Question
Find the equation for the least squares regression line of the data described below.
Alana teaches a standardized test prep course. Students take a diagnostic test at the beginning of the course and then take the real test at the end. Alana wanted to know how well the diagnostic test predicts a student's score on the real test.
She recorded each student's score on the diagnostic test, x, and on the real test, y. Both tests were graded on a scale from 120 to 180.
Diagnostic test score Real test score
152 129
154 126
156 135
169 151
178 172
Round your answers to the nearest thousandth.
y=
x+
Answers
GPT-4o mini
To find the least squares regression line for the given data, we first need to calculate the necessary statistics. Specifically, we need the means (\( \bar{x} \) and \( \bar{y} \)), the sums of squares, and the sums of products needed to find the slope (\( b \)) and intercept (\( a \)) of the regression line.
The least squares regression line is given in the form:
\[ y = bx + a \]
where
- \( b \) is the slope,
- \( a \) is the y-intercept.
Let's summarize the data:
| Diagnostic Test Score (x) | Real Test Score (y) |
|----------------------------|----------------------|
| 152 | 129 |
| 154 | 126 |
| 156 | 135 |
| 169 | 151 |
| 178 | 172 |
### Step 1: Calculate the means of x and y
\[
\bar{x} = \frac{152 + 154 + 156 + 169 + 178}{5} = \frac{809}{5} = 161.8
\]
\[
\bar{y} = \frac{129 + 126 + 135 + 151 + 172}{5} = \frac{713}{5} = 142.6
\]
### Step 2: Calculate the slope (b)
To find \( b \), we use the formula:
\[
b = \frac{\sum (x_i - \bar{x})(y_i - \bar{y})}{\sum (x_i - \bar{x})^2}
\]
#### Calculate \( \sum (x_i - \bar{x})(y_i - \bar{y}) \)
\[
\begin{align*}
152: & \quad (152 - 161.8)(129 - 142.6) = (-9.8)(-13.6) = 133.28 \\
154: & \quad (154 - 161.8)(126 - 142.6) = (-7.8)(-16.6) = 129.48 \\
156: & \quad (156 - 161.8)(135 - 142.6) = (-5.8)(-7.6) = 44.08 \\
169: & \quad (169 - 161.8)(151 - 142.6) = (7.2)(8.4) = 60.48 \\
178: & \quad (178 - 161.8)(172 - 142.6) = (16.2)(29.4) = 476.28 \\
\end{align*}
\]
Now sum these products:
\[
\sum (x_i - \bar{x})(y_i - \bar{y}) = 133.28 + 129.48 + 44.08 + 60.48 + 476.28 = 843.6
\]
#### Calculate \( \sum (x_i - \bar{x})^2 \)
\[
\begin{align*}
152: & \quad (152 - 161.8)^2 = (-9.8)^2 = 96.04 \\
154: & \quad (154 - 161.8)^2 = (-7.8)^2 = 60.84 \\
156: & \quad (156 - 161.8)^2 = (-5.8)^2 = 33.64 \\
169: & \quad (169 - 161.8)^2 = (7.2)^2 = 51.84 \\
178: & \quad (178 - 161.8)^2 = (16.2)^2 = 262.44 \\
\end{align*}
\]
Now sum these squares:
\[
\sum (x_i - \bar{x})^2 = 96.04 + 60.84 + 33.64 + 51.84 + 262.44 = 504.80
\]
Now we can calculate \( b \):
\[
b = \frac{843.6}{504.80} \approx 1.670
\]
### Step 3: Calculate the intercept (a)
The intercept \( a \) is given by:
\[
a = \bar{y} - b\bar{x}
\]
Calculating \( a \):
\[
a = 142.6 - (1.670)(161.8)
\]
\[
a \approx 142.6 - 270.366 = -127.766
\]
### Final Equation
Round \( a \) and \( b \) to the nearest thousandth:
- \( a \approx -127.766 \) rounds to \(-127.766\),
- \( b \approx 1.670 \) rounds to \(1.670\).
Thus, the least squares regression line is:
\[
y = 1.670x - 127.766
\]
The least squares regression line is given in the form:
\[ y = bx + a \]
where
- \( b \) is the slope,
- \( a \) is the y-intercept.
Let's summarize the data:
| Diagnostic Test Score (x) | Real Test Score (y) |
|----------------------------|----------------------|
| 152 | 129 |
| 154 | 126 |
| 156 | 135 |
| 169 | 151 |
| 178 | 172 |
### Step 1: Calculate the means of x and y
\[
\bar{x} = \frac{152 + 154 + 156 + 169 + 178}{5} = \frac{809}{5} = 161.8
\]
\[
\bar{y} = \frac{129 + 126 + 135 + 151 + 172}{5} = \frac{713}{5} = 142.6
\]
### Step 2: Calculate the slope (b)
To find \( b \), we use the formula:
\[
b = \frac{\sum (x_i - \bar{x})(y_i - \bar{y})}{\sum (x_i - \bar{x})^2}
\]
#### Calculate \( \sum (x_i - \bar{x})(y_i - \bar{y}) \)
\[
\begin{align*}
152: & \quad (152 - 161.8)(129 - 142.6) = (-9.8)(-13.6) = 133.28 \\
154: & \quad (154 - 161.8)(126 - 142.6) = (-7.8)(-16.6) = 129.48 \\
156: & \quad (156 - 161.8)(135 - 142.6) = (-5.8)(-7.6) = 44.08 \\
169: & \quad (169 - 161.8)(151 - 142.6) = (7.2)(8.4) = 60.48 \\
178: & \quad (178 - 161.8)(172 - 142.6) = (16.2)(29.4) = 476.28 \\
\end{align*}
\]
Now sum these products:
\[
\sum (x_i - \bar{x})(y_i - \bar{y}) = 133.28 + 129.48 + 44.08 + 60.48 + 476.28 = 843.6
\]
#### Calculate \( \sum (x_i - \bar{x})^2 \)
\[
\begin{align*}
152: & \quad (152 - 161.8)^2 = (-9.8)^2 = 96.04 \\
154: & \quad (154 - 161.8)^2 = (-7.8)^2 = 60.84 \\
156: & \quad (156 - 161.8)^2 = (-5.8)^2 = 33.64 \\
169: & \quad (169 - 161.8)^2 = (7.2)^2 = 51.84 \\
178: & \quad (178 - 161.8)^2 = (16.2)^2 = 262.44 \\
\end{align*}
\]
Now sum these squares:
\[
\sum (x_i - \bar{x})^2 = 96.04 + 60.84 + 33.64 + 51.84 + 262.44 = 504.80
\]
Now we can calculate \( b \):
\[
b = \frac{843.6}{504.80} \approx 1.670
\]
### Step 3: Calculate the intercept (a)
The intercept \( a \) is given by:
\[
a = \bar{y} - b\bar{x}
\]
Calculating \( a \):
\[
a = 142.6 - (1.670)(161.8)
\]
\[
a \approx 142.6 - 270.366 = -127.766
\]
### Final Equation
Round \( a \) and \( b \) to the nearest thousandth:
- \( a \approx -127.766 \) rounds to \(-127.766\),
- \( b \approx 1.670 \) rounds to \(1.670\).
Thus, the least squares regression line is:
\[
y = 1.670x - 127.766
\]