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In: Statistics and Probability

What does a descriptive and bivariate analyses followed by logistic regression to evaluate risks and analysis...


What does a descriptive and bivariate analyses followed by logistic regression to evaluate risks and analysis of covariance testing to determine outcome differences mean?

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Expert Solution

Analysis of covariance (ANCOVA) is a general linear model which blends ANOVA and regression. ANCOVA evaluates whether the means of a dependent variable (DV) are equal across levels of a categorical independent variable (IV) often called a treatment, while statistically controlling for the effects of other continuous variables that are not of primary interest, known as covariates (CV) or nuisance variables. Mathematically, ANCOVA decomposes the variance in the DV into variance explained by the CV(s), variance explained by the categorical IV, and residual variance. Intuitively, ANCOVA can be thought of as 'adjusting' the DV by the group means of the CV(s).[1]

The ANCOVA model assumes a linear relationship between the response (DV) and covariate (CV):

y i j = μ + τ i + B ( x i j − x ¯ ) + ϵ i j . {\displaystyle y_{ij}=\mu +\tau _{i}+\mathrm {B} (x_{ij}-{\overline {x}})+\epsilon _{ij}.}

In this equation, the DV, y i j {\displaystyle y_{ij}} is the jth observation under the ith categorical group; the CV, x i j {\displaystyle x_{ij}} is the jth observation of the covariate under the ith group. Variables in the model that are derived from the observed data are μ {\displaystyle \mu } (the grand mean) and x ¯ {\displaystyle {\overline {x}}} (the global mean for covariate x {\displaystyle x} ). The variables to be fitted are τ i {\displaystyle \tau _{i}} (the effect of the ith level of the IV), B {\displaystyle B} (the slope of the line) and ϵ i j {\displaystyle \epsilon _{ij}} (the associated unobserved error term for the jth observation in the ith group).

Under this specification, the a categorical treatment effects sum to zero ( ∑ i a τ i = 0 ) . {\displaystyle \left(\sum _{i}^{a}\tau _{i}=0\right).}


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