Build the analysis of variance table for the following 2 k factorial using: a) Y
ID: 3269823 • Letter: B
Question
Build the analysis of variance table for the following 2k factorial using:
a) Yates Method
b) Standard computational formulas
c) Software
A completely randomized experiment is run to determine the influence of three factors, A, B, and C, in a plant on the ppm of a particular byproduct. Each of the three factors are considered at two levels. Analyze according to the instructions above.
A1 A2 B1 B2 B3 B4 C1 C2 C1 C2 C1 C2 C1 C2 1595 1745 1835 1838 1573 2184 1700 1717 1578 1689 1823 1614 1592 1538 1815 1806Explanation / Answer
a) Yates Method
Yates method gives a systematic approach to find the sum of squares.
In this experiment, we have three factors A, B, C and two levels each 1 and 2. In a 2k factorial method, k is the number of factors which is 2 in this case. The number of observations for each combination of factors are 2 which is denoted by n subsequently.
To find the sum of squares for each effect, following method is used -
1. First write the treatment combinations in the standard order in the column at the beginning of table, called as treatment column.
2. Find the total yield for each treatment. Write this as second column of the table, called as yield column.
3. Obtain columns (1), (2),…,(n) successively (i) obtain column (1) from yield column
a) upper half is obtained by adding yields in pairs.
b) second half is obtained by taking differences in pairs, the difference obtained by subtracting the first term of pairs from the second term.
(ii) The columns (2), (3),…, (n) are obtained from preceding ones in the same manner as used for getting (1) from the yield columns. 4.
This process of finding columns is repeated K times in 2K factorial experiment.
Hence, with the above method, the following table is generated. The last column gives the sum of squares for each effect.
Total SS = Total of all the SS due to main and interaction effects.
b) Standard Computation formulas
The sums of squares SST and SSE are computed for the ANOVA are used to form two mean squares, one for treatments and the second for error. These mean squares are denoted by MST and MSE, respectively. These are typically displayed in a tabular form, known as an ANOVA Table. The ANOVA table also shows the statistics used to test hypotheses about the population means.
The mean squares are formed by dividing the sum of squares by the associated degrees of freedom.
Let N=total number of observations, k = number of columns, n = number of entries in one column
The degrees of freedom for treatment are -
DFT=k1,
The degrees of freedom for error are -
DFE=Nk.
The corresponding mean squares are:
MST=SST/DFT
MSE=SSE/DFE
The F-test statistic, used in testing the equality of treatment means is: F=MST/MSE.
The calculations are displayed in an ANOVA table, as follows:
Source
SS
DF
MS
F
Treatments
SST
k1
SST/(k1)
MST/MSE
Error
SSE
Nk
SSE/(Nk)
Total (corrected)
SS
N1
Some authors prefer to use "between" and "within" instead of "treatments" and "error", respectively.
In the given question, N = 16, k = 2, n = 8
Therefore, table of calculation is as follows -
Anova table is as follows -
Treatment Notation Experiment 1 Experiment 2 Yield from n experiments Column 1 Column 2 Column 3 Effect = Column 3 SS(Effect) = ((Effect)^2)/(n*8) A1B1C1 1 1595 1578 3173 6607 13717 27642 [M] 47755010.25 A1B1C2 a 1745 1689 3434 7110 13925 620 [A] 24025 A1B1C3 b 1835 1823 3658 6887 55 654 [B] 26732.25 A1B1C4 ab 1838 1614 3452 7038 565 -1016 [AB] 64516 A1B1C5 c 1573 1592 3165 261 503 208 [C] 2704 A1B1C6 ac 2184 1538 3722 -206 151 510 [AC] 16256.25 A1B1C7 bc 1700 1815 3515 557 -467 -352 [BC] 7744 A1B1C8 abc 1717 1806 3523 8 -549 -82 [ABC] 420.25Related Questions
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