Problem 5 - Differential Equations and Linear Algebra

Question

Find the equation of the least squares line associated with the given set of data points. (2,5),(0,-1),(5,3),(1,-3).

Step-by-Step Solution

Answer

The equation of the least squares line associated with the given set of data points is \(y = x - 1\).

Step 1: Find the mean of x values and y values

To find the mean of the x values, add up the x values and divide by the total number of points: x_mean = (2 + 0 + 5 + 1) / 4 x_mean = 8 / 4 x_mean = 2 For the y values: y_mean = (5 + (-1) + 3 + (-3)) / 4 y_mean = 4 / 4 y_mean = 1

Step 2: Find the slope of the least squares line (m)

Using the formula for the slope: m = Σ((x-x_mean)*(y-y_mean))/Σ((x-x_mean)^2) m = ((2-2)*(5-1) + (0-2)*(-1-1) + (5-2)*(3-1) + (1-2)*(-3-1))/((2-2)^2 + (0-2)^2 + (5-2)^2 + (1-2)^2) m = ((0*4) + (-2*(-2)) + (3*2) + (-1*(-4)))/(0 + 4 + 9 + 1) m = (0 + 4 + 6 + 4)/(14) m = 14/14 m = 1

Step 3: Find the y-intercept (b)

Using the formula for the y-intercept: b = y_mean - m * x_mean b = 1 - 1 * 2 b = 1 - 2 b = -1

Step 4: Write the equation of the least squares line

Now, we have the slope (m = 1) and the y-intercept (b = -1). So, the equation of the least squares line is: y = mx + b y = 1x - 1 y = x - 1 The equation of the least squares line associated with the given set of data points is \(y = x - 1\).

Key Concepts

Mean of x and y values

The first step in finding the least squares line is to calculate the mean of the x and y values from the given data points. This forms the foundation for further calculations in our analysis.

To find the mean of the x values, we take each x-coordinate from the data points and add them together. Once we have the total, we divide it by the number of data points. For the given points

• (2,5)
• (0,-1)
• (5,3)
• (1,-3)

we find:

• Mean of x = \( \frac{2 + 0 + 5 + 1}{4} = \frac{8}{4} = 2 \)

We follow a similar process for finding the mean of the y values, adding them up and dividing by the number of points:

• Mean of y = \( \frac{5 + (-1) + 3 + (-3)}{4} = \frac{4}{4} = 1 \)

Knowing these means helps position the regression line correctly within the data set.

Slope Calculation

After determining the means of x and y, we proceed to calculate the slope of the least squares line. The slope, often represented as \(m\), describes how much y changes for a unit change in x. We use the formula:

• \( m = \frac{\Sigma((x-x_{\text{mean}})\cdot(y-y_{\text{mean}}))}{\Sigma((x-x_{\text{mean}})^2)} \)

With the means already calculated, substitute and find:

• Each term in the numerator: - \((x-x_{\text{mean}})(y-y_{\text{mean}})\) for each point: - \((2-2)(5-1) = 0\) - \((0-2)(-1-1) = 4\) - \((5-2)(3-1) = 6\) - \((1-2)(-3-1) = 4\)

• The denominator sums: - \((2-2)^2 = 0\) - \((0-2)^2 = 4\) - \((5-2)^2 = 9\) - \((1-2)^2 = 1\)

Adding these up:

• \( m = \frac{0 + 4 + 6 + 4}{0 + 4 + 9 + 1} = \frac{14}{14} = 1 \)

This 1 tells us that for every unit increase in x, y increases by 1.

Y-intercept Calculation

With the slope calculated, finding the y-intercept is the next step. The y-intercept is the point where the line crosses the y-axis, and it can be found using the formula:

• \( b = y_{\text{mean}} - m \cdot x_{\text{mean}} \)

Plug in the values we have:

• \( b = 1 - 1 \cdot 2 \)
• \( b = 1 - 2 \)
• \( b = -1 \)

The y-intercept of -1 indicates where the line would touch the y-axis when x is zero. This provides another crucial piece of information for the equation of the line.

Data Points Analysis

Once we have both the slope and y-intercept, it's insightful to analyze how well our line approximates or captures the trend within the data. A least squares line minimizes the sum of the squares of the vertical distances of the points from the line – a fancy way to say it best fits those {x, y} coordinates.Let's recall:

• The slope \( m = 1 \) means for every increase in x, y also increases by 1.
• The y-intercept \( b = -1 \) anchors the line downward, initiating from below the origin point when measured against our axes.

Now, considering each point:

• (2, 5) gives approximate equation value: \( 2 \times 1 - 1 = 1 \). This is lower than 5, confirming small offset handled by square sizing.


• (0, -1) fits snug at \( 0 \times 1 - 1 = -1 \), showing perfect alignment.
• (5, 3) comparison: \( 5 \times 1 - 1 = 4 \). A tiny difference again refreshing confirmation of overall fit.
• (1, -3) at \( 1 \times 1 - 1 = 0 \) suggests a notable degree apart, resubmitting adjustments typical in fitting models.

Reviewing these helps students appreciate the blend of statistical models and real-world data limitations.