The following table gives the production costs, in cents per pound, of broilers
ID: 3159226 • Letter: T
Question
The following table gives the production costs, in cents per pound, of broilers produced in three production methods, A, B, C, as represented by by ten producers randomly selected from each method. Do these data provide sufficient evidence to indicate a difference in mean cost among three methods? Let alpha = 5%. The following table shows the results, in miles per gallon, of an experiment conducted to compare three brands of gasoline. Each brand was used with seven different cars of the same weight and engine size, driven under similar conditions. Do these data provide sufficient evidence at the 0.01 level of significance to indicate a difference between brands of gasoline?Explanation / Answer
I ) H0: Mean cost among three methods are equal with one another
H1: Atleast one of the Mean cost is not equal to rest of the methods.
Linear Model for One way ANOVA Yij = µ + ti + eij
Correction factor (CF) = (GT)2/n
Total sum of squares (Total SS) = ( Y112+Y122 + …+ Y3,102)-CF
Treatment sum of squares (TSS)
= [ First Column Total2 /10+ Second Column Total2 /10 + Third Column Total2 /10 ] – CF
Error sum of squares(ESS)
= Total SS – TSS
Calculation:
GT = 352
CF = 3522 /30 = 4130.13
Total Sum of Square = 112 + 102+ 122+...+ 142 +142 +152 - CF
Total Sum of Square = 4256-4130.13
TSS = 125.87
Treatment sum of square = (1082 + 1072 +1372 )/10 - CF
Tr.SS = 4188.2-4130.13
Tr.SS = 58.07
Error Sum of Square = TSS-Tr.SS
ESS = 125.87-58.07
ESS= 67.8
ANOVA TABLE
Conclusion:
From the ANOVA table, it is clear that the F critical is lesser than the F calculated, so we have reasonable evidence to reject null hypothesis at 5% level of significance. So the mean cost for three methods are not equal to one another.
II) H0: The effect of different brands of gasoline in giving mileage is same.
H1: The effect of different brands of gasoline in giving mileage is different with atleast one of the brand.
Linear Model for One way ANOVA Yij = µ + ti + eij
Correction factor (CF) = (GT)2/n
Total sum of squares (Total SS) = ( Y112+Y122 + …+ Y3,102)-CF
Treatment sum of squares (TSS)
= [ First Column Total2 /10+ Second Column Total2 /10 + Third Column Total2 /10 ] – CF
Error sum of squares(ESS)
= Total SS – TSS
Calculation:
GT = 419
CF = 4192 /21 = 8360.048
Total Sum of Square = 142 + 192+ 192+...+ 232 +252 +232 - CF
Total Sum of Square = 8563-8360.048
TSS = 202.95
Treatment sum of square = (1202 + 1352 +1642 )/7 - CF
Tr.SS = 8503-8360.048
Tr.SS = 142.95
Error Sum of Square = TSS-Tr.SS
ESS = 202.95-142.95
ESS= 60.00
ANOVA TABLE
Conclusion:
From the ANOVA table, it is clear that the F critical is lesser than the F calculated, so we have reasonable evidence to reject null hypothesis at 1% level of significance. So the mean mileage of three brands of gasoline is different with one another.
S.No A B C 1 11 12 12 2 10 10 13 3 12 9 15 4 10 11 14 5 11 10 14 6 9 12 11 7 8 12 15 8 13 14 14 9 12 8 14 10 12 9 15 Total 108 107 137 Mean 10.8 10.7 13.7Related Questions
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