Estimation of Parameters in Complete Randomized Design lecture - 04

 

Estimation of Parameters

in

Complete Randomized Design

Consider the linear statistical model of CR design:

Yij = μ + τj + ϵij         i=1, 2, ..., r

                                         j=1, 2, ..., t   

Where:

μ is the true mean effect, τj represents the effects of jth treatment and ϵij denoted random error. Yij is the yield of j^th treatment in the i^th experimental unit.

Further;

ϵij  ~ NIID (0, σ2)

that,s 

E ( ϵij ) = 0 and Var ( ϵij ) = σ2

Let

• Read More: Mathematical Expectation in CRD

Post Hock Tests in Complete Randomized Design lecture - 03

 

Post Hock Tests in Complete Randomized Design

Example

Determinations are made on the yield using three methods of catalyzing a chemical process.

Methods	observations
1	47.2	49.8	48.5	48.7
2	50.1	49.3	51.5	50.9
3	49.1	53.2	51.2	52.8	52.3

Write the appropriate statistical model and test the hypothesis that the three methods differ significantly at 5 % level of significance? Further examine which methods are significant by using Post Hock tests.

Solution:

Let Yij of three catalyzing process is considered the treatments and nothing is stated about the other variation in the statement of the problem, so it is pure CRD Problem and represent by the following linear statistical model.

Yij = μ + τj + ϵij

Statement of hypothesis

i. H0 : μ1 = μ2 = μ3    Vs.   H1 : μ1 ≠ μ2 ≠ μ3

ii. The significance level; α = 0.05

iii. The test statistic: CR Design

F =MST / MSE ~ F (2, 10)

vi. Reject H0, when F ≥ F0.05 (2, 10) = 4.10

v. Computation:

Observation	Method 1	Method 2	Method 3
1	47.2	50.1	49.1
2	49.8	49.3	53.2
3	48.5	51.5	51.2
4	48.7	50.9	52.8
5			52.3
T.j	194.2	201.8	258.6	654.6
T.j ^2	37713.64	40723.24	66873.96

ANOVA Table

Source of Variation	d.f	Sum of Squares	Mean Squares	F - ratio
Treatment	2	22.392	11.196	6.59
Error	10	16.988	1.6988
Total	12	39.38

F = MST / MSE = 6.59

vi. Remarks: 

As F calculated value falls in the rejection region, so we have not sufficient evidence to accept  thus we conclude that the three methods are significant

Post Hock Tests

1.    LSD Test

The method 1 and method 2 have identical effects but the method 3 effect is significantly different from method 1. 

Now if it is desired to construct (1 -  α ) % confidence interval for μi  -  μj

Then it is given by:

 

In the same manner construct confidence interval for other pairs of treatment. 

DMR Test

(Duncan’s Multiple Range Test)

The same result is provide by DMRT. 

• Read More: Estimation of Parameters

Complete Randomized Design lecture - 02

 Complete Randomized Design

continue.......

Example:

In order to study the storage conditions on the moisture content of white pine lumber, five storage methods were investigated with vary number of experimental units being stored under each condition. The data thus obtained are given below:

Storage conditions
1	2	3	4	5
7.3	5.4	8.1	7.9	7.1
8.3	7.4	6.4	9.5
7.6	7.1		10.5
8.4
8.3

Write an appropriate statistical model and state your null hypothesis about the storage conditions and stat your comments.

Solution: To compare 5 storage conditions only, we use CRD model.

$$

Yij = μ + τj + ϵij

 τj: Represents the storage condition considered as treatments,

Statement of hypothesis:

i.        H0 : τ1 = τ2 = τ3 = τ4 = τ5   Vs.  H1 : τ1 ≠ τ2 ≠ τ3 ≠ τ4 ≠ τ5 

ii. The significance level; α = 0.05

iii. The test statistic: CR Design

F = MST / MSE ~ F(4, 10)

iv.  Reject H0, when F >= F(4, 10) = 3.71 

v. Computation:

	Storage conditions
	1	2	3	4	5
	7.3	5.4	8.1	7.9	7.1
	8.3	7.4	6.4	9.5
	7.6	7.1		10.5
	8.4
	8.3
T . j	39.9	19.9	14.5	27.9	7.1	109.3
Sqr T . j	1592.01	396.01	210.25	778.41	50.41

ANOVA Table:

Source of variation	d.f	Sum of squares	Mean squares	F - ratio
Treatment	4	12.08	12.08 / 4 = 3.02	3.32
Error	9	8.21	8.21 / 9 = 0.91
Total	14	20.29

vi.  Remarks: The Calculated F value does not fall in the rejection region. We cannot reject the null hypothesis and conclude that the 5 storage conditions are identical.

Estimation of Missing value in CR Design

To estimate the single missing value in CR design as;

i.                   Substitute x for missing value.

ii.                 Calculate the sum of squares error for CRD in the usual manners.

iii.              Differentiate the sum of squares error with respect to x and equate to zero.

iv.               Solve for x.

 Example:

The data for 3 treatments are obtained by using CRD model. The last value for 3^rd treatment is not recorded. The data tabulated as:

Observation
1	2	4	4
2	5	2	5
3	3	2	1
4	1	3	?

Estimate the missing value.

Solution:

	A	B	C
	2	4	4
	5	2	5
	3	2	1
	1	3	x
T . j	11	11	10 + x	32 + x
Sqr T.j	121	121	(10 + x)^2

• Read More: POST HOCK tests in CRD