In: Statistics and Probability
5.18 The shear strength of an adhesive is thought to be affected by the application pressure and temperature. A factorial experiment is performed in which both factors are assumed to be fixed. Analyze the data and draw conclusions. Perform a test for nonadditivity on minitab.
| Temperature (°F) | ||||
| Pressure (lb/in2) | 250 | 260 | 270 | |
| 120 | 9.60 | 11.28 | 9.00 | |
| 130 | 9.69 | 10.10 | 9.57 | |
| 140 | 8.43 | 11.01 | 9.03 | |
| 150 | 9.98 | 10.44 | 9.80 | |
| Strength | Temperature | Pressure | 
| 9.6 | 250 | 120 | 
| 9.69 | 250 | 130 | 
| 8.43 | 250 | 140 | 
| 9.98 | 250 | 150 | 
| 11.28 | 260 | 120 | 
| 10.1 | 260 | 130 | 
| 11.01 | 260 | 140 | 
| 10.44 | 260 | 150 | 
| 9 | 270 | 120 | 
| 9.57 | 270 | 130 | 
| 9.03 | 270 | 140 | 
| 9.8 | 270 | 150 | 
Temperature:
Null Hypothesis: The mean shear strength of an adhesive is the same for all levels of the temperature.
Alternative Hypothesis: The mean shear strength of an adhesive is not the same for all levels of the temperature.
Pressure
Null Hypothesis: The mean shear strength of an adhesive is the same for all levels of the pressure.
Alternative Hypothesis: The mean shear strength of an adhesive is not the same for all levels of the pressure.

Temperature: The p-value for the treatment Temperature is 0.032 and less than 0.05. Hence, we can reject the null hypothesis and conclude that there is a significant effect of temperature on the shear strength of an adhesive.
Pressure:
The p-value for the treatment Pressure is 0.673 and greater than 0.05. Hence, we can not reject the null hypothesis and conclude that there is not a significant effect of pressure on the shear strength of an adhesive.