In: Statistics and Probability
A manufacturer of rubber doors sealing accessories wants to test how closely the hardness of their door sealing products is correlated with temperature. It is known that the higher the temperature, the lower the hardness, beginning at around 35˚ C. (95˚ F.) and up. The data from a standardized test they performed is shown below.
Use the Excel correlation and the Regression tool to determine the regression equation and the R2 value. There is no template for this problem; you will have to use the Regression tool to determine the answers. (2 points)
Temperature |
Hardness |
35 |
97 |
40 |
94 |
45 |
95 |
50 |
93 |
55 |
90 |
60 |
86 |
65 |
87 |
70 |
82 |
75 |
79 |
80 |
81 |
85 |
75 |
90 |
69 |
95 |
60 |
100 |
52 |
105 |
42 |
Determine the regression equation and explain the strength of the relationship.
If the oven is increased to 60˚ C, what would you predict that the hardness measure will be?
a. Regression Equation is given as:
Hardness = 127.35 - 0.69357 * Temperature
b. R-Square = 0.8809 88.1%
That is, 88.1% of the total variation in the hardness measure is explained by the Temperature. Thus model fits the data adequately
c. If Temperature is 60˚ C then the predicted hardness measure is given as:
Hardness = 127.35 - 0.69357 * 60 = 85.7358
Excel Output