Table 4: Cost index values evaluated at sample mean values and at selected percentiles of
food safety technology use and performance of sanitation and process control tasks.1
|
Industry__________ |
Variable________ |
-----Percentile----- |
--Mean--- |
------Percentile------ | ||
|
95____ |
75________ |
25______ |
5_________ | |||
|
Model: Food safety technology use and performance of sanitation and process control | ||||||
|
Meat Slaughter |
Sanitation and Process Control |
0.99 |
0.99 |
1.00 |
1.02 |
1.11 |
|
Meat processing |
Sanitation and Process Control |
0.90 |
0.90 |
1.00 |
1.00 |
1.15 |
|
Poultry Slaughter |
Sanitation and Process Control |
0.89 |
0.90 |
1.00 |
1.04 |
1.29 |
|
Meat Slaughter |
Technology |
0.97 |
0.98 |
1.00 |
1.05 |
1.24 |
|
Meat processing |
Technology |
0.99 |
TÔ1 |
1.00 |
Tôî |
1.18 |
|
Poultry Slaughter |
Technology |
0.76 |
0.96 |
1.00 |
1.28 |
0.62 |
|
Model: Performance of sanitation and process control tasks only, no food safety | ||||||
|
Meat Slaughter |
Sanitation and Process Control |
0.92 |
0.92 |
1.00 |
1.04 |
1.12 |
|
Meat processing |
Sanitation and Process Control |
0.89 |
0.89 |
1.00 |
1.00 |
0.99 |
|
Poultry Slaughter |
Sanitation and Process Control |
0.87 |
0.90 |
1.00 |
1.03 |
1.02 |
1Costs were estimated for sanitation and process control task and food safety technology by
allowing S and T to vary and setting all other values at their sample means. Most terms drop out
and we are left with: ln C= Intercept + βsln S + βS2 ln S * ln S and
ln C= Intercept + βTln T + βT2 ln T * ln T.
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