Chapter 3 Problems: 3-6myplace.frontier.com/~stevebrainerd1/STATISTICS/ECE-580-DOE WEEK 6...Chapter...

Chapter 3 Homework Steve Brainerd 1

Homework Chapter 3 Problem: 3-6


Chapter 3 Problems: 3-6 EXCELDesign Expert Format EXCEL Format

Coating Type ConductivityCoating type

1Coating type 2

Coating type 3

Coating type 4

1.00 143.00 Conductivity 143.00 152.00 134.00 129.001.00 141.00 141.00 149.00 136.00 127.001.00 150.00 150.00 137.00 132.00 132.001.00 146.00 146.00 143.00 127.00 129.002.00 152.002.00 149.00 Anova: Single Factor2.00 137.002.00 143.00 SUMMARY3.00 134.00 Groups Count Sum Average Variance3.00 136.00 Coating type 1 4 580 145 15.333333.00 132.00 Coating type 2 4 581 145.25 44.253.00 127.00 Coating type 3 4 529 132.25 14.916674.00 129.00 Coating type 4 4 517 129.25 4.254.00 127.004.00 132.004.00 129.00 ANOVA

Source of Variation SS df MS F P-value F critBetween Groups 844.6875 3 281.5625 14.302 0.00029 3.4903Within Groups 236.25 12 19.6875

Total 1080.9375 15


Chapter 3 Problems: 3-6 EXCELConductivity Problem 3-6

143.00141.00

150.00

146.00

152.00

149.00

137.00

143.00

134.00136.00

132.00

127.00129.00127.00

132.00

129.00y = -0.8125x2 - 1.9625x + 148.94R2 = 0.6814

125.00

130.00

135.00

140.00

145.00

150.00

155.00

0 1 2 3 4 5Coating Type

Con

duct

ivity

ConductivityPoly. (Conductivity)


Chapter 3 Problems: 3-6

• Design Expert Analysis 3-6 a


Chapter 3 Problems: 3-6• Design Expert Analysis 3-6a


Chapter 3 Problems: 3-6• Design Expert Analysis 3-6b


Chapter 3 Problems: 3-6• Design Expert Analysis and EXCEL Analysis 3-6c

Total N = 16a = # levels = 4 TINV(0.05,df) TINV(0.01,df)

0.05 0.01

Mean COATING 4 df =N-a ta/2,N-a ta/2.N-a MSe

n # Replicates

SE Standard Error:

sqrt(Mse/n)

129 12 2.178812792 3.05454 19.69 4 2.218670773

Confidence Interval low highmean-((tα/2,N-a)*sqrt(Mse/n))

mean-((tα/2,N-a)*sqrt(Mse/n))

95% 124 13499% 122 136


Chapter 3 Problems: 3-6• Design Expert Analysis and EXCEL: 3-6c second part Mean

difference Coating 1 - 4

Total N = 16a = # factors = 4 TINV(0.05,df) TINV(0.01,df)

0.05 0.01

Mean Difference COATING 1 - 4 df =N-a ta/2,N-a ta/2.N-a MSe

n # Replicates

SE Standard Error:

sqrt(2Mse/n)

15.75 12 2.178812792 3.05454 19.69 4 3.137674298

Confidence Interval low highmean-((tα/2,N-a)*sqrt(Mse/n))

mean-((tα/2,N-a)*sqrt(Mse/n))

95% 9 2399% 6 25


Chapter 3 Problems: 3-6• 3-6d Fishers LSD ( page 99)• Calculate the LSD term as:

• LSD = tα/2,n-a*Sqrt(2MSe/n)

• tα/2,n-a = t0.05,12 = 2.18• MSe = 19.69• n= 4• LSD = 2.18*3.14 = 6.66• So any absolute value mean difference > 6.66 = signifiance

Mse n sqrt(2Mse/n) t(0.05,12) t(0.01,15) LSD= t0.05,12 (sqrt(2Mse/n))

19.6875 4 3.14 2.18 3.05 6.66


Chapter 3 Problems: 3-6• 3-6d Fishers LSD EXCEL If Absolute difference between

means > LSD = significance!

Mse n sqrt(2Mse/n) t(0.05,12) t(0.01,15) LSD= t0.05,12

(sqrt(2Mse/n))19.6875 4 3.14 2.18 3.05 6.66

Coating Coating Value A value BDifference A-

BLSD Significance at

α = 0.05

1 2 145 145.25 0.25 NO1 3 145 132.25 12.75 YES1 4 145 129.25 15.75 YES2 3 145.25 132.25 13 YES2 4 145.25 129.25 16 YES3 4 132.25 129.25 3 NO


Chapter 3 Problems: 3-6• 3-6d Fishers LSD 1–3, 1-4, 2-3 ,and 2-4 are significantly

different from JUMP


Chapter 3 Problems: 3-6• Graphical comparison of Means 3-6e : t df for error = 12

Graphical Comparison of Means Example 3-6e

0.00%

5.00%

10.00%

15.00%

20.00%

25.00%

30.00%

35.00%

40.00%

45.00%

50.00%

125 130 135 140 145 150 155

Mean +/- SE*t

P(X)

Stu

dent

t di

strib

utio

n

Student t P(X) (df = , 1 tail)

ACTUAL DATA TreatmentMeans


Chapter 3 Problems: 3-6• Design Expert Analysis 3-6


Chapter 3 Problems: 3-6• JUMP JMP Analysis 3-6


Chapter 3 Problems: 3-6• JUMP JMP Analysis 3-6 Bartlett’s test


Chapter 3 Problems: 3-6• JUMP JMP Analysis 3-6 JUMP Starter


Chapter 3 Problems: 3-6• JUMP JMP Analysis 3-6 JUMP Starter

Index LSD


Chapter 3 Problems: 3-6• JUMP JMP Analysis 3-6 d LSD

Chapter 3 Problems: 3-6myplace.frontier.com/~stevebrainerd1/STATISTICS/ECE-580-DOE WEEK 6...Chapter...

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