Formulate Design of Experiments and ANOVA

Formulate Design of Experiments and ANOVA Student’s Name Institutional Affiliation Formulate Design of Experiments and ANOVA Assignment 9 1 What are the treatments for a designed experiment with two factors, one qualitative with two levels (A and B) and one quantitative with five levels (50, 60, 70, 80, and 90)? There are 10 treatments which are 50A, 60A, 70A, 80A, 90A, 50B, 60B, 70B, 80B, and 90B. 2 A quality control supervisor measures the quality of a steel ingot on a scale of 0 to 10. He designs an experiment in which three different temperatures ranging from 1,100 to 1,200 degrees Fahrenheit) and five different pressures (ranging from 500 to 600 psi) are utilized, with 20 ingots produced at each Temperature – Pressure combination. Identify the following elements of the experiment: 1 Response…………… the quality of steel ingot. 2 Factor(s) and factor type(s)……………factors are 3 different temperatures and 5 different pressures, qualitative and quantitative respectively. 3 Treatment…………….15 treatments. 4 Experimental units…………20 ingots. 3 A partially completed ANOVA table for a completely randomized design is shown here: Source df SS MS F Treatments 6 17.5 2.9167 3.52 Error 35 29.0 .8286 Total 41 46.5 5 Complete the ANOVA table. 6 How many treatments are involved in the experiment? V=7. 7 Do the data provide sufficient evidence to indicate a difference among the population means? Test using α = .10. Test for the hypothesis: H0: t1=t2=t3=t4=t5=t6=t7 vs H1: anything different. F (6, 35, 3.52) = 1.955. So 1.955<3.52. Reject H0 and conclude that the data provide sufficient evidence that there is a significant difference among the population means. 8 Find the approximate observed significance level for the test in part c, and interpret it. Read the corresponding values from the F tables that give 3.52. At α = 0.05, the f calculated value 3.52 lies at 2, 19. The observed significance level is 0.05. 9 Suppose that x1=3.7 and x2=4.1. Do the data provide sufficient evidence to indicate a difference between µ1 and µ2? Assume that there are six observations for each treatment. Test using α = .10 There are 42 observations (n) and 7 treatments. The mean difference is -0.4. The critical value at t 0.10/2 (n-v) i.e t 0.05 35= 1.6896. The critical difference is 1.6896*sqrt 2/7 mss(e) = 0.822096. So -0.4<0.822096. There is no significant difference between the two means (Antony & Antony, 2003). 10 Refer to part b. Find a 90% confidence interval for (µ1 – µ2). C.I= (-0.6153,-0.18469) Refer to part b. Find a 90% confidence interval for µ1. C.I= (3.4847, 3.9153) 4 The data in the table below resulted from an experiment that utilized a completely randomized design. Treatment 1 Treatment 2 Treatment 3 3.8 5.4 1.3 1.2 2.0 0.7 4.1 4.8 2.2 5.5 3.8 2.3 Use SPSS to generate an ANOVA analysis and attach a copy of the output Check the attachment for Question 4. Test the null hypothesis that µ1 = µ2 = µ3, where µi represents the true mean for treatment I, against the alternative that at least tw...