Question

A healthy child’s systolic blood pressure \(p\) (in millimetres of mercury) and weight \(w\) (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.

\(\begin{array}{*{20}{c}}w& & {44}&{61}&{81}&{113}&{131} \\ \hline{\ln w}& & {3.78}&{4.11}&{4.39}&{4.73}&{4.88} \\ \hline p& & {91}&{98}&{103}&{110}&{112} \end{array}\)

Short answer

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

Step-by-step solution

The General Linear Model

The equation of the general linear model is defined as:

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

Here, \({\bf{y}} = \left( {\begin{array}{*{20}{c}}{{y_1}}\\{{y_2}}\\ \vdots \\{{y_n}}\end{array}} \right)\) is an observational vector, \(X = \left( {\begin{array}{*{20}{c}}1&{{x_1}}& \cdots &{x_1^n}\\1&{{x_2}}& \cdots &{x_2^n}\\ \vdots & \vdots & \ddots & \vdots \\1&{{x_n}}& \cdots &{x_n^n}\end{array}} \right)\) is the design matrix, \(\beta = \left( {\begin{array}{*{20}{c}}{{\beta _1}}\\{{\beta _2}}\\ \vdots \\{{\beta _n}}\end{array}} \right)\) is the parameter vector, and \( \in = \left( {\begin{array}{*{20}{c}}{{ \in _1}}\\{{ \in _2}}\\ \vdots \\{{ \in _n}}\end{array}} \right)\) 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:

\(\begin{array}{*{20}{c}}w& & {44}&{61}&{81}&{113}&{131} \\ \hline{\ln w}& & {3.78}&{4.11}&{4.39}&{4.73}&{4.88} \\ \hline p& & {91}&{98}&{103}&{110}&{112} \end{array}\)

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:

\(X = \left( {\begin{array}{*{20}{c}}1&{3.78}\\1&{4.11}\\1&{4.39}\\1&{4.73}\\1&{4.88}\end{array}} \right)\)

Observational vector:

\({\bf{y}} = \left( {\begin{array}{*{20}{c}}{91}\\{98}\\{103}\\{110}\\{112}\end{array}} \right)\)

And the parameter vectorfor the given equation is,

\({\bf{\beta }} = \left( {\begin{array}{*{20}{c}}{{\beta _0}}\\{{\beta _1}}\end{array}} \right)\)

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:

\(\left( {\begin{array}{*{20}{c}}{{\beta _0}}\\{{\beta _1}}\end{array}} \right) = {\left( {{{\left( {\begin{array}{*{20}{c}}1&{3.78}\\1&{4.11}\\1&{4.39}\\1&{4.73}\\1&{4.88}\end{array}} \right)}^T}\left( {\begin{array}{*{20}{c}}1&{3.78}\\1&{4.11}\\1&{4.39}\\1&{4.73}\\1&{4.88}\end{array}} \right)} \right)^{ - 1}}{\left( {\begin{array}{*{20}{c}}1&{3.78}\\1&{4.11}\\1&{4.39}\\1&{4.73}\\1&{4.88}\end{array}} \right)^T}\left( {\begin{array}{*{20}{c}}{91}\\{98}\\{103}\\{110}\\{112}\end{array}} \right)\)

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

1. Enter the data \(\left( {\begin{array}{*{20}{c}}{{\beta _0}}\\{{\beta _1}}\end{array}} \right) = {\left( {{{\left( {\begin{array}{*{20}{c}}1&{3.78}\\1&{4.11}\\1&{4.39}\\1&{4.73}\\1&{4.88}\end{array}} \right)}^T}\left( {\begin{array}{*{20}{c}}1&{3.78}\\1&{4.11}\\1&{4.39}\\1&{4.73}\\1&{4.88}\end{array}} \right)} \right)^{ - 1}}{\left( {\begin{array}{*{20}{c}}1&{3.78}\\1&{4.11}\\1&{4.39}\\1&{4.73}\\1&{4.88}\end{array}} \right)^T}\left( {\begin{array}{*{20}{c}}{91}\\{98}\\{103}\\{110}\\{112}\end{array}} \right)\) in the tab in the form of \({\left( {{{\left( {\begin{array}{*{20}{c}}1&{3.78}\\1&{4.11}\\1&{4.39}\\1&{4.73}\\1&{4.88}\end{array}} \right)}^T}\left( {\begin{array}{*{20}{c}}1&{3.78}\\1&{4.11}\\1&{4.39}\\1&{4.73}\\1&{4.88}\end{array}} \right)} \right)^{ - 1}}{\left( {\begin{array}{*{20}{c}}1&{3.78}\\1&{4.11}\\1&{4.39}\\1&{4.73}\\1&{4.88}\end{array}} \right)^T}\left( {\begin{array}{*{20}{c}}{91}\\{98}\\{103}\\{110}\\{112}\end{array}} \right)\).
2. Use colons after that and press ENTER.

So, the value of \(\left( {\begin{array}{*{20}{c}}{{\beta _0}}\\{{\beta _1}}\end{array}} \right)\) is: \(\left( {\begin{array}{*{20}{c}}{18.56}\\{19.24}\end{array}} \right)\).

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 valueof\(p\) for \(w = 100\) is \(107\)millimeters of mercury.