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Q12E

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Linear Algebra and its Applications
Found in: Page 331
Linear Algebra and its Applications

Linear Algebra and its Applications

Book edition 5th
Author(s) David C. Lay, Steven R. Lay and Judi J. McDonald
Pages 483 pages
ISBN 978-03219822384

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Short Answer

A healthy child’s systolic blood pressure (in millimetres of mercury) and weight (in pounds) are approximately related by the equation

\({\beta _0} + {\beta _1}\ln w = p\)

Use the following experimental data to estimate the systolic blood pressure of healthy child weighing 100 pounds.

\(\begin{array} w&\\ & {44}&{61}&{81}&{113}&{131} \\ \hline {\ln w}&\\vline & {3.78}&{4.11}&{4.39}&{4.73}&{4.88} \\ \hline p&\\vline & {91}&{98}&{103}&{110}&{112} \end{array}\)

The value of for \(w = 100\) is \(107\) millimeters of mercury.

See the step by step solution

Step by Step Solution

TABLE OF CONTENTS :
TABLE OF CONTENTS

The General Linear Model

The equation of the general linear model is defined as:

\({\bf{y}} = X\beta + \in \)

Here, \({\bf{y}} = \left( {\begin{aligned}{{y_1}}\\{{y_2}}\\ \vdots \\{{y_n}}\end{aligned}} \right)\) is an observational vector, \(X = \left( {\begin{aligned}1&{{x_1}}& \cdots &{x_1^n}\\1&{{x_2}}& \cdots &{x_2^n}\\ \vdots & \vdots & \ddots & \vdots \\1&{{x_n}}& \cdots &{x_n^n}\end{aligned}} \right)\) is the design matrix, \(\beta = \left( {\begin{aligned}{{\beta _1}}\\{{\beta _2}}\\ \vdots \\{{\beta _n}}\end{aligned}} \right)\) is the parameter vector, and \( \in = \left( {\begin{aligned}{{ \in _1}}\\{{ \in _2}}\\ \vdots \\{{ \in _n}}\end{aligned}} \right)\) is the residual vector.

Find design matrix, observation vector, parameter vector for given data

The given equation is \({\beta _0} + {\beta _1}\ln w = p\) , and the given table is as shown:

\(\begin{aligned} w&\vline & {44}&{61}&{81}&{113}&{131} \\

\hline {\ln w}&\vline & {3.78}&{4.11}&{4.39}&{4.73}&{4.88} \\

\hline p&\vline & {91}&{98}&{103}&{110}&{112} \end{aligned}\)

Write the Design matrix, observational vector, and the parameter vector for the given equation and data set by using the information provided in step 1.

Design matrix:

\(X = \left( {\begin{aligned}1&{3.78}\\1&{4.11}\\1&{4.39}\\1&{4.73}\\1&{4.88}\end{aligned}} \right)\)

Observational vector:

\({\bf{y}} = \left( {\begin{aligned}{91}\\{98}\\{103}\\{110}\\{112}\end{aligned}} \right)\)

And the parameter vector for the given equation is,

\({\bf{\beta }} = \left( {\begin{aligned}{{\beta _0}}\\{{\beta _1}}\end{aligned}} \right)\)

These are the best fit for the given data set and equation.

Normal equation

The normal equation is given as:

\({X^T}X\beta = {X^T}{\bf{y}}\)

Find the least-squares curve

The general least-squares equation is given by \({\beta _0} + {\beta _1}\ln w = p\), and to find the associated least-squares curve, the values of \({\beta _0},{\beta _1}\) are required, so find the values of \({\beta _0},{\beta _1}\) by using normal equation.

By using the obtained information from step 2, the normal equation will be:

\(\beta = {\left( {{X^T}X} \right)^{ - 1}}{X^T}{\bf{y}}\)

That implies:

\(\left( {\begin{aligned}{{\beta _0}}\\{{\beta _1}}\end{aligned}} \right) = {\left( {{{\left( {\begin{aligned}1&{3.78}\\1&{4.11}\\1&{4.39}\\1&{4.73}\\1&{4.88}\end{aligned}} \right)}^T}\left( {\begin{aligned}1&{3.78}\\1&{4.11}\\1&{4.39}\\1&{4.73}\\1&{4.88}\end{aligned}} \right)} \right)^{ - 1}}{\left( {\begin{aligned}1&{3.78}\\1&{4.11}\\1&{4.39}\\1&{4.73}\\1&{4.88}\end{aligned}} \right)^T}\left( {\begin{aligned}{91}\\{98}\\{103}\\{110}\\{112}\end{aligned}} \right)\)

Use the following steps to find the associated values for the obtained data in MATLAB.

  1. Enter the data \(\left( {\begin{aligned}{{\beta _0}}\\{{\beta _1}}\end{aligned}} \right) = {\left( {{{\left( {\begin{aligned}1&{3.78}\\1&{4.11}\\1&{4.39}\\1&{4.73}\\1&{4.88}\end{aligned}} \right)}^T}\left( {\begin{aligned}1&{3.78}\\1&{4.11}\\1&{4.39}\\1&{4.73}\\1&{4.88}\end{aligned}} \right)} \right)^{ - 1}}{\left( {\begin{aligned}1&{3.78}\\1&{4.11}\\1&{4.39}\\1&{4.73}\\1&{4.88}\end{aligned}} \right)^T}\left( {\begin{aligned}{91}\\{98}\\{103}\\{110}\\{112}\end{aligned}} \right)\) in the tab in the form of \({\left( {{{\left( {\begin{aligned}1&{3.78}\\1&{4.11}\\1&{4.39}\\1&{4.73}\\1&{4.88}\end{aligned}} \right)}^T}\left( {\begin{aligned}1&{3.78}\\1&{4.11}\\1&{4.39}\\1&{4.73}\\1&{4.88}\end{aligned}} \right)} \right)^{ - 1}}{\left( {\begin{aligned}1&{3.78}\\1&{4.11}\\1&{4.39}\\1&{4.73}\\1&{4.88}\end{aligned}} \right)^T}\left( {\begin{aligned}{91}\\{98}\\{103}\\{110}\\{112}\end{aligned}} \right)\).
  2. Use colons after that and press ENTER.

So, the value of \(\left( {\begin{aligned}{{\beta _0}}\\{{\beta _1}}\end{aligned}} \right)\) is: \(\left( {\begin{aligned}{18.56}\\{19.24}\end{aligned}} \right)\).

Now, substitute the obtained values into \({\beta _0} + {\beta _1}\ln w = p\).

\(p = 18.56 + 19.24\ln w\)

For \(w = 100\), \(p \approx 107\)

So, the required value of for \(w = 100\) is \(107\) millimeters of mercury.

Most popular questions for Math Textbooks

Show that if an \(n \times n\) matrix satisfies \(\left( {U{\bf{x}}} \right) \cdot \left( {U{\bf{y}}} \right) = {\bf{x}} \cdot {\bf{y}}\) for all x and y in \({\mathbb{R}^n}\), then \(U\) is an orthogonal matrix.

Given data for a least-squares problem, \(\left( {{x_1},{y_1}} \right), \ldots ,\left( {{x_n},{y_n}} \right)\), the following abbreviations are helpful:

\(\begin{aligned}{l}\sum x = \sum\nolimits_{i = 1}^n {{x_i}} ,{\rm{ }}\sum {{x^2}} = \sum\nolimits_{i = 1}^n {x_i^2} ,\\\sum y = \sum\nolimits_{i = 1}^n {{y_i}} ,{\rm{ }}\sum {xy} = \sum\nolimits_{i = 1}^n {{x_i}{y_i}} \end{aligned}\)

The normal equations for a least-squares line \(y = {\hat \beta _0} + {\hat \beta _1}x\)may be written in the form

\(\begin{aligned}{{\hat \beta }_0} + {{\hat \beta }_1}\sum x = \sum y \\{{\hat \beta }_0}\sum x + {{\hat \beta }_1}\sum {{x^2}} = \sum {xy} {\rm{ (7)}}\end{aligned}\)

16. Use a matrix inverse to solve the system of equations in (7) and thereby obtain formulas for \({\hat \beta _0}\) , and that appear in many statistics texts.

In exercises 1-6, determine which sets of vectors are orthogonal.

\(\left[ {\begin{array}{*{20}{c}}5\\{ - 4}\\0\\3\end{array}} \right]\), \(\left[ {\begin{array}{*{20}{c}}{ - 4}\\1\\{ - 3}\\8\end{array}} \right]\), \(\left[ {\begin{array}{*{20}{c}}3\\3\\5\\{ - 1}\end{array}} \right]\)

Let \(X\) be the design matrix used to find the least square line of fit data \(\left( {{x_1},{y_1}} \right), \ldots ,\left( {{x_n},{y_n}} \right)\). Use a theorem in Section 6.5 to show that the normal equations have a unique solution if and only if the data include at least two data points with different \(x\)-coordinates.

Suppose \(A = QR\), where \(Q\) is \(m \times n\) and R is \(n \times n\). Show that if the columns of \(A\) are linearly independent, then \(R\) must be invertible.
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