To find the least squares linear regression line, we need to determine the linear relationship between the size of the group (independent variable [tex]\( x \)[/tex]) and the number of hours spent on the job (dependent variable [tex]\( y \)[/tex]). The equation of the line can be written as:
[tex]\[ y = b_0 + b_1 x \][/tex]
where [tex]\( b_0 \)[/tex] is the y-intercept and [tex]\( b_1 \)[/tex] is the slope of the line.
We proceed with the following steps:
1. Calculate the mean of [tex]\( x \)[/tex] and [tex]\( y \)[/tex]:
[tex]\[
\text{mean\_size} = \frac{1}{n} \sum_{i=1}^{n} x_i
\][/tex]
[tex]\[
\text{mean\_hours} = \frac{1}{n} \sum_{i=1}^{n} y_i
\][/tex]
Given the data:
[tex]\[
\text{sizes} = [5, 8, 4, 6, 10, 3, 5, 9, 10, 7]
\][/tex]
[tex]\[
\text{hours} = [10, 7, 13, 4, 8, 1, 7, 3, 2, 5]
\][/tex]
From the result provided, we know:
[tex]\[
\text{mean\_size} = 6.7
\][/tex]
[tex]\[
\text{mean\_hours} = 6.0
\][/tex]
2. Calculate the slope [tex]\( b_1 \)[/tex]:
[tex]\[
b_1 = \frac{\sum{(x_i - \text{mean\_size})(y_i - \text{mean\_hours})}}{\sum{(x_i - \text{mean\_size})^2}}
\][/tex]
The numerator for [tex]\( b_1 \)[/tex] is:
[tex]\[
\text{b1\_numerator} = -20.0
\][/tex]
The denominator for [tex]\( b_1 \)[/tex] is:
[tex]\[
\text{b1\_denominator} = 56.1
\][/tex]
Therefore, the slope [tex]\( b_1 \)[/tex] is:
[tex]\[
b_1 = \frac{-20.0}{56.1} \approx -0.3565
\][/tex]
3. Calculate the intercept [tex]\( b_0 \)[/tex]:
[tex]\[
b_0 = \text{mean\_hours} - b_1 \cdot \text{mean\_size}
\][/tex]
Using the calculated values:
[tex]\[
b_0 = 6.0 - (-0.3565 \cdot 6.7) \approx 8.389
\][/tex]
Therefore, the least squares linear regression line is:
[tex]\[
y = 8.389 - 0.357x
\][/tex]
In conclusion:
- The slope [tex]\( b_1 \)[/tex] is approximately -0.357.
- The intercept [tex]\( b_0 \)[/tex] is approximately 8.389.
