Biostatistics for Animal Science

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32 Biostatistics for Animal Science Example: Determine the probability distribution of the number of female calves in three consecutive calvings. Assume that only a single calf is possible at each calving, and that the probability of having a female in a single calving is p = 0.5. The random variable y is defined as the number of female calves in three consecutive calvings. Possible outcomes are 0, 1, 2 and 3. The distribution is binomial with p = 0.5 and n = 3: ⎛ 3⎞ y 3− y P( y) = ⎜ ⎟( 0. 5) ( 0. 5) y = 0,1,2,3 ⎝ y⎠ Possible values with corresponding probabilities are presented in the following table: y p(y) 0 1 2 3 ⎛3⎞ 0 3 ⎜ ⎟( 0. 5) ( 0. 5) = 0. 125 ⎝0⎠ ⎛3⎞ 1 2 ⎜ ⎟( 0. 5) ( 0. 5) = 0. 375 ⎝1⎠ ⎛3⎞ 2 1 ⎜ ⎟( 0. 5) ( 0. 5) = 0. 375 ⎝2⎠ ⎛3⎞ 3 0 ⎜ ⎟( 0. 5) ( 0. 5) = 0. 125 ⎝3⎠ The sum of the probabilities of all possible values is: Σi p(yi) = 1 The expectation and variance are: µ = E(y) = np = (3)(0.5) = 1.5 σ 2 = var(y) = npq = (3)(0.5)(0.5) = 0.75 Another example: In a swine population susceptibility to a disease is genetically determined at a single locus. This gene has two alleles: B and b. The disease is associated with the recessive allele b, animals with the genotype bb will have the disease, while animals with Bb are only carriers. The frequency of the b allele is equal to 0.5. If a boar and sow both with Bb genotypes are mated and produce a litter of 10 piglets: a) how many piglets are expected to have the disease; b) what is the probability that none of the piglets has the disease; c) what is the probability that at least one piglet has the disease; d) what is the probability that exactly a half of the litter has the disease. The frequency of the b allele is 0.5. The probability that a piglet has the disease (has the bb genotype) is equal to (0.5)(0.5) = 0.25. Further, the probability that a piglet is healthy is 1 - 0.25 = 0.75. Thus, a binomial distribution with p = 0.25 and n = 10 can be used. a) Expectation = np = 2.5, that is, between two and three piglets can be expected to have the disease.
⎛10⎞ 0 10 0 10 b) P ( y = 0) = ⎜ ⎟ p q = 1( 0. 25) ( 0. 75) = 0. 056 ⎝ 0 ⎠ c) P(y ≥1) = 1 – P(y = 0) = 1 – 0.056 = 0.944 ⎛10⎞ 5 5 10! 5 5 d) P ( y = 5) = ⎜ ⎟ p q = ( 0. 25) ( 0. 75) = 0. 058 ⎝ 5 ⎠ 5! 5! Chapter 3 Random Variables and their Distributions 33 Third example: A farmer buys an expensive cow with hopes that she will produce a future elite bull. How many calves must that cow produce such that the probability of having at least one male calf is greater than 0.99. Solution: Assume that the probability of having a male calf in a single calving is 0.5. For at least one male calf the probability must be greater than 0.99: P(y ≥ 1) > 0.99 Using a binomial distribution, the probability that at least one calf is male is equal to one minus the probability that n calves are female: Thus: n 0 ⎟ ⎛ ⎞ ⎝ ⎠ P(y ≥ 1) = 1 – P(y < 1) = 1 – P(y = 0) = 1 – ()() n 1 0 1 ⎜ 2 2 ⎛n ⎞ − ⎜ ⎟ ⎝0 ⎠ 1 2 2 1 0 1 n ( ) ( ) > 0. 99 Solving for n in this inequality: n > 6.64 Or rounded to an integer: n = 7 = 1 2 3.2.4 Hyper-geometric Distribution n < 0.01 Assume a set of size N with R successes and N – R failures. A single trial has only two outcomes, but the set is finite, and each trial depends on the outcomes of previous trials. The random variable y is the number of successes in a sample of size n drawn from the source set of size N. Such a variable has a hyper-geometric probability distribution: where: ⎛ R⎞⎛ N − R⎞ ⎜ ⎟⎜ ⎟ ⎝ y ⎠⎝ n − y P ( y) = ⎠ ⎛ N ⎞ ⎜ ⎟ ⎝ n ⎠ y = random variable, the number of successful trials in the sample n = size of the sample
Page 1: Biostatistics for Animal Science
Page 4 and 5: CABI Publishing is a division of CA
Page 6 and 7: vi Biostatistics for Animal Science
Page 8 and 9: viii Biostatistics for Animal Scien
Page 10 and 11: x Biostatistics for Animal Science
Page 12 and 13: Preface This book was written to se
Page 15 and 16: Chapter 1 Presenting and Summarizin
Page 17 and 18: Brown 5% Simmental 76% Holstein 19%
Page 19 and 20: Number of calves 18 16 14 12 10 8 6
Page 21 and 22: Σi y 2 i = 1 2 + 3 2 + 6 2 = 46 Co
Page 25 and 26: 1.4.5 Measures of Relative Position
Page 29 and 30: Chapter 2 Probability The word prob
Page 31 and 32: 2.2.1 Multiplicative Rule Chapter 2
Page 33 and 34: Chapter 2 Probability 19 event. As
Page 35 and 36: Chapter 2 Probability 21 P(A ∩ B)
Page 37 and 38: 2.4 Bayes Theorem First calf Second
Page 39 and 40: We define: P(A1) = 0.6 is the proba
Page 41 and 42: Chapter 3 Random Variables and thei
Page 45: 3.2.3 Binomial Distribution Chapter
Page 49 and 50: E(y) = µ = λ and Var(y) = σ 2 =
Page 53 and 54: 2.5% µ−1.96σ µ−σ µ µ+σ F
Page 61 and 62: e f ( y) = − 1 2 −1 ( y −µ )
Page 65 and 66: The expectation and variance of the
Page 67 and 68: Chapter 4 Population and Sample A p
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Page 71 and 72: 5.2 Maximum Likelihood Estimation C
Page 73 and 74: Chapter 5 Estimation of Parameters
Page 75 and 76: By setting both terms to zero the e
Page 77 and 78: Chapter 5 Estimation of Parameters
Page 79 and 80: Chapter 6 Hypothesis Testing The fo
Page 81 and 82: µ 0 y − z ≈ s n Chapter 6 Hypo
Page 83 and 84: Known values are: y = 4000 kg σ =
Page 85 and 86: α -zα Figure 6.7 The critical val
Page 87 and 88: s( y 1− y2 ) = 2 s1 s + n n 1 2 2
Page 89 and 90: The estimated pooled variance is: s
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Page 93 and 94: ( y − y2 ) − 0 ( 11. 857 − 21
Page 95 and 96: Normal Approximation Z 2.0626 One-S
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pˆ − pˆ 1 The estimator has var
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2 χ [ yi E( yi ) ] E( y ) ∑ −
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∑ ∑ y i i p0 = i = 1,..., k n i
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Chapter 6 Hypothesis Testing 89 Exp
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Chapter 6 Hypothesis Testing 91 An
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Decision of a statistical test H 0
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Chapter 6 Hypothesis Testing 95 nor
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Chapter 6 Hypothesis Testing 97 One
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0.45 0.40 0.35 0.30 0.25 0.20 0.15
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SAS output: alpha n mi0 mi1 stdev d
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6.12 Sample Size Chapter 6 Hypothes
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SAS program: DATA a; DO n = 2 TO 10
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Chapter 6 Hypothesis Testing 107 PR
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Chapter 7 Simple Linear Regression
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The measurements of the dependent v
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2 SSRES s = = MSRES = 115. 826 n
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Confidence intervals follow the cla
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spread yi about y A) y y * * * * *
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SS REG ( SSxy ) = SS xx 2 Chapter 7
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E(MSREG) = σ 2 + β1 2 SSxx Chapte
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The likelihood function when H0 is
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2 R = SS SS REG TOT Chapter 7 Simpl
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= (X'X) -1 X'y A maximum likelihood
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DATA a; alpha=0.05; n=6; b=7.52941;
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Chapter 8 Correlation 147 of correl
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has a t-distribution with (n - 2) d
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Chapter 8 Correlation 151 under H0:
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Exercises Chapter 8 Correlation 153
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The assumptions of the model are: 1
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where: Chapter 9 Multiple Linear Re
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The residual mean square, which is
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The null and alternative hypotheses
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Chapter 9 Multiple Linear Regressio
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yi = the weight of bull i x1i = the
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SAS program: DATA bulls; INPUT weig
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9.6.1 Analysis of Residuals Chapter
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y * * 1 * * * * * * * * * 2 * * * 4
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where: Chapter 9 Multiple Linear Re
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SAS program: DATA bull; INPUT weigh
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where: SSRES = residual mean square
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Chapter 10 Curvilinear Regression I
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Chapter 10 Curvilinear Regression 1
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The variance estimate is s 2 = 210.
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SAS program: DATA a; INPUT age weig
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Weight (g) 1000 800 600 400 200 0 0
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Chapter 11 One-way Analysis of Vari
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From the assumptions it follows: Ch
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By defining: SS SS SS TOT TRT RES =
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1) Total sum: Σi Σj yij = (270 +
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2 L( µ , σ | y) = The log likelih
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And as shown previously: Thus: 2
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The hypotheses can be tested using
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ANOVA table: Chapter 11 One-way Ana
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F 3 = 2890 2 296. 67 = 4. 871 Chapt
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SAS output: Chapter 11 One-way Anal
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The effect of sow 1 is: ˆ 1 τ = 0
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The log likelihood that is to be ma
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Solution for Random Effects Chapter
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11.3.1.2 Estimating Parameters Chap
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The sums of squares needed for test
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The Xr'Xr matrix and its inverse ar
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11.3.2 The Random Effects Model 11.
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11.3.2.4 Restricted Maximum Likelih
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Chapter 12 Concepts of Experimental
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12.5 Required Number of Replication
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SAS output: alfa df1 df2 n power 0.
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Chapter 13 Blocking 273 environment
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1) Total sum: Σi Σj yij 2) Correc
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Chapter 13 Blocking 277 Example: Th
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13.1.3 SAS Example for Block Design
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Animal No. (Treatment) Blocks I II
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Mean square for treatments: Mean sq
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= s y − y MS ij . i' j' . RES ⎛
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1347. 90 F = = 7. 67 175. 83 F test
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Dependent Variable: d_gain i/j 1 2
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Chapter 13 Blocking 291 the adjuste
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Exercises Chapter 13 Blocking 293 1
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µ = the overall mean τi = the fix
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BLOCK II The hypotheses are: Chapte
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The results are shown in the ANOVA
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Chapter 14 Change-over Designs 301
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3) Total (corrected) sum of squares
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3) Total (corrected) sum of squares
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Least Squares Means Adjustment for
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Chapter 15 Factorial Experiments A
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1) Total sum: Σi Σj Σk yijk 2) C
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Chapter 15 Factorial Experiments 31
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5) Sum of squares for vitamin II: (
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Chapter 15 Factorial Experiments 32
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Chapter 16 Hierarchical or Nested D
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6) Sum of squares within factor B (
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3) Total sum of squares: Chapter 16
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SAS output of the NESTED procedure:
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Chapter 17 More about Blocking If t
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Chapter 17 More about Blocking 333
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SAS program: DATA steer; INPUT pen
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SAS output of the GLM procedure: De
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Chapter 17 More about Blocking 339
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The ANOVA table is: Source Degrees
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The model for this design is: where
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Chapter 18 Split-plot Design 345 su
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Least Squares Means Chapter 18 Spli
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Source Degrees of freedom Factor A
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Effects Means Estimators Standard e
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Chapter 18 Split-plot Design 353 Ex
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Chapter 19 Analysis of Covariance A
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DATA gain; INPUT treatment $ initia
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Chapter 19 Analysis of Covariance 3
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For males (M) the model is: E(yi) =
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Chapter 19 Analysis of Covariance 3
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Chapter 20 Repeated Measures Experi
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Chapter 20 Repeated Measures 367 Ex
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8 9 1 1.3 8 10 1 1.3 8 11 1 1.3 8 1
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SAS output: Covariance Parameter Es
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where: Chapter 20 Repeated Measures
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Chapter 20 Repeated Measures 375 Ex
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σ ⎡ σ b σ ⎡ ⎤ 0 b0b1 [ 1 t
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⎡1 ⎢ ⎢ 1 R ˆ = ⎢1 ⎢ ⎣1
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⎡1 ⎢ ˆR ⎢ 1 = ⎢1 ⎢ ⎣1
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Chapter 20 Repeated Measures 383
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Chapter 21 Analysis of Numerical Tr
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21.2.1 SAS Example for Polynomial C
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Chapter 22 Discrete Dependent Varia
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where: pi = the proportion with mas
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Analysis Of Parameter Estimates Cha
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SAS program: DATA a; INPUT n y farm
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Farm The model is: where: Total no.
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LR Statistics For Type 1 Analysis C
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Solutions of Exercises 1.1. Mean =
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Appendix A: Vectors and Matrices A
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⎡a A + B = ⎢ ⎢ a ⎢⎣ a A +
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a = (X'X) -1 X'y Appendix A: Vector
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Critical Values of Student t Distri
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Critical Values of Chi-square Distr
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Critical Values of F Distributions,
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Critical Value of F Distributions,
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Critical Values of the Studentized
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References 437 Gianola, D. and Hamm
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Subject Index accuracy, 266 Akaike
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y hypothesis testing, 56 by paramet
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hypothesis test, 186 t test, 186 ra
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of regression estimators, 119, 159
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