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Expert-verified Found in: Page 331 ### 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 # 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

## 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. ### Want to see more solutions like these? 