/*! 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 cetane number is a critical ... [FREE SOLUTION] | 91影视

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The cetane number is a critical property in specifying the ignition quality of a fuel used in a diesel engine. Determination of this number for a biodiesel fuel is expensive and time-consuming. The article "Relating the Cetane Number of Biodiesel Fuels to Their Fatty Acid Composition: A Critical Study" (J. Automobile Engr., 2009: 565-583) included the following data on \(x=\) iodine value \((\mathrm{g})\) and \(y=\) cetane number for a sample of 14 biofuels. The iodine value is the amount of iodine necessary to saturate a sample of \(100 \mathrm{~g}\) of oil. The article's authors fit the simple linear regression model to this data, so let's follow their lead. $$ \begin{aligned} &\begin{array}{l|rrrrrrr} x & 132.0 & 129.0 & 120.0 & 113.2 & 105.0 & 92.0 & 84.0 \\ \hline y & 46.0 & 48.0 & 51.0 & 52.1 & 54.0 & 52.0 & 59.0 \end{array}\\\ &\begin{array}{l|rrrrrrr} x & 83.2 & 88.4 & 59.0 & 80.0 & 81.5 & 71.0 & 69.2 \\ \hline y & 58.7 & 61.6 & 64.0 & 61.4 & 54.6 & 58.8 & 58.0 \end{array} \end{aligned} $$ $$ \begin{aligned} &\sum x_{i}=1307.5, \quad \sum y_{i}=779.2 \\ &\sum x_{i}^{2}=128,913.93, \quad \sum x_{i} y_{i}=71,347.30, \\ &\sum y_{i}^{2}=43,745.22 \end{aligned} $$ a. Obtain the equation of the least squares line, and then calculate a point prediction of the cetane number that would result from a single observation with an iodine value of 100 . b. Calculate and interpret the coefficient of determination. c. Calculate and interpret a point estimate of the model standard deviation \(\sigma\).

Short Answer

Expert verified
a. Regression line equation: \( y = \beta_0 + \beta_1 x \). b. Coefficient of determination \( R^2 \) explains how much of \( y \) is explained by \( x \). c. \( \sigma \) indicates the typical error of the model's predictions.

Step by step solution

01

Calculate the slope (尾鈧)

To find the slope, we use the formula: \( \beta_1 = \frac{n(\sum xy) - (\sum x)(\sum y)}{n(\sum x^2) - (\sum x)^2} \). With our sums and \( n = 14 \), we substitute these values: \( \beta_1 = \frac{14(71347.3) - (1307.5)(779.2)}{14(128913.93) - (1307.5)^2} \). Calculate to get \( \beta_1 \).
02

Calculate the intercept (尾鈧)

Using \( \beta_0 = \bar{y} - \beta_1 \bar{x} \), find the means \( \bar{x} = \frac{1307.5}{14} \) and \( \bar{y} = \frac{779.2}{14} \). Substitute \( \beta_1 \) from Step 1 into this formula to find \( \beta_0 \).
03

Write the equation of the regression line

Combine \( \beta_0 \) and \( \beta_1 \) into the equation \( y = \beta_0 + \beta_1 x \). This is the least squares regression line.
04

Predict value at x = 100

Substitute \( x = 100 \) into the regression equation derived in Step 3 to find the predicted cetane number.
05

Calculate Coefficient of Determination (R虏)

Use the formula: \( R^2 = \frac{[n(\sum xy) - (\sum x)(\sum y)]^2}{(n \sum x^2 - (\sum x)^2)(n \sum y^2 - (\sum y)^2)} \). Plug in the sums to calculate \( R^2 \) which expresses the proportion of the variance in \( y \) explained by \( x \).
06

Calculate model standard deviation (蟽)

The model standard deviation is found using \( \sigma = \sqrt{\frac{1}{n-2} \left( \sum y^2 - \beta_0 \sum y - \beta_1 \sum xy \right)} \). Substitute \( \beta_0 \) and \( \beta_1 \) alongside the sums to compute \( \sigma \).

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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 approach in regression analysis for finding the best-fitting line through a set of points. This technique minimizes the sum of the squares of the vertical distances of the points from the line. By doing so, it ensures the errors between the observed values and the values predicted by the model are as small as possible.

To derive the least squares line, we calculate the slope (\beta_1) and the intercept (\beta_0) of the regression line using specific formulas. The slope is calculated by:\[\beta_1 = \frac{n(\sum xy) - (\sum x)(\sum y)}{n(\sum x^2) - (\sum x)^2}\]where \( n \) is the number of data points. Once the slope is found, the intercept is determined by:\[\beta_0 = \bar{y} - \beta_1 \bar{x}\]where \( \bar{x} \) and \( \bar{y} \) are the means of the x and y variables, respectively.

With the slope and intercept, we can express the least squares line as \( y = \beta_0 + \beta_1 x \). This line represents the linear relationship between two variables in the dataset, helping us make predictions by substituting different values of \( x \) into the equation.
Coefficient of Determination
The Coefficient of Determination, denoted as \( R^2 \), is a statistical measure that assesses how well a regression line fits the data. It indicates the proportion of variance in the dependent variable (\( y \)) that can be explained by the independent variable (\( x \)). A higher \( R^2 \) value suggests a better fit of the model.

The formula for calculating \( R^2 \) is:\[R^2 = \frac{[n(\sum xy) - (\sum x)(\sum y)]^2}{(n \sum x^2 - (\sum x)^2)(n \sum y^2 - (\sum y)^2)}\]where each component involves sums and squared sums of the x and y data.

An \( R^2 \) value of 1 signifies a perfect fit, meaning 100% of the variation in \( y \) is explained by the relationship between \( x \) and \( y \). If \( R^2 \) is 0, it means the model does not explain any of the variation in \( y \).

Understanding \( R^2 \) helps in evaluating the effectiveness of the regression model and deciding whether it's a suitable predictive tool based on how much it explains the observed data.
Model Standard Deviation
Model Standard Deviation, often represented as \( \sigma \), quantifies the variation or spread of the residuals in a regression model. Residuals are the differences between the observed values and the predicted values. A small standard deviation indicates that the data points are close to the fitted regression line, while a large standard deviation suggests more variability.

To calculate the model standard deviation, we use the formula:\[\sigma = \sqrt{\frac{1}{n-2} \left( \sum y^2 - \beta_0 \sum y - \beta_1 \sum xy \right)}\]Here, \( n \) is the number of observations, and \( n - 2 \) accounts for the degrees of freedom, adjusting for the estimation of the slope and intercept.

Having a precise value of the model standard deviation is crucial for further statistical inference, like constructing confidence intervals and hypothesis tests. It provides insights into the reliability and precision of the regression estimates.
Linear Regression
Linear Regression is a popular statistical method for modeling the relationship between a dependent variable and one or more independent variables. In cases with a single independent variable, it's termed simple linear regression, as seen in the exercise with the cetane number data.

Through linear regression, we aim to find a linear equation, \( y = \beta_0 + \beta_1 x \), that best predicts the dependent variable based on the independent variable. This line is often called the regression line.

Linear regression involves several important steps:
  • Determine the least squares line by calculating the line's slope and intercept.
  • Assess the model's fit using \( R^2 \) to quantify how well the independent variable explains the dependent variable's variance.
  • Calculate the model's standard deviation to evaluate the spread of residuals and the precision of predictions.
By using linear regression, analysts and scientists can make informed predictions, explore relationships between variables, and gain deeper insights into the data.

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Most popular questions from this chapter

The article "Analysis of the Modeling Methodologies for Predicting the Strength of Air-Jet Spun Yarns" (Textile Res. \(J ., 1997: 39-44\) ) reported on a study carried out to relate yarn tenacity \((y\), in \(\mathrm{g} /\) tex) to yarn count ( \(x_{1}\), in tex), percentage polyester \(\left(x_{2}\right)\), first nozzle pressure \(\left(x_{3}\right.\), in \(\left.\mathrm{kg} / \mathrm{cm}^{2}\right)\), and second nozzle pressure \(\left(x_{4}\right.\), in \(\left.\mathrm{kg} / \mathrm{cm}^{2}\right) .\) The estimate of the constant term in the corresponding multiple regression equation was \(6.121\). The estimated coefficients for the four predictors were \(-.082, .113, .256\), and \(-.219\), respectively, and the coefficient of multiple determination was .946. Assume that \(n=25\). a. State and test the appropriate hypotheses to decide whether the fitted model specifies a useful linear relationship between the dependent variable and at least one of the four model predictors. b. Calculate the value of adjusted \(R^{2}\) and comment. c. Calculate a \(99 \%\) confidence interval for true mean yarn tenacity when yarn count is 16.5, yarn contains \(50 \%\) polyester, first nozzle pressure is 3 , and second nozzle pressure is 5 if the estimated standard deviation of predicted tenacity under these circumstances is \(.350\).

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