Problem 64 - A Graphical Approach to College Algebra

Question

The average hourly wage (adjusted to 1982 dollars) was \(\$ 7.66\) in 1990 and \(\$ 8.27\) in 2009 (Source: U.S. Census Bureau.) (a) Find a point-slope form of a line that passes through the points \((1990,7.66)\) and \((2009,8.27)\) (b) Interpret the slope. (c) Use the equation from part (a) to approximate the hourly wage in \(2005 .\) Compare it with the actual value of \(\$ 8.18\)

Step-by-Step Solution

Answer

(a) Equation: \(y - 7.66 = 0.0321(x - 1990)\). (b) The slope represents an annual wage increase of \(\$0.0321\). (c) Estimated 2005 wage is \$8.14, slightly lower than the actual \$8.18.

Step 1: Find the Slope

The first step is to find the slope of the line that passes through the points \((1990, 7.66)\) and \((2009, 8.27)\). The slope \(m\) is calculated using the slope formula: \[m = \frac{y_2 - y_1}{x_2 - x_1} = \frac{8.27 - 7.66}{2009 - 1990} = \frac{0.61}{19} \approx 0.0321\]

Step 2: Write the Point-Slope Form Equation

Next, use the point-slope form of the equation of a line \(y - y_1 = m(x - x_1)\) with the slope \(m = 0.0321\) and one of the points. Using the point \((1990, 7.66)\), the equation is:\[y - 7.66 = 0.0321(x - 1990) \]

Step 3: Interpret the Slope

The slope of \(0.0321\) represents the average annual increase in hourly wage (in dollars) between 1990 and 2009 when adjusted to 1982 dollars. Therefore, for each year, the hourly wage increased by approximately \(0.0321\) dollars.

Step 4: Use the Equation to Estimate 2005 Wage

Substitute \(x = 2005\) into the equation from Step 2 to find the approximate wage for 2005:\[y - 7.66 = 0.0321(2005 - 1990) \y - 7.66 = 0.0321(15) \y - 7.66 = 0.4815 \y = 7.66 + 0.4815 = 8.1415\]So, the estimated wage for 2005 is \$8.14.

Step 5: Compare Estimated Wage to Actual Wage

The estimated wage for 2005 is \\(8.14, while the actual wage was \\)8.18. This means the estimation using the point-slope form is slightly lower than the actual observed wage by approximately \$0.04.

Key Concepts

Slope

In mathematics, the **slope** of a line provides a measure of how steep the line is. When we talk about slope, we're referring to the change in the vertical direction (y-values) relative to the change in the horizontal direction (x-values). Essentially, it's the rise over run.

To calculate slope, the formula used is:

• \[ m = \frac{y_2 - y_1}{x_2 - x_1} \]

In the context of our original problem, we calculated the slope as approximately \(0.0321\). This suggests that for each year from 1990 to 2009, the average hourly wage increased by \(0.0321\) dollars. This slight increase indicates that the wages are rising gradually, reflecting economic changes and inflation adjustments over time.

Equation of a Line

The **equation of a line** can be expressed in various forms, and one of the most common is the point-slope form. This is particularly useful when you have a point on the line and the slope.

The point-slope form is given by:

• \( y - y_1 = m(x - x_1) \)

Here, \(m\) is the slope, and \((x_1, y_1)\) is a known point on the line. From our exercise, using the point \((1990, 7.66)\) and the slope \(0.0321\), we formed the equation:

• \( y - 7.66 = 0.0321(x - 1990) \)

This equation allows us to find the y-value (hourly wage) for any given x-value (year). It's a simple way to model linear growth over time.

Approximation

**Approximation** involves finding a value that is close to the exact value, often used when working with predictions or estimates. In our exercise, we used the equation derived from the point-slope form to approximate the hourly wage for the year 2005.

By substituting \(x = 2005\) into the equation, we calculated an estimated wage of \\(8.14. Though this doesn't match the actual wage of \\)8.18, it's important to note that approximations are bound to have slight discrepancies.

This is because approximations depend on a simple linear model, while real-world data can have more irregular patterns.

Linear Regression

**Linear regression** is a statistical technique used to model and study the relationships between variables. It provides a way to understand how one variable affects another, often using a straight line to summarize this relationship.

In our case, we're using a form of linear regression when applying the point-slope equation to predict the wage. By analyzing the increase in wage from 1990 to 2009, the slope we calculated represents the "best fit" line through our points. This is useful for both making predictions and understanding past trends.

• Linear regression helps identify trends over time.
• It accounts for variability in data and finds the most accurate straight line fit.
• This makes it an invaluable tool in data analysis and forecasting.

While simple models provide valuable insights, keep in mind that they may ignore other complex factors, capturing primarily linear relationships.