Question
Find the equation for the least squares regression line of the data described below.
The math teachers at Springdale High School had never golfed before. They set up a friendly competition to see who would have the best round on a full course. The teachers practiced on a driving range the week before they played.
The statistics teacher tracked the number of hours practiced, x, and the final score, y, of each player.
Hours practiced Score
0.88 115
4.35 116
4.46 110
4.58 108
4.59 105
9.95 101
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 will use the formula for the line of best fit, which is given by:
\[
y = mx + b
\]
where \( m \) is the slope and \( b \) is the y-intercept. We need to first compute the necessary statistical values to find \( m \) and \( b \).
Here are the data points:
- (0.88, 115)
- (4.35, 116)
- (4.46, 110)
- (4.58, 108)
- (4.59, 105)
- (9.95, 101)
Let's compute the required sums:
1. \( n \) = number of data points = 6
2. \( \sum x \) = \( 0.88 + 4.35 + 4.46 + 4.58 + 4.59 + 9.95 = 29.81 \)
3. \( \sum y \) = \( 115 + 116 + 110 + 108 + 105 + 101 = 655 \)
4. \( \sum xy \) = \( (0.88 \cdot 115) + (4.35 \cdot 116) + (4.46 \cdot 110) + (4.58 \cdot 108) + (4.59 \cdot 105) + (9.95 \cdot 101) \)
- \( = 101.2 + 505.6 + 490.6 + 494.64 + 482.95 + 1004.95 = 3079.99 \)
5. \( \sum x^2 \) = \( (0.88^2) + (4.35^2) + (4.46^2) + (4.58^2) + (4.59^2) + (9.95^2) \)
- \( = 0.7744 + 18.9225 + 19.7616 + 20.9764 + 20.0681 + 99.0025 = 179.5055 \)
Using these sums, we can calculate \( m \) and \( b \):
\[
m = \frac{n(\sum xy) - (\sum x)(\sum y)}{n(\sum x^2) - (\sum x)^2}
\]
Substituting in our values:
\[
m = \frac{6(3079.99) - (29.81)(655)}{6(179.5055) - (29.81)^2}
\]
Calculating each term:
- \( n(\sum xy) = 6 \times 3079.99 = 18479.94 \)
- \( (\sum x)(\sum y) = 29.81 \times 655 = 19544.55 \)
- \( n(\sum x^2) = 6 \times 179.5055 = 1077.033 \)
- \( (\sum x)^2 = (29.81)^2 = 888.6561 \)
Thus:
\[
m = \frac{18479.94 - 19544.55}{1077.033 - 888.6561} = \frac{-1064.61}{188.3769} \approx -5.655
\]
Now, calculating \( b \):
\[
b = \frac{\sum y - m(\sum x)}{n}
\]
Substituting the values:
\[
b = \frac{655 - (-5.655)(29.81)}{6}
\]
Calculating \( -5.655 \times 29.81 \approx -168.50 \):
\[
b = \frac{655 + 168.50}{6} = \frac{823.50}{6} \approx 137.25
\]
Now we can write the regression line:
\[
y \approx -5.655x + 137.250
\]
Therefore, rounding to the nearest thousandth, the least squares regression line is:
\[
\boxed{y = -5.655x + 137.250}
\]
\[
y = mx + b
\]
where \( m \) is the slope and \( b \) is the y-intercept. We need to first compute the necessary statistical values to find \( m \) and \( b \).
Here are the data points:
- (0.88, 115)
- (4.35, 116)
- (4.46, 110)
- (4.58, 108)
- (4.59, 105)
- (9.95, 101)
Let's compute the required sums:
1. \( n \) = number of data points = 6
2. \( \sum x \) = \( 0.88 + 4.35 + 4.46 + 4.58 + 4.59 + 9.95 = 29.81 \)
3. \( \sum y \) = \( 115 + 116 + 110 + 108 + 105 + 101 = 655 \)
4. \( \sum xy \) = \( (0.88 \cdot 115) + (4.35 \cdot 116) + (4.46 \cdot 110) + (4.58 \cdot 108) + (4.59 \cdot 105) + (9.95 \cdot 101) \)
- \( = 101.2 + 505.6 + 490.6 + 494.64 + 482.95 + 1004.95 = 3079.99 \)
5. \( \sum x^2 \) = \( (0.88^2) + (4.35^2) + (4.46^2) + (4.58^2) + (4.59^2) + (9.95^2) \)
- \( = 0.7744 + 18.9225 + 19.7616 + 20.9764 + 20.0681 + 99.0025 = 179.5055 \)
Using these sums, we can calculate \( m \) and \( b \):
\[
m = \frac{n(\sum xy) - (\sum x)(\sum y)}{n(\sum x^2) - (\sum x)^2}
\]
Substituting in our values:
\[
m = \frac{6(3079.99) - (29.81)(655)}{6(179.5055) - (29.81)^2}
\]
Calculating each term:
- \( n(\sum xy) = 6 \times 3079.99 = 18479.94 \)
- \( (\sum x)(\sum y) = 29.81 \times 655 = 19544.55 \)
- \( n(\sum x^2) = 6 \times 179.5055 = 1077.033 \)
- \( (\sum x)^2 = (29.81)^2 = 888.6561 \)
Thus:
\[
m = \frac{18479.94 - 19544.55}{1077.033 - 888.6561} = \frac{-1064.61}{188.3769} \approx -5.655
\]
Now, calculating \( b \):
\[
b = \frac{\sum y - m(\sum x)}{n}
\]
Substituting the values:
\[
b = \frac{655 - (-5.655)(29.81)}{6}
\]
Calculating \( -5.655 \times 29.81 \approx -168.50 \):
\[
b = \frac{655 + 168.50}{6} = \frac{823.50}{6} \approx 137.25
\]
Now we can write the regression line:
\[
y \approx -5.655x + 137.250
\]
Therefore, rounding to the nearest thousandth, the least squares regression line is:
\[
\boxed{y = -5.655x + 137.250}
\]