# FF

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