/*! This file is auto-generated */ .wp-block-button__link{color:#fff;background-color:#32373c;border-radius:9999px;box-shadow:none;text-decoration:none;padding:calc(.667em + 2px) calc(1.333em + 2px);font-size:1.125em}.wp-block-file__button{background:#32373c;color:#fff;text-decoration:none} Problem 19 The following data is representa... [FREE SOLUTION] | 91Ó°ÊÓ

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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. Eng. for Power, July 1973: 165-170), with \(x=\) burner area liberation rate \(\left(\mathrm{MBtu} / \mathrm{hr}-\mathrm{ft}^{2}\right)\) and \(y=\mathrm{NO}_{X}\) emission rate (ppm): $$ \begin{array}{l|ccccccc} x & 100 & 125 & 125 & 150 & 150 & 200 & 200 \\ \hline y & 150 & 140 & 180 & 210 & 190 & 320 & 280 \\ x & 250 & 250 & 300 & 300 & 350 & 400 & 400 \\ \hline y & 400 & 430 & 440 & 390 & 600 & 610 & 670 \end{array} $$ 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 \(\mathrm{NO}_{\mathrm{X}}\) emission rate when burner area liberation rate equals 225 ? c. Estimate the amount by which you expect \(\mathrm{NO}_{\mathrm{X}}\) 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?

Short Answer

Expert verified
a. Regression line: \(\hat{y} = 118.43 + 1.207x\). b. Estimate: 390.08 ppm. c. Decrease by 60.35 ppm. d. No, it's extrapolation.

Step by step solution

01

Organize the Data

The data consists of two variables: \(x\) (burner area liberation rate) and \(y\) (NO\(_X\) emission rate). We have fourteen data points: \([100, 150], [125, 140], [125, 180], [150, 210], [150, 190], [200, 320], [200, 280], [250, 400], [250, 430], [300, 440], [300, 390], [350, 600], [400, 610], [400, 670]\).
02

Compute Means

Calculate the means of \(x\) and \(y\):\[\bar{x} = \frac{\sum x}{n} = \frac{2750}{14} = 196.43\]\[\bar{y} = \frac{\sum y}{n} = \frac{5000}{14} = 357.14\]
03

Calculate the Slope

Calculate the slope \(b_1\) using the formula:\[b_1 = \frac{\sum (x_i - \bar{x})(y_i - \bar{y})}{\sum (x_i - \bar{x})^2}\]Compute \(\sum (x_i - \bar{x})(y_i - \bar{y}) = 26857.14\) and \(\sum (x_i - \bar{x})^2 = 22232.14\).So, \[b_1 = \frac{26857.14}{22232.14} = 1.207\]
04

Calculate the Intercept

Calculate the intercept \(b_0\) using the formula:\[b_0 = \bar{y} - b_1 \times \bar{x}\]Substitute the known values:\[b_0 = 357.14 - 1.207 \times 196.43 = 118.43\]
05

Write the Regression Equation

The least squares estimate of the true regression line is:\[\hat{y} = 118.43 + 1.207x\]
06

Estimate NO_X for 225 Burner Rate

Substitute \(x = 225\) into the regression equation:\[\hat{y} = 118.43 + 1.207 \times 225 = 390.08\]Expected NO\(_X\) emission rate is approximately 390.08 ppm.
07

Estimate Change When Rate Decreases by 50

The change in \(y\) for a change in \(x\) is given by the slope \(b_1\): \(-1.207 \times 50 = -60.35\). Thus, NO\(_X\) emission rate decreases by approximately 60.35 ppm.
08

Assess Prediction for Liberation Rate of 500

Extrapolating beyond the range of the observed data (100 to 400) is risky because it may not follow the linear trend established. Thus, using the regression line for \(x = 500\) is not advisable without additional data.

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Key Concepts

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

Least Squares Method
The Least Squares Method is a fundamental technique used in linear regression to find the best-fitting line through a set of points. The goal is to minimize the sum of the squares of the vertical distances between the observed values and the values predicted by the linear model. This method ensures that the differences between the observed data points and the predicted values are as small as possible, thereby producing the most accurate line of fit.

To apply the least squares method in the context of the given exercise, one computes two important components: the slope and the intercept of the regression line. First, the slope (\(b_1\)) is calculated using the formula:\[b_1 = \frac{\sum (x_i - \bar{x})(y_i - \bar{y})}{\sum (x_i - \bar{x})^2}\]This involves summing up the products of the deviations of each point from their respective means, both for the \(x\) and \(y\) variables. Similarly, the intercept (\(b_0\)) is determined by extrapolating the point where the regression line crosses the \(y\)-axis, using the formula:\[b_0 = \bar{y} - b_1 \times \bar{x}\]This helps us define the regression equation, which predicts the value of \(y\) for any given \(x\). In our example, the resulting equation is:\[\hat{y} = 118.43 + 1.207x\]
Emission Rate Estimation
Estimating the emission rate involves using the regression model to predict future values. In the exercise, you have a model that allows you to estimate the expected NO\(_X\) emission rate for any given burner area liberation rate.

For instance, if you need to find the estimated emission rate for a liberation rate of 225, substitute this value into the regression equation:\[\hat{y} = 118.43 + 1.207 \times 225\]Carrying out this calculation provides an estimated emission rate of approximately 390.08 ppm. This indicates that with the provided burner rate, we can expect the NO\(_X\) level to be around 390.08 parts per million.

Moreover, by evaluating how changing the burner rate affects emissions, you can understand the sensitivity of emissions to burner rate fluctuations. For instance, the slope (\(b_1 = 1.207\)) tells us that for every 1-unit increase in burner rate, NO\(_X\) emissions increase by 1.207 ppm.
Extrapolation Risks
Extrapolation involves making predictions outside the range of the data used to build the model. It can be risky because the assumptions about the linear relationship may not hold outside of the observed \(x\) values.

In this exercise, using the regression line to predict emission rates for a burner area liberation rate of 500 falls under extrapolation. Since our data ranges from 100 to 400, using the model to estimate values for \(x = 500\) is speculative. The linear trend detected within the observed data may not continue beyond this range, and thus, predictions at \(x = 500\) could be inaccurate.

The best practice is to gather more data within the extended range to verify whether the linear model remains valid or if adjustments are required. This practice safeguards against the misapplication of the model, preventing overconfidence in predictions that venture far from the data's safe limits.
Statistical Analysis
Statistical analysis provides insight into data patterns and relationships, aiding decision-making based on empirical evidence. In linear regression, statistical tools help us quantify the relationship between variables, like burner area liberation rate and NO\(_X\) emission rate.

By calculating key statistics such as means, variances, and covariances, one can better understand the data's underlying structure. This often involves thorough computation that leads to identifying the line best representing the connection between \(x\) and \(y\).

Another crucial aspect is the model's goodness-of-fit, usually measured using the coefficient of determination (R²). This statistic tells us how much of the variation in \(y\) is explained by the model. A high R² value indicates that the model explains a significant portion of the variance in the emission rates, suggesting reliable predictions. However, it's important to remember that statistical significance does not imply real-world significance. Critical thinking should always accompany statistical findings to ensure practical applicability.

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Most popular questions from this chapter

The accompanying data on \(x=\) diesel oil consumption rate measured by the drain-weigh method and \(y=\) rate measured by the CI-trace method, both in \(\mathrm{g} / \mathrm{hr}\), was read from a graph in the article "A New Measurement Method of Diesel Engine Oil Consumption Rate" (J. Society Auto Engr., 1985: 28-33). $$ \begin{array}{l|ccccccccccccc} x & 4 & 5 & 8 & 11 & 12 & 16 & 17 & 20 & 22 & 28 & 30 & 31 & 39 \\ \hline y & 5 & 7 & 10 & 10 & 14 & 15 & 13 & 25 & 20 & 24 & 31 & 28 & 39 \end{array} $$ a. Assuming that \(x\) and \(y\) are related by the simple linear regression model, carry out a test to decide whether it is plausible that on average the change in the rate measured by the CI-trace method is identical to the change in the rate measured by the drain-weigh method. b. Calculate and interpret the value of the sample correlation coefficient.

Suppose that \(x\) and \(y\) are positive variables and that a sample of \(n\) pairs results in \(r \approx 1\). If the sample correlation coefficient is computed for the \(\left(x, y^{2}\right)\) pairs, will the resulting value also be approximately 1 ? Explain.

Show that the "point of averages" \((\bar{x}, \bar{y})\) lies on the estimated regression line.

The article "Chronological Trend in Blood Lead Levels" (N. Engl. J. Med., 1983: 1373-1377) gives the following data on \(y=\) average blood lead level of white children age 6 months to 5 years and \(x=\) amount of lead used in gasoline production (in 1000 tons) for ten 6-month periods: $$ \begin{array}{l|ccccc} x & 48 & 59 & 79 & 80 & 95 \\ \hline y & 9.3 & 11.0 & 12.8 & 14.1 & 13.6 \\ x & 95 & 97 & 102 & 102 & 107 \\ \hline y & 13.8 & 14.6 & 14.6 & 16.0 & 18.2 \end{array} $$ a. Construct separate normal probability plots for \(x\) and \(y\). Do you think it is reasonable to assume that the \((x, y)\) pairs are from a bivariate normal population? b. Does the data provide sufficient evidence to conclude that there is a linear relationship between blood lead level and the amount of lead used in gasoline production? Use \(\alpha=.01\).

Suppose an investigator has data on the amount of shelf space \(x\) devoted to display of a particular product and sales revenue \(y\) for that product. The investigator may wish to fit a model for which the true regression line passes through \((0,0)\). The appropriate model is \(Y=\beta_{1} x+\epsilon\). Assume that \(\left(x_{1}, y_{1}\right), \ldots\), \(\left(x_{n}, y_{n}\right)\) are observed pairs generated from this model, and derive the least squares estimator of \(\beta_{1}\). [Hint: Write the sum of squared deviations as a function of \(b_{1}\), a trial value, and use calculus to find the minimizing value of \(b_{1}\).]

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