Question

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

Short answer

a) 1.4408, b) 630.95, c) 5.7783, d) 1.6733, e) 0.1276.

Step-by-step solution

Model a) Substitute x in the Equation

Given the model \(\hat{y}=1.2+0.8 \log x\), we need to substitute \(x=2\). The equation becomes \(\hat{y}=1.2 + 0.8 \log 2\).

Calculate Logarithm

Using the logarithm base 10, calculate \(\log 2\). This is approximately 0.3010. Now the equation becomes \(\hat{y}=1.2 + 0.8 \times 0.3010\).

Solve for y in Model a)

Multiply 0.8 by 0.3010 to get 0.2408. Now, add 0.2408 to 1.2 to get \(\hat{y} = 1.4408\).

Model d) Setup Equation

For model d, we are given \(\hat{y}^2=1.2+0.8x\). Substitute \(x=2\) to get \(\hat{y}^2=1.2+0.8(2)\).

Solve for y in Model d)

Calculate the right side: \(1.2 + 1.6 = 2.8\). Therefore, \(\hat{y}^2 = 2.8\). Solve for \(\hat{y}\) by taking the square root: \(\hat{y} = \sqrt{2.8}\), which is approximately 1.6733.

Model b) Setup Equation

We have \(\log \hat{y} = 1.2 + 0.8x\). Plug in \(x=2\) to get \(\log \hat{y} = 1.2 + 0.8(2)\).

Solve for y in Model b)

Calculate the right side: \(1.2 + 1.6 = 2.8\). Raise 10 to the power of both sides to get \(\hat{y} = 10^{2.8}\). Compute \(10^{2.8}\) to get \(\hat{y} \approx 630.95\).

Model e) Setup Equation

The model is \(\frac{1}{\sqrt{\hat{y}}} = 1.2 + 0.8x\). Substitute \(x=2\): \(\frac{1}{\sqrt{\hat{y}}} = 1.2 + 0.8(2)\).

Solve for y in Model e)

Perform the operations: \(1.2 + 1.6 = 2.8\). Therefore, \(\frac{1}{\sqrt{\hat{y}}} = 2.8\). Invert both sides to find \(\sqrt{\hat{y}} = \frac{1}{2.8}\), then square both sides to find that \(\hat{y} \approx 0.1276\).

Model c) Setup Equation

We have \(\ln \hat{y} = 1.2 + 0.8 \ln x\). Substitute \(x=2\) to get \(\ln \hat{y} = 1.2 + 0.8 \ln 2\).

Solve for y in Model c)

Calculate \(\ln 2\), which is approximately 0.6931. Then substitute: \(\ln \hat{y} = 1.2 + 0.8 \times 0.6931 = 1.75448\). Raise \(e\) to the power of both sides to solve for \(\hat{y}\), \(\hat{y} \approx e^{1.75448} \approx 5.7783\).

Key concepts

Logarithmic Transformation

Logarithmic transformation is a technique used to stabilize the variance and make data more normally distributed. It is particularly useful when dealing with exponential growth and helps linearize data by transforming a multiplicative relationship into an additive one. In our regression model example, we have the equation \( \hat{y} = 1.2 + 0.8 \log x \). Here, \( \log x \) implies using base 10 logarithms, making it easier to handle changes in \( x \) over large scales.

• The logarithmic function is denoted as \( \log(x) \), and it compresses the larger values of \( x \) more than the smaller values.
• This transformation is beneficial when data exhibits exponential growth or follows a power law, as it helps in achieving a linear relationship.

For example, substituting \( x = 2 \) in our equation and applying the transformation gives us a more understandable form to predict outcomes by reducing skewness in the data.

Square Root Transformation

The square root transformation is another method used to stabilize variance and make a dataset closer to meeting the assumptions of normality. It reduces positive skewness, making it valuable in data with a Poisson distribution. For the model \( \hat{y}^2 = 1.2 + 0.8x \), this concept comes into play by later solving for \( \hat{y} \).

• By taking the square root of both sides of the equation, the transformed model makes the variance more constant over different levels of \( x \).
• Unlike the logarithmic transformation, the square root is less aggressive on data values, making small values more prominent than larger ones.

In the step-by-step solution provided, the equation is simplified by taking the square root to get \( \hat{y} = \sqrt{2.8} \), allowing us to find the predicted value more easily.

Exponential Functions

Exponential functions are mathematical expressions in which a variable appears in the exponent. In regression models, they are used when the change in the dependent variable is proportional to the value of the independent variable. In our example, the equation \( \log \hat{y} = 1.2 + 0.8 x \) features an exponential relationship after transforming back from the logarithm.

• An exponential function is typically represented as \( a^x \), where \( a \) is a constant.
• When taking the logarithm of an exponential function, the equation transforms from a multiplicative form to an additive form.

Solving the equation involves reversing the logarithmic transformation by raising 10 to the power of the calculated result, yielding \( \hat{y} = 10^{2.8} \). This method reflects how exponential growth can be analyzed in regression models.

Equation Solving

Equation solving is a fundamental aspect of handling regression models. It involves manipulating algebraic expressions to find unknown values. In the context of our regression equations, such as \( \ln \hat{y} = 1.2 + 0.8 \ln x \), solving equations is crucial for predicting outcomes accurately.

• The main goal is to isolate the dependent variable, \( \hat{y} \), using mathematical operations such as addition, subtraction, multiplication, division, and applying inverse transformations like exponentiation.
• Each step aims to simplify the equation; for example, converting logarithmic equations back into their exponential form.

For the given model \( \ln \hat{y} = 1.2 + 0.8 \ln 2 \), solution steps involve calculating \( \ln 2 \), performing arithmetic, and finally exponentiating, adding clarity in processing the final predicted values. Understanding how to methodically solve such equations enhances analytical skills and enables more accurate data predictions.