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 $ \sigma0 $ 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).

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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).

cdquestions Admin · Accepted Answer

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} $.

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