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When we take measurements of the same general type, a power law of the form \(y=\alpha x^{\beta}\) often gives an excellent fit to the data. A lot of research has been conducted as to why power laws work so well in business, economics, biology, ecology, medicine, engineering, social science, and so on. Let us just say that if you do not have a good straight-line fit to data pairs \((x, y)\), and the scatter plot does not rise dramatically (as in exponential growth), then a power law is often a good choice. College algebra can be used to show that power law models become linear when we apply logarithmic transformations to both variables. To see how this is done, please read on. Note: For power law models, we assume all \(x>0\) and all \(y>0\). Suppose we have data pairs \((x, y)\) and we want to find constants \(\alpha\) and \(\beta\) such that \(y=\alpha x^{\beta}\) is a good fit to the data. First, make the logarithmic transformations \(x^{\prime}=\log x\) and \(y^{\prime}=\log y .\) Next, use the \(\left(x^{\prime}, y^{\prime}\right)\) data pairs and a calculator with linear regression keys to obtain the least-squares equation \(y^{\prime}=a+b x^{\prime} .\) Note that the equation \(y^{\prime}=a+b x^{\prime}\) is the same as \(\log y=a+b(\log x)\). If we raise both sides of this equation to the power 10 and use some college algebra, we get \(y=10^{a}(x)^{b} .\) In other words, for the power law model \(y=\alpha x^{\beta}\), we have \(\alpha \approx 10^{a}\) and \(\beta \approx b\). In the electronic design of a cell phone circuit, the buildup of electric current (Amps) is an important function of time (microseconds). Let \(x=\) time in microseconds and let \(y=\) Amps built up in the circuit at time \(x\). $$ \begin{array}{c|ccccc} \hline x & 2 & 4 & 6 & 8 & 10 \\ \hline y & 1.81 & 2.90 & 3.20 & 3.68 & 4.11 \\ \hline \end{array} $$ (a) Make the logarithmic transformations \(x^{\prime}=\log x\) and \(y^{\prime}=\log y .\) Then make a scatter plot of the \(\left(x^{\prime}, y^{\prime}\right)\) values. Does a linear equation seem to be a good fit to this plot? (b) Use the \(\left(x^{\prime}, y^{\prime}\right)\) data points and a calculator with regression keys to find the least-squares equation \(y^{\prime}=a+b x^{\prime} .\) What is the correlation coefficient? (c) Use the results of part (b) to find estimates for \(\alpha\) and \(\beta\) in the power law \(y=\) \(\alpha x^{\beta .}\) Write the power law giving the relationship between time and Amp buildup. Note: The TI-84Plus calculator fully supports the power law model. Place the original \(x\) data in list \(\mathrm{L} 1\) and the corresponding \(y\) data in list \(\mathrm{L} 2 .\) Then press STAT, followed by CALC, and scroll down to option A: PwrReg. The output gives values for \(\alpha, \beta\), and the correlation coefficient \(r\).

Step by step solution

01

- Perform Logarithmic Transformations

First, apply the transformations to the given measurements: \(x' = \log x\) and \(y' = \log y\). This means, you'll need to calculate the logarithm (base 10) for each value of \(x\) and \(y\). The data pairs \(x = \{2, 4, 6, 8, 10\}\) and \(y = \{1.81, 2.90, 3.20, 3.68, 4.11\}\) are transformed into \(x' = \{0.301, 0.602, 0.778, 0.903, 1.000\}\) and \(y' = \{0.258, 0.463, 0.505, 0.566, 0.614\}\).

02

- Create Scatter Plot

Using the transformed \((x', y')\) pairs, create a scatter plot. Look at the plot to evaluate whether the relationship appears linear. In this case, the plot indicates a linear trend, suggesting that a linear equation will be a good fit.

03

- Perform Linear Regression

Use a calculator with linear regression features to find the least-squares line from the \((x', y')\) pairs. The regression will provide you with an equation of the form \(y' = a + bx'\). For the given data, the line is approximately \(y' = 0.184 + 0.429x'\). The correlation coefficient for this regression is \(r \approx 0.991\), indicating a very strong linear relationship.

04

- Derive Power Law Parameters

Convert the linear equation back to the power law form. Since \(y' = a + bx'\) corresponds to \(\log y = a + b\log x\), the power law form is \(y = 10^a \cdot x^b\). Substitute the values \(a \approx 0.184\) and \(b \approx 0.429\) to find \(\alpha = 10^{0.184} \approx 1.524\) and \(\beta \approx 0.429\).

05

- Construct Power Law Equation

Use the derived values of \(\alpha\) and \(\beta\) to write the power law equation: \(y \approx 1.524x^{0.429}\). This equation models the relationship between time (in microseconds) and current buildup (in Amps).

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Key Concepts

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

Logarithmic Transformation

A logarithmic transformation is a mathematical tool used to simplify exponential relationships, which are not linear, into linear ones. This transformation modifies the scale of data and is particularly effective in analyzing datasets that adhere to a power law model. In our exercise, we started with data pairs \((x, y)\) that fit a power law model of the form \(y = \alpha x^{\beta}\).

By taking logarithms of both variables, turning them into \(x' = \log x\) and \(y' = \log y\), we transformed the non-linear relationship into a linear form: \(y' = a + bx'\). This transformation enables us to use linear regression techniques to estimate parameters that describe the original power law model: \(\alpha\) and \(\beta\).

Logarithmic transformations are crucial in making non-linear relationships easier to analyze, allowing us to apply linear statistical methods. It effectively straightens exponential curves, making it easier to interpret and build relationships.

Linear Regression

Linear regression is a fundamental statistical method used to model the relationship between two variables by fitting a linear equation to observed data. In the context of logarithmic transformations, linear regression helps us find the best-fitting line for the transformed data.

Once we have transformed our original data pairs into \((x', y')\) as discussed, we apply linear regression to these new pairs. This involves plotting these points on a graph and using tools like a calculator or software to derive the line that minimizes the difference between predicted \(y'\) values and actual \(y'\) values. The resulting equation is usually of the form: \(y' = a + bx'\).

In our example problem, the equation derived was \(y' = 0.184 + 0.429x'\). This represents the straight-line relationship between the transformed data. The coefficients \(a\) and \(b\) are key in transforming back to the power law model, where \(\alpha \approx 10^a\) and \(\beta \approx b\).

Linear regression provides an efficient way to analyze data with linear trends, assisting us in understanding complex datasets by simplifying the underlying patterns.

Correlation Coefficient

The correlation coefficient, often denoted as \(r\), measures the strength and direction of a linear relationship between two variables. It ranges from -1 to 1, where -1 indicates a perfect negative linear relationship, 0 no linear relationship, and 1 a perfect positive linear relationship.

When applying linear regression to transformed \((x', y')\) data pairs, the correlation coefficient aids in determining how well the regression line fits the data. A high absolute value of \(r\), close to 1 or -1, indicates a strong relationship between the variables.

• In our exercise, the correlation coefficient was approximately \(r = 0.991\), suggesting a very strong positive linear relationship between \(x'\) and \(y'\).
• This high value confirms the linear regression's accuracy and appropriateness for our transformed data, and it reassures us that the transformation and subsequent modeling faithfully represent the underlying pattern of the original dataset.

The correlation coefficient therefore plays a vital role in evaluating the validity of our linear regression model.