# Stat 8120 Assignment 3 Solutions

```					                                             Stat 8120
Spring 2011
Assignment 3 Solutions

1: Answer the first question in problem 13.18 on page 540 of the textbook.

As we discussed in class, the F test statistics for testing the interaction effects are
relatively straightforward.

MSABC         MSAB          MSAC            MSBC
FABC          , FAB        , FAC        and FBC 
MSE           MSABC         MSABC           MSABC

The test statistics for testing the main effects are more complicated. There are no simple
ratios of mean squares that allow for testing the main effects. The approximate F test
   statistics for testing the main effects are

MSA  MSABC        MSB  MSABC            MSC  MSABC
FA                , FB              , and FC 
MSAB  MSAC        MSAB  MSBC            MSAC  MSBC

Questions 2 through 5 all refer to the following experiment: The first four columns in the
data set are the results of a factorial experiment run by Starbucks to identify the best
   setting to reduce leakage in the bags the coffee is packed in for retail sales. The
experiment is real, but the data have been changed to avoid disclosing any proprietary
information. The predictor variables are the viscosity of the adhesive used to seal the
bag, the pressure at which the bag is sealed and the spacing (gap) of the sealing
apparatus. The gap is measured in millimeters larger (positive) or smaller (negative) than
the nominal setting.

For this experiment, 20 bags are produced at each combination of settings and the bags
are then tested for air leakage. The response variable is the proportion of the twenty bags
that leaked.

2: Analyze the data and identify any significant effects. Also identify the best
combination of viscosity, pressure and gap - the combination that minimizes the
proportion of bags that leak.

The ANOVA table below indicates the presence of a significant interaction between the
viscosity and gap variables. There is no significant effect of pressure at all.

The Tukey pairwise comparisons find the treatment combinations with the lowest mean
leakage to be those with gap = 2 and viscosity equal to 320 or 380. Pressure can be set at
either of its values with no significant effect.
Analysis of Variance for Leakage, using Adjusted SS for Tests

Viscosity                      1   0.002500   0.002500    0.002500     0.50   0.500
Pressure                       1   0.010000   0.010000    0.010000     2.00   0.195
Gap                            1   0.140625   0.140625    0.140625    28.13   0.001
Viscosity*Pressure             1   0.000625   0.000625    0.000625     0.12   0.733
Viscosity*Gap                  1   0.090000   0.090000    0.090000    18.00   0.003
Pressure*Gap                   1   0.010000   0.010000    0.010000     2.00   0.195
Viscosity*Pressure*Gap         1   0.005625   0.005625    0.005625     1.13   0.320
Error                          8   0.040000   0.040000    0.005000
Total                         15   0.299375

Grouping Information Using Tukey Method and 95.0% Confidence

Viscosity    Gap   N   Mean   Grouping
380          -2    4    0.5   A
320          -2    4    0.3     B
320           2    4    0.2     B C
380           2    4    0.1       C

The residual plots indicate reasonable Normality and one treatment combination with
rather larger variance than the others.
The treatment combination with the largest variability is the pressure = 170,
viscosity = 380, gap = 2 combination. Because the viscosity = 380, gap = 2 combination
is one of those that yields the lowest mean leakage value, I might suggest using the
pressure = 190 for producing the coffee bags. It appears that even though pressure does
not affect the mean value of leakage, pressure may affect the variability.

3: There had to be 320 total bags produced for the entire experiment (20 at each of the 16
combinations). Suppose it was impossible to produce all 320 bags in one day. Each
replicate of the experiment was run on a different day identified by the variable “Day” in
the data set.

Analyze the data again taking into account the fact that the experiment took two days to
run. Comment on whether it was important to account for the different days. Comment
on whether the results differ from those in question 2.

While the values of the sums of squares, F statistics and p-values changed slightly, there
are no substantial differences between the results here and those of question 2. The F
value for the blocking variable, Day, is 4.49, indicating that blocking was worthwhile.

The Normal probability plot of the residuals shows the tails to be a bit longer than in
question 2. The pressure = 170, viscosity = 380, gap = 2 results (circled on the graph)
still have rather large variability, so I would suggest using pressure = 190 for producing
the coffee bags.
4: Suppose the pressure settings in the original experiment had been chosen randomly
from a large possible number of settings. Analyze the data again, estimating the variance
components of all random effects. Comment on any differences between these results
and those of question 2.

The results are now quite different from those in question 2. The following ANOVA
table shows there are no significant effects at all.
Factor        Type       Levels      Values
Viscosity     fixed           2      320, 380
Pressure      random          2      170, 190
Gap           fixed           2      -2, 2

Analysis of Variance for Leakage, using Adjusted SS for Tests

Viscosity                        1   0.002500      0.002500     0.002500       4.00   0.295
Pressure                         1   0.010000      0.010000     0.010000       2.00   0.707x
Gap                              1   0.140625      0.140625     0.140625      14.06   0.166
Viscosity*Pressure               1   0.000625      0.000625     0.000625       0.11   0.795
Viscosity*Gap                    1   0.090000      0.090000     0.090000      16.00   0.156
Pressure*Gap                     1   0.010000      0.010000     0.010000       1.78   0.410
Viscosity*Pressure*Gap           1   0.005625      0.005625     0.005625       1.13   0.320
Error                            8   0.040000      0.040000     0.005000
Total                           15   0.299375

x Not an exact F-test.

The residual plots are identical to those in question 2.

The reasons for the loss of significant of the viscosity*gap interaction term lies in the test
statistic now that the model has a random effect. In question 2, the F test statistic for the
viscosity*gap interaction was

MSVG
FVG         with 8 degrees of freedom for the denominator.
MSE

In this question, the F test statistic for the viscosity*gap interaction is
          MSVG
FVG           with only 1 degree of freedom for the denominator. Not only have we lost
MSVPG
7 degrees of freedom for the denominator, which equates to a substantial loss of power,
but the denominator has increased from MSE=0.005 to MSVPG=0.005625.
   The table below shows that none of the random terms contribute much to the overall
variability.

Estimated
Source                              Value
Pressure                          0.00063
Viscosity*Pressure               -0.00125
Pressure*Gap                      0.00109
Viscosity*Pressure*Gap            0.00031
Error                             0.00500
5: Suppose that in the original experiment the gap had been hard to change. The
engineers are willing to change the gap three times throughout the experiment (that is one
original setting and three changes). Design the appropriate experiment and perform the
analysis. Compare the results with those of question 2.

The ANOVA table indicates a significant viscosity*gap interaction, as in question 2.

Fixed Effect Tests
Source                      DF DFDen F Ratio Prob > F
Viscosity                    1     6 0.8000    0.4055
Pressure                     1     6 3.2000    0.1238
Gap                          1     2 13.2353   0.0679
Viscosity*Pressure           1     6 0.2000    0.6704
Viscosity*Gap                1     6 28.8000 0.0017*
Pressure*Gap                 1     6 3.2000    0.1238
Viscosity*Pressure*Gap       1     6 1.8000    0.2283

The Fisher’s LSD pairwise tests have an unusual pattern. It appears the tests can
distinguish a statistically significant difference between the means (actually the least
squares means, i.e. the fitted means) of the 320, 2 and 380, 2 treatment combinations
(with least squares means of 0.2375 and 0.1125), but not the 320, -2 and 380, 2 treatment
combinations (with least squares means of 0.275 and 0.1125).

Viscosity*Gap

LSMeans Differences Student's t

α=0.050

Level                    Least Sq Mean
380,-2    A                 0.45000000
320,-2         B   C        0.27500000
320,2          B            0.23750000
380,2              C        0.11250000

The reason for the unusual pattern can be seen in the pairwise tests that make up the
Fisher’s LSD. In the comparisons below, we are first comparing across levels of the
whole plot factor (from a gap = 2 setting to gap = -2 setting). Note the degrees of
freedom of the t-test (3.255605). Because the whole plots are a random effect, this
comparison is done with an approximate test.

Comparing 320,2 with 320,-2

Difference               -0.03750   t Ratio        -0.6396
Std Err Dif               0.05863   DF           3.255605
Upper CL Dif              0.14107   Prob > |t|      0.5646
Lower CL Dif             -0.21607   Prob > t        0.7177
Confidence                   0.95   Prob < t        0.2823
The next test is done comparing treatment combinations within the same level of the
whole plot factor (gap = -2). Note the degrees of freedom of the t-test (df = 6). This is an
“exact” t-test.

Comparing 380,-2 with 320,-2

Difference             0.175000   t Ratio           4.427189
Std Err Dif            0.039528   DF                        6
Upper CL Dif           0.271723   Prob > |t|          0.0044*
Lower CL Dif           0.078277   Prob > t            0.0022*
Confidence                 0.95   Prob < t             0.9978

The test that led to the unusual pattern in the Fisher’s LSD is a test done across levels of
the whole pot factor (from gap = -2 to gap = 2). The reduced degrees of freedom of the
test is a reflection of the fact that we imposed a restriction on the randomization of the
whole plot factor. Because we did not allow the gap variable to be changed as often as it
might have been under complete randomization, we have lost some power in tests across
different levels of gap.

Comparing 380,2 with 320,-2

Difference             -0.16250   t Ratio           -2.77161
Std Err Dif             0.05863   DF                3.255605
Upper CL Dif            0.01607   Prob > |t|           0.0633
Lower CL Dif           -0.34107   Prob > t             0.9683
Confidence                 0.95   Prob < t            0.0317*

You can see the comparison above is “nearly” significant (p-value = 0.0633). If we are
willing to be a little flexible with our  level, we can decide that the 380, 2 combination
yields results that are, on average, significantly different from the results of the 320, -2
combination.

Our conclusion is that the 380, 2 combination of viscosity and gap gives the best overall
result. Using the profiler, we can see the margin of error of the 95% confidence interval
for the predicted leakage percentage to be ± 0.124, a rather large margin of error.

The whole plots contribute 37.5% of the total variance.

Random Effect    Var Ratio   Var Component      Std Error   95% Lower    95% Upper    Pct of Total
Whole Plots            0.6         0.001875    0.0026943     -0.003406    0.0071558        37.500
Residual                           0.003125    0.0018042     0.0012976    0.0151534        62.500
Total                                 0.005                                               100.000

The residual lots were much like those in question 2, including the larger variability for
the pressure = 170, viscosity = 380, gap = 2 combination.

6: The next three columns in the data set are from an experiment to test how well the
temperature of an industrial oven is controlled. The target temperature for the oven is
475° F. The experiment is run in four plants using four operators within each plant. For
the purposes of this experiment, the operators are fixed effects.
Analyze the data from this experiment and identify which plant(s), if any are doing a
satisfactory job of maintaining a furnace temperature near the target of 475° F.

The ANOVA table indicates a significant plant effect and a significant effect of the
operators nested in the plants.

Analysis of Variance for Temp, using Adjusted SS for Tests

Plant                  3    731.52      731.52    243.84    13.74    0.000
Operator(Plant)       12    499.81      499.81     41.65     2.35    0.008
Error                176   3122.92     3122.92     17.74
Total                191   4354.24

The Tukey pairwise comparisons show that plants 1 and 4 are doing the best job of
holding the mean temperature at 475° F. It could even be argued that plant 2 is not doing
badly. Plant 3 is clearly the doing the worst job.

Grouping Information Using Tukey Method and 95.0% Confidence

Plant    N    Mean    Grouping
2       48   477.3    A
4       48   475.8    A B
1       48   474.5      B
3       48   472.0        C

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 views: 86 posted: 5/27/2011 language: English pages: 8