Question

the value of a bicycle was recorded over a period of five years. based on how the plot of the residuals might look, would a linear function that passes through (0,250) and (4,20) be a good fit for the data shown?

chart: bicycle prices, x-axis: age of bicycle (in years), y-axis: value of bicycle (in dollars), points at (1, ~150), (2, ~100), (3, ~80), (4, ~80)

options:
no, the residuals would tend to be one - sided
yes, the residuals would be randomly distributed
no, the residuals would be randomly distributed
yes, the residuals would tend to be one - sided

Explanation:

To determine if a linear function through (0, 250) and (4, 20) is a good fit, we analyze residuals. Residuals are the differences between observed and predicted values. The linear function \( y = mx + b \) (with \( b = 250 \), slope \( m=\frac{20 - 250}{4-0}=\frac{-230}{4}=-57.5 \)) predicts values. The observed data points (from the scatter plot) have lower values than the linear model's predictions initially (e.g., at \( x = 1 \), predicted \( y=250 - 57.5(1)=192.5 \), but observed is ~125) and closer later, but overall, the residuals (observed - predicted) would be negative (since observed < predicted) initially and maybe less negative later, but they are not randomly distributed. Wait, no—actually, the linear model overestimates the value (since the actual points are below the line from (0,250) to (4,20) initially and maybe closer later, but the key is residual distribution. Wait, the correct reasoning: A good linear fit has residuals randomly distributed around zero. Here, the linear function from (0,250) to (4,20) is a steep line. The actual data points are below this line (since at x=0, value is 250? No, the y - axis starts at 0, and the first point is around x=1, y=125, which is below 250 - 57.5(1)=192.5. So the residuals (observed - predicted) are negative (since observed < predicted) for most points, meaning they are not randomly distributed—they tend to be one - sided (negative, so one - sided below the line). Wait, no, the options: Let's re - evaluate. The linear function passes through (0,250) and (4,20). The scatter plot shows that the actual values are much lower than the linear model's predictions (except maybe at x = 4, where the predicted is 20 and the observed is around 20? Wait, the last point at x=4 is around y=20, which matches. But at x=1, predicted is 250 - 57.5(1)=192.5, observed is ~125 (residual = 125 - 192.5=-67.5). At x=2, predicted is 250 - 57.5(2)=250 - 115 = 135, observed is ~80 (residual=80 - 135=-55). At x=3, predicted is 250 - 57.5(3)=250 - 172.5 = 77.5, observed is ~60 (residual=60 - 77.5=-17.5). At x=4, predicted is 20, observed is ~20 (residual=0). So the residuals start very negative and become less negative, but they are all negative (or mostly negative) and not randomly distributed (they have a pattern: decreasing in magnitude as x increases). So the residuals are not randomly distributed, meaning the linear function is not a good fit because residuals are one - sided (mostly negative, so tending to be one - sided). Wait, but the options: The first option: "No, the residuals would tend to be one - sided". Wait, let's check the options again:

Options:

1. No, the residuals would tend to be one - sided

1. Yes, the residuals would be randomly distributed

1. No, the residuals would be randomly distributed

1. Yes, the residuals would tend to be one - sided

Wait, the linear model is a bad fit because residuals are not random. So the answer is "No, the residuals would tend to be one - sided" (first option). Wait, why? Because the linear function overestimates the value (predicts higher than actual) for most points (residuals are negative, so one - sided), so they are not randomly distributed. So the correct option is "No, the residuals would tend to be one - sided".

Answer:

No, the residuals would tend to be one - sided