For k = 3 groups with equal n, express the Tukey-Kramer denominator simplification algebraically starting from sqrt(MSE/2 * (1/n + 1/n)).

• A. Denominator simplifies to sqrt(MSE/2n) because the two terms halve the MSE and n together.
• B. Denominator simplifies to MSE/n because the square root cancels when n is equal across groups.
• C. Denominator simplifies to sqrt(2*MSE/n) because the 2 in the numerator doubles when you combine the fractions.
• D. Denominator simplifies to sqrt(MSE/n).

Answer: (D)