048 Cor) a=2*(-0.5048) = 0.9904 (100-99.04) -10.96-9.) Hence proved.