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### Transcript of 4 .. 5 4 .. 5 4 .. 5 6 7 6 7 6 7 Y 1 s f s f 2 3 2 3 2 ...

• F tests and the Extra Sum of Squares

Example:

Y = plaster hardness

s = sand content

f = fibre content

Model:

Yi = β0 + β1si + β2s 2 i + β3fi + β4f

2 i + β5si fi + ǫi

In matrix form: Y1... Yn

 =  1 s1 s

2 1 f1 f

2 1 s1f1

... ...

... ...

... ...

1 sn s 2 n fn f

2 n snfn

  β0...

β5



Richard Lockhart STAT 350: F tests

• Hypotheses to test

Questions:

◮ Do we need the S ∗ F term? ◮ Do we need the F or F 2 terms? ◮ Do we need the S or S2 terms?

To answer these questions we test

◮ Ho : β5 = 0 ◮ Ho : β3 = β4 = 0 (or perhaps Ho : β3 = β4 = β5 = 0) ◮ Ho : β1 = β2 = 0

Richard Lockhart STAT 350: F tests

• Technique: we fit a sequence of models:

(a) Original “full” model.

(b) The model with no interactions:

µi = β0 + β1si + β2s 2 i + β3fi + β4f

2 i

(c) The Sand only model:

µi = β0 + β1si + β2s 2 i

(d) The Fibre only model:

µi = β0 + β3fi + β4f 2 i

(e) The “Empty” model: µi = β0

Richard Lockhart STAT 350: F tests

• Each model has design matrix which is submatrix of full design matrix:

Y = [1|X1|X2|X3]



β0 β1 β2 β3 β4 β5

 + ǫ

The design matrices for the models a, b, c, d and e are given by

Xa = X

Xb = [1|X1|X2] Xc = [1|X1] Xd = [1|X2] Xe = [1]

Note that 1 is a column vector of n 1s.

Richard Lockhart STAT 350: F tests

• F tests

◮ Can compare two models easily if one is a special case of the other.

◮ Example: design matrix of first model is submatrix of second obtained by selecting subcolumns.

◮ Model (b) is a special case of (a), model (c) is a special case of (a) or (b) but models (c) and (d) are not comparable.

Richard Lockhart STAT 350: F tests

• Comparing two models: General Theory

◮ Consider matrix X partitioned into pieces X1 and X2.

X = [X1|X2] ◮ The Full Model is

Y = Xβ + ǫ

= [X1|X2] [

β1 β2

] + ǫ

= X1β1 + X2β2 + ǫ

◮ The Reduced model is

Y = X1β1 + ǫ

◮ Difference is term X2β2 which is 0 IF the null hypothesis Ho : β2 = 0 is true.

Richard Lockhart STAT 350: F tests

• Dimensions

◮ β has p parameters. ◮ βi has pi parameters with p1 + p2 = p.

Richard Lockhart STAT 350: F tests

• Testing Ho : β2 = 0

◮ Fit both full and reduced models; get:

Y = µ̂F + ǫ̂F

= µ̂R + ǫ̂R

◮ Subscript F refers to full model and R to reduced model. ◮ Get decomposition of data vector Y into sum of three

perpendicular vectors:

Y = µ̂R + (µ̂F − µ̂R) + ǫ̂F ◮ I showed

µ̂R ⊥ µ̂F − µ̂R µ̂R ⊥ ǫ̂F

µ̂F − µ̂R ⊥ ǫ̂F

Richard Lockhart STAT 350: F tests

• Resulting ANOVA table

Source Sum of Squares Degrees of Freedom

X1 ||µ̂R ||2 p1

X2|X1 ||µ̂F − µ̂R ||2 p2

Error ||ǫ̂||2 n − p

Richard Lockhart STAT 350: F tests

• F tests again

◮ In this table the notation X2|X1 means X2 adjusted for X1 or X2 after fitting X1.

◮ This table can now be used to test Ho : β2 = 0 by computing

F = MS(X2|X1)

MSE = ||µ̂F − µ̂R ||2/p2 ||ǫ̂||2/(n − p)

◮ Get P value from Fp2,n−p distribution. ◮ P value usually computed by software. ◮ Do level α test by comparing P to α.

Richard Lockhart STAT 350: F tests

• Another Formula for this F statistic

◮ Recall that ||µ̂R ||2 + ||ǫ̂R ||2 = ||Y ||2

and ||µ̂R ||2 + ||µ̂F − µ̂R ||2 + ||ǫ̂F ||2 = ||Y ||2

so that ||ǫ̂R ||2 = ||ǫ̂F ||2 + ||µ̂F − µ̂R ||2

◮ This makes

F = (ESSR − ESSF )/p2

ESSF/(n − p) =

ExtraSS/p2 ESSF/(n − p)

Richard Lockhart STAT 350: F tests

• Remarks

1. If the errors are normal then

ESSF σ2

∼ χ2n−p

2. If the errors are normal and Ho : β2 = 0 is true then

ExtraSS σ2

∼ χ2p2 3. ESSF is independent of the Extra SS. 4. SO:

ExtraSS/(σ2p2) ESSF /(σ2(n − p)) = F ∼ Fp2,n−p

Richard Lockhart STAT 350: F tests

• Example of the above: Multiple Regression

◮ Hardness, Yi , of plaster measured for each of 9 combinations of sand content and fibre content.

◮ Sand content si was set at one of 3 levels as was fibre content fi .

◮ All possible combinations tried, each on two batches of plaster.

◮ Here is an excerpt of the data: Sand Fibre Hardness Strength

0 0 61 34 0 0 63 16

15 0 67 36 15 0 69 19 30 0 65 28

...

Richard Lockhart STAT 350: F tests

• Models Fitted

◮ I fit submodels of the following ”Full” model:

Yi = β0 + β1si + β2s 2 i + β3fi + β4f

2 i + β5si fI + ǫi

◮ There are 25 = 32 possible submodels of the full model ◮ Many of these 32 models are not sensible, such as

Yi = β0 + β4f 2 i + ǫi

or Yi = β0 + β5si fi + ǫi

◮ Assume interaction term (β5si fi) is negligible unless each of S and F have some effect .

◮ Assume that quadratic terms will probably not be present unless linear terms are present.

◮ This limits the set of potential reasonable models. ◮ Fit each; error sum of squares in following table:

Richard Lockhart STAT 350: F tests

• Fitting various models

Model for µ Error SS Error df Full 81.264 12

β0 + β1si + β2s 2 i + β3fi + β4f

2 i 82.389 13

β0 + β1si + β2s 2 i + β3fi 104.167 14

β0 + β1si + β2s 2 i 169.500 15

β0 + β1si 174.194 16 β0 + β1si + β3fi + β4f

2 i 87.083 14

β0 + β1fi + β2f 2 i 189.167 15

β0 + β1fi 210.944 16 β0 + β1si + β3fi 108.861 15

Richard Lockhart STAT 350: F tests

• Hypotheses tested

◮ First question: do 2nd degree polynomial terms, that is, those involving β2, β4 and β5 need to be included?

◮ Compare top line with model containing only β0 + β1si + β3fi . ◮ The extra SS is 108.861-81.264 on 3 degrees of freedom

which gives a mean square of (108.861-81.264)/3= 9.199.

◮ The MSE is 81.264/12 = 6.772. ◮ Gives an F -statistic of 9.199/6.772=1.358 on 3 numerator

and 12 denominator degrees of freedom.

◮ P-value is 0.30 which is not significant. ◮ So we delete quadratic terms and consider the coefficients of

S and F .

Richard Lockhart STAT 350: F tests

• A t test and an F test

Q : Are the effects of S and F additive?

A : Test Ho : β5 = 0. ◮ There are two methods to carry out such a test:

1. A t test 2. A F test.

Fact : the F test is equivalent to a two sided t test.

◮ The t test uses

t = β̂5 − 0

σ̂β̂5 =

β̂5√ MSE

√ (XT X )−166

∼ t1,n−6

Richard Lockhart STAT 350: F tests

• t tests

◮ See Distribution Theory section for linear combinations:

β5 = [0, 0, 0, 0, 0, 1]︸ ︷︷ ︸ xT

 β0 β1 ...

β5

 and xT (XT X )−1x is the lower right hand corner entry in (XTX )−1, that is, (XTX )−166 .

◮ The F test uses

F = (ESSR − ESSF )/1 ESSFULL/(n − 6) ∼ F1,n−6 (= t

2)

Richard Lockhart STAT 350: F tests

• Testing for Main Effects

The Data

◮ Y = hardness of plaster. n = 18 batches. ◮ S = sand content. Values used 0%, 15% 30%. ◮ F = fibre content. Values used 0%, 25% 50%. ◮ Factorial design with 2 replicates.

Richard Lockhart STAT 350: F tests

• Comparison of Models “Full” model

Yi = β0 + β1si + β2s 2 i + β3fi + β4f

2 i + β5si fI + ǫi

Fitted Models, ESS, df for error

Model for µ ESS Error df Full 81.264 12 β0 + β1si + β2s

2 i + β3fi + β4f

2 i 82.389 13

β0 + β1si + β2s 2 i + β3fi 104.167 14

β0 + β1si + β2s 2 i 169.500 15

β0 + β1si 174.194 16 β0 + β1si + β3fi + β4f

2 i 87.083 14

β0 + β1fi + β2f 2 i 189.167 15

β0 + β1fi 210.944 16 β0 + β1si + β3fi 108.861 15 β0(empty model) 276.278 17

Richard Lockhart STAT 350: F tests

• Hypothesis tests:

1. Quadratic terms needed? Ho : β2 = β4 = β5 = 0.

◮ Extra SS = 108.861-81.264. ◮ F = [(108.861− 81.264)/3]/[81.264/12] = 1.358. ◮ Degrees of freedom are 3, 12 so P = 0.30, not significant.

Richard Lockhart STAT 350: F tests

• 3. Linear terms needed? There are several possible F -tests.

3.1 Compare full model to empty model.

F = (276.278− 81.264)/5/(81.264/12) = 5.76

so P is about .006. 3.2 Assume full model is now additive, linear model

β0 + β1si + β3fi .

Then

F = [(276.278− 108.861)/2]/[108.861/15] = 11.53

and P is about 0.0009. 3.3 Use estimate of σ2 from full model 3.4 But get extra SS from last comparison:

F =