91Ó°ÊÓ

The following data is representative of that reported in the article "An Experimental Correlation of Oxides of Nitrogen Emissions from Power Boilers Based on Field Data" (J. of Engr: for Power, July 1973: 165-170), with emission rate \((\mathrm{ppm})\) : $$ \begin{array}{l|lllllll} x & 100 & 125 & 125 & 150 & 150 & 200 & 200 \\ \hline y & 150 & 140 & 180 & 210 & 190 & 320 & 280 \\ x & 250 & 250 & 300 & 300 & 350 & 400 & 400 \\ \hline y & 400 & 430 & 440 & 390 & 600 & 610 & 670 \end{array} $$ a. Assuming that the simple linear regression model is valid, obtain the least squares estimate of the true regression line. b. What is the estimate of expected \(\mathrm{NO}_{x}\) emission rate when burner area liberation rate equals 225 ? c. Estimate the amount by which you expect \(\mathrm{NO}_{x}\) emission rate to change when burner area liberation rate is decreased by 50 . d. Would you use the estimated regression line to predict emission rate for a liberation rate of 500 ? Why or why not?

Step by step solution

01

Calculate the Mean of x and y

To find the mean values for the predictor variable \( x \) and response variable \( y \), sum the values in each data set and divide by the number of observations.\[\bar{x} = \frac{100 + 125 + \ldots + 400}{14} = 237.5\]\[\bar{y} = \frac{150 + 140 + \ldots + 670}{14} = 360\]

02

Calculate the Slope (b1) and Intercept (b0)

To find the least squares estimates, calculate the slope \( b_1 \) using the formula:\[b_1 = \frac{\sum (x_i - \bar{x})(y_i - \bar{y})}{\sum (x_i - \bar{x})^2}\]Calculate the intercept \( b_0 \):\[b_0 = \bar{y} - b_1\bar{x}\]Plugging in the calculations:\( b_1 = \frac{42000}{46962.5} = 0.8946 \)\( b_0 = 360 - 0.8946 \times 237.5 = 146.33 \) So, the regression equation is:\[y = 146.33 + 0.8946x\]

03

Predict NOx Emission Rate for x=225

To predict \( NO_x \) emission for \( x = 225 \), substitute \( x = 225 \) into the regression equation:\[y = 146.33 + 0.8946 \times 225 \]Calculate:\[y = 146.33 + 201.285 = 347.615\]

04

Estimate the Change in NOx Emission if x Decreases by 50

The slope of the regression line, \( b_1 = 0.8946 \), indicates how much the predicted \( NO_x \) emission rate changes with a unit change in \( x \). Therefore, decreasing \( x \) by 50 changes \( y \) by:\[0.8946 \times (-50) = -44.73\]

05

Decide on Using the Regression Line for a Liberation Rate of 500

The extrapolation beyond the observed x-values (100 to 400) to predict \( NO_x \) at \( x = 500 \) is risky because it falls outside the range of data used to fit the regression line. Thus, it's not recommended to use this model for \( x = 500 \).

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Key Concepts

These are the key concepts you need to understand to accurately answer the question.

Least Squares Estimate

Least squares estimation is a core technique in linear regression, used to find the line that best fits a set of data points on a graph. This technique minimizes the sum of the squared differences, or residuals, between the observed values and those predicted by the linear model.
The regression equation is typically written as \( y = b_0 + b_1x \), where:

• \( y \) is the dependent variable we're trying to predict.
• \( x \) is the independent variable, which influences \( y \).
• \( b_0 \) is the y-intercept of the line.
• \( b_1 \) is the slope of the line, indicating the change in \( y \) for a one-unit change in \( x \).

The goal of least squares estimation is to calculate \( b_0 \) and \( b_1 \) such that the sum of the squares of the differences between the observed values and those predicted by the model is minimized. This method effectively finds the line that fits the data best by reducing prediction errors.

Slope and Intercept Calculation

Calculating the slope and intercept is crucial in determining the linear relationship between two variables. This involves finding \( b_1 \) for the slope and \( b_0 \) for the intercept using specific formulas.
The formula for the slope \( b_1 \) is:\[ b_1 = \frac{\sum (x_i - \bar{x})(y_i - \bar{y})}{\sum (x_i - \bar{x})^2} \]where:

• \( x_i \) and \( y_i \) are individual data points.
• \( \bar{x} \) and \( \bar{y} \) are the means of \( x \) and \( y \) respectively.

Once the slope is calculated, the intercept \( b_0 \) can be determined using:\[ b_0 = \bar{y} - b_1\bar{x} \]This intercept is the point where the line crosses the y-axis, which represents the value of \( y \) when \( x \) is zero.
Understanding and calculating these components help in forming the regression equation, which can then be used for making predictions based on the relationship between the variables.

Extrapolation

Extrapolation involves extending a model beyond the range of the observed data. It is typically used to make predictions about unknown values based on the trends established by the current data. While it can be useful, it also carries risks and uncertainties, particularly if the extrapolated values are far outside the data range.
In linear regression analysis, extrapolation can lead to questionable predictions because the relationship established within the data range may not hold outside of it.

• If we have data ranging from 100 to 400 and want to make predictions at 500, the result can be unreliable because we are assuming the same linear trend holds true beyond the observed data.
• It’s essential to be cautious while extrapolating, as the model accuracy diminishes when predicting values that weren't part of the initial model fit.

To mitigate the risks of extrapolation, it is advisable to collect more data covering a broader range, ensuring any predictions lie within this range, thereby improving the prediction accuracy and reliability.