Question

Consider the multi-linear regression model \[ y_i = \beta_0 + \beta_1 x_{1i} + \beta_2 x_{2i} + \beta_3 x_{3i} + \beta_4 x_{4i} + \epsilon_i, \quad i = 1, 2, \dots, 25, \] {where } \( \beta_i, i = 0, 1, 2, 3, 4 \) 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 that the value of the coefficient of determination \( R^2 \) is obtained as \( \frac{5}{6} \). Then the value of adjusted \( R^2 \) is _________ (rounded off to two decimal places).

Hint:

The adjusted \( R^2 \) accounts for the number of predictors in the model and adjusts the coefficient of determination accordingly. Use the formula: \[ R_{{adj}}^2 = 1 - \frac{(1 - R^2)(n - 1)}{n - k - 1}. \]

Answer 1

The formula for the adjusted \( R^2 \) is given by: \[ R_{{adj}}^2 = 1 - \left( \frac{(1 - R^2)(n - 1)}{n - k - 1} \right), \] where:
\( R^2 \) is the coefficient of determination,
\( n \) is the number of observations,
\( k \) is the number of independent variables (excluding the intercept).
Given:
\( R^2 = \frac{5}{6} \),
\( n = 25 \) (since there are 25 observations),
\( k = 4 \) (since there are 4 independent variables, \( x_1, x_2, x_3, x_4 \)).
Substitute these values into the formula: \[ R_{{adj}}^2 = 1 - \left( \frac{(1 - \frac{5}{6})(25 - 1)}{25 - 4 - 1} \right). \] Simplify the expression: \[ R_{{adj}}^2 = 1 - \left( \frac{(\frac{1}{6})(24)}{20} \right) = 1 - \left( \frac{4}{20} \right) = 1 - 0.2 = 0.8. \] Thus, the adjusted \( R^2 \) is \( \boxed{0.80} \).