Box's test may be non-significant because of violation of which assumption?

Box's test may be non-significant because of violation of which assumption?

• A. Linearity
• B. Normality of univariate distributions
• C. Homogeneity of variance across groups
• D. Multivariate normality

Answer: (D)

Box's M test is used to assess whether the covariance matrices across groups are equal in MANOVA, and it relies on the assumption that the dependent variables are multivariate normally distributed within each group. When multivariate normality holds, the test’s chi-square approximation is valid and a significant result suggests different covariance structures across groups. If multivariate normality is violated, that approximation becomes unreliable, and the test can fail to detect differences in covariance matrices, yielding a non-significant result even when the covariances differ. So the reason a Box’s test may be non-significant is a violation of multivariate normality. The other assumptions—linearity, univariate normality, and homogeneity of variance across groups—do not directly underpin Box’s M in the same way.

Box's M test is used to assess whether the covariance matrices across groups are equal in MANOVA, and it relies on the assumption that the dependent variables are multivariate normally distributed within each group. When multivariate normality holds, the test’s chi-square approximation is valid and a significant result suggests different covariance structures across groups. If multivariate normality is violated, that approximation becomes unreliable, and the test can fail to detect differences in covariance matrices, yielding a non-significant result even when the covariances differ. So the reason a Box’s test may be non-significant is a violation of multivariate normality. The other assumptions—linearity, univariate normality, and homogeneity of variance across groups—do not directly underpin Box’s M in the same way.