Question

Solve using the Neyman-Pearson Theorem

63. The error X in a measurement has a normal distri- bution with mean value 0 and variance o2. Con- sider testing Ho: a = 2 versus H: = 3 based on a random sample X1, . . . , X, of errors a. Show that a most powerful test rejects Ho when Σ>. b. For n 10, find the value of c for the test in (a) that results in a = .05

Answer 1

63. @ Solution :- - e art ? with zero & variance mean 2) on form of By Neyman - Pearson Theorem, the the critical region is given by (with to ² = 2 and on = 3) C = $(x – an) EIR" ALL x. - - sen Llr²4 . sen) ..L(6²) = Likelilood function /27 To - x2 VAX2 e 2x3 Similarly, V2.7 x3 (i 1 Now, C = f(x - xn) EIR 1 = SC 2012 Prie (2-3) Exiscl c = foc – kn) 20 c sen) FIRM ) Elec EIRAL

It is knopen thad, IO f (x un Elk"I niz y iz Hence, the most powerful test is of a the form rejeet Housb Exi2zc (6) To find the value of c, we need the probability distribution of xpt 410 (for n= 10). It is knosen then, 2 (x:3² ~ x 10 [:squere of standerd normal variate is x² distribusion] 0.05 = 2 = P(Reject Hol Ho is true) = P 2 (xi) zc102= 2) = P(x) = & from the chi- quere table, we get Xhos = 18.30 =) = 18.30 - - 36.60