**The following table shows the relationship between the weight (in hundreds of pounds) and the miles per gallon (mpg) for five cars.



**Weight (in hundreds of pounds)**

**Miles per gallon (mpg)**

5

32

9

27

13

15

17

14

20

11

**Mean**

12.8

19.8

**Standard Deviation**

6.02

9.15


Weight is the explanatory variable and has a mean of 12.8 and a standard deviation of 6.02. Miles per gallon is the response variable and has a mean of 19.8 and a standard deviation of 9.15.

The correlation was found to be -0.959.

**Select the correct slope and y-intercept for the least-squares line (answer choices are rounded to the hundredths place).**

A.)

slope = 1.46
y-intercept = 14.71

B.)

slope = 0.63
y-intercept = 25.27

C.)

slope = -1.46
y-intercept = 38.49

D.)

slope = -0.63
y-intercept = 27.86

To determine the slope (\( b_1 \)) and y-intercept (\( b_0 \)) of the least-squares regression line, we use the following formulas:

1. **Slope** (\( b_1 \)):
\[
b_1 = r \cdot \frac{s_y}{s_x}
\]
Where:
- \( r \) is the correlation coefficient
- \( s_y \) is the standard deviation of the response variable (mpg)
- \( s_x \) is the standard deviation of the explanatory variable (weight)

Given:
- \( r = -0.959 \)
- \( s_y = 9.15 \)
- \( s_x = 6.02 \)

So, the slope is:
\[
b_1 = -0.959 \cdot \frac{9.15}{6.02}
\]

2. **Y-intercept** (\( b_0 \)):
\[
b_0 = \bar{y} - b_1 \cdot \bar{x}
\]
Where:
- \( \bar{y} \) is the mean of the response variable (mpg)
- \( \bar{x} \) is the mean of the explanatory variable (weight)

Given:
- \( \bar{y} = 19.8 \)
- \( \bar{x} = 12.8 \)

Let's first calculate the slope and then use it to find the y-intercept.

### Step 1: Calculate the slope
\[
b_1 = -0.959 \cdot \frac{9.15}{6.02}
\]

Let's calculate this value.The calculated slope is approximately \( b_1 = -1.46 \).

### Step 2: Calculate the y-intercept
Using the formula for the y-intercept:
\[
b_0 = \bar{y} - b_1 \cdot \bar{x}
\]
Substitute the values:
\[
b_0 = 19.8 - (-1.46) \cdot 12.8
\]

Now, let's compute the y-intercept.The calculated y-intercept is approximately \( b_0 = 38.49 \).

### Conclusion:
The correct slope and y-intercept for the least-squares regression line are:
- Slope = \(-1.46\)
- Y-intercept = \(38.49\)

Therefore, the correct answer is **C**.

Would you like further details or have any other questions? Here are some related questions for deeper understanding:

1. How is the correlation coefficient interpreted in the context of weight and miles per gallon?
2. What is the significance of a negative slope in a regression line?
3. How would changing the units of weight or mpg affect the regression equation?
4. Can the least-squares regression line be used for prediction? If so, how?
5. How does the value of the correlation coefficient affect the accuracy of the regression model?

**Tip:** Always check the signs of the slope and intercept to ensure they align with the relationship (positive or negative) observed in the data.

The following table shows the relationship between the weight (in hundreds of pounds) and the miles per gallon (mpg) for five cars.

Weight (in hundreds of pounds)
Miles per gallon (mpg)
5 - 32
9 - 27
13 - 15
17 - 14
20 - 11
Mean: 12.8, 19.8
Standard Deviation: 6.02, 9.15
Correlation: -0.959

Select the correct slope and y-intercept for the least-squares line. Choices: 
A) slope = 1.46, y-intercept = 14.71
B) slope = 0.63, y-intercept = 25.27
C) slope = -1.46, y-intercept = 38.49
D) slope = -0.63, y-intercept = 27.86

Calculate the slope and y-intercept for the least-squares regression line relating car weight to miles per gallon (mpg). Given a correlation of -0.959, this example demonstrates the relationship between these variables, suitable for high school statistics students.

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