In: Statistics and Probability
A food processing operation consists of mixing ingredients, forming and cooking. Figure below shows the cooking process where the formed material travels through a cooking oven on a conveyer belt. The raw material is subject to shrinkage due to the cooking process. The specification limits for shrinkage is between 5% and 9%. There are two machines operating two shifts daily. A technician selects samples before and after cooking process and calculates percent shrinkage. Table below shows the data set for several cooked samples taken from the upper and lower machines. The weight is in pounds.
Upper Cooked |
Lower Cooked |
1.020 |
1.024 |
1.034 |
1.018 |
1.034 |
0.974 |
1.016 |
1.032 |
1.028 |
1.034 |
1.012 |
1.012 |
0.998 |
0.998 |
1.016 |
1.000 |
1.018 |
1.016 |
1.018 |
1.020 |
1.028 |
0.974 |
Answer: Refer Following tables with Diagram for control chart:
i) Quality Control chart for Upper Machine using below table of data
Using this table to create the Control chart below:
Interpretation: Looking at control chart of Upper Machine we can see no data points are below or above UCL and LCL lines which means that quality of Upper machine is good to use for food processing operations.
ii) Quality Control chart for Lower Machine using below table of data
Using this table to create the Control chart below:
Interpretation: Looking at control chart of Lower Machine we can see no data points are below or above UCL and LCL lines which means that quality of Lower machine is good to use for food processing operations.
iii) to compare both machines, as Is supervisor’s claiming that both machines are the same, we need to use following testing of equality of two population means using Exact test.
Here using test statistic we got Calculated value as 1.6418 and table value (at 5% LOS) is 1.96
Which says, since Calculated value less than table value so we may accept Ho here.
It concludes that both machines are same in process which support claim of supervisor.