Question

We want to evaluate a ferrous material in terms of its resistance. The factors that we want to evaluate are the effect of temperature and the Cooling Method in the resistance of the material. The temperatures evaluated were 850 and 1000 Celsius degrees, while the cooling method consisted of cooling: exposed to air, exposed to water, and water with ice. Four plates of appoximately 1 squared inch were cutted of the ferrous material. Each one of these plates were subdivided in 6 parts. Three parts of plate 1 were at a temperature ramdomoly selected for 20 minutes and then cooled down using one plate for each cooling method. Then the temperature is changed in the oven and the other 3 parts of plate 1 are processed. The process is repeated similarly for the other 3 plates.

Indicate how would be conducted an F-test for the sources of interest. (You may write the equations).

Answer 1

Solution

We have 4 plates each cut into 6 cut into 6 pieces. Thus, there are 24 pieces of plates.

Let Pij represent the jth piece cut off from plate i, i = 1, 2, 3, 4; j = 1, 2, 3, 4, 5, 6.

Then, allotment of these 24 pieces to the six treatment combinations can be represented as follows:

Cooling Method	Temperature
	850   Celsius degrees				1000 Celsius degrees
Air	P11	P21	P31	P41	P14	P24	P34	P44
Water	P12	P22	P32	P42	P15	P25	P35	P45
Water + Ice	P13	P23	P33	P43	P16	P26	P36	P46

  The resistance of these allotted plates form the response variables.

A two-way ANOVA with equal number of replications per treatment combination (4 per each of temperature-cooling method combination) would bring out significance of temperature. Cooling method and their interaction.

DONE