Question

2. Suppose that X1, X2, ..., Xn " N(41,01) and Yı,Y2,...,Ym * N(H2;02) are two independent random samples. (a) What is E[X - Ÿ]? (b) Find a general expression for Var[X – Ý), and use this to find an expression for the standard error ox-ý = StDev(X – Ỹ). (c) Suppose that of = 2 and o = 2.5, and also that n = 10 and m = 15. Determine the probability P(|X – Ý - (µ1 – 42)| < 0.5) by standardizing X - Y. You can either use a z-table (such as in the back of your textbook), or R or Mathematica (or other software, as you like). (d) Suppose still that of = 2 and o = 2.5, but that m = n (call this a balanced experimental design when you have equal sample sizes). Determine the smallest value of n such that X - Ỹ lies within 0.5 units of H1 - 42 with a probability of at least 0.95; i.e. P(|X - Y- (1- 2)| < 0.5) > 0.95. Make sure you round your answer up to a whole number.

Answer 1

Sop": Xi , Xxj. .· ,Xn HoN(A,, G;?). and Y,%,.., Ym A NI Ab, G?) • ') X1 , Xx,. , Xn and Y, Y, , . »,Ym are andependent IEEIX) n it .: ElX) = Mi nd ī .. E(7) = Me My-Mq E (X) – E♡) = > E[x-7] = %3D arnd YNNLAS, E) (G) Since, X,X2,- -, Xn and y,,- - Ym ome endefondant * and ī are endependent Cov (8,7)=0 = Var (x)+ Varl7)-2ov(8,9) - 2*0 Vr [ §-7]= =>

Var (x-Ÿ) = Now, 675 = stder.(x-9) = Var lx-Ÿ) ² O77 = Stderl 8-7) = =-> © Greven that , 6,²=2; 6²= 2.5 6= 2,5 n=10, m = 15 x-7-(24-2) NNlo,1) 6x-y S0.5 = P[ 1 3-3-1A4) <0s7 = P[ ]zl < ] <0.5 Mut Ox-y 2.5 = 0.605 10 15 P) Iz1 5_0.5 0.605]= PL 121s 0:8267

>' P[ 1*-7 - (44-42) Z 0.5] PL IZI so.826] $ (0.826) - I fo.826) ewhere $) Cs the cdif. f zNN = O (0.826) – ( 1- 0 c0.621) O-3) = i- O(3) since = 2E (0.826) -1 i from the z-table = 2 * 0.7967 -1 1.5934 -1 0•5934 %3| > P[ 1X-7-(14-4)| s 0.5] = 0-5934

@Geiren that; 0,² = 2 , 6,2 = 2,5 and we need to find Amallest value af n such that PE IX-7-(4-14)|so5],0.95 S0.5 2+2.5 4.5 Ox-y = %3D → P[ 1포7-14- <>095 OR-9 0:5 Izl 3 o15 VY5n,0:95 P => 9f x is a random vartable with mean u gnd variance 62 then fo any +ive integor P[ 1x-41<Ko] 1- (By chebychevs 9nequaliy) Tustn 0:5 K *1 0° ZN N(0, 1)

and O•95 -1- K2 = 0.05 K2 Too (00 = 20 0•25. K² = 0.5 -> 20 = 0.25n Tos 4. 45*2* J = 4.5* P0 0:25 n n = 45 * 8 = 360 o: m-n=360