BCABCDADBDABDCDACDBCDABCDErrorTotalTest all hypotheses of interest using a = 0.01. Find the critical value. (Round your answer to two decimal places.)F₁2=Which main effects are significant at level 0.01? (Select all that apply.)A (nature of roof)B (power setting)OC (scrap used)D (charge)There are no significant factors at level 0.01.Which interactions are significant at level 0.01? (Select all that apply.)AB (nature of roof and power setting)AC (nature of roof and scrap used)AD (nature of roof and charge)BC (power setting and scrap used)BD (power setting and charge)CD (scrap used and charge)ABC (nature of roof, power setting, and scrap used)pers Calculator The following data on power consumption in electric furnace heats (kW consumed per ton of melted product) resulted from a 24 factorial experiment with three replicates. The factors were nature ofroof (A, low/high), power setting (B, low/high), scrap used (C, tube/plate), and charge (D, 700 lb/1000lb).TreatmentTreatment(1)ababсacbcabcThe hypotheses are as follows:HOA: all a's = 0Ha4: at least one a, * 0HOAB: all y ABS = 0H₂4B: at least one y, AB #0BD₁s = 0HaBD: at least one y, BD 0HOBD: all yHOABC: all yik's = 0HaABC at least one Yijk * 0AXijklm867, 860, 799946, 800, 840776, 834, 746709, 789, 6461017, 990, 9541028, 906, 977817, 783, 771829, 806, 691SourceBABCACdfdadbdabdcdacdbcdabcdHos: all Bjs = 0Has: at least one ß, *0Xijklm996, 808, 650966, 976, 876SS702, 659, 650784, 700, 595922, 808, 8681055, 870, 905AC # 0HOAC: all YAC'S = 0HAC: at least one YikHocp: all YCD's = 0Hacp: at least one yCD * 0798, 725, 700751, 713, 715HOABD: all y's 0HaABD: at least one y* 0HOABCD: all Yiki's = 0HaABCD: at least one Yijkl 0Construct the ANOVA table. (Round your answers to two decimal places.)MSHoc: all 8's = 0Hac: at least one 8 0HOBC: all Y₁BC¹s = 0HOAD: all y, AD's = 0HaBC: at least one YBC #0 HaAD: at least one y,AD + 0HOACD: all yik's=0HACD: at least one Yikl 0Hop: all cjs = 0Hap: at least one , # 0HOBCD: all Yiki's = 0HaBCD: at least one Yjkl # 0

TreatmentxijklmTreatmentxijklm(1)867, 860, 799d996, 808, 650a946, 800, 840ad966, 976, 876b776, 834, 746bd702, 659, 650ab709, 789, 646abd784, 700, 595c1017, 990, 954cd922, 808, 868ac1028, 906, 977acd1055, 870, 905bc817, 783, 771bcd798, 725, 700abc829, 806, 691abcd751, 713, 715Consider a 24 factorial experiment with 4 factors; nature of roof (A), power setting (B), scrap used (C) and charge (D) each of at 2 levels i.e. 16 treatment combinations (1), a, b, c, d , ab, ac, ad,, ab, ac, ad, bc, bd, cd, abc, abd, acd, bcd and abcd with three replications.ABCDTreatmentObservations1111(1)867, 860, 7992111a946, 800, 8401211b776, 834, 7462211ab709, 789, 6461121c1017, 990, 9542121ac1028, 906, 9771221bc817, 783, 7712221abc829, 806, 6911112d996, 808, 6502112ad966, 976, 8761212bd702, 659, 6502212abd784, 700, 5951122cd922, 808, 8682122acd1055, 870, 9051222bcd798, 725, 7002222abcd751, 713, 715a)Use the Minitab Technology to perform the ANOVA for 24 factorial experiment with 3 replications with the following steps:Step 1) Open Minitab and Go to Stats-> DOE-> Create the 24 factorial experiment with 3 replications.Step 2) Now the respective observation under the given combinations of factors.Step 3) Go to Stats-> DOE-> Factorial-> Analyze the factorial experiment.Step 4) Select yield on power consumption under Response Box and Press OK.Output:Thus, the obtained ANOVA table is:b) Determine the critical value using the excel formula, "=F.INV.RT(0.01,1,32)":Fα,v1,v2=7.499281Fα,v1,v2=7.50The critical value at 1% significance level is 7.50.c)The p-value corresponding to main effect B and C is and 8 respectively which is less than the significance level((α=0.01). Thus, the corresponding null hypothesis is rejected.It can be concluded that main effects B(Power setting) and C(Scrap used) are significant at 1% significance level as their respective p-values are less than the significant level(α=0.01).d) The p-value corresponding to all the interaction effects are greater than the significance level(α=0.01). Thus, the corresponding null hypothesis is not rejected.It can be concluded that all the interaction effects are not significant at 1% significance level as their respective p-values are greater than the significant level(α=0.01).a)Use the Minitab Technology to perform the ANOVA for 24 factorial experiment with 3 replications with the following steps:Step 1) Open Minitab and Go to Stats-> DOE-> Create the 24 factorial experiment with 3 replications.Step 2) Now the respective observation under the given combinations of factors.Step 3) Go to Stats-> DOE-> Factorial-> Analyze the factorial experiment.Step 4) Select yield on power consumption under Response Box and Press OK.Output:Thus, the obtained ANOVA table is:b) The critical value at 1% significance level is 7.50.c) The p-value corresponding to main effect B and C is and 8 respectively which is less than the significance level((α=0.01). Thus, the corresponding null hypothesis is rejected.It can be concluded that main effects B(Power setting) and C(Scrap used) are significant at 1% significance level as their respective p-values are less than the significant level(α=0.01).d)The p-value corresponding to all the interaction effects are greater than the significance level(α=0.01). Thus, the corresponding null hypothesis is not rejected.It can be concluded that all the interaction effects are not significant at 1% significance level as their respective p-values are greater than the significant level(α=0.01).

The following data on power consumption in electric furnace heats (kW consumed per ton of melted product) resulted from a 24 factorial experiment with three replicates. The factors were nature of roof (A, low/high), power setting (B, low/high), scrap used (C, tube/plate), and charge (D, 700 lb/1000lb). Treatment (1) a b ab с ac bc abc Xijklm 867, 860, 799 946, 800, 840 776, 834, 746 709, 789, 646 1017, 990, 954 The hypotheses are as follows: HOA: all a's = 0 H₂4: at least one a, # 0 B 1028, 906, 977 817, 783, 771 829, 806, 691 HOAB: all y ABS = 0 HAB: at least one y, AB = 0 HOBD: all y BD¹s = 0 HaBD: at least one y AB C AC HOABC: all Yk's=0 HaABC at least one Yk # 0 BD # 0 Treatment d ad bd abd cd acd bcd abcd Hos: all B;'s = 0 Has: at least one 8, +0 SS Xijklm 996, 808, 650 HOAC: all y ACS=0 HAC: at least one y#0 HOCD: all YCD's=0 HacD: at least one YCD = 0 966, 976, 876 702, 659, 650 784, 700, 595 922, 808, 868 1055, 870, 905 798, 725, 700 751, 713, 715 HOABD: all y's=0 HABD: at least one Yijl * 0 MS HOABCD: all Yk's = 0 HABCD: at least one Yijk * 0 Construct the ANOVA table. (Round your answers to two decimal places.) Source df Hoc: all 8's = 0 Hac: at least one 8 = 0 HOBC: all y BC¹s = 0 Hasc: at least one y₁BC0 Hop: all es = 0 Hap: at least one , # 0 HOACD: all y's = 0 HOBCD: all Yjk's=0 HACD: at least one Yikl 0 HaBCD: at least one Yjkl 0 f HOAD: all y AD'S = 0 HAD: at least one y AD + 0

A First Course in Probability (10th Edition)

10th Edition

ISBN:9780134753119

Author:Sheldon Ross

Publisher:Sheldon Ross

Chapter1: Combinatorial Analysis

Section: Chapter Questions

Problem 1.1P: a. How many different 7-place license plates are possible if the first 2 places are for letters and...

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