# Step-by-Step Solution Let's solve the problem step by step. --- ## **Step 1: Understanding the Problem** We need to predict the **sepal width** for **Iris setosa** when the **sepal length** is **7.79 cm** using the least squares line. - Predictor (independent variable, \( x \)): Sepal length - Response (dependent variable, \( y \)): Sepal width --- ## **Step 2: Get the Data (Iris Setosa Only)** The Iris dataset contains 50 samples for each species. The summary statistics for **Iris setosa** are: - Sepal length (mean): 5.006 cm - Sepal width (mean): 3.428 cm Let’s denote: - \( \bar{x} = \) mean sepal length = **5.006** - \( \bar{y} = \) mean sepal width = **3.428** The sample regression line is: \[ y = a + bx \] Where: - \( b = \frac{\sum (x_i - \bar{x})(y_i - \bar{y})}{\sum (x_i - \bar{x})^2} \) - \( a = \bar{y} - b\bar{x} \) For **Iris setosa** from the original dataset: - \( S_{x} = .352 \) (std dev of sepal length) - \( S_{y} = .379 \) (std dev of sepal width) - \( r = .742 \) (correlation between sepal length and width for setosa) --- ## **Step 3: Calculate the Slope (\( b \))** \[ b = r \frac{S_{y}}{S_{x}} \] \[ b = .742 \times \frac{.379}{.352} = .742 \times 1.0767 \approx .799 \] --- ## **Step 4: Calculate the Intercept (\( a \))** \[ a = \bar{y} - b \bar{x} \] \[ a = 3.428 - .799 \times 5.006 \] \[ a = 3.428 - 4.003 \approx -.575 \] --- ## **Step 5: Regression Line Equation** \[ \text{Predicted sepal width} = a + b \times (\text{sepal length}) \] \[ \text{Predicted sepal width} = -.575 + .799 \times (\text{sepal length}) \] --- ## **Step 6: Predict for Sepal Length 7.79** \[ \text{Predicted sepal width} = -.575 + .799 \times 7.79 \] \[ = -.575 + 6.223 \] \[ = 5.648 \text{ cm} \] --- # **Final Answer** **The predicted sepal width for Iris setosa with a sepal length of 7.79 cm is approximately:** \[ \boxed{5.65 \text{ cm}} \] --- ## **Summary Table** | Step | Calculation | Result | |------------------------|----------------------------------|-------------| | Mean Sepal Length | \( \bar{x} \) | 5.006 | | Mean Sepal Width | \( \bar{y} \) | 3.428 | | Slope (\( b \)) | \( .742 \times \frac{.379}{.352} \) | .799 | | Intercept (\( a \)) | \( 3.428 - .799 \times 5.006 \) | -.575 | | Regression Line | \( y = -.575 + .799x \) | | | Prediction at 7.79 | \( -.575 + .799 \times 7.79 \) | 5.65 cm | --- If you need the calculations for other species or more explanation, let me know!

# Detailed Calculations and Explanation Let's break down the calculations and provide more detailed explanations for each step. --- ## **Step 1: Understand the Variables** In this analysis, we are working with the **Iris setosa** species, focusing on: - **Independent Variable** \( (x) \): Sepal Length - **Dependent Variable** \( (y) \): Sepal Width We aim to find the relationship between these two variables using the least squares regression method. --- ## **Step 2: Data for Iris Setosa** ### Key Statistics - **Mean Sepal Length** \( (\bar{x}) \): 5.006 cm - **Mean Sepal Width** \( (\bar{y}) \): 3.428 cm - **Standard Deviation of Sepal Length** \( (S_x) \): 0.352 cm - **Standard Deviation of Sepal Width** \( (S_y) \): 0.379 cm - **Correlation Coefficient** \( (r) \): 0.742 These statistics are essential for calculating the slope and intercept of the regression line. --- ## **Step 3: Calculate the Slope (\( b \))** The slope \( b \) of the regression line is calculated using the formula: \[ b = r \frac{S_{y}}{S_{x}} \] ### Calculation Substituting the known values: \[ b = 0.742 \times \frac{0.379}{0.352} \] Calculating the ratio: \[ \frac{0.379}{0.352} \approx 1.0767 \] Now calculating \( b \): \[ b \approx 0.742 \times 1.0767 \approx 0.799 \] --- ## **Step 4: Calculate the Intercept (\( a \))** The intercept \( a \) can be calculated using the formula: \[ a = \bar{y} - b\bar{x} \] ### Calculation Substituting the known values: \[ a = 3.428 - 0.799 \times 5.006 \] Calculating \( 0.799 \times 5.006 \): \[ 0.799 \times 5.006 \approx 4.003 \] Now substituting back to find \( a \): \[ a \approx 3.428 - 4.003 \approx -0.575 \] --- ## **Step 5: Formulate the Regression Equation** The regression line equation is: \[ y = a + bx \] Substituting \( a \) and \( b \): \[ y = -0.575 + 0.799x \] --- ## **Step 6: Predict Sepal Width for Sepal Length 7.79** Now we can utilize the regression equation to predict the sepal width for a sepal length of 7.79 cm: \[ \text{Predicted sepal width} = -0.575 + 0.799 \times 7.79 \] ### Calculation First, calculate \( 0.799 \times 7.79 \): \[ 0.799 \times 7.79 \approx 6.223 \] Now substitute back into the equation: \[ \text{Predicted sepal width} \approx -0.575 + 6.223 \approx 5.648 \text{ cm} \] --- # **Final Answer** The predicted sepal width for **Iris setosa** with a sepal length of **7.79 cm** is approximately: \[ \boxed{5.65 \text{ cm}} \] --- ## **Summary of Key Steps and Calculations** | Step | Formula/Calculation | Result | |------------------------|----------------------------------------------|-------------| | Mean Sepal Length | \( \bar{x} \) | 5.006 | | Mean Sepal Width | \( \bar{y} \) | 3.428 | | Slope (\( b \)) | \( 0.742 \times \frac{0.379}{0.352} \) | 0.799 | | Intercept (\( a \)) | \( 3.428 - 0.799 \times 5.006 \) | -0.575 | | Regression Equation | \( y = -0.575 + 0.799x \) | | | Prediction at 7.79 | \( -0.575 + 0.799 \times 7.79 \) | 5.65 cm | --- If you require further details or have additional questions, feel free to ask!