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Answer :
We are given the following data for burner area liberation rate, [tex]$x$[/tex], and NOₓ emission rate, [tex]$y$[/tex]:
[tex]$$
\begin{array}{|c|cccccccccccccc|}
\hline
x & 100 & 125 & 125 & 150 & 150 & 200 & 200 & 250 & 250 & 300 & 300 & 350 & 400 & 400 \\
\hline
y & 160 & 140 & 170 & 200 & 180 & 310 & 280 & 400 & 430 & 450 & 390 & 590 & 610 & 660 \\
\hline
\end{array}
$$[/tex]
We assume that the simple linear regression model
[tex]$$
y = \beta_0 + \beta_1 x + \epsilon
$$[/tex]
is valid. The following steps show how we obtain the estimates and answer each part of the question.
––––––––––– Part (a): Estimate the True Regression Line ––––––––––––––––
The least squares method finds the best fitting line by estimating the slope, [tex]$\beta_1$[/tex], and intercept, [tex]$\beta_0$[/tex], using the formulas
[tex]$$
\hat{\beta}_1 = \frac{\sum_{i=1}^n (x_i - \bar{x})(y_i - \bar{y})}{\sum_{i=1}^n (x_i - \bar{x})^2}
$$[/tex]
and
[tex]$$
\hat{\beta}_0 = \bar{y} - \hat{\beta}_1 \bar{x},
$$[/tex]
where [tex]$\bar{x}$[/tex] and [tex]$\bar{y}$[/tex] are the sample means of [tex]$x$[/tex] and [tex]$y$[/tex], respectively.
After the computations are carried out, we obtain
[tex]$$
\hat{\beta}_0 \approx -47.8515 \quad \text{and} \quad \hat{\beta}_1 \approx 1.7091.
$$[/tex]
Thus, the estimated regression line is
[tex]$$
y = -47.8515 + 1.7091\, x.
$$[/tex]
––––––––––– Part (b): Estimate Expected NOₓ Emission Rate When [tex]$x=200$[/tex] ––––––––––––––––
To predict the emission rate for [tex]$x = 200$[/tex], substitute [tex]$x=200$[/tex] in the estimated line:
[tex]$$
\hat{y} = -47.8515 + 1.7091(200).
$$[/tex]
Carrying out the calculation:
[tex]$$
\hat{y} \approx -47.8515 + 341.82 \approx 293.96.
$$[/tex]
Hence, the expected NOₓ emission rate when the burner area liberation rate is 200 (MBtu/hr-ft²) is approximately [tex]$293.96$[/tex] ppm.
––––––––––– Part (c): Estimate the Change in NOₓ Emission Rate for a Decrease of 50 in [tex]$x$[/tex] ––––––––––––––––
A decrease in [tex]$x$[/tex] by 50 units corresponds to a change in [tex]$y$[/tex] equal to the slope times the change in [tex]$x$[/tex]. Since the change in [tex]$x$[/tex] is [tex]$-50$[/tex], we have
[tex]$$
\Delta y = \hat{\beta}_1 \times (-50).
$$[/tex]
Substitute the value of [tex]$\hat{\beta}_1$[/tex]:
[tex]$$
\Delta y \approx 1.7091 \times (-50) \approx -85.45.
$$[/tex]
Thus, we expect the NOₓ emission rate to decrease by approximately [tex]$85.45$[/tex] ppm when the burner area liberation rate decreases by 50 units.
––––––––––– Part (d): Prediction at a Liberation Rate of [tex]$x=500$[/tex] ––––––––––––––––
The available data for [tex]$x$[/tex] range from 100 to 400. Predicting the emission rate when [tex]$x = 500$[/tex] would require extrapolation well beyond the observed data range. Extrapolation can be unreliable because the relationship observed within the data may not hold outside that range. Therefore, it is not advisable to use the estimated regression line to predict emission rates for [tex]$x = 500$[/tex]. The correct choice is:
[tex]$$\textbf{No, this value is too far away from the known values for useful extrapolation.}$$[/tex]
––––––––––– Final Answers Summary ––––––––––––––––
(a) The estimated regression line is
[tex]$$
y = -47.8515 + 1.7091\, x.
$$[/tex]
(b) The estimated NOₓ emission rate when [tex]$x = 200$[/tex] is approximately [tex]$293.96$[/tex] ppm.
(c) When the burner area liberation rate decreases by 50, the NOₓ emission rate is expected to decrease by approximately [tex]$85.45$[/tex] ppm.
(d) It is not appropriate to predict the emission rate for a burner area liberation rate of 500 because this value is outside the range of the observed data, making the prediction unreliable.
[tex]$$
\begin{array}{|c|cccccccccccccc|}
\hline
x & 100 & 125 & 125 & 150 & 150 & 200 & 200 & 250 & 250 & 300 & 300 & 350 & 400 & 400 \\
\hline
y & 160 & 140 & 170 & 200 & 180 & 310 & 280 & 400 & 430 & 450 & 390 & 590 & 610 & 660 \\
\hline
\end{array}
$$[/tex]
We assume that the simple linear regression model
[tex]$$
y = \beta_0 + \beta_1 x + \epsilon
$$[/tex]
is valid. The following steps show how we obtain the estimates and answer each part of the question.
––––––––––– Part (a): Estimate the True Regression Line ––––––––––––––––
The least squares method finds the best fitting line by estimating the slope, [tex]$\beta_1$[/tex], and intercept, [tex]$\beta_0$[/tex], using the formulas
[tex]$$
\hat{\beta}_1 = \frac{\sum_{i=1}^n (x_i - \bar{x})(y_i - \bar{y})}{\sum_{i=1}^n (x_i - \bar{x})^2}
$$[/tex]
and
[tex]$$
\hat{\beta}_0 = \bar{y} - \hat{\beta}_1 \bar{x},
$$[/tex]
where [tex]$\bar{x}$[/tex] and [tex]$\bar{y}$[/tex] are the sample means of [tex]$x$[/tex] and [tex]$y$[/tex], respectively.
After the computations are carried out, we obtain
[tex]$$
\hat{\beta}_0 \approx -47.8515 \quad \text{and} \quad \hat{\beta}_1 \approx 1.7091.
$$[/tex]
Thus, the estimated regression line is
[tex]$$
y = -47.8515 + 1.7091\, x.
$$[/tex]
––––––––––– Part (b): Estimate Expected NOₓ Emission Rate When [tex]$x=200$[/tex] ––––––––––––––––
To predict the emission rate for [tex]$x = 200$[/tex], substitute [tex]$x=200$[/tex] in the estimated line:
[tex]$$
\hat{y} = -47.8515 + 1.7091(200).
$$[/tex]
Carrying out the calculation:
[tex]$$
\hat{y} \approx -47.8515 + 341.82 \approx 293.96.
$$[/tex]
Hence, the expected NOₓ emission rate when the burner area liberation rate is 200 (MBtu/hr-ft²) is approximately [tex]$293.96$[/tex] ppm.
––––––––––– Part (c): Estimate the Change in NOₓ Emission Rate for a Decrease of 50 in [tex]$x$[/tex] ––––––––––––––––
A decrease in [tex]$x$[/tex] by 50 units corresponds to a change in [tex]$y$[/tex] equal to the slope times the change in [tex]$x$[/tex]. Since the change in [tex]$x$[/tex] is [tex]$-50$[/tex], we have
[tex]$$
\Delta y = \hat{\beta}_1 \times (-50).
$$[/tex]
Substitute the value of [tex]$\hat{\beta}_1$[/tex]:
[tex]$$
\Delta y \approx 1.7091 \times (-50) \approx -85.45.
$$[/tex]
Thus, we expect the NOₓ emission rate to decrease by approximately [tex]$85.45$[/tex] ppm when the burner area liberation rate decreases by 50 units.
––––––––––– Part (d): Prediction at a Liberation Rate of [tex]$x=500$[/tex] ––––––––––––––––
The available data for [tex]$x$[/tex] range from 100 to 400. Predicting the emission rate when [tex]$x = 500$[/tex] would require extrapolation well beyond the observed data range. Extrapolation can be unreliable because the relationship observed within the data may not hold outside that range. Therefore, it is not advisable to use the estimated regression line to predict emission rates for [tex]$x = 500$[/tex]. The correct choice is:
[tex]$$\textbf{No, this value is too far away from the known values for useful extrapolation.}$$[/tex]
––––––––––– Final Answers Summary ––––––––––––––––
(a) The estimated regression line is
[tex]$$
y = -47.8515 + 1.7091\, x.
$$[/tex]
(b) The estimated NOₓ emission rate when [tex]$x = 200$[/tex] is approximately [tex]$293.96$[/tex] ppm.
(c) When the burner area liberation rate decreases by 50, the NOₓ emission rate is expected to decrease by approximately [tex]$85.45$[/tex] ppm.
(d) It is not appropriate to predict the emission rate for a burner area liberation rate of 500 because this value is outside the range of the observed data, making the prediction unreliable.
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