Question

The following data is representative of that reported in an article on nitrogen emissions, with x = burner area liberation rate (MBtu/hr-ft²) and y = NO_x emission rate (ppm):

x	100	125	125	150	150	200	200	250	250	300	300	350	400	400
y	150	130	190	220	190	330	280	400	420	430	400	590	610	680

(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 225? (Round your answer to two decimal places.)
ppm _______

(c) Estimate the amount by which you expect NO_x 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.

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.

Answer 1

Solution:

use lm function in R to fit linear regression of y on x

Rcode:

          
x   <- c(100,   125,   125,   150   ,150,   200   ,200,   250,   250   ,300   ,300,   350,   400,   400)
y <- c(   150,   130,   190   ,220,   190   ,330,   280,   400,   420,   430   ,400,   590   ,610,   680)

linmod <- lm(y~x)
summary(linmod)
coefficients(linmod)

summary(linmod)

Call:
lm(formula = y ~ x)

Residuals:
Min 1Q Median 3Q Max
-68.19 -27.13 12.33 29.95 41.30

Coefficients:
Estimate Std. Error t value Pr(>|t|)
(Intercept) -43.35085 25.36085 -1.709 0.113
x 1.70512 0.09927 17.177 8.18e-10 ***
---
Signif. codes: 0 ‘***’ 0.001 ‘**’ 0.01 ‘*’ 0.05 ‘.’ 0.1 ‘ ’ 1

Residual standard error: 36.59 on 12 degrees of freedom
Multiple R-squared: 0.9609,   Adjusted R-squared: 0.9577
F-statistic: 295 on 1 and 12 DF, p-value: 8.177e-10

> coefficients(linmod)
(Intercept) x
-43.350854 1.705125

y =-43.350854 +1.705125*225

y=340.3023

when change is 50

y =-43.350854 +1.705125*x1

y =-43.350854 +1.705125*x2

adding 2 equations

=1.705125(x1-x2)

=1.705125*50

= 85.25625

we cannot fit x-values which are out of x range.it leads to extrapolation

ANSWER(a)

y =-43.3509+1.7051x

(b)340.30

(c)85.26

(d)No, this value is too far away from the known values for useful extrapolation