Math 558 Lecture #18

Sum of Squares for the Contrasts

The quantity discussed in the previous lecture

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∑ki=1 lwiyi.2 r∑ki=1 lw2i

S ,where Sw=r(lw1+lw2+….+lwk) w

is also called Sum of the Squares for Contrasts. We can use the notation SSlw for the contrast lw. where lw = lw1τ1 + lw2τ2 + … + lwkτk. Here lwi’s are coefficients of the given contrast. The contrasts can also be written in terms of means as

lw =lw1μ1+lw2μ2+…+lwkμk =∑lwiμi

Sum of Squares for Contrasts can be computed by using the following

SSlw = Where yi. is the ith treatment total.

Example2: Computing contrast sum of squares

A product development engineer is interested in the tensile strength1 of a new synthetic fibre. The engineer knows from her previous experience that the strength is affected by the weight percent of cotton used in the blend of materials for the fibre. Furthermore, she suspects that increasing the cotton content will increase the strength atleast initially. . She also knows that the cotton should range between 10-40 percent if the final product is to have other characteristics that are desired (such as ability to take a permanent press finishing treatment). The engineer decided to test specimens at five levels of cotton weight percent:15, 20, 25, 30 and 35 percent. She also tests five specimens for each factor level. There are five factor levels (5 treatments)i.e. t = 5 each with five replications i.e. r=5. She runs a completely randomized design with 25 runs.

1The tensile strength of a material is the maximum amount of tensile stress that it can take before failure. Tensile strength is measured in units of force per unit area.

Example: Computing contrast sum of squares

Cotton OBS Total AVG

Weight 12 3 45 Percentage

15 77 15 20 12 17 12 25 14 18 18 30 19 25 22 35 7 10 11

119499.8 18 18 77 15.4 19 19 88 17.6 19 23 108 21.6 15 11 54 10.8

Example: Computing contrast sum of squares

The engineer rejects the null hypothesis under F-tests and concludes some cotton weight percentages produce different tensile strengths than others. In order to further investigate which cotton weight percentages cause this difference she decides to test the following hypothesis.

H0 :μ4 = μ5

H0 :μ1 + μ3 = μ4 + μ5

H0 :μ1 = μ3

H0 :4μ2 = μ1 + μ3 + μ4 + μ5

l1 = − y4. + y5.

l2 =y1. +y2. −y4. −y5.

l3 =y1. − y3.

l2 =−y1. +4y2. −y3. −y4. −y5.

From the table We can find the values of the contrasts and the corresponding sum of squares.

l1 = − 1(108) + 1(154) = −54

l2 = + 1(149) + 1(188) − 1(154) = −25

l3 = + 1(149) − 1(188) = −39

l4 =−1(149)+4(77)−1(88)−1(108)−1(54) = 9

Example: Computing contrast sum of squares

SSl1 = 5(2)

SSl2 = 5(4)

SSl1 = 5(2)

= 291.60 = 31.25 = 152.10

SSl1 =5(20)=0.81

Example: ANOVA Table for the Tensile Strength Data

Source of SS df variation

Treatments 475.76 4

118.94 291.60 31.25 152.10 0.81 8.06

14.76 36.18 3.88 18.87 0.10

P-value ≈ 0.001

≈ 0.001 0.06 ≈ 0.001 0.76

l1 l2 l3 l4 Error

291.60 1 31.25 1 152.10 1 0.81 1 161.20 20

Results Discussion

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