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Question

In a completely randomized design, with 4 treatments replicated 5 times, the following information is obtained. SST = 26234.95, SSE = 11558.80, the Fcal is equal to:

This question was previously asked in
SSC CGL 2024 (Tier-I) Previous Year Paper (17-Sep-2024) (Shift 3)
The correct answer is

12.11

The F-statistic in an ANOVA test is calculated as the ratio of the mean square between treatments (MST) to the mean square error (MSE). The formulas are:
MST = (SST - SSE) / (df between), where df between = number of treatments - 1 = 4 - 1 = 3
MSE = SSE / df error, where df error = number of treatments * number of replications - number of treatments = 4 * 5 - 4 = 16
Now, MST = (26234.95 - 11558.80) / 3 = 4892.05, and MSE = 11558.80 / 16 = 722.425
Thus, Fcal = MST / MSE = 4892.05 / 722.425 ≈ 12.11.

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