Question

1) You have a process to detect surface flaws on an orthopedic device. Your process is correct 70% of the time with current employees, and you are hoping that newer employees will perform better when executing your process due to improved training methods. a. Specify a null and alternative process to determine whether your newer employees perform as well as your experienced employees. b. You decide to reject the null hypothesis Ho: p= 0.7 if newer employees correctly identify flaws in at least 13 out of 16 tries. What is the type I error probability a? c. Find the minimum number of correct identifications out of 16 tries that would lead you to reject the null hypothesis Ho: p = 0.7 if you want a type I error probability a below 5%. d. You hope your test is powerful enough to detect a 20% increase in performance, so you specify the alternative to be Hı: p=0.9. What is the type II error probability B if you decide to reject the null hypothesis if newer employees correctly identify flaws in at least 13 out of 16 tries? e. What is the minimum number of correct identifications out of 16 attempts that would lead you to reject the null hypothesis and accept the alternative hypothesis Hi: p = 0.9 if you want a type II error probability ß below 5%?

Answer 1

(d) ß = Type I esroo= pq XE W l H, is love? = 1-Pfx6wl Hi es true} where w is the region of rejection Wa { xx | 22,13? So, ß= 1- P1 2213 Þ=0.93 51-|(13(0-9)%0.-117 (46(0.9)"(0-13°46%)(0-2)*(0-1) +(14)(0-2)56 0.06871 1- 100142 +0.2745 +0.3294 +0.1851 (e) we want a probability like as follows: Let's be the minimum attempts "!" that will enable us to reject Ho We know that. Bel-Pf 24W 14,3 T 7 or 1-84 (16) pr qlb | H:p=0.9) <0.05 7) (46) (0.4)*(0-1)!6–, 0105 =12] T

(0) 4.: Neuer employees berform as well as the oleler employees - Hii Neuner employees don't berform better than the older employees inferior to denotes the number of flaws en ni a Hows in 'n outcomes (6) Let X then kn Bln, p) Type I error = d = pf Reject Ho Ho is true} 5) Q = P{*713/6=0.7]. = (11) 60-360-3)+(14)62)*(0-3)?(1967)*60-3) + (54) (0.7)" T: X~ B116,0.7)}. =) = 0.0156 +0.0732+0.022 +0.00332 = 0.11412 Let 'n be the minimum number correct esentification sut of 16 trials that would enable to reject null hypothesis then x = pq reject Ho Ho is true? < 0.05 = {(16)b” 26* | p = 071 <0.05 (0) = (%) 6-73" (0.3)6-n soos + (19)C-3€ (0.3)” <005 < () (0-7)(0-337