Math 558 Lecture #5
Statistical Analysis for CRD description of some concepts from lecture 4
We considered the following model for the response variable Yiw in lecture 4.
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Yiw = μi + εiw
Here μi is the mean response for the ith treatment. The responses are normally distributed. i. e Yiw ∼ N(μi, σ2). Equivalently εiw ∼ N(0, σ2). Let us consider the case of equal replication (r1 = r2 = ..ri = ..rt). The sample mean of the data for the ith treatment is
The overall mean is
1t y ̄=t y ̄
1 ri y ̄=r y
i. ∑ iw i w=1
.. ∑ i. i=1
RCD: Statistical Analysis
We want to estimate the model parameters in β. One commonly used estimation technique under the model assumptions given in lecture 4 is called ‘Least squares estimation”. In this technique we minimize the sum of the squared errors. We can use the following expression for this purpose,
SSE = ∑ ∑(yiw −μi)2
1The least squares estimator for the population parameters can be
obtained by solving the following matrix equation. (X′ X)−1X′ y = βˆ
The resulting estimators are μˆ = y ̄ for i = 1, 2, ..t i i.
1I am skipping the further details
Statistical Analysis
We also discussed an example of a CRD with 3 treatments replicated twice. There are six plots and we have the following equations for the rersponse on each plot.
Y11 = μ1 + ε11 Y12 = μ1 + ε12 Y21 = μ2 + ε21 Y22 = μ2 + ε21 Y31 = μ3 + ε31 Y32 = μ3 + ε32
Statistical Analysis
y11 1 y12 1
0 0 ε11
0 0 μ1 ε12
1 0 ε21
1 0 × μ2 + ε22 (1)
μ3 0 1 ε31
y21 0 y22 = 0
y31 0
y32 001 ε32 Y = Xβ + ε
Statistical Analysis
The values in Y come from the observed responses. The product
(X′ X)−1X′ y will give you βˆ which is the matrix of estimators. From the previous slide
1 0 0 110000 100 200
0 0 1 1 0 0×0 1 0=0 2 0
000011 002 0 0 1
Statistical Analysis
1/2 0 0 (X′X)−1= 0 1/2 0
Note: The inverse of a diagonal matrix is obtained by replacing each
element in the diagonal with its reciprocal. y11
1 1 0 0 0 0 y12 y11+y12 y21
0 0 1 1 0 0×y22 = y21 +y22 (2) 0 0 0 0 1 1 y21+y22
y31 y32
Statistical Analysis
(X X)−1X y = y2. = βˆ
Linear Model for CRD
The above model can be written as
Yiw = μ+τi + εiw
Here τi = μi − μ is the effect of the ith treatment. The responses are normally distributed. i. e Yiw ∼ N(μ + τi, σ2). Equivalently
εiw ∼ N(0, σ2). In the matrix form the model is
Y = Xβ + ε
Let us consider a CRD with three treatments, each with four replications. Then the effects model in the matrix form can be written as
y11 y
12 y13
y14 y 21
y22 1 = y23 1
y24 y31 y32 y33
0 τ1 ε22
× + (3)
ε31 ε32 ε33
1100 ε11 1100 ε
12 1100 ε13
1100 ε14 1010 μ ε
1 1 1 1
0 τ2 ε23 0 τ3 ε24
y34 1001 ε34
Statistical Analysis
The matrix β is the matrix of model parameters. The least squares estimators of the model parameters are obtained by solving the normal equations X′ Xβ = X′ y The matrix X′ X is singular therefore, cannot be inverted. The R command lm solves this problem by dropping the column that corresponds to the first treatment. The matrix that will be
1 0 0 1 0 0
1 0 0 1 0 0
1 1 0
1 1 0 used in the calculations is therefore, X1 = 1 1 0
1 1 0 1 0 1 1 0 1 1 0 1
Statistical Analysis
Solving the above system of equations with X1 we have (X′ X)−1X′ y = βˆ
μˆ + τˆ 1
Where βˆ = τˆ − τˆ 21
τˆ − τˆ 31
The first treatment is treated as a control and all other treatments are compared to it.
Statistical Analysis
Rise Time 35
Loaf Heights 4.4, 5.0,5.5,6.75 6.5, 6.5, 10.5,9.5 9.75,8.75,6.5,8.25
Table: Bread Rise Experiment
Statistical Analysis
lm(formula = height time, data = bread) Coefficients:
Estimate Std. Error t value
Pr(>|t|) 5.65e-05 *** 0.0288 * 0.0262 *
(Intercept) time40 time45
5.4375 2.8125 2.8750
0.7655 1.0825 1.0825
7.104 2.598 2.656
— Signif. codes: 0 ‘***’ 0.001 ‘**’ 0.01 ‘*’ 0.05 ‘.’ 0.1 ‘ ’ 1
Residual standard error: 1.531 on 9 degrees of freedom Multiple R-squared: 0.5056, Adjusted R-squared: 0.3958 F-statistic: 4.602 on 2 and 9 DF, p-value: 0.042
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