5. A research team wished to specify a manufacturing process so that Y, the area
ID: 3325600 • Letter: 5
Question
5. A research team wished to specify a manufacturing process so that Y, the area in a product affected by surface flaws was as small as possible. They had four levels of concentration of a chemical used to wash the product before the final manufacturing step and wanted to determine whether different concentration levels caused a change in E(). They ran a balanced one-way layout with4 observations at each concentration with level 1 set at 5%, level 2 set at 10%. level 3 set at 15%, and level 4 set at 20%. They observed that y, = 21.5, ½172, y,-18. 8, and y,-193, where y, was the average of the observations taken on the ith level. They also observed that s-14 19, s-1 52.2, s 13 18, and s-1619, where s, was the unbiased estimate of the variance for the observations taken at the ith soncentraton, The four parts of this problem are worth a total of 80 points. +' Complete the analysis of variance table for these results; that is, be sure to specify the degrees of freedom, sum of squares mean square, and F-test. Use significance levels set to 0.10 0.05, and 0.01. (20 points) Find the sum of squares due to the linear, quadratic, and cubic contrasts. The coefficients of the linear contrast are -3.-11.3 the coefficients of the quadratic contrast are 1.-1.-1,1; and the coefficients of the cubic contrast are -13,-3,1. (30 points)- Find the 95% Scheffe confidence interval for the linear and quadratic contrasts. (20 points) What is the optimal setting of the concentration level and how do you document it. (10 points) a. b. c. d. Analysis of variance tablev MS 12.61* 146.95* Source 37.84p 1763.4* 1801.242 | Treat 0.09 Error' Total 15- Accept at the 0.10, 0.05, and 0.01 levels of significance. +' The estimated contrasts 5.0.1 -4.8Ac =-7.0, and ss, = 5.0, sse = 23.04, ss, 9.8-, Note that the three sums of squares add up to the son of squares for treatments. +' The 95% Sche econ idence interval for 41s-50487 8 . The 99% Scheffe confidence interval for is 4.8±39.3·There is not an optimal setting since all of the treatment means appear equal +'Explanation / Answer
Answer:
Group
Mean
SD
n
Total
SSGroups
SSError
1
21.5
9.049862
6
129
8.64
409.5
2
11.2
9.602083
6
67.2
793.5
461
3
18.8
8.473488
6
112.8
91.26
359
4
39.3
10.09455
6
235.8
1653.36
509.5
ANOVA
Source
df
SS
MS
F
P-value
Groups
3
2546.76
848.92
9.763312
0.000356
Error
20
1739
86.95
Total
23
4285.76
H_0: mu_1 = mu_2= mu_3 = mu_4
H1: At least one pair of population means are different
Calculated F=9.76, P=0.0004 which is < 0.01 level.
Ho is rejected at all the 0.10, 0.05 and 0.01 levels.
Group
Mean
SD
n
Total
SSGroups
SSError
1
21.5
9.049862
6
129
8.64
409.5
2
11.2
9.602083
6
67.2
793.5
461
3
18.8
8.473488
6
112.8
91.26
359
4
39.3
10.09455
6
235.8
1653.36
509.5
Related Questions
Hire Me For All Your Tutoring Needs
Integrity-first tutoring: clear explanations, guidance, and feedback.
Drop an Email at
drjack9650@gmail.com
drjack9650@gmail.com
Navigate
Integrity-first tutoring: explanations and feedback only — we do not complete graded work. Learn more.