Question

Again consider the automobile parts supplier situation. Remember that a problem-solving team will be assigned to rectify the process producing the cylindrical engine parts if the null hypothesis Ho: u= 3 is rejected in favor of Ha: u#3. In this exercise we calculate probabilities of various Type Il errors in the context of this situation. (a) Suppose that the parts supplier's hypothesis test is based on a sample of n= 100 diameters and that o equals .023. If the parts supplier sets a = .05, calculate the probability of a Type Il error for each of the following alternative values of 2.990, 2.995, 3.005, 3.010. (Round answers to 4 decimal places.) 0.0084 2.990, B 2.995, B 3.005, B 3.010, B 0.0084 (b) If we want both the probabilities of making a Type Il error when u equals 2.995 and when u equals 3.005 to be very small, is the parts supplier's hypothesis test adequate? If not, and if we wish to maintain the value of a at .05, what must be done? (Round answer to 4 decimal places.) ncrease the sample size. No when u 2.995 and when u 3.005.

I'm sure sure why, but it kept saying my answers for the two middle parts were wrong. Please explain

Answer 1

a. Probability of type II error are 0.3336, 0.4129, 0.5871 and 0.6664

b. Probability of type II error are: 0.5879 and 0.4121

0.63 0.023 2.0 0.4347 0.023 -8.2013 0.2173 Hoid-3 vs He: M43 22100, 0.0.023, 420.05 a ka prob. (Type- II Essar) ap (Accept He / His tur) = P(Z < 2.99 / 423) =P(und (2.99-3) =P (Z < 0.02 ) = P (2 20.4347) = 0.50 P(022 20.43) = 0.50 0.1664 zettty 2. is 4-2.995 RoPln 42.995 / 4 = 3.) -P(Z < -0.005) -P(Z <=0,2173) = 0.5-Pfo<2<21.93) = 0.5 – 1.0871 -0.4129 3. il 4-3.005 P (53,005) 423) = P(%5 000) - P(Z < 0.2173) 2015+ P(OLZ60.2173) = 0.5 + 0.0874 -0.5871 M = 3.000 B = P(#4 3.010/123) - P(%50.00%) P(014342) =0.5+ P(O<Zc0.4347) 2015 +0J664 = 0.6664 A P(x33/422.995) - P(Z < 0.05)=P(Z50.2173) 20.5879 2. ß-P(733/423.00s) - P(%6-0.2193) 0.5-0.0878 = 0.4121 0. 2/93 34