one a graph Are point points : ﻿(6, 10)﻿(5, 7)(4, 3)(3, 6)(2, 2) and a straight line with x intercept on 1 and these pints : (2, 2)°(6, 10) What is the observed value of at x=5 2. What is the predicted value of at x=3 3. Find the sum of the residual values of the five points from line. 4. Find the sum of the squares of the deviations of the five points from line.

To solve this problem, let's break down the steps:

### 1. The Equation of the Line
You mentioned a straight line with:
- \( x \)-intercept at \( x = 1 \)
- Passing through points \( (2, 2) \) and \( (6, 10) \)

#### Step 1: Find the slope of the line.
The slope \( m \) is calculated using the two points \( (2, 2) \) and \( (6, 10) \):
\[
m = \frac{y_2 - y_1}{x_2 - x_1} = \frac{10 - 2}{6 - 2} = \frac{8}{4} = 2
\]

#### Step 2: Find the equation of the line.
Using the point-slope form \( y - y_1 = m(x - x_1) \) with \( (2, 2) \) and slope \( m = 2 \):
\[
y - 2 = 2(x - 2)
\]
\[
y - 2 = 2x - 4
\]
\[
y = 2x - 2
\]

Thus, the equation of the line is:
\[
y = 2x - 2
\]

### 2. Observed and Predicted Values
Next, let's calculate the observed and predicted values at the given points.

#### 2.1. Observed Value at \( x = 5 \):
From the table of given points, the observed value at \( x = 5 \) is \( y = 7 \).

#### 2.2. Predicted Value at \( x = 3 \):
Using the line equation \( y = 2x - 2 \), substitute \( x = 3 \):
\[
y = 2(3) - 2 = 6 - 2 = 4
\]
So, the predicted value at \( x = 3 \) is \( y = 4 \).

### 3. Sum of the Residuals
The residual for each point is the difference between the observed \( y \)-value and the predicted \( y \)-value from the line.

| \( x \) | Observed \( y \) | Predicted \( y = 2x - 2 \) | Residual (Observed - Predicted) |
|--------|------------------|----------------------------|---------------------------------|
| 6      | 10               | \( 2(6) - 2 = 10 \)        | \( 10 - 10 = 0 \)              |
| 5      | 7                | \( 2(5) - 2 = 8 \)         | \( 7 - 8 = -1 \)               |
| 4      | 3                | \( 2(4) - 2 = 6 \)         | \( 3 - 6 = -3 \)               |
| 3      | 6                | \( 2(3) - 2 = 4 \)         | \( 6 - 4 = 2 \)                |
| 2      | 2                | \( 2(2) - 2 = 2 \)         | \( 2 - 2 = 0 \)                |

Sum of residuals:
\[
0 + (-1) + (-3) + 2 + 0 = -2
\]
So, the sum of the residuals is \( -2 \).

### 4. Sum of the Squares of the Deviations
To find the sum of the squares of the residuals, we square each residual value.

| Residual | Squared Residual |
|----------|------------------|
| 0        | 0                |
| -1       | 1                |
| -3       | 9                |
| 2        | 4                |
| 0        | 0                |

Sum of squared residuals:
\[
0 + 1 + 9 + 4 + 0 = 14
\]
So, the sum of the squares of the deviations is \( 14 \).

### Summary of Results:
1. **Observed value at \( x = 5 \)**: \( 7 \)
2. **Predicted value at \( x = 3 \)**: \( 4 \)
3. **Sum of the residuals**: \( -2 \)
4. **Sum of the squares of the deviations**: \( 14 \)

---

Would you like further clarification on any of the steps? Here are some related questions:

1. How do you find the equation of a line given two points?
2. What is the significance of the residual in regression analysis?
3. How are residuals used to measure the accuracy of a model?
4. How do we interpret the sum of residuals when it’s non-zero?
5. Why do we square residuals when calculating deviations?

**Tip:** When summing residuals, a value close to zero indicates that the line is a good fit for the data.

One a graph are point points: (6, 10), (5, 7), (4, 3), (3, 6), (2, 2) and a straight line with x intercept on 1 and these points: (2, 2), (6, 10). What is the observed value of y at x=5? What is the predicted value of y at x=3? Find the sum of the residual values of the five points from the line. Find the sum of the squares of the deviations of the five points from the line.

This problem involves finding the equation of a line passing through specific points, calculating observed and predicted values, and determining the sum of residuals and squared deviations. Learn to apply slope-intercept formulas and residual analysis for a set of five points.

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