The following data is representative of that reported in an article on nitrogen
ID: 3230321 • Letter: T
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 235? (Round your answer to two decanal places.) ______ ppm (c) Estimate the amount by winch you expect NO_x. emission rate to change when burner area liberation rate is decreased by 50. ______ 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 n 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
X
Y
XY
X2
100
160
1600
10000
125
140
17500
15625
125
170
21250
12525
150
210
31500
22500
150
180
27000
22500
200
320
64000
40000
200
270
54000
40000
250
390
97500
62500
250
440
1100000
62500
300
430
129000
90000
300
380
114000
90000
350
590
206500
122500
400
620
248000
160000
400
680
272000
160000
3300
4980
1408250
913750
m = ( n xy – (x).(y)) / nx2 –(x)2
= (14 *1408250 – 3300*4980) / 14*913750 –(3300*3300)
=(19715500 -1643400) /(12792500 – 10890000)
=1.724836
b= (y –m(x)) /n
= (4980 -1.724836*3300) /14
=-50.8541
The linear regression => y=mx +b
y = 1.724836x – 50.8541
b). when liberation rate x= 235
y=1.724836(235) -50.8541
= 354.48236
c) Expect NO emission rate to change when burner area liberation rate is decreased by 50
=-50m
=-50*1.724836.
=-86.2418 ( - indicate decreased value)
=86.2418
d). 500 is not within the range, so option c is correct
No, this value is too far away from the known values for useful extrapolation
X
Y
XY
X2
100
160
1600
10000
125
140
17500
15625
125
170
21250
12525
150
210
31500
22500
150
180
27000
22500
200
320
64000
40000
200
270
54000
40000
250
390
97500
62500
250
440
1100000
62500
300
430
129000
90000
300
380
114000
90000
350
590
206500
122500
400
620
248000
160000
400
680
272000
160000
3300
4980
1408250
913750
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