Biostatistics for Animal Science

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22 Biostatistics for Animal Science Find the following probabilities: a) the first calf is spotted, b) the first calf is either black or red, c) the second calf is black if the first was spotted, d) the first calf is spotted and the second black, e) the first two calves are spotted and black, regardless of order. Solutions: a) There is a total of 10 calves, and 5 are spotted. The number of favorable outcomes are m = 5 and the total number of outcomes is n = 10. Thus, the probability that a calf is spotted is: P(spotted) = 5 /10 = 1 /2 b) The probability that the first calf is either black or red is an example of union. P(black or red) = P(black) + P(red) = 2 /10 + 3 /10 = 5 /10 = ½. Also, this is equal to the probability that the first calf is not spotted, the complement of the event described in a): P(black ∪ red ) = 1 - P(spotted) = 1 - 1 /2 = 1 /2 c) This is an example of conditional probability. The probability that the second calf is black if we know that the first one was spotted is the number of black calves (2) divided by the number of calves remaining after removing a spotted one from the pen (9): P(black | spotted) = 2 /9 d) This is an example of the probability of an intersection of events. The probability that the first calf is spotted is P(spotted) = 0.5. The probability that the second calf is black when the first was spotted is: P(black | spotted) = 2 /9 The probability that the first calf is spotted and the second is black is the intersection: P[spotted ∩ (black | spotted)] = ( 5 /10) ( 2 /9) = 1 /9 e) We have already seen that the probability that the first calf is spotted and the second is black is: P[spotted ∩ (black | spotted)] = 1 /9. Similarly, the probability that the first is black and the second is spotted is: P[black ∩ (spotted | black)] = ( 2 /10) ( 5 /9) = 1 /9 Since we are looking for a pair (black, spotted) regardless of the order, then we have either (spotted, black) or (black, spotted) event. This is an example of union, so the probability is: P{[spotted ∩ (black | spotted)] ∪ [black ∩ (spotted | black)]} = ( 1 /9) + ( 1 /9) = 2 /9 We can illustrate the previous examples using a tree diagram:
2.4 Bayes Theorem First calf Second calf 1 black ( 2 /10) ( 1 /9) 2 black ( 2 /10) 3 red ( 2 /10) ( 3 /9) 5 spotted ( 2 /10) ( 5 /9) 2 black ( 3 /10) ( 2 /9) 3 red ( 3 /10) 2 red ( 3 /10) ( 2 /9) 5 spotted ( 3 /10) ( 5 /9) 2 black ( 5 /10) ( 2 /9) 5 spotted ( 5 /10) 3 red ( 5 /10) ( 3 /9) 4 spotted ( 5 /10) ( 4 /9) Chapter 2 Probability 23 Bayes theorem is useful for stating the probability of some event A if there is information about the probability of some event E that happened after the event A. Bayes theorem is applied to an experiment that occurs in two or more steps. Consider two cages K1 and K2, in the first cage there are three mice, two brown and one white, and in the second there are two brown and two white mice. Each brown mouse is designated with the letter B, and each white mouse with the letter W. Cage K1 Cage K2 B,B,W B,B,W,W A cage is randomly chosen and then a mouse is randomly chosen from that cage. If the chosen mouse is brown, what is the probability that it is from the first cage? The first step of the experiment is choosing a cage. Since it is chosen randomly, the probability of choosing the first cage is P(K1) = ( 1 /2). The second step is choosing a mouse from the cage. The probability of choosing a brown mouse from the first cage is P(B|K1) = ( 2 /3), and of choosing a brown mouse from the second cage is P(B|K2) = ( 2 /4). The probability that the first cage is chosen if it is known that the mouse is brown is an example of conditional probability: P( K1 ∩ B) ( K |B) = P( B) P 1 The probability that the mouse is from the first cage and that it is brown is: P(K1 ∩ B) = P(K1) P(B | K1) = ( 1 /2) ( 2 /3) = ( 1 /3)
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
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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: Chapter 2 Probability 21 P(A ∩ B)
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 and 46: 3.2.3 Binomial Distribution Chapter
Page 47 and 48: ⎛10⎞ 0 10 0 10 b) P ( y = 0) =
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
Page 69 and 70: Chapter 4 Population and Sample 55
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
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s( y 1− y2 ) = 2 s1 s + n n 1 2 2
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The estimated pooled variance is: s
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Chapter 6 Hypothesis Testing 77 tha
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( y − y2 ) − 0 ( 11. 857 − 21
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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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