Answer: 0.0599

The White test for heteroskedasticity uses the test statistic nR², where n is the sample size and R² is from the auxiliary regression of squared residuals on the explanatory variables. The test statistic follows a chi-square distribution with degrees of freedom equal to the number of explanatory variables in the auxiliary regression. For a model with one explanatory variable, the auxiliary regression includes the explanatory variable and its square, so the degrees of freedom = 2. At 5% significance level, the critical value for χ²(2) is approximately 5.991. We need to find R² such that: nR² = 100 × R² ≥ 5.991 Solving for R²: R² ≥ 5.991/100 = 0.05991 Therefore, the null hypothesis of homoskedasticity will be rejected when R² ≥ 0.0599. Option D (0.0599) is the correct answer as it represents the threshold R² value at which the test statistic would exceed the critical value at the 5% significance level.

Author: Nikitesh Somanthe