We want to evaluate a ferrous material in terms of its resistance. The factors t
ID: 3350901 • Letter: W
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).
Explanation / Answer
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
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
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