# Given Information Data follows the model: \( Y = \beta_ + \beta_1 x + \epsilon \) Given summary statistics: - \( \sum x = 20 \) - \( \sum Y = 50 \) - \( \sum x^2 = 120 \) - \( \sum xY = 180 \) - \( n = 5 \) # What to Find Find the least-squares estimates of \( \beta_ \) and \( \beta_1 \). # Definition / Concept Used For a simple linear regression \( Y = \beta_ + \beta_1 x + \epsilon \), the least-squares estimates are: - \( \hat{\beta}_1 = \dfrac{n\sum xY - (\sum x)(\sum Y)}{n\sum x^2 - (\sum x)^2} \) - \( \hat{\beta}_ = \overline{Y} - \hat{\beta}_1 \overline{x} \) where \( \overline{x} = \dfrac{\sum x}{n} \), \( \overline{Y} = \dfrac{\sum Y}{n} \). # Step-by-Step Solution Calculate sample means: - \( \overline{x} = \dfrac{20}{5} = 4 \) - \( \overline{Y} = \dfrac{50}{5} = 10 \) Find \( \hat{\beta}_1 \): \[ \hat{\beta}_1 = \frac{5 \times 180 - 20 \times 50}{5 \times 120 - 20^2} \] \[ \hat{\beta}_1 = \frac{900 - 100}{600 - 400} \] \[ \hat{\beta}_1 = \frac{-100}{200} = -.5 \] Find \( \hat{\beta}_ \): \[ \hat{\beta}_ = 10 - (-.5 \times 4) \] \[ \hat{\beta}_ = 10 + 2 = 12 \] # Summary - \( \hat{\beta}_ = 12 \) - \( \hat{\beta}_1 = -.5 \)

# Given Information The data follows the linear model: \( Y = \beta_0 + \beta_1 x + \epsilon \) Provided summary statistics: - \( \sum x = 20 \) - \( \sum Y = 50 \) - \( \sum x^2 = 120 \) - \( \sum xY = 180 \) - \( n = 5 \) # What to Find Determine the least-squares estimates of \( \beta_0 \) and \( \beta_1 \). # Definition / Concept Used In simple linear regression, the least-squares estimates are defined as follows: - \( \hat{\beta}_1 = \dfrac{n\sum xY - (\sum x)(\sum Y)}{n\sum x^2 - (\sum x)^2} \) - \( \hat{\beta}_0 = \overline{Y} - \hat{\beta}_1 \overline{x} \) where \( \overline{x} = \dfrac{\sum x}{n} \) and \( \overline{Y} = \dfrac{\sum Y}{n} \). # Step-by-Step Solution Calculate the sample means: - \( \overline{x} = \dfrac{20}{5} = 4 \) - \( \overline{Y} = \dfrac{50}{5} = 10 \) Compute \( \hat{\beta}_1 \): \[ \hat{\beta}_1 = \frac{5 \times 180 - 20 \times 50}{5 \times 120 - 20^2} \] \[ = \frac{900 - 100}{600 - 400} \] \[ = \frac{800}{200} = 4 \] Calculate \( \hat{\beta}_0 \): \[ \hat{\beta}_0 = 10 - (4 \times 4) \] \[ = 10 - 16 = -6 \] # Summary - \( \hat{\beta}_0 = -6 \) - \( \hat{\beta}_1 = 4 \)

# Given Information The data follows the linear model: \( Y = \beta_0 + \beta_1 x + \epsilon \) Provided summary statistics: - \( \sum x = 20 \) - \( \sum Y = 50 \) - \( \sum x^2 = 120 \) - \( \sum xY = 180 \) - \( n = 5 \) # What to Find Determine the least-squares estimates of \( \beta_0 \) and \( \beta_1 \). # Definition / Concept Used In simple linear regression, the least-squares estimates are calculated as follows: - \( \hat{\beta}_1 = \frac{n\sum xY - \sum x \sum Y}{n\sum x^2 - (\sum x)^2} \) - \( \hat{\beta}_0 = \overline{Y} - \hat{\beta}_1 \overline{x} \) where \( \overline{x} = \frac{\sum x}{n} \) and \( \overline{Y} = \frac{\sum Y}{n} \). # Step-by-Step Solution Calculate the sample means: - \( \overline{x} = \frac{20}{5} = 4 \) - \( \overline{Y} = \frac{50}{5} = 10 \) Compute \( \hat{\beta}_1 \): \[ \hat{\beta}_1 = \frac{5 \times 180 - 20 \times 50}{5 \times 120 - 20^2} \] Breaking it down: - Numerator: \( 900 - 100 = 800 \) - Denominator: \( 600 - 400 = 200 \) Thus, \[ \hat{\beta}_1 = \frac{800}{200} = 4 \] Calculate \( \hat{\beta}_0 \): \[ \hat{\beta}_0 = 10 - (4 \times 4) \] \[ = 10 - 16 = -6 \] # Summary - \( \hat{\beta}_0 = -6 \) - \( \hat{\beta}_1 = 4 \)

# Given Information The model follows the equation: \( Y = \beta_0 + \beta_1 X + \epsilon \) Provided data: - \( \sum X = 20 \) - \( \sum Y = 50 \) - \( \sum X^2 = 120 \) - \( \sum XY = 180 \) - \( n = 5 \) # What to Find Calculate the least-squares estimates of \( \beta_0 \) and \( \beta_1 \). # Definition / Concept Used The least-squares estimates for simple linear regression are computed using the following formulas: - Slope: \[ \hat{\beta}_1 = \frac{S_{xy}}{S_{xx}} \] - Intercept: \[ \hat{\beta}_0 = \overline{Y} - \hat{\beta}_1 \overline{X} \] Where: - \( S_{xx} = \sum X^2 - \frac{(\sum X)^2}{n} \) - \( S_{xy} = \sum XY - \frac{(\sum X)(\sum Y)}{n} \) - \( \overline{X} = \frac{\sum X}{n} \) - \( \overline{Y} = \frac{\sum Y}{n} \) # Step-by-Step Solution Calculate the sample means: - \( \overline{X} = \frac{20}{5} = 4 \) - \( \overline{Y} = \frac{50}{5} = 10 \) Compute \( S_{xx} \): \[ S_{xx} = \sum X^2 - \frac{(\sum X)^2}{n} \] \[ = 120 - \frac{20^2}{5} = 120 - \frac{400}{5} = 120 - 80 = 40 \] Compute \( S_{xy} \): \[ S_{xy} = \sum XY - \frac{(\sum X)(\sum Y)}{n} \] \[ = 180 - \frac{20 \times 50}{5} = 180 - \frac{1000}{5} = 180 - 200 = -20 \] Calculate the slope \( \hat{\beta}_1 \): \[ \hat{\beta}_1 = \frac{S_{xy}}{S_{xx}} = \frac{-20}{40} = -0.5 \] Calculate the intercept \( \hat{\beta}_0 \): \[ \hat{\beta}_0 = \overline{Y} - \hat{\beta}_1 \overline{X} \] \[ = 10 - (-0.5 \times 4) = 10 + 2 = 12 \] # Summary - \( \hat{\beta}_0 = 12 \) - \( \hat{\beta}_1 = -0.5 \)

# Given Information The data follows the model: \( Y = \beta_0 + \beta_1 X + \epsilon \) Provided summary statistics: - \( \sum X = 20 \) - \( \sum Y = 50 \) - \( \sum X^2 = 120 \) - \( \sum XY = 180 \) - \( n = 5 \) # What to Find Calculate the least-squares estimates of \( \beta_0 \) and \( \beta_1 \). # Definition / Concept Used In simple linear regression, the estimates are calculated using: - Slope: \[ \hat{\beta}_1 = \frac{S_{xy}}{S_{xx}} \] - Intercept: \[ \hat{\beta}_0 = \overline{Y} - \hat{\beta}_1 \overline{X} \] Where: - \( S_{xx} = \sum X^2 - \frac{(\sum X)^2}{n} \) - \( S_{xy} = \sum XY - \frac{(\sum X)(\sum Y)}{n} \) - \( \overline{X} = \frac{\sum X}{n} \) - \( \overline{Y} = \frac{\sum Y}{n} \) # Step-by-Step Solution Calculate the sample means: - \( \overline{X} = \frac{20}{5} = 4 \) - \( \overline{Y} = \frac{50}{5} = 10 \) Compute \( S_{xx} \): \[ S_{xx} = \sum X^2 - \frac{(\sum X)^2}{n} \] \[ = 120 - \frac{20^2}{5} = 120 - \frac{400}{5} = 120 - 80 = 40 \] Compute \( S_{xy} \): \[ S_{xy} = \sum XY - \frac{(\sum X)(\sum Y)}{n} \] \[ = 180 - \frac{20 \times 50}{5} = 180 - \frac{1000}{5} = 180 - 200 = -20 \] Calculate the slope \( \hat{\beta}_1 \): \[ \hat{\beta}_1 = \frac{S_{xy}}{S_{xx}} = \frac{-20}{40} = -0.5 \] Calculate the intercept \( \hat{\beta}_0 \): \[ \hat{\beta}_0 = \overline{Y} - \hat{\beta}_1 \overline{X} \] \[ = 10 - (-0.5 \times 4) = 10 + 2 = 12 \] # Summary - \( \hat{\beta}_0 = 12 \) - \( \hat{\beta}_1 = -0.5 \)

# Given Information The model is defined as: \( Y = \beta_0 + \beta_1 X + \epsilon \) Provided statistics: - \( \sum X = 20 \) - \( \sum Y = 50 \) - \( \sum X^2 = 120 \) - \( \sum XY = 180 \) - \( n = 5 \) # Objective Determine the least-squares estimates for \( \beta_0 \) and \( \beta_1 \). # Definitions / Concepts Used The least-squares estimates are given by: - Slope: \[ \hat{\beta}_1 = \frac{S_{xy}}{S_{xx}} \] - Intercept: \[ \hat{\beta}_0 = \overline{Y} - \hat{\beta}_1 \overline{X} \] Where: - \( S_{xx} = \sum X^2 - \frac{(\sum X)^2}{n} \) - \( S_{xy} = \sum XY - \frac{(\sum X)(\sum Y)}{n} \) - \( \overline{X} = \frac{\sum X}{n} \) - \( \overline{Y} = \frac{\sum Y}{n} \) # Detailed Solution Steps First, calculate the averages: - \( \overline{X} = \frac{20}{5} = 4 \) - \( \overline{Y} = \frac{50}{5} = 10 \) Next, compute \( S_{xx} \): \[ S_{xx} = \sum X^2 - \frac{(\sum X)^2}{n} \] \[ = 120 - \frac{20^2}{5} = 120 - \frac{400}{5} = 120 - 80 = 40 \] Now, calculate \( S_{xy} \): \[ S_{xy} = \sum XY - \frac{(\sum X)(\sum Y)}{n} \] \[ = 180 - \frac{20 \times 50}{5} = 180 - \frac{1000}{5} = 180 - 200 = -20 \] With \( S_{xy} \) and \( S_{xx} \) determined, find the slope \( \hat{\beta}_1 \): \[ \hat{\beta}_1 = \frac{S_{xy}}{S_{xx}} = \frac{-20}{40} = -0.5 \] Finally, calculate the intercept \( \hat{\beta}_0 \): \[ \hat{\beta}_0 = \overline{Y} - \hat{\beta}_1 \overline{X} \] \[ = 10 - (-0.5 \times 4) = 10 + 2 = 12 \] # Summary of Results - \( \hat{\beta}_0 = 12 \) - \( \hat{\beta}_1 = -0.5 \)

# Given Information The linear model is represented as: \( Y = \beta_0 + \beta_1 X + \epsilon \) Provided data includes: - \( \sum X = 20 \) - \( \sum Y = 50 \) - \( \sum X^2 = 120 \) - \( \sum XY = 180 \) - \( n = 5 \) # Objective Find the least-squares estimates for \( \beta_0 \) and \( \beta_1 \). # Definitions / Concepts Used To find the least-squares estimates, the following equations are utilized: - Slope (\( \hat{\beta}_1 \)): \[ \hat{\beta}_1 = \frac{S_{xy}}{S_{xx}} \] - Intercept (\( \hat{\beta}_0 \)): \[ \hat{\beta}_0 = \overline{Y} - \hat{\beta}_1 \overline{X} \] Where: - \( S_{xx} = \sum X^2 - \frac{(\sum X)^2}{n} \) - \( S_{xy} = \sum XY - \frac{(\sum X)(\sum Y)}{n} \) - \( \overline{X} = \frac{\sum X}{n} \) - \( \overline{Y} = \frac{\sum Y}{n} \) # Step-by-Step Solution First, calculate the averages: - \( \overline{X} = \frac{20}{5} = 4 \) - \( \overline{Y} = \frac{50}{5} = 10 \) Next, compute \( S_{xx} \): \[ S_{xx} = \sum X^2 - \frac{(\sum X)^2}{n} \] \[ = 120 - \frac{20^2}{5} = 120 - \frac{400}{5} = 120 - 80 = 40 \] Now, calculate \( S_{xy} \): \[ S_{xy} = \sum XY - \frac{(\sum X)(\sum Y)}{n} \] \[ = 180 - \frac{20 \times 50}{5} = 180 - \frac{1000}{5} = 180 - 200 = -20 \] With \( S_{xy} \) and \( S_{xx} \) calculated, determine the slope \( \hat{\beta}_1 \): \[ \hat{\beta}_1 = \frac{S_{xy}}{S_{xx}} = \frac{-20}{40} = -0.5 \] Finally, compute the intercept \( \hat{\beta}_0 \): \[ \hat{\beta}_0 = \overline{Y} - \hat{\beta}_1 \overline{X} \] \[ = 10 - (-0.5 \times 4) = 10 + 2 = 12 \] # Summary - \( \hat{\beta}_0 = 12 \) - \( \hat{\beta}_1 = -0.5 \)

# Given Information The linear regression model is described by the equation: \( Y = \beta_0 + \beta_1 X + \epsilon \) Provided statistical values: - \( \sum X = 20 \) - \( \sum Y = 50 \) - \( \sum X^2 = 120 \) - \( \sum XY = 180 \) - \( n = 5 \) # Objective To derive the least-squares estimates for \( \beta_0 \) and \( \beta_1 \). # Definitions / Concepts Used The least-squares estimates are derived using: - **Slope (\( \hat{\beta}_1 \))**: \[ \hat{\beta}_1 = \frac{S_{xy}}{S_{xx}} \] - **Intercept (\( \hat{\beta}_0 \))**: \[ \hat{\beta}_0 = \overline{Y} - \hat{\beta}_1 \overline{X} \] Where: - \( S_{xx} = \sum X^2 - \frac{(\sum X)^2}{n} \) - \( S_{xy} = \sum XY - \frac{(\sum X)(\sum Y)}{n} \) - \( \overline{X} = \frac{\sum X}{n} \) - \( \overline{Y} = \frac{\sum Y}{n} \) # Detailed Solution Steps 1. **Calculate the Means**: - \( \overline{X} = \frac{20}{5} = 4 \) - \( \overline{Y} = \frac{50}{5} = 10 \) 2. **Compute \( S_{xx} \)**: \[ S_{xx} = \sum X^2 - \frac{(\sum X)^2}{n} = 120 - \frac{20^2}{5} \] \[ = 120 - \frac{400}{5} = 120 - 80 = 40 \] 3. **Compute \( S_{xy} \)**: \[ S_{xy} = \sum XY - \frac{(\sum X)(\sum Y)}{n} = 180 - \frac{20 \times 50}{5} \] \[ = 180 - \frac{1000}{5} = 180 - 200 = -20 \] 4. **Calculate the Slope \( \hat{\beta}_1 \)**: \[ \hat{\beta}_1 = \frac{S_{xy}}{S_{xx}} = \frac{-20}{40} = -0.5 \] 5. **Calculate the Intercept \( \hat{\beta}_0 \)**: \[ \hat{\beta}_0 = \overline{Y} - \hat{\beta}_1 \overline{X} \] \[ = 10 - (-0.5 \times 4) = 10 + 2 = 12 \] # Summary - \( \hat{\beta}_0 = 12 \) - \( \hat{\beta}_1 = -0.5 \)

# Given Information The linear model is expressed as: \( Y = \beta_0 + \beta_1 X + \epsilon \) Provided data: - \( \sum X = 20 \) - \( \sum Y = 50 \) - \( \sum X^2 = 120 \) - \( \sum XY = 180 \) - \( n = 5 \) # Objective Calculate the least-squares estimates for \( \beta_0 \) and \( \beta_1 \). # Definitions / Concepts Used Using the direct computational formula method, the estimates are calculated as follows: - **Slope (\( \hat{\beta}_1 \))**: \[ \hat{\beta}_1 = \frac{n\sum XY - (\sum X)(\sum Y)}{n\sum X^2 - (\sum X)^2} \] - **Intercept (\( \hat{\beta}_0 \))**: \[ \hat{\beta}_0 = \overline{Y} - \hat{\beta}_1 \overline{X} \] Where: - \( \overline{X} = \frac{\sum X}{n} \) - \( \overline{Y} = \frac{\sum Y}{n} \) # Step-by-Step Solution ### Calculate \( \hat{\beta}_1 \) First, compute the numerator: \[ n\sum XY = 5 \times 180 = 900 \] \[ (\sum X)(\sum Y) = 20 \times 50 = 1000 \] Thus, the numerator is: \[ 900 - 1000 = -100 \] Next, compute the denominator: \[ n\sum X^2 = 5 \times 120 = 600 \] \[ (\sum X)^2 = 20^2 = 400 \] Thus, the denominator is: \[ 600 - 400 = 200 \] Now calculate the slope: \[ \hat{\beta}_1 = \frac{-100}{200} = -0.5 \] ### Calculate \( \hat{\beta}_0 \) First, find the means: \[ \overline{X} = \frac{20}{5} = 4 \] \[ \overline{Y} = \frac{50}{5} = 10 \] Now calculate the intercept: \[ \hat{\beta}_0 = \overline{Y} - \hat{\beta}_1 \overline{X} \] Substituting the values: \[ \hat{\beta}_0 = 10 - (-0.5 \times 4) = 10 + 2 = 12 \] # Summary - \( \hat{\beta}_0 = 12 \) - \( \hat{\beta}_1 = -0.5 \)

# Given Information The linear regression model is defined as: \( Y = \beta_0 + \beta_1 X + \epsilon \) The provided statistics are: - \( \sum X = 20 \) - \( \sum Y = 50 \) - \( \sum X^2 = 120 \) - \( \sum XY = 180 \) - \( n = 5 \) # Objective To compute the least-squares estimates of \( \beta_0 \) and \( \beta_1 \). # Definitions / Concepts Used The least-squares estimates can be calculated using the following formulas: - **Slope (\( \hat{\beta}_1 \))**: \[ \hat{\beta}_1 = \frac{n\sum XY - (\sum X)(\sum Y)}{n\sum X^2 - (\sum X)^2} \] - **Intercept (\( \hat{\beta}_0 \))**: \[ \hat{\beta}_0 = \overline{Y} - \hat{\beta}_1 \overline{X} \] Where: - \( \overline{X} = \frac{\sum X}{n} \) - \( \overline{Y} = \frac{\sum Y}{n} \) # Step-by-Step Solution ### Calculate \( \hat{\beta}_1 \) **Numerator Calculation**: - Compute \( n\sum XY \): \[ n\sum XY = 5 \times 180 = 900 \] - Compute \( (\sum X)(\sum Y) \): \[ (\sum X)(\sum Y) = 20 \times 50 = 1000 \] - Therefore, the numerator is: \[ 900 - 1000 = -100 \] **Denominator Calculation**: - Compute \( n\sum X^2 \): \[ n\sum X^2 = 5 \times 120 = 600 \] - Compute \( (\sum X)^2 \): \[ (\sum X)^2 = 20^2 = 400 \] - Thus, the denominator is: \[ 600 - 400 = 200 \] **Slope Calculation**: Now, calculate the slope: \[ \hat{\beta}_1 = \frac{-100}{200} = -0.5 \] ### Calculate \( \hat{\beta}_0 \) **Means Calculation**: - Find the means: \[ \overline{X} = \frac{20}{5} = 4 \] \[ \overline{Y} = \frac{50}{5} = 10 \] **Intercept Calculation**: Now calculate the intercept: \[ \hat{\beta}_0 = \overline{Y} - \hat{\beta}_1 \overline{X} \] Substituting the values: \[ \hat{\beta}_0 = 10 - (-0.5 \times 4) = 10 + 2 = 12 \] # Summary - \( \hat{\beta}_0 = 12 \) - \( \hat{\beta}_1 = -0.5 \)

# Given Information The linear regression model is expressed as: \[ Y = \beta_0 + \beta_1 X + \epsilon \] Provided statistics: - \( \sum X = 20 \) - \( \sum Y = 50 \) - \( \sum X^2 = 120 \) - \( \sum XY = 180 \) - \( n = 5 \) # Objective To compute the least-squares estimates of \( \beta_0 \) and \( \beta_1 \). # Definitions / Concepts Used The least-squares estimates can be calculated using the following formulas: - **Slope (\( \hat{\beta}_1 \))**: \[ \hat{\beta}_1 = \frac{n\sum XY - (\sum X)(\sum Y)}{n\sum X^2 - (\sum X)^2} \] - **Intercept (\( \hat{\beta}_0 \))**: \[ \hat{\beta}_0 = \overline{Y} - \hat{\beta}_1 \overline{X} \] Where: - \( \overline{X} = \frac{\sum X}{n} \) - \( \overline{Y} = \frac{\sum Y}{n} \) # Step-by-Step Solution ### Calculate \( \hat{\beta}_1 \) **Numerator Calculation**: - Compute \( n\sum XY \): \[ n\sum XY = 5 \times 180 = 900 \] - Compute \( (\sum X)(\sum Y) \): \[ (\sum X)(\sum Y) = 20 \times 50 = 1000 \] - Therefore, the numerator is: \[ 900 - 1000 = -100 \] **Denominator Calculation**: - Compute \( n\sum X^2 \): \[ n\sum X^2 = 5 \times 120 = 600 \] - Compute \( (\sum X)^2 \): \[ (\sum X)^2 = 20^2 = 400 \] - Thus, the denominator is: \[ 600 - 400 = 200 \] **Slope Calculation**: Now, calculate the slope: \[ \hat{\beta}_1 = \frac{-100}{200} = -0.5 \] ### Calculate \( \hat{\beta}_0 \) **Means Calculation**: - Find the means: \[ \overline{X} = \frac{20}{5} = 4 \] \[ \overline{Y} = \frac{50}{5} = 10 \] **Intercept Calculation**: Now calculate the intercept: \[ \hat{\beta}_0 = \overline{Y} - \hat{\beta}_1 \overline{X} \] Substituting the values: \[ \hat{\beta}_0 = 10 - (-0.5 \times 4) = 10 + 2 = 12 \] # Summary - \( \hat{\beta}_0 = 12 \) - \( \hat{\beta}_1 = -0.5 \)

# Given Information The linear regression model is represented as: \[ Y = \beta_0 + \beta_1 X + \epsilon \] Provided data: - \( \sum X = 20 \) - \( \sum Y = 50 \) - \( \sum X^2 = 120 \) - \( \sum XY = 180 \) - \( n = 5 \) # Objective To compute the least-squares estimates for \( \beta_0 \) and \( \beta_1 \). # Definitions / Concepts Used The least-squares estimates are determined using the following formulas: - **Slope (\( \hat{\beta}_1 \))**: \[ \hat{\beta}_1 = \frac{n\sum XY - (\sum X)(\sum Y)}{n\sum X^2 - (\sum X)^2} \] - **Intercept (\( \hat{\beta}_0 \))**: \[ \hat{\beta}_0 = \overline{Y} - \hat{\beta}_1 \overline{X} \] Where: - \( \overline{X} = \frac{\sum X}{n} \) - \( \overline{Y} = \frac{\sum Y}{n} \) # Step-by-Step Solution ### Calculate \( \hat{\beta}_1 \) **Numerator Calculation**: 1. Compute \( n\sum XY \): \[ n\sum XY = 5 \times 180 = 900 \] 2. Compute \( (\sum X)(\sum Y) \): \[ (\sum X)(\sum Y) = 20 \times 50 = 1000 \] 3. Therefore, the numerator is: \[ 900 - 1000 = -100 \] **Denominator Calculation**: 1. Compute \( n\sum X^2 \): \[ n\sum X^2 = 5 \times 120 = 600 \] 2. Compute \( (\sum X)^2 \): \[ (\sum X)^2 = 20^2 = 400 \] 3. Thus, the denominator is: \[ 600 - 400 = 200 \] **Slope Calculation**: Now, calculate the slope: \[ \hat{\beta}_1 = \frac{-100}{200} = -0.5 \] ### Calculate \( \hat{\beta}_0 \) **Means Calculation**: 1. Find the means: \[ \overline{X} = \frac{20}{5} = 4 \] \[ \overline{Y} = \frac{50}{5} = 10 \] **Intercept Calculation**: Now calculate the intercept: \[ \hat{\beta}_0 = \overline{Y} - \hat{\beta}_1 \overline{X} \] Substituting the values: \[ \hat{\beta}_0 = 10 - (-0.5 \times 4) = 10 + 2 = 12 \] # Summary - \( \hat{\beta}_0 = 12 \) - \( \hat{\beta}_1 = -0.5 \)

# Given Information The linear regression model is defined as: \[ Y = \beta_0 + \beta_1 X + \epsilon \] The provided statistics are: - \( \sum X = 20 \) - \( \sum Y = 50 \) - \( \sum X^2 = 120 \) - \( \sum XY = 180 \) - \( n = 5 \) # Objective Calculate the least-squares estimates for \( \beta_0 \) and \( \beta_1 \). # Definitions / Concepts Used The least-squares estimates can be calculated using: - **Slope (\( \hat{\beta}_1 \))**: \[ \hat{\beta}_1 = \frac{n\sum XY - (\sum X)(\sum Y)}{n\sum X^2 - (\sum X)^2} \] - **Intercept (\( \hat{\beta}_0 \))**: \[ \hat{\beta}_0 = \overline{Y} - \hat{\beta}_1 \overline{X} \] Where: - \( \overline{X} = \frac{\sum X}{n} \) - \( \overline{Y} = \frac{\sum Y}{n} \) # Step-by-Step Solution ### Calculate \( \hat{\beta}_1 \) **Numerator Calculation**: 1. Calculate \( n\sum XY \): \[ n\sum XY = 5 \times 180 = 900 \] 2. Calculate \( (\sum X)(\sum Y) \): \[ (\sum X)(\sum Y) = 20 \times 50 = 1000 \] 3. The numerator then is: \[ 900 - 1000 = -100 \] **Denominator Calculation**: 1. Calculate \( n\sum X^2 \): \[ n\sum X^2 = 5 \times 120 = 600 \] 2. Calculate \( (\sum X)^2 \): \[ (\sum X)^2 = 20^2 = 400 \] 3. The denominator is: \[ 600 - 400 = 200 \] **Slope Calculation**: Now, calculate the slope: \[ \hat{\beta}_1 = \frac{-100}{200} = -0.5 \] ### Calculate \( \hat{\beta}_0 \) **Means Calculation**: 1. Calculate the means: \[ \overline{X} = \frac{20}{5} = 4 \] \[ \overline{Y} = \frac{50}{5} = 10 \] **Intercept Calculation**: Now calculate the intercept: \[ \hat{\beta}_0 = \overline{Y} - \hat{\beta}_1 \overline{X} \] Substituting the values: \[ \hat{\beta}_0 = 10 - (-0.5 \times 4) = 10 + 2 = 12 \] # Summary - \( \hat{\beta}_0 = 12 \) - \( \hat{\beta}_1 = -0.5 \)