The data alongside gives the rate of emission of Oxides of Nitrogen (in ppm) cor
ID: 3232016 • Letter: T
Question
The data alongside gives the rate of emission of Oxides of Nitrogen (in ppm) corresponding to the burner-area liberation rate of a certain type of boiler (in MBTU/hr-ft^2). a) Taking burner-area liberation rate as the independent variable and emission rate as the dependent variable, fit a regression line to the given data. (i.e. find the coefficients of regression and give the regression equation). b) Determine the correlation coefficient and coefficient of determination. State the level of the strength of linear relationship between the given variables (weak, moderate, or strong?). What will be the Emission Rate if Burner Area Liberation Rate is 225 units?Explanation / Answer
PART-A
FROM FOLLOWING RESUTLS WE HAVE REGRESSION LINE AS FOLLOWS:
Regression Equation
Emission Rate = -45.6 + 1.7114 Burner Area Liberation Rate
SO CORRESPONDING TO unit increase in burner Area Liberation Rate there is on an average an incease of 1.7114 in emission rate.
Model Summary
S R-sq R-sq(adj) R-sq(pred)
36.7485 96.09% 95.76% 94.73%
Coefficients
Term Coef SE Coef T-Value P-Value VIF
Constant -45.6 25.5 -1.79 0.099
Burner Area Liberation Rate 1.7114 0.0997 17.17 0.000 1.00
Regression Equation
Emission Rate = -45.6 + 1.7114 Burner Area Liberation Rate
PArt-B
From following results correlaion coefficient is r=0.98 which is very stron, positive and signicant with p-valueless than 0.05.
So, coefficient of detemriantion is R2=0.98*0.98=96.04% which means that burner Area Liberation Rate explained 96.04% variations in emission rate.
For burner Area Liberation Rate=225,
emission rate=-45.6 + 1.7114 *225=339.465
Correlation: Burner Area Liberation Rate, Emission Rate
Pearson correlation of Burner Area Liberation Rate and Emission Rate = 0.980
P-Value = 0.000
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