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 (4,25) be a good fit for the data shown? graph: bicycle prices, x-axis: age of bicycle (in years), y-axis: value of bicycle (in dollars). points at (0,250), (1, ~130), (2, ~80), (3, ~40), (4, ~30). options: no, the residuals would be randomly distributed. no, the residuals would tend to be one-sided. yes, the residuals would tend to be one-sided. yes, the residuals would be randomly distributed.

Explanation:

Step1: Understand Residuals for Linear Fit

Residuals are the differences between observed and predicted values. For a good linear fit, residuals should be randomly distributed (no pattern). If residuals are one - sided (all positive or all negative in a pattern), the linear model is not a good fit.

Step2: Analyze the Data and Linear Function

The linear function passes through \((0,250)\) and \((4,25)\). The slope \(m=\frac{25 - 250}{4-0}=\frac{- 225}{4}=- 56.25\). The equation is \(y=-56.25x + 250\).
Looking at the scatter plot, the observed points are below the line at some points and above? Wait, no. Wait, the line goes from \((0,250)\) to \((4,25)\). The actual data points (at \(x = 1\), \(x = 2\), \(x = 3\), \(x = 4\)) are below the line? Wait, no, at \(x = 0\) the line is at 250 (matches the point). At \(x = 1\), the line would predict \(y=-56.25(1)+250 = 193.75\), but the actual point is around 125 - 150 (below the line). At \(x = 2\), predicted \(y=-56.25(2)+250=137.5\), actual point is around 75 - 100 (below the line). At \(x = 3\), predicted \(y=-56.25(3)+250 = 81.25\), actual point is around 25 - 50 (below the line). At \(x = 4\), predicted \(y = 25\), actual point is around 25 - 50 (above the line? Wait, no, the actual point at \(x = 4\) is below 25? Wait, the plot shows at \(x = 4\) the point is around 25 - 50? Wait, maybe I misread. Wait, the key is about residuals. If the linear model is used, the residuals (observed - predicted) would have a pattern. Wait, the data points seem to be decreasing in a non - linear way (maybe exponential decay). So the linear model would have residuals that are not random. Wait, the options: A good linear fit has randomly distributed residuals. But here, the residuals would be one - sided (all positive or all negative in a trend). Wait, let's think again. The line is a straight line, but the data points are curving down (decreasing at a decreasing rate? Or increasing rate of decrease?). So when we fit a straight line, the residuals (observed - predicted) would be, for example, at the start (x = 1,2,3) the observed values are below the line (residuals negative) and at x = 4 maybe above? No, the line at x = 4 is 25, and the actual point is around 25 - 50? Wait, maybe the actual point at x = 4 is below 25? Wait, the plot: x - axis is age (0 - 5), y - axis is value. At x = 0, y = 250. At x = 1, a point around 125. At x = 2, around 75. At x = 3, around 25 - 50. At x = 4, around 25 - 50. The line from (0,250) to (4,25) is a straight line. So for x = 1, predicted y=250 - 56.25*1 = 193.75, observed is ~125, residual=125 - 193.75=- 68.75 (negative). For x = 2, predicted=250 - 56.25*2 = 137.5, observed ~75, residual=75 - 137.5=- 62.5 (negative). For x = 3, predicted=250 - 56.25*3 = 81.25, observed ~25 - 50, residual (say 40 - 81.25=- 41.25) (negative). For x = 4, predicted=25, observed ~25 - 50 (say 30), residual=30 - 25 = 5 (positive). Wait, but the pattern of residuals: from x = 0 to x = 3, residuals are negative, and at x = 4, positive. But maybe the overall trend is that the residuals are not random. Wait, the correct reasoning: A linear model is a good fit if residuals are randomly distributed (no pattern). If the residuals tend to be one - sided (e.g., all positive or all negative in a way that shows a pattern), then the linear model is not a good fit. Looking at the data, the bicycle's value is decreasing, but the rate of decrease is changing (the points are curving), so a linear model would not fit well. The residuals would tend to be one - sided (either mostly positive or mostly negative in a pattern). So the answer is "No, the residuals would tend to be one - sided."

Answer:

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