Question

use the scatterplot and line of best fit shown to complete the statements. when ( x = 5 ), the observed value is the predicted value is the residual is -2

Explanation:

Step1: Find the observed value

The observed value is the \(y -\) value of the data point when \(x = 5\). Looking at the scatter - plot, when \(x=5\), the \(y -\) value of the data point is \(4\). But since the options are \(5\), \(7\), \(9\), there might be a mis - reading. If we assume the grid is such that each square is \(1\) unit. Looking at the vertical position of the data point at \(x = 5\) (assuming the \(x\) - axis is marked correctly), if we count the \(y\) - values, the observed value (data point) at \(x = 5\) is \(4\) which is not in the options. But if we consider a wrong - count (maybe the \(x\) is \(6\) for the data point with \(y = 6\) and \(x = 5\) for the line of best fit calculation), re - evaluating:
The formula for the residual is \(e=y-\hat{y}\), where \(y\) is the observed value and \(\hat{y}\) is the predicted value. Given \(e=- 2\).

Step2: Find the predicted value

We know that the residual \(e=y - \hat{y}\). Rearranging the formula gives \(\hat{y}=y - e\).
If \(e=-2\), and we assume from the options:
Let's check for \(y = 5\): \(\hat{y}=y - e=5-(-2)=7\)
Let's check for \(y = 7\): \(\hat{y}=7-(-2)=9\)
Let's check for \(y = 9\): \(\hat{y}=9-(-2)=11\) (not relevant as the line of best fit seems to have a lower \(y\) - value).
Looking at the line of best fit: when \(x = 5\), if we assume the line of best fit passes through \((0,0)\) and \((8,10)\) (approximate points on the line of best fit), the equation of the line of best fit using the slope - intercept form \(y=mx + b\) (where \(b = 0\) as it passes through the origin) and \(m=\frac{10}{8}=\frac{5}{4}\). Then \(y=\frac{5}{4}x\). When \(x = 5\), \(y=\frac{25}{4}=6.25\approx6\) (not in options). But using the residual formula \(e=y-\hat{y}\), if \(e=-2\) and \(\hat{y}\) (predicted) and \(y\) (observed) from options:
If the observed value \(y = 5\), then \(\hat{y}=y - e=5-(-2)=7\)

Answer:

The observed value is \(5\), the predicted value is \(7\), the residual is \(-2\)