Complete Randomized Design lecture - 02

 Complete Randomized Design

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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

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