The 2k full factorial designs of the experiment is a specific case of the full factorial where each k factor can assume only two values. In this case, 2k is all the possible combinations of factors.


For the execution of this test, we can use the Yates algorithm. For setup, this test, the first step is write the combination of factor in this way:

  • Assuming that each factor of the test can assume high or low value, so we are + for high and – for low;
  • Now write all the combination of value, for example for a three factor (a,b,c) we have:
    1. – – – that means a=low, b=low, c=low;
    2. + – – that means a=high, b=low, c=low. It can be also rappresented as a;
    3. – + – that means a=low, b=high, c=low. It can be also rappresented as b;
    4. + + – that means a=high, b=high, c=low. It can be also rappresented as a,b;
    5. – – + that means a=low, b=low, c=high. It can be also rappresented as c;
    6. + – + that means a=high, b=low, c=high. It can be also rappresented as a,c;
    7. – + + that means a=low, b=high, c=high. It can be also rappresented as b,c;
    8. + + + that means a=high, b=high, c=high. It can be also rappresented as a,b,c;


Now we need to make the calc in image1:

image1 - Yates algorithm calc
image1 – Yates algorithm calc

where:

  • Effect: represent the combination of high value of the n=3 independent variable as explained first;
  • Run1 and Run2: is two-run (r=2) of test and represent the value observed at the end of the test (our Y);
  • Sum: is the sum of run1 and run2 for each effet;
  • Col1: for the first 4 value is the sum of adjacent pair of summed responses, for example (E2) 82 = (D2) 41 + (D3) 41; for the last 4 value is the difference of adjacent pair of response, for example (E9) 11 = (D9) 37 – (D8) 26;
  • Col2: is made in the same way of Col1 but on the value of col1;
  • Col3: is made in the same way of Col1 but on the value of col2;
  • Estimate (of impact): is Col3 / r*2^n, where n is the number of indipendent variable, r is the number of run, 2 is the possibile value (+ or -)


In this example, it seems that C has a high estimated impact on the dependant variable. To assess if this impact has a statistical significance, we can make an ANOVA hypothesis test for each combination of values.

In image2 we can look at the calc for the ANOVA test:

image2 - Anova test
image2 – Anova test

where:

  • SS is the sum of squared for all combinations. Is calculated as (COL3^2)/(r*(2^n));
  • Response squared run 1 is simply the square of Run1, the same for the run2;
  • Df is the degree of freedom, that is 1 for each vale except for the error that is 2^n *(r-1)
  • Sum of response squarred is only the sum of the two respons squared column (it’s a single value in A15);
  • MS = SS / DF exepct for the MS of the error that is MSerror= sum of all SS / DF
  • F (the statistics) if finally calculated as: F=MS/MSError


Finally we have our F statistics and we need to get the critical value from the F statistics table. In this case we have df1 =1 and df2 = 8 that is 2^n*(r-1) and we select alpha = 0,05. So F(df=1; df2=8; alpha= 0,05) = 5,3177 so the test denote that only the factor C have a significant impact.

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