Question

Q2. [20] A random sample of size 36 is taken from X which follow N . 64). [5] Show that to test the null hypothesis .. = 80 against W -76, the best critical region for a 0.05 is of the forms [3] Show that to test the null hypothesis = 80 against M: > 80, the uniformly most powerful critical region for -0.05 is of the form M [10] Show that to test the null hypothesis M -80 against 8 0, the likelihood ratio test for a = 0.05 is of the form - 80 C () [3] Can you explain the reasons of considering three different tests for (a), (b), and (c)?

Answer 1

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DIIL. PAGE NO: 70 1622 497), X~. NlM264) ! Ho: H=90 Uls. H=76; d =0.05. But critical Region :- (We); - Wc = f lo sk . I 2, where; . 1 din Sanx8 TX-4² Lī_728 - 8 Janu = (8 Jes)" Hp ECX-Me)? lo X2 i Elxi - Mo) zo &C X2-Mo) We = fetteri)? - ZOXE=Rod 7 ek : 122 Tating log con foth Sider

PAGE NO We = L 1 [[ (x2-1) - {(x - 2) ) e lege :: لا شروع 12 = (ربما 3x - (رب )) We feli't muilmy, – Tri -niha + fxa&mi s pohyb We = { mun) + Exi (240-241) 5 128 loge Of Exi (Re-drive) is Ipilog kaming = {{ xi ( 2 Mo -2M,) e Co . Here: Mo = 80; 4= 76 s & Exi ( 160152) s to SEXi çdo a & ł s. 4, CHeme Proved). where; e = lo = 128 lg k-In Mind

M WANAWIENNUWUN LAUW DATE: PAGE NO: Some s Now, from egration (1) m part (ft we had Deaf Exi (Atle=&AD) 5. rasloj k = (mmente) Note that informly most powerful critical region - is some as Best critical Region with o parameter) - can be any value, but it should be best for all values of o. - fare; Wc = {Xi (QHo - dpi) s e He=80; 4, 380 it t o ✓ Vuo-HD) 20 ..! (2 Mo - 241) co * We = f s.Xi > es (Ho-lhe 3 we = f 7.7. M. 6 40 2412 " we = ł 71 M 2 Hrnce Proved)

. Likelihood Ratio Trst is t h e or (X) = 27 ) | nk) where î (he) - supremum 10a). Oche Heren [ N = zcz (Wi-ps? (2D62)?, Likelihood Patio tut :- d(x) = ? (2) Ľl dk) za En et 13 (xi – ? XS (7- min Leu?) {(x1280) s(x1=0) 4-80 s n {u? px x-go 2, 4, 5х8 (depending on es).

DATE: PAGE NO: IX-801), us. Hence Provid? 22 The search fos Comidering 3 different that is that live have to consider hypothesis and according to that only we can apply different Since Mo ?Mi in fait part; we applied Alsts in Best vietical region method a Because of Ho Chi ( always in part (f.), We applied uniformly most powerful region and time we don't know any relation between Me and Mi in part (e); we applied - Likelihood Patio test: Please LIKE' Thanks ☺