Answer Happy

Please answer Question 3 only. Example 8.4.2 is for
reference.
Example 8.4.2. Let X be N(0, 100). To find the sequential probability ratio test for testing Ho 0 = 75 against H₁0 = 78 such that each of a and 3 is approximately equal to 0.10, take 0.10 1 10.10 ko = k₁ = = 9. 10.10 9' 0.10 Since L(75, n) exp[-(x-75)2²/2(100)] 6 Συ; – 459n = L(78, n) exp[-(x-78)²/2(100)] 200 the inequality 1 L(75, n) ko = < 9 L(78, n) 8.4. *The Sequential Probability Ratio Test can be rewritten, by taking logarithms, as 6 Σx; – 459η - log 9 < < log 9. 200 This inequality is equivalent to the inequality n 100 153 co (n) 153 -n- 2 log9 < Σxi < 100 -n + -log 9 = c₁ (n). 3 3 2 Xi Moreover, L(75, n)/L(78, n) ≤ ko and L(75, n)/L(78, n) ≥ k₁ are equivalent to the inequalities x ≥ c₁(n) and 1 x ≤ co(n), respectively. Thus the observation of outcomes is discontinued with the first value of n of N for which either x ≥ ci(n) or xico(n). The inequality Ex c₁(n) leads to the rejection of Ho 0= 75, and the inequality co(n) leads to the acceptance of 0.10 when Ho is true, and Ho 075. The power of the test is approximately approximately 0.90 when H₁ is true. ■ = exp < 9 = k₁ 505
3. (Exercise 8.4.1 from edition 8 of the textbook) Let X be N(0,0) and it is desired to test Ho: 0 = 0' = 4 versus H₁ : 0 = 0″ = 9. Let aa = 0.05, and a = 0.10. (a). Show that the sequential probability ratio test can be based upon the statistic Σ=1 X². (b). Determine the two constant co(n) and c₁ (n). [Hint: For the two notations co(n) and c₁(n), read and follow the notations in Example 8.4.1 and Example 8.4.2 of the textbook.]