The following data is representative of that reported in an article on nitrogen

Question

The following data is representative of that reported in an article on nitrogen emissions, with x = burner area liberation rate (MBtu/hr-ft^2) and y = NO_x emission rate (ppm): (a) Assuming that the simple linear regression model is valid, obtain the least squares estimate of the true regression line. (Round all numerical values to four decimal places.) y = (b) What is the estimate of expected NO_x emission rate when burner area liberation rate equals 250? () ppm (c) Estimate the amount by which you expect NO_x emission rate to change when burner area liberation rate is decreased by 60. () ppm (d) Would you use the estimated regression line to predict emission rate for a liberation rate of 500? Why or why not? Yes, the data is perfectly linear, thus lending to accurate predictions. Yes, this value is between two existing values. No, this value is too far away from the known values for useful extrapolation. No, the data near this point deviates from the overall regression model.

Explanation / Answer

The statistical software output for this problem is:

Simple linear regression results:
Dependent Variable: y
Independent Variable: x
y = -46.018397 + 1.7103811 x
Sample size: 14
R (correlation coefficient) = 0.97793198
R-sq = 0.95635096
Estimate of error standard deviation: 38.884748

Parameter estimates:

Analysis of variance table for regression model:

Predicted values:

a) The regression equation will be:

y = -46.0184 + 1.7104x

b) For x = 250,

y = -46.0184 + 1.7104(250)

y = 381.58

So,

The expected NOx emission for area liberation rate of 250 is 381.58 ppm.

c) If we decrease the liberation rate by 60 then NOx emission rate will decrease by:

1.7104(60) = 102.62 ppm

Parameter Estimate Std. Err. Alternative DF T-Stat P-value Intercept -46.018397 26.948255 0 12 -1.7076577 0.1134 Slope 1.7103811 0.10548264 0 12 16.214812 <0.0001

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