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  • #124254

    FF
    Participant

    The following statements are equivalent if you assume Zshift=1.5:
    Zst=6, Zlt = 4.5, Cpk=2, Ppk = 1.5, DPMOlt = 3.4
     

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    #114207

    FF
    Participant

    …… I must have some problems in pasting data from Minitab. I hope this time it works:

    Analysis of Variance (Balanced Designs)

    Factor     Type Levels Values
    Temp      fixed      4     1     2     3     4
    Oven     random      4     A     B     C     D

    Analysis of Variance for Sigma  

    Source       DF         SS         MS       F           P
    Temp          3    4107.09    1369.03   20.48      0.000
    Oven          3     142.59      47.53    0.71          0.569
    Temp*Oven     9     601.53      66.84   23.00    0.000
    Error        16      46.50       2.91
    Total        31    4897.72

     
     
     

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    #114206

    FF
    Participant

    Brookiep,
    you are right, I erroneously posted a wrong analysis where I had removed some data. This is the correct analysis with the Oven considered as a random factor:
      

    Analysis of Variance (Balanced Designs)
    Factor Type Levels Values
    Temp fixed 4 1 2 3 4
    Oven random 4 A B C D
    Analysis of Variance for Sigma
    Source DF SS   MS       F       P
    Temp 3 4107.09 1369.03 20.48   0.000
    Oven 3 142.59  47.53   0.71    0.569
    Temp*Oven 9 601.53 66.84 23.00 0.000
    Error 16 46.50 2.91
    Total 31 4897.72

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    #114133

    FF
    Participant

    Sorry for the bad format in the previous post
     

    Analysis of Variance (Balanced Designs)

    Factor     Type Levels Values
    Temper    fixed      3   350   400   450
    Oven typ random      4     A     B     C     D

    Analysis of Variance for Sigma Y

    Source             DF         SS         MS       F      P
    Temper              2    1035.75     517.87    8.48  0.018
    Oven typ            3     319.17     106.39    1.74  0.258
    Temper*Oven typ     6     366.58      61.10   25.28  0.000
    Error              12      29.00       2.42
    Total              23    1750.50

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    #114132

    FF
    Participant

    Thank you all for your replies,
    considering that the Oven has a small effect compared to Temperature, I’m wondering if the factor Oven can ve considered just a blocking factor. In fact the different ovens simply heat up the components under analysis, and maybe there souldn’t be any reasonable difference among them.
    Using a Balanced anova where the Oven is considered a random factor I get these results:
    Analysis of Variance (Balanced Designs)
     
    Factor Type Levels Values
    Temper fixed 3 350 400 450
    Oven typ random 4 A B C D
     
    Analysis of Variance for Sigma Y
    Source DF SS MS F P
    Temper 2 1035.75 517.87 8.48 0.018
    Oven typ 3 319.17 106.39 1.74 0.258
    Temper*Oven typ 6 366.58 61.10 25.28 0.000
    Error 12 29.00 2.42
    Total 23 1750.50
     
    In this case the Oven is not significant

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    #113838

    FF
    Participant

    Thanks K.Subbiah,
    these are the F values and the MSE in case the interaction is included
    Temp          3    4107.09    1369.03  471.06  0.000
    Oven          3     142.59      47.53   16.35  0.000
    Temp*Oven     9     601.53      66.84   23.00  0.000
    Error        16      46.50       2.91
    Total        31    4897.72
     
    This is the result if interaction term is NOT included
    Source      DF         SS         MS       F      P
    Temp         3    4107.09    1369.03   52.82  0.000
    Oven         3     142.59      47.53    1.83  0.167
    Error       25     648.03      25.92
    Total       31    4897.72

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    #100666

    FF
    Participant

    This is how I would organize data to compare processing time between regions (ri) and center types (cj)
    For example: i=3, j=4
     

    r1
    r2
    r3

    c1
    12
    8
    15

    c2
    7
    8
    13

    c3
    13
    12
    15

    c4
    7
    8
    6
    p=0.848

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    #100582

    FF
    Participant

    CHI SQUARE

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Viewing 8 posts - 1 through 8 (of 8 total)