Question

For each of the models listed below, predict \(y\) when \(x=2\). a) \(\ln \hat{y}=1.2+0.8 x\) d) \(\hat{y}=1.2+0.8 \ln x\) b) \(\sqrt{\hat{y}}=1.2+0.8 x\) e) \(\log \hat{y}=1.2+0.8 \log x\) c) \(\frac{1}{\hat{y}}=1.2+0.8 x\)

Short answer

Model (a): \(\hat{y} \approx 16.44\), Model (d): \(\hat{y} \approx 1.754\), Model (b): \(\hat{y} = 7.84\), Model (e): \(\hat{y} \approx 27.53\), Model (c): \(\hat{y} \approx 0.357\).

Step-by-step solution

Identify Variables

We need to find specific predictions for each model equation given. The value we need to use for each equation is \(x = 2\).

Model a: Transform and solve

For model (a), the equation is \(\ln \hat{y} = 1.2 + 0.8x\). Substitute \(x = 2\): \[ \ln \hat{y} = 1.2 + 0.8(2) = 1.2 + 1.6 = 2.8 \] To find \(\hat{y}\), use the exponential function: \[ \hat{y} = e^{2.8} \approx 16.44 \]

Model d: Solve directly

For model (d), the equation is \(\hat{y} = 1.2 + 0.8 \ln x\). Substitute \(x = 2\): \[ \ln 2 \approx 0.693 \] \[ \hat{y} = 1.2 + 0.8(0.693) \approx 1.2 + 0.554 = 1.754 \]

Model b: Transform and solve

In model (b), the equation is \(\sqrt{\hat{y}} = 1.2 + 0.8x\). Substitute \(x = 2\): \[ \sqrt{\hat{y}} = 1.2 + 0.8(2) = 1.2 + 1.6 = 2.8 \] Squaring both sides to solve for \(\hat{y}\): \[ \hat{y} = (2.8)^2 = 7.84 \]

Model e: Transform and solve

Model (e)'s equation is \(\log \hat{y} = 1.2 + 0.8 \log x\). Substitute \(x = 2\): \[ \log 2 \approx 0.301 \] \[ \log \hat{y} = 1.2 + 0.8(0.301) \approx 1.2 + 0.241 = 1.441 \] To find \(\hat{y}\), use: \[ \hat{y} = 10^{1.441} \approx 27.53 \]

Model c: Transform and solve

For model (c), the equation is \(\frac{1}{\hat{y}} = 1.2 + 0.8x\). Substitute \(x = 2\): \[ \frac{1}{\hat{y}} = 1.2 + 0.8(2) = 1.2 + 1.6 = 2.8 \] To find \(\hat{y}\): \[ \hat{y} = \frac{1}{2.8} \approx 0.357 \]

Key concepts

Exponential Function

An exponential function is a mathematical expression where the variable appears as an exponent. In mathematical terms, it can be expressed as \( y = a \cdot b^{x} \), where \( a \) is a constant, \( b \) is the base of the exponential, and \( x \) is the exponent or power. This type of function is essential in many real-world applications due to its nature of representing rapid growth or decay.
One of the most critical aspects of working with exponential functions is understanding the base of the natural logarithm, \( e \), which is approximately equal to 2.71828. The exponential function with this base, often written as \( e^x \), models continuous growth, commonly used in fields such as finance for compound interest calculations or biological sciences to describe population dynamics.
In model (a) of our solution, we solved \( \ln \hat{y} = 2.8 \) by taking the exponential function of both sides. This gave us \( \hat{y} = e^{2.8} \). The solution demonstrated how to deal with the natural logarithm by reversing it with its corresponding exponential function to find \( \hat{y} \approx 16.44 \).
Exponential functions are powerful tools in statistical modeling, offering an efficient way to handle scenarios involving multiplicative growth patterns.

• Used for modeling growth processes.
• Involves the mathematical constant \( e \).
• Complementary to logarithmic functions in mathematical operations.

Logarithmic Transformation

Logarithmic transformation is a mathematical operation used to convert multiplicative relationships into additive ones. It helps in handling a wide range of data where the variance increases with the mean. This transformation is particularly useful in data analysis and regression modeling.
The transformation involves changing a variable, \( y \), into \( \log(y) \) or \( \ln(y) \), depending on the context and base used (common logarithm for base 10 or natural logarithm for base \( e \)). This process is advantageous for linearizing exponential data, stabilizing variance, and often making the data more "normal" in distribution.
In our problem, model (e) required us to use a logarithmic transformation with base 10. By substituting \( x = 2 \): \( \log 2 \approx 0.301 \), then applying the equation \( \log \hat{y} = 1.2 + 0.8 \times 0.301 \), it allowed us to solve for \( \hat{y} \), giving \( \hat{y} = 10^{1.441} \approx 27.53 \).
With logarithmic transformations:

• Data distributions can be normalized.
• Variance is stabilized for more robust analyses.
• Exponential trends can be linearized for easier interpretation.

Predictive Model

Predictive models are mathematical tools used to forecast future data based on historical patterns. These models use various statistical and machine learning algorithms to identify relationships amongst variables.
In predictive modeling, it's crucial to choose a model that best represents the relationships in the data. Factors like linearity, distribution, and variance play significant roles in the selection process. For example, choosing the right transformation (like logarithmic or square root) can improve model accuracy and interpretability.
In our context, solving models (a-d) involved different transformations and their direct solutions to predict \( y \) at \( x = 2 \). Each model equation represents a distinctive approach to prediction, leveraging transformations to adjust for different data patterns.
Strategies in predictive modeling:

• Select appropriate transformations based on data characteristics.
• Evaluate different model equations for accuracy and goodness of fit.
• Continuous refinement of models is key to improved predictions.

Square Root Transformation

Square root transformation is a mathematical technique where the square root of a variable is taken, often used to address skewness in data. This transformation is effective in stabilizing variance and is frequently applied in the presence of Poisson and gamma-type data distributions.
The transformation is expressed as \( y' = \sqrt{y} \), and it is particularly useful when dealing with right-skewed data, allowing for a closer approximation to normality and thus more valid statistical analyses.
In our exercise, model (b) applied a square root transformation: \( \sqrt{\hat{y}} = 1.2 + 0.8 \times 2 = 2.8 \). Squaring both sides, \( \hat{y} = (2.8)^2 \), resulted in \( \hat{y} = 7.84 \). This shows how the transformation reduces extreme values' impact, leading to a moremanageable range for analysis.
Benefits of square root transformation:

• Addresses right-skewness in data distributions.
• Stabilizes variance for more accurate predictions.
• Simplifies the process of achieving normal distribution of residuals.