Question

Least-squares idea The table below gives a small set of data. Which of the following two lines fits the data better: y = 1 - x or y = 3 - 2x? Use the least-squares criterion to justify your answer. (Note: Neither of these two lines is the least-squares regression line for these data.)

Answer 1

To determine which line fits the data better, we use the least-squares criterion to compare the sum of squared errors (SSE) for both lines.

Step-by-step explanation:

To determine which of the two lines fits the data better, we need to use the least-squares criterion. The least-squares criterion states that the line with the smallest sum of squared errors (SSE) is the best fit for the data. SSE measures the distance between each data point and the corresponding point on the line.

Let's calculate the SSE for both lines and compare them. For the line y = 1 - x, we find the SSE by calculating the squared difference between the actual y-values and the predicted y-values on the line. For example, if one data point is (2, 3), the predicted y-value on the line would be y = 1 - 2 = -1. The squared difference is (-1 - 3)^2 = 16.

We repeat this calculation for all data points and sum up the squared differences. We then repeat the process for the line y = 3 - 2x. The line with the smaller SSE is the one that fits the data better.

Answer 2

The line
`\( \hat{y}=1-x \)` is the better fit for the data based on the least-squares criterion because it has the smaller sum of squared residuals.

The predicted values
`\(\hat{y}\)` for each x using the equation
`\(\hat{y}=1-x\)` are:

For
`\(x = -1\), \(\hat{y} = 1 - (-1) = 2\)`

For
`\(x = 1\), \(\hat{y} = 1 - 1 = 0\)`

For
`\(x = 1\) (again), \(\hat{y} = 1 - 1 = 0\)`

For
`\(x = 3\), \(\hat{y} = 1 - 3 = -2\)`

For x = 5,
`\(\hat{y} = 1 - 5 = -4\)`

The squared residuals for the first line equation
`\(\hat{y}=1-x\)` are:

For x = -1, residual squared is
`\((2 - 2)^2 = 0\)`

For x = 1, residual squared is
`\((0 - 0)^2 = 0\)`

For x = 1 (again), residual squared is
`\((1 - 0)^2 = 1\)`

For x = 3, residual squared is
`\((-1 - (-2))^2 = 1\)`

For x = 5, residual squared is
`\((-5 - (-4))^2 = 1\)`

The sum of the squared residuals for the first line equation
`\(\hat{y}=1-x\)` is 0 + 0 + 1 + 1 + 1 = 3.

predicted values
`\(\hat{y}\)` for each x using the equation
`\(\hat{y}=3-2x\)` are:

For
`\(x = -1\), \(\hat{y} = 3 - 2(-1) = 5\)`

For
`\(x = 1\), \(\hat{y} = 3 - 2(1) = 1\)`

For
`\(x = 1\) (again), \(\hat{y} = 3 - 2(1) = 1\)`

For
`\(x = 3\), \(\hat{y} = 3 - 2(3) = -3\)`

For
`\(x = 5\), \(\hat{y} = 3 - 2(5) = -7\)`

The squared residuals for the second line equation
`\(\hat{y}=3-2x\)` are:

For x = -1, residual squared is
`\((2 - 5)^2 = 9\)`

For x = 1, residual squared is
`\((0 - 1)^2 = 1\)`

For x = 1 (again), residual squared is
`\((1 - 1)^2 = 0\)`

For x = 3, residual squared is
`\((-1 - (-3))^2 = 4\)`

For x = 5, residual squared is
`\((-5 - (-7))^2 = 4\)`

The sum of the squared residuals for the second line equation
`\(\hat{y}=3-2x\)` is 9 + 1 + 0 + 4 + 4 = 18.

The sum of squared residuals for the first line equation
`\(\hat{y}=1-x\)` is smaller (3) than the sum for the second line equation
`\(\hat{y}=3-2x\) (18)`, so the line
`\(\hat{y}=1-x\)` better fits the data according to the least-squares criterion.