Question

find the residual values, and use the graphing calculator tool to make a residual plot. does the residual plot show that the line of best fit is appropriate for the data? yes, the points have no pattern. no, the points are evenly distributed about the x - axis. no, the points are in a linear pattern. yes, the points are in a curved pattern.

Answer:

Part 1: Calculating Residual Values

The formula for the residual is \( \text{Residual} = \text{Given} - \text{Predicted} \).

Step 1: Residual for \( x = 1 \)

Given \( \text{Given} = -3.5 \), \( \text{Predicted} = -1.1 \)

$$ \text{Residual} = -3.5 - (-1.1) = -3.5 + 1.1 = -2.4 $$

Step 2: Residual for \( x = 2 \)

Given \( \text{Given} = -2.9 \), \( \text{Predicted} = 2 \)

$$ \text{Residual} = -2.9 - 2 = -4.9 $$

Step 3: Residual for \( x = 3 \)

Given \( \text{Given} = -1.1 \), \( \text{Predicted} = 5.1 \)

$$ \text{Residual} = -1.1 - 5.1 = -6.2 $$

Step 4: Residual for \( x = 4 \)

Given \( \text{Given} = 2.2 \), \( \text{Predicted} = 8.2 \)

$$ \text{Residual} = 2.2 - 8.2 = -6.0 $$

Step 5: Residual for \( x = 5 \)

Given \( \text{Given} = 3.4 \), \( \text{Predicted} = 11.3 \)

$$ \text{Residual} = 3.4 - 11.3 = -7.9 $$

Part 2: Analyzing the Residual Plot

To determine if the line of best fit is appropriate, we check the residual plot. A good line of best fit has residuals with no pattern (randomly scattered). Let's analyze the residuals we calculated: \(-2.4, -4.9, -6.2, -6.0, -7.9\). Wait, actually, let's re - evaluate the residual pattern. Wait, maybe I made a mistake in the residual calculation interpretation. Wait, no, the key is:

For a line of best fit to be appropriate, the residual plot should have points with no pattern (random). Let's check the options:

• Option 1: "Yes, the points have no pattern." If the residuals are randomly scattered, this is correct. But wait, let's check the other options.
• Option 2: "No, the points are evenly distributed about the x - axis." Even distribution about the x - axis is a sign of a good fit, but the wording "No" is wrong.
• Option 3: "No, the points are in a linear pattern." If residuals form a linear pattern, the line of best fit is not appropriate. Let's see our residuals: \(-2.4, -4.9, -6.2, -6.0, -7.9\). Let's plot them (mentally) against x (1,2,3,4,5). The residuals seem to have a pattern? Wait, no, maybe I miscalculated the residuals. Wait, no, the formula is \( \text{Residual}=\text{Observed}-\text{Predicted} \), which is what I used.

Wait, maybe I made a mistake in the residual calculation. Let's recalculate:

For \( x = 1 \): Observed = - 3.5, Predicted=-1.1. Residual = - 3.5-(-1.1)=-3.5 + 1.1=-2.4 (correct)

For \( x = 2 \): Observed=-2.9, Predicted = 2. Residual=-2.9 - 2=-4.9 (correct)

For \( x = 3 \): Observed=-1.1, Predicted = 5.1. Residual=-1.1-5.1=-6.2 (correct)

For \( x = 4 \): Observed = 2.2, Predicted = 8.2. Residual=2.2 - 8.2=-6.0 (correct)

For \( x = 5 \): Observed = 3.4, Predicted = 11.3. Residual=3.4 - 11.3=-7.9 (correct)

Now, let's check the residual pattern. The residuals are: - 2.4 (x = 1), - 4.9 (x = 2), - 6.2 (x = 3), - 6.0 (x = 4), - 7.9 (x = 5). If we plot x (1 - 5) against residuals, we can see if there is a pattern. Wait, maybe the original data's predicted values are from a linear model. Let's check the slope between predicted values: from x = 1 to x = 2, predicted goes from - 1.1 to 2, slope=(2 - (-1.1))/(2 - 1)=3.1. From x = 2 to x = 3, (5.1 - 2)/(3 - 2)=3.1. From x = 3 to x = 4, (8.2 - 5.1)/(4 - 3)=3.1. From x = 4 to x = 5, (11.3 - 8.2)/(5 - 4)=3.1. So the predicted values are from a linear model with slope 3.1. Now, the observed values: at x = 1, - 3.5; x = 2, - 2.9 (increase by 0.6); x = 3, - 1.1 (increase by 1.8); x = 4, 2.2 (increase by 3.3); x = 5, 3.4 (increase by 1.2). The observed values do not have a linear pattern, but the predicted values are linear. Now, the residuals: as x increases, the residuals (observed - predicted) are - 2.4, - 4.9, - 6.2, - 6.0, - 7.9. Wait, the residuals seem to have a pattern? No, wait, maybe I messed up. Wait, the correct way: if the residual plot has no pattern (random), the line of best fit is good. Let's check the options again.

Option 1: "Yes, the points have no pattern." If the residuals are randomly scattered, this is correct. But let's check the other options. Option 3: "No, the points are in a linear pattern." Let's see the residuals: - 2.4, - 4.9, - 6.2, - 6.0, - 7.9. Let's find the slope between (1, - 2.4) and (2, - 4.9): slope=(-4.9+2.4)/(2 - 1)=-2.5. Between (2, - 4.9) and (3, - 6.2): slope=(-6.2 + 4.9)/(3 - 2)=-1.3. Between (3, - 6.2) and (4, - 6.0): slope=(-6.0+6.2)/(4 - 3)=0.2. Between (4, - 6.0) and (5, - 7.9): slope=(-7.9 + 6.0)/(5 - 4)=-1.9. These slopes are not consistent, so maybe the residuals do not form a linear pattern. Wait, maybe the correct answer for the second part is Option 1: "Yes, the points have no pattern." But wait, let's check the residual calculation again. Wait, maybe the formula is \( \text{Residual}=\text{Predicted}-\text{Observed} \)? No, the standard formula is \( \text{Residual}=\text{Observed}-\text{Predicted} \).

Final Residual Table

x	Given	Predicted	Residual
2	-2.9	2	-4.9
3	-1.1	5.1	-6.2
4	2.2	8.2	-6.0
5	3.4	11.3	-7.9

Part 2 Answer

For the second part, a line of best fit is appropriate when the residual plot has no pattern (randomly scattered points).

• Option 1: "Yes, the points have no pattern." - This is correct because if residuals have no pattern, the linear model is appropriate.
• Option 2: "No, the points are evenly distributed about the x - axis." - Even distribution about the x - axis is a good sign, but the "No" is incorrect.
• Option 3: "No, the points are in a linear pattern." - Our residual slopes are not consistent, so no linear pattern.
• Option 4: "Yes, the points are in a curved pattern." - A curved pattern in residuals means the linear model is not appropriate.

Part 1 Answer (Residuals):

For \( x = 1 \): \(-2.4\); \( x = 2 \): \(-4.9\); \( x = 3 \): \(-6.2\); \( x = 4 \): \(-6.0\); \( x = 5 \): \(-7.9\)

Part 2 Answer:

A. Yes, the points have no pattern.