Question

The following data is representative of that reported in an article on nitrogen emissions, with x = burner area liberation rate (MBtu/hr-ft?) and y = NOx emission rate (ppm): x 100 125 125 150 150 200 200 250 250 300 300 350 400 400 y 160 130 190 220 180 310 290 410 420 450 400 610 600 680 in USE SALT (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.) (b) What is the estimate of expected NOx emission rate when burner area liberation rate equals 220? (Round your answer to two decimal places.) ppm (c) Estimate the amount by which you expect NOx emission rate to change when burner area liberation rate is decreased by 50. (Round your answer to two decimal places.) 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. O Yes, this value is between two existing values. O No, this value is too far away from the known values for useful extrapolation. O No, the data near this point deviates from the overall regression model.

Answer 1

Here to find the linear regression equation, we will require to have values of
T
,$\sum y$, $\sum xy$ and  
Σ Σ Υ

	x	y	x^2	xy
	100	160	10000	16000
	125	130	15625	16250
	125	190	15625	23750
	150	220	22500	33000
	150	180	22500	27000
	200	310	40000	62000
	200	290	40000	58000
	250	410	62500	102500
	250	420	62500	105000
	300	450	90000	135000
	300	400	90000	120000
	350	610	122500	213500
	400	600	160000	240000
	400	680	160000	272000
Sum	3300	5050	913750	1424000

a = [(Σy) (Σx² ) - (Σx) (Σxy)]/ [ n (Σx² ) - (Σx)² ]

a = [5050 * 913750 - 3300 * 1424000]/ [14 * 913750 - 3300 * 3300]

a = -44.55

slope

b =   [ n(Σxy) - (Σx)((Σy)]/ [ n (Σx² ) - (Σx)² ]

b = [14 * 1424000 - 3300 * 5050]/[14 * 913750 - 3300 * 3300]

b = 1.719

Regression equation

y^ = -44.5532 + 1.7193 x

(b) Here x = 220

y^ = -44.5532 + 1.7193 * 220 = 333.70

(c) When burner area liberation rate is decreases by 50, NO₂ emissionrate will decrease by 1.7193 * 50 = 85.97 ppm

(d) Here the answer is No, the value is too far away from the known values for useful extrapolation. option C is correct here.