Question

Consider the simple linear regression model \[ y_i = \alpha + \beta x_i + \epsilon_i, \quad i = 1, 2, \dots, 24, \] where \( \alpha \in \mathbb{R} \) and \( \beta \in \mathbb{R} \) are unknown parameters, the errors \( \epsilon_i \)'s are i.i.d. random variables having \( N(0, \sigma^2) \) distribution, where \( \sigma>0 \) is unknown. Suppose the following summary statistics are obtained from a data set of 24 observations \( (x_1, y_1), \dots, (x_{24}, y_{24}) \): \[ S_{xx} = \sum_{i=1}^{24} (x_i - \bar{x})^2 = 22.82, \quad S_{yy} = \sum_{i=1}^{24} (y_i - \bar{y})^2 = 43.62, \quad S_{xy} = \sum_{i=1}^{24} (x_i - \bar{x})(y_i - \bar{y}) = 15.48, \] where \( \bar{x} = \frac{1}{24} \sum_{i=1}^{24} x_i \) and \( \bar{y} = \frac{1}{24} \sum_{i=1}^{24} y_i \). Then, for testing \( H_0: \beta = 0 \) against \( H_1: \beta \neq 0 \), the value of the \( F \)-test statistic based on the least squares estimator of \( \beta \), whose distribution is \( F_{1,22} \), equals (rounded off to two decimal places):

• A. 2.54
• B. 2.98
• C. 3.17
• D. 6.98

Hint:

When performing hypothesis testing in simple linear regression, the \( F \)-statistic is used to test the significance of the regression model. It compares the variation explained by the model to the unexplained variation.

Answer 1

The Correct Option is D

Step 1: Formula for the \( F \)-test statistic.
The \( F \)-test statistic for testing \( H_0: \beta = 0 \) against \( H_1: \beta \neq 0 \) in a simple linear regression model is given by: \[ F = \frac{\frac{(S_{xy})^2}{S_{xx}}}{\frac{S_{yy} - \frac{(S_{xy})^2}{S_{xx}}}{22}}. \] Step 2: Plug in the values.
From the problem statement, we are given the following values: \[ S_{xx} = 22.82, \quad S_{yy} = 43.62, \quad S_{xy} = 15.48. \] Substitute these values into the formula for the \( F \)-test statistic: \[ F = \frac{\frac{(15.48)^2}{22.82}}{\frac{43.62 - \frac{(15.48)^2}{22.82}}{22}}. \] Step 3: Simplify the expression.
First, calculate \( \frac{(S_{xy})^2}{S_{xx}} \): \[ \frac{(15.48)^2}{22.82} = \frac{239.0304}{22.82} = 10.47. \] Next, calculate \( S_{yy} - \frac{(S_{xy})^2}{S_{xx}} \): \[ 43.62 - 10.47 = 33.15. \] Now, calculate the denominator: \[ \frac{33.15}{22} = 1.51. \] Finally, compute the \( F \)-statistic: \[ F = \frac{10.47}{1.51} = 6.98. \] Thus, the value of the \( F \)-test statistic is approximately \( \boxed{6.98} \) (rounded to two decimal places).