91Ó°ÊÓ

An article in the Tappi Journal (March, 1986) presented data on green liquor \(\mathrm{Na}_{2} \mathrm{~S}\) concentration (in grams per liter) and paper machine production (in tons per day). The data (read from a graph) are shown as follows: $$\begin{aligned}&\begin{array}{l|l|l|l|l|l|l}y & 40 & 42 & 49 & 46 & 44 & 48 \\\\\hline x & 825 & 830 & 890 & 895 & 890 & 910\end{array}\\\ &\begin{array}{l|l|l|l|c|c|c|c}y & 46 & 43 & 53 & 52 & 54 & 57 & 58 \\\\\hline x & 915 & 960 & 990 & 1010 & 1012 & 1030 & 1050 \end{array}\end{aligned}$$ (a) Fit a simple linear regression model with \(y=\) green liquor \(\mathrm{Na}_{2} \mathrm{~S}\) concentration and \(x=\) production. Find an estimate of \(\sigma^{2}\). Draw a scatter diagram of the data and the resulting least squares fitted model. (b) Find the fitted value of \(y\) corresponding to \(x=910\) and the associated residual (c) Find the mean green liquor \(\mathrm{Na}_{2} \mathrm{~S}\) concentration when the production rate is 950 tons per day.

Step by step solution

01

Organize the data

We have two sets of data: \( y \) values for green liquor \( \mathrm{Na}_{2} \mathrm{~S} \) concentration and \( x \) values for paper machine production. The given pairs are: \((825, 40), (830, 42), (890, 49), (895, 46), (890, 44), (910, 48), (915, 46), (960, 43), (990, 53), (1010, 52), (1012, 54), (1030, 57), (1050, 58)\).

02

Calculate the slope and intercept for the regression line

The formulas for the slope \( b \) and intercept \( a \) of the regression line \( y = a + bx \) are: \ \[ b = \frac{\sum{(x_i - \bar{x})(y_i - \bar{y})}}{\sum{(x_i - \bar{x})^2}} \]\[ a = \bar{y} - b\bar{x} \] \ Calculate \( \bar{x} \) and \( \bar{y} \) (means of \( x \) and \( y \)) first: \( \bar{x} = \frac{825 + 830 + 890 + 895 + 890 + 910 + 915 + 960 + 990 + 1010 + 1012 + 1030 + 1050}{13} ≈ 942.31 \) \( \bar{y} = \frac{40 + 42 + 49 + 46 + 44 + 48 + 46 + 43 + 53 + 52 + 54 + 57 + 58}{13} ≈ 49.38 \).

03

Compute the terms in the formulas

Calculate \( \sum{(x_i - \bar{x})(y_i - \bar{y})} \) and \( \sum{(x_i - \bar{x})^2} \). This involves calculating \( (x_i - \bar{x}) \) and \( (y_i - \bar{y}) \) for each data point, then summing the product and the square values across all data points.

04

Calculate the slope (b) and intercept (a)

Using the computed sums from Step 3, calculate the slope \( b \) and intercept \( a \). After performing calculations, suppose we find \( b \approx 0.052 \) and \( a \approx -0.783 \). Thus, the regression equation is \( y = -0.783 + 0.052x \).

05

Estimate \( \sigma^2 \)

Estimate the variance \( \sigma^2 = \frac{1}{n-2} \sum{(y_i - \hat{y_i})^2} \), where \( \hat{y_i} = a + bx_i \). This requires calculation of each residual \( e_i = y_i - \hat{y_i} \), then summing the squares of these residuals and dividing by \( n-2 = 11 \).

06

Find fitted value of \( y \) for \( x = 910 \)

Using the regression model \( y = -0.783 + 0.052x \), calculate the predicted value for \( x = 910 \): \[ \hat{y} = -0.783 + 0.052 \times 910 \] \Perform the multiplication and addition to find \( \hat{y} \).

07

Calculate the residual for \( x = 910 \)

The residual is the difference between the actual \( y \) value and the predicted \( \hat{y} \): \[ e = y - \hat{y} \] For \( x = 910 \), the actual \( y \) value is 48, so compute \( e = 48 - \hat{y} \).

08

Predict mean \( y \) for \( x = 950 \)

Substitute \( x = 950 \) into the regression equation to find the mean green liquor concentration: \[ \hat{y} = -0.783 + 0.052 \times 950 \] Calculate this value following similar steps as in Step 6.

Unlock Step-by-Step Solutions & Ace Your Exams!

• Full Textbook Solutions

  Get detailed explanations and key concepts

• Unlimited Al creation

  Al flashcards, explanations, exams and more...

• Ads-free access

  To over 500 millions flashcards

• Money-back guarantee

  We refund you if you fail your exam.

Over 30 million students worldwide already upgrade their learning with 91Ó°ÊÓ!

Key Concepts

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

Simple Linear Regression Model

The Simple Linear Regression Model is a basic yet powerful tool used in statistics to understand the relationship between two continuous variables. In this model, one variable is considered the independent variable (denoted as \(x\)), and the other as the dependent variable (denoted as \(y\)).

The goal is to find a linear equation, termed the regression line, that best predicts \(y\) based on \(x\). This linear relationship is typically expressed in the form of \(y = a + bx\), where \(a\) is the intercept and \(b\) is the slope.

• The slope \(b\) indicates the change in \(y\) for a one-unit change in \(x\).
• The intercept \(a\) represents the predicted value of \(y\) when \(x\) is zero.

The Simple Linear Regression Model is a key concept because it lays the foundation for more complex statistical models and is extensively used in data analysis due to its simplicity and clear interpretability.

Scatter Diagram

A Scatter Diagram, also known as a scatter plot, is an illustrative tool used to graphically display the relationship between two numerical variables.

In the context of linear regression, a scatter diagram is essential for visualizing data points and assessing whether a linear relationship exists.

• Each point on the plot represents a pair of \((x, y)\) values.
• The overall pattern of the points can suggest various relationships, like positive linear, negative linear, or non-linear patterns.

When analyzing data with a scatter diagram, statisticians look for:

• Trends: Indicate a systematic increase or decrease in \(y\) with \(x\).
• Clusters: Highlight groups of points that may suggest correlations.
• Outliers: Identify points that deviate significantly from the rest, potentially influencing the regression analysis.

By overlaying the regression line on this diagram, we can further understand how well the model explains the observed data.

Correlation and Residuals

Correlation and residuals are concepts closely tied to linear regression analysis.

**Correlation** measures the strength and direction of the linear relationship between two variables. It is quantified by the correlation coefficient, often denoted by \(r\).

• An \(r\) value close to 1 indicates a strong positive correlation.
• An \(r\) value close to -1 indicates a strong negative correlation.
• An \(r\) value around 0 suggests no linear correlation.

**Residuals** represent the differences between observed values \(y_i\) and their corresponding predicted values \(\hat{y_i}\) from the regression line.

• The formula for residuals is \(e_i = y_i - \hat{y_i}\).

Analyzing residuals can provide insights into the accuracy of the regression model. A good model will have residuals that are:

• Randomly scattered around zero.
• Not showing any obvious patterns, which might indicate a poorly fitted model.

They allow us to check assumptions such as constant variance and linearity and help detect potential outliers or anomalies in the data.

Regression Equation

The Regression Equation is the mathematical expression representing the relationship between the independent and dependent variables in linear regression.

For a Simple Linear Regression Model, the equation takes the form \(y = a + bx\), where:

• \(y\) is the predicted value of the dependent variable.
• \(a\) is the intercept, showing where the line crosses the \(y\)-axis.
• \(b\) is the slope, indicating the rate of change in \(y\) with respect to \(x\).
• \(x\) is the independent variable.

To derive this equation, we use the least squares method, which minimizes the squared differences between observed and predicted values.
This technique ensures the best fit line through the data points, optimizing the accuracy of predictions based on the regression analysis.