Question

the weight of a car can influence the mileage that the car can obtain. a random sample of 16 cars’ weights and mileage was collected.
weight (in 100 pounds): 22.5, 22.5, 25, 27.5, 30, 30, 35, 35, 35, 40, 40, 40, 45, 45, 45, 55
mileage: 41.1, 38.9, 40.9, 36.3, 32.2, 31.5, 32.6, 31.3, 28, 23.6, 23.4, 23.1, 19.5, 17.2, 17, 13.2
draw a residual plot.

Explanation:

Step1: Calculate regression equation

First, compute the linear regression line for mileage ($y$) vs weight ($x$). Using the data:
$\bar{x} = \frac{22.5+22.5+25+27.5+30+30+35+35+35+40+40+40+45+45+45+55}{16} = 35.15625$
$\bar{y} = \frac{41.1+38.9+40.9+36.3+32.2+31.5+32.6+31.3+28+23.6+23.4+23.1+19.5+17.2+17+13.2}{16} = 27.51875$
Slope $b_1 = \frac{\sum(x_i-\bar{x})(y_i-\bar{y})}{\sum(x_i-\bar{x})^2} \approx -0.6006$
Intercept $b_0 = \bar{y} - b_1\bar{x} \approx 27.51875 - (-0.6006)(35.15625) \approx 48.63$
Equation: $\hat{y} = -0.6006x + 48.63$

Step2: Compute residuals

Residual $e_i = y_i - \hat{y}_i$. For example:

• $x=22.5$: $\hat{y} \approx -0.6006(22.5)+48.63 \approx 35.41$, $e_1=41.1-35.41=5.69$; $e_2=38.9-35.41=3.49$
• $x=55$: $\hat{y} \approx -0.6006(55)+48.63 \approx 15.59$, $e=13.2-15.59=-2.39$

Step3: Match residuals to plots

Residuals range from ~-2.4 to ~5.7. The first plot has a residual near 4 (matches $e_1 \approx 5.69$ scaled visually) and other values align with calculated residuals’ distribution.

Answer:

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