VIPSolutionsGive step-by-step solution with explanation and final answer: What is the predicted sepal width for iris setosa for a flower with a sepal length of ? The famous iris dataset ( the first sheet of the spreadsheet linked above ) was first published in 1 9 3 6 by Ronald Fisher. The dataset contains 5 0 samples from 3 iris species: setosa, virginia, and versicolor. Four features are measured, all in cm: sepal length, sepal width, petal length, and petal width.What is the equation for the least square regression line where the independent or predictor variable is sepal length and the dependent or response variable is sepal width for iris setosa for a flower with a sepal length of 7 . 7 9
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Give step-by-step solution with explanation and final answer: What is the predicted sepal width for iris setosa for a flower with a sepal length of ? The famous iris dataset ( the first sheet of the spreadsheet linked above ) was first published in 1 9 3 6 by Ronald Fisher. The dataset contains 5 0 samples from 3 iris species: setosa, virginia, and versicolor. Four features are measured, all in cm: sepal length, sepal width, petal length, and petal width.What is the equation for the least square regression line where the independent or predictor variable is sepal length and the dependent or response variable is sepal width for iris setosa for a flower with a sepal length of 7 . 7 9
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Asked by: anil
Created at: 2025-09-27 22:09:10
Asked by: anil
Created at: 2025-09-27 22:09:10
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# Step-by-Step Solution
We are asked:
- What is the **predicted sepal** for **iris setosa** **sepal length = 7.9 cm**?
- What is the **least squares regression equation** for sepal width (dependent variable) vs. sepal length (independent variable) for iris setosa?
Let's proceed step by step.
---
## 1. **Regression Line Equation**
The simple linear regression equation is:
\[
\hat{y} = b_ + b_1 x
\]
Where:
- \( \hat{y} \) = predicted sepal width
- \( x \) = sepal length
- \( b_ \) = intercept
- \( b_1 \) = slope
---
## 2. **Extracting Setosa Data from Iris Dataset**
For **iris setosa** (first 50 rows in the classic iris dataset):
- **Mean sepal length** (\( \bar{x} \)) = 5.006 cm
- **Mean sepal width** (\( \bar{y} \)) = 3.428 cm
- **Sample standard deviations**:
- \( s_x = .3525 \)
- \( s_y = .3791 \)
- **Pearson correlation** (\( r \)) between sepal length and sepal width for setosa:
- \( r = -.1176 \)
---
## 3. **Finding Slope (\( b_1 \)) and Intercept (\( b_ \))**
\[
b_1 = r \frac{s_y}{s_x}
\]
\[
b_1 = -.1176 \times \frac{.3791}{.3525}
\approx -.1176 \times 1.0754
\approx -.1265
\]
\[
b_ = \bar{y} - b_1 \bar{x}
\]
\[
b_ = 3.428 - (-.1265) \times 5.006
\]
\[
b_ = 3.428 + .6335
\approx 4.0615
\]
**Regression equation:**
\[
\boxed{
\hat{y} = 4.0615 - .1265x
}
\]
---
## 4. **Predict Sepal Width at Sepal Length = 7.9**
Plug in \( x = 7.9 \):
\[
\hat{y} = 4.0615 - .1265 \times 7.9
\]
\[
= 4.0615 - 1.0004
\]
\[
= 3.0611
\]
---
## 5. **Final Answers**
### Regression Line Equation
\[
\boxed{
\hat{y} = 4.0615 - .1265x
}
\]
Where \( x \) = sepal length (cm), \( \hat{y} \) = predicted sepal width (cm).
### Predicted Sepal Width for Sepal Length = 7.9 cm
\[
\boxed{
\text{Predicted sepal width} = 3.06\ \text{cm} \ (\text{rounded to 2 decimal places})
}
\]
---
## **Summary Table**
| Sepal Length (cm) | Predicted Sepal Width (cm) |
|-------------------|---------------------------|
| 7.9 | 3.06 |
---
### **Note**
- The negative slope (\( b_1 \)) reflects the weak negative correlation in setosa between sepal length and width.
- This calculation uses classic iris dataset summary statistics. If you use your own data, recalculate means, standard deviations, and correlation.
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