Let's go through a detailed, step-by-step approach to solve this problem using linear regression.
### Step 1: Understanding the Dataset
The dataset contains measurements for three species of iris flowers: setosa, virginia, and versicolor. Each species has 50 samples, and four features are measured:
- Sepal length
- Sepal width
- Petal length
- Petal width
We are specifically interested in the Iris virginica species and need to find the least squares regression line where the predictor variable is sepal length ([tex]\(x\)[/tex]) and the response variable is sepal width ([tex]\(y\)[/tex]).
### Step 2: Filter for Iris Virginica
We extract the data for the species "Iris-virginica".
### Step 3: Formulating the Linear Regression Problem
The linear regression model can be represented by the equation:
[tex]\[
\hat{y} = b_0 + b_1 x
\][/tex]
where:
- [tex]\(\hat{y}\)[/tex] is the predicted sepal width
- [tex]\(b_0\)[/tex] is the y-intercept
- [tex]\(b_1\)[/tex] is the slope of the regression line
### Step 4: Calculate Means of [tex]\(x\)[/tex] and [tex]\(y\)[/tex]
To begin solving for [tex]\(b_0\)[/tex] and [tex]\(b_1\)[/tex], calculate the average (mean) of the sepal lengths ([tex]\(\bar{x}\)[/tex]) and the average (mean) of the sepal widths ([tex]\(\bar{y}\)[/tex]).
### Step 5: Calculate the Slope ([tex]\(b_1\)[/tex])
The slope [tex]\(b_1\)[/tex] is determined by the following formula:
[tex]\[
b_1 = \frac{\sum (x_i - \bar{x})(y_i - \bar{y})}{\sum (x_i - \bar{x})^2}
\][/tex]
### Step 6: Calculate the Y-Intercept ([tex]\(b_0\)[/tex])
The y-intercept [tex]\(b_0\)[/tex] can be calculated by:
[tex]\[
b_0 = \bar{y} - b_1 \bar{x}
\][/tex]
### Step 7: Formulating the Regression Line
Once we have [tex]\(b_0\)[/tex] and [tex]\(b_1\)[/tex], the regression equation [tex]\(\hat{y}\)[/tex] can be written.
### Step 8: Predict Sepal Width for Sepal Length of 5.57 cm
After deriving the regression equation, we substitute [tex]\(x = 5.57\)[/tex] into the equation to predict the corresponding sepal width.
### Example Calculations (Hypothetical Data for Illustrative Purposes)
Assume for Iris virginica:
- Average sepal length ([tex]\(\bar{x}\)[/tex]) = 6.59 cm
- Average sepal width ([tex]\(\bar{y}\)[/tex]) = 2.97 cm
- Sum of products of the deviations: [tex]\(\sum (x_i - \bar{x})(y_i - \bar{y}) = 15.02\)[/tex]
- Sum of squared deviations: [tex]\(\sum (x_i - \bar{x})^2 = 7.81\)[/tex]
So,
[tex]\[ b_1 = \frac{15.02}{7.81} = 1.922 \][/tex]
[tex]\[ b_0 = 2.97 - (1.922 \times 6.59) = 2.97 - 12.664 = -9.694 \][/tex]
The regression equation is:
[tex]\[ \hat{y} = 1.922 x - 9.694 \][/tex]
### Predicting Sepal Width for Sepal Length of 5.57 cm:
[tex]\[ \hat{y} = 1.922 \times 5.57 - 9.694 \][/tex]
[tex]\[ \hat{y} = 10.70434 - 9.694 \][/tex]
[tex]\[ \hat{y} = 1.010 \][/tex]
### Final Answers:
1. The least square regression line equation is:
[tex]\[ \hat{y} = 1.922 x - 9.694 \][/tex]
2. The predicted sepal width for a sepal length of 5.57 cm is:
[tex]\[ \hat{y} = 1.010 \, \text{cm} \][/tex]
Please note: The numerical values used here for illustration are arbitrary. Using the actual data from the Iris dataset will yield precise numbers, which should be computed using dedicated statistical software or programming tools.
