This mathematics problem involves applying core mathematical principles and formulas. Below you will find a complete step-by-step solution with detailed explanations for each step, helping you understand not just the answer but the method behind it.
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Here's the continuation of the solution to your question: II. Calculate the F-value. Step 1: Calculate the sum, mean, and sum of squares for each group. Agric. Education (G_1): Scores (8, 5, 5, 9, 7, 8, 9, 5, 6, 8) n_1 = 10 x_1 = 70 x_1 = (70)/(10) = 7 x_1^2 = 514 Business Education (G_2): Scores (7, 6, 6, 4, 8, 3, 6, 5, 7, 4) n_2 = 10 x_2 = 56 x_2 = (56)/(10) = 5.6 x_2^2 = 336 Industrial Technical Education (G_3): Scores (5, 4, 5, 6, 3, 5, 4, 7, 3, 4) n_3 = 10 x_3 = 46 x_3 = (46)/(10) = 4.6 x_3^2 = 230 Overall: N = n_1 + n_2 + n_3 = 10 + 10 + 10 = 30 Grand Sum ( x) = 70 + 56 + 46 = 172 Grand Sum of Squares ( x^2) = 514 + 336 + 230 = 1080 Step 2: Calculate the Correction Factor (CF), Total Sum of Squares (SST), Sum of Squares Between Groups (SSB), and Sum of Squares Within Groups (SSW). Correction Factor: CF = (( x)^2)/(N) = ((172)^2)/(30) = (29584)/(30) = 986.1333 Total Sum of Squares: SST = x^2 - CF = 1080 - 986.1333 = 93.8667 Sum of Squares Between Groups: SSB = ((( x_1)^2)/(n_1) + (( x_2)^2)/(n_2) + (( x_3)^2)/(n_3)) - CF SSB = ((70^2)/(10) + (56^2)/(10) + (46^2)/(10)) - 986.1333 SSB = (490 + 313.6 + 211.6) - 986.1333 = 1015.2 - 986.1333 = 29.0667 Sum of Squares Within Groups: SSW = SST - SSB = 93.8667 - 29.0667 = 64.8 Step 3: Calculate the degrees of freedom. Degrees of freedom between groups (df_B) = k - 1 = 3 - 1 = 2 Degrees of freedom within groups (df_W) = N - k = 30 - 3 = 27 Step 4: Calculate the Mean Square Between Groups (MSB) and Mean Square Within Groups (MSW). Mean Square Between Groups: MSB = (SSB)/(df_B) = (29.0667)/(2) = 14.53335 Mean Square Within Groups: MSW = (SSW)/(df_W) = (64.8)/(27) = 2.4 Step 5: Calculate the F-value. F = (MSB)/(MSW) = (14.53335)/(2.4) = 6.0555625 The calculated F-value is 6.06. III. What is the F-table value? Step 6: Determine the critical F-value from the F-distribution table. Using = 0.05, df_1 = df_B = 2, and df_2 = df_W = 27: The F-table value is 3.35. IV. What is your decision based on the results? Step 7: Compare the calculated F-value with the F-table value. Since the calculated F-value (6.06) is greater than the critical F-table value (3.35), we reject the null hypothesis. V. What is your conclusion? Step 8: State the conclusion. There is sufficient evidence to conclude that there is a significant difference in the mean quiz scores among the students in Agric. Education, Business Education, and Industrial Technical Education departments at the 0.05 level of significance. Drop the next question.