91Ó°ÊÓ

The article "Bank Full Discharge of Rivers" (Water 91Ó°ÊÓ \(J ., 1978: 1141-1154)\) reports data on discharge amount \(\left(q\right.\), in \(\left.\mathrm{m}^{3} / \mathrm{sec}\right)\), flow area \(\left(a\right.\), in \(\left.\mathrm{m}^{2}\right)\), and slope of the water surface \((b\), in \(\mathrm{m} / \mathrm{m})\) obtained at a number of floodplain stations. A subset of the data follows. The article proposed a multiplicative power model \(Q=\alpha a^{\beta} b^{\gamma} \epsilon\). \begin{tabular}{l|rrrrr} \(q\) & \(17.6\) & \(23.8\) & \(5.7\) & \(3.0\) & \(7.5\) \\ \hline\(a\) & \(8.4\) & \(31.6\) & \(5.7\) & \(1.0\) & \(3.3\) \\ \hline\(b\) & \(.0048\) & \(.0073\) & 0037 & \(.0412\) & \(.0416\) \\ \(q\) & \(89.2\) & \(60.9\) & \(27.5\) & \(13.2\) & \(12.2\) \\ \hline\(a\) & \(41.1\) & \(26.2\) & \(16.4\) & \(6.7\) & \(9.7\) \\ \hline\(b\) & \(.0063\) & \(.0061\) & \(.0036\) & \(.0039\) & \(.0025\) \end{tabular} a. Use an appropriate transformation to make the model linear, and then estimate the regression parameters for the transformed model. Finally, estimate \(\alpha, \beta\), and \(\gamma\) (the parameters of the original model). What would be your prediction of discharge amount when flow area is 10 and slope is 01 ? b. Without actually doing any analysis, how would you fit a multiplicative exponential model \(Q=\alpha e^{\beta a} e^{\gamma b} \boldsymbol{\epsilon}\) ? c. After the transformation to linearity in part (a), a \(95 \%\) CI for the value of the transformed regression function when \(a=3.3\) and \(b=.0046\) was obtained from computer output as \((.217,1.755)\). Obtain a \(95 \%\) Cl for \(\alpha a^{\beta} b\) when \(a=3.3\) and \(b=.0046\).

Step by step solution

01

Transform the Multiplicative Model

The given multiplicative model is \( Q = \alpha a^{\beta} b^{\gamma} \epsilon \). Taking the natural log on both sides, we linearize it: \( \ln(Q) = \ln(\alpha) + \beta \ln(a) + \gamma \ln(b) + \ln(\epsilon) \). This equation is now in the form \( Y = \beta_0 + \beta_1 X_1 + \beta_2 X_2 + e \), where \( Y = \ln(Q) \), \( \beta_0 = \ln(\alpha) \), \( X_1 = \ln(a) \), and \( X_2 = \ln(b) \).

02

Perform Regression on Transformed Data

Using the transformed variables, perform a multiple linear regression analysis. Calculate \( Y = \ln(q) \), \( X_1 = \ln(a) \), and \( X_2 = \ln(b) \) for each data entry given. With these regression inputs, estimate the coefficients \( \beta_0 \), \( \beta_1 \), and \( \beta_2 \). These coefficients will allow us to determine \( \alpha, \beta, \) and \( \gamma \) for the original model.

03

Estimate Parameters \( \alpha, \beta, \gamma \)

From the regression output, \( \beta_0 = \ln(\alpha) \), thus \( \alpha = e^{\beta_0} \). Similarly, \( \beta = \beta_1 \) and \( \gamma = \beta_2 \). Use these values for any predictions in the original model.

04

Predict Discharge for Given Values

To predict the value of \( Q \) when \( a = 10 \) and \( b = 0.01 \), use the original model with the estimated parameters: \( Q = \alpha \cdot 10^{\beta} \cdot 0.01^{\gamma} \). Substitute the values of \( \alpha, \beta, \) and \( \gamma \) obtained from Step 3.

05

Fit a Multiplicative Exponential Model

For part (b), without calculations, to fit \( Q = \alpha e^{\beta a} e^{\gamma b} \epsilon \), take the natural logarithm again: \( \ln(Q) = \ln(\alpha) + \beta a + \gamma b + \ln(\epsilon) \). This is an exponential form directly relating \( Q \) to inputs, which implies using nonlinear regression techniques.

06

Calculate 95% Confidence Interval for Original Model

Convert the confidence interval for the transformed model to the original model by exponentiating the interval bounds. For the provided CI \((0.217, 1.755)\), use \( (e^{0.217}, e^{1.755}) \) to obtain the CI for \( \alpha a^{\beta} b \).

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Key Concepts

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

Multiplicative Model

A multiplicative model is one where the independent variables are multiplied together. In this problem, the model is given as \( Q = \alpha a^{\beta} b^{\gamma} \epsilon \). This implies that the discharge amount \( Q \) depends on the product of the flow area \( a \), the slope of the water surface \( b \), and some error term \( \epsilon \). The parameters \( \alpha, \beta, \) and \( \gamma \) are constants that need to be estimated from the data.

These models are powerful for their ability to describe relationships where changes in one variable could compound with others, making them suitable for various scientific and engineering fields. They are especially useful in contexts where the effect of one variable on the outcome changes depending on the level of another variable. This is common in ecological models, economics, and other applied sciences.

• \( \alpha \) acts as a scaling factor.
• \( \beta \) and \( \gamma \) determine how strong the influence of \( a \) and \( b \) are, respectively.
• \( \epsilon \) accounts for unobserved factors that affect the output.

Natural Logarithm

The natural logarithm (\( \ln \)) is a logarithm to the base \( e \), where \( e \approx 2.718 \). It's a powerful mathematical tool used to linearize multiplicative models. When you take the natural log of both sides of the multiplicative model, it simplifies multiplicative relationships into additive ones. For instance,

\[ Q = \alpha a^{\beta} b^{\gamma} \epsilon \]

becomes

\[ \ln(Q) = \ln(\alpha) + \beta \ln(a) + \gamma \ln(b) + \ln(\epsilon) \].
This transformation, therefore, converts the problem into a multiple linear regression form.

Logarithms such as these are extensively used in solving real-world problems because they can simplify complex equations. They also make it easier to interpret the multiplicative effects of variables in terms of percentage changes because an increase in the log form approximates percentage changes in linear space.

Regression Parameters

In the transformed linear model \( Y = \ln(Q) \), the regression parameters correspond to the coefficients of \( \ln(a) \) and \( \ln(b) \). By performing a multiple linear regression, you estimate these coefficients, which are:\[ \beta_0 = \ln(\alpha), \ \beta_1 = \beta, \ \text{and} \ \beta_2 = \gamma \].

These parameters are crucial as they represent the strength and direction of the effect each variable \( a \) and \( b \) has on \( Q \). A positive \( \beta \) implies \( a \) is positively correlated with \( Q \); the same logic applies to \( \gamma \) and \( b \).

• \( \beta_0 \) is the intercept, equivalent to the log of the multiplicative constant \( \alpha \).
• \( \beta_1 \) and \( \beta_2 \) are the slopes for \( \ln(a) \) and \( \ln(b) \) respectively.
• They help predict the natural log of \( Q \) using the input variables.

Performing this regression allows you to derive the necessary parameters for predicting discharge in situations beyond your existing data.

Confidence Interval

A confidence interval (CI) provides a range of values that is likely to contain the parameter of interest. In regression analysis, this helps us to understand the reliability of the estimated parameters. In our context, after transforming the multiplicative model, a \(95\%\) CI for the transformed regression function was given.

For example, if we have a CI as \((0.217, 1.755)\) for the transformed model, we can convert this to a CI for \( \alpha a^{\beta} b \) by exponentiating the bounds:
\[ (e^{0.217}, e^{1.755}) \].
This transformation is crucial because it gives a CI for the original scale of the model, allowing us to predict the discharge more accurately when specific parameter values are plugged into the equation.

• Exponentiating converts the CI back from the log scale to the original scale.
• It tells us with \(95\%\) confidence that the true parameter lies within this range.
• This supports more reliable predictions and decisions in practical applications.

Using confidence intervals in this process ensures that we account for the degree of uncertainty in our parameter estimates.