Consider the following four simples A, B, C, D. Are there significant different between the mean of the treatment of the samples on the basis of the 5% of level of significant?

Question

Asked on August 28, 2026Mathematics

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.

Accepted Answer · 2026-08-28T20:23:01.555Z

To determine if there are significant differences between the means of the four samples, we will perform a One-Way Analysis of Variance (ANOVA) at a 5% level of significance. Step 1: State the Hypotheses Null Hypothesis (H_0): There is no significant difference between the means of the four samples. H_0: _A = _B = _C = _D Alternative Hypothesis (H_1): At least one sample mean is significantly different from the others. H_1: At least one _j is different The significance level is = 0.05. Step 2: Calculate Descriptive Statistics for Each Sample Let n_j be the number of observations in sample j, x_j be the sum of observations, and x_j^2 be the sum of squared observations. Sample A: n_A = 5 x_A = 9+7+12+14+13 = 55 x_A = 11 x_A^2 = 9^2+7^2+12^2+14^2+13^2 = 81+49+144+196+169 = 639 Sample B: n_B = 6 x_B = 18+12+15+13+16+16 = 90 x_B = 15 x_B^2 = 18^2+12^2+15^2+13^2+16^2+16^2 = 324+144+225+169+256+256 = 1374 Sample C: n_C = 6 x_C = 10+9+11+10+7+10 = 57 x_C = 9.5 x_C^2 = 10^2+9^2+11^2+10^2+7^2+10^2 = 100+81+121+100+49+100 = 551 Sample D: n_D = 6 x_D = 13+8+9+12+11+10 = 63 x_D = 10.5 x_D^2 = 13^2+8^2+9^2+12^2+11^2+10^2 = 169+64+81+144+121+100 = 679 Step 3: Calculate Overall Statistics Total number of observations N = n_A+n_B+n_C+n_D = 5+6+6+6 = 23. Grand Sum X = 55+90+57+63 = 265. Grand Sum of Squares X^2 = 639+1374+551+679 = 3243. Correction Factor CF = (( X)^2)/(N) = (265^2)/(23) = (70225)/(23) ≈ 3053.2609. Step 4: Calculate Sum of Squares Sum of Squares Total (SST): SST = X^2 - CF = 3243 - 3053.2609 = 189.7391 Sum of Squares Between Treatments (SSTreatment): SSTreatment = _j=1^k (( x_j)^2)/(n_j) - CF SSTreatment = ( (55^2)/(5) + (90^2)/(6) + (57^2)/(6) + (63^2)/(6) ) - 3053.2609 SSTreatment = ( (3025)/(5) + (8100)/(6) + (3249)/(6) + (3969)/(6) ) - 3053.2609 SSTreatment = (605 + 1350 + 541.5 + 661.5) - 3053.2609 SSTreatment = 3158 - 3053.2609 = 104.7391 Sum of Squares Within Treatments (SSError): SSError = SST - SSTreatment = 189.7391 - 104.7391 = 85.0000 Step 5: Calculate Degrees of Freedom (df) df for Treatments (df_1 = k-1): 4-1 = 3 df for Error (df_2 = N-k): 23-4 = 19 df Total (N-1): 23-1 = 22 Step 6: Calculate Mean Squares (MS) Mean Square for Treatments (MSTreatment): MSTreatment = (SSTreatment)/(df_1) = (104.7391)/(3) ≈ 34.9130 Mean Square for Error (MSError): MSError = (SSError)/(df_2) = (85.0000)/(19) ≈ 4.4737 Step 7: Calculate the F-statistic F = (MSTreatment)/(MSError) = (34.9130)/(4.4737) ≈ 7.8041 Step 8: Determine the Critical F-value For = 0.05, df_1 = 3, and df_2 = 19, the critical F-value from the F-distribution table is approximately F_critical = 3.127. Step 9: Make a Decision Since the calculated F-statistic (7.8041) is greater than the critical F-value (3.127), we reject the null hypothesis (H_0). Step 10: State the Conclusion There is sufficient evidence to conclude that there are significant differences between the means of the treatments (samples A, B, C, D) at the 5% level of significance. The final answer is Yes, there are significant differences between the means of the treatments. Send me the next one 📸

Accepted Answer · 2026-08-28T20:23:01.555Z

To determine if there are significant differences between the means of the four samples, we will perform a One-Way Analysis of Variance (ANOVA) at a 5% level of significance. Step 1: State the Hypotheses Null Hypothesis (H_0): There is no significant difference between the means of the four samples. H_0: _A = _B = _C = _D Alternative Hypothesis (H_1): At least one sample mean is significantly different from the others. H_1: At least one _j is different The significance level is = 0.05. Step 2: Calculate Descriptive Statistics for Each Sample Let n_j be the number of observations in sample j, x_j be the sum of observations, and x_j^2 be the sum of squared observations. Sample A: n_A = 5 x_A = 9+7+12+14+13 = 55 x_A = 11 x_A^2 = 9^2+7^2+12^2+14^2+13^2 = 81+49+144+196+169 = 639 Sample B: n_B = 6 x_B = 18+12+15+13+16+16 = 90 x_B = 15 x_B^2 = 18^2+12^2+15^2+13^2+16^2+16^2 = 324+144+225+169+256+256 = 1374 Sample C: n_C = 6 x_C = 10+9+11+10+7+10 = 57 x_C = 9.5 x_C^2 = 10^2+9^2+11^2+10^2+7^2+10^2 = 100+81+121+100+49+100 = 551 Sample D: n_D = 6 x_D = 13+8+9+12+11+10 = 63 x_D = 10.5 x_D^2 = 13^2+8^2+9^2+12^2+11^2+10^2 = 169+64+81+144+121+100 = 679 Step 3: Calculate Overall Statistics Total number of observations N = n_A+n_B+n_C+n_D = 5+6+6+6 = 23. Grand Sum X = 55+90+57+63 = 265. Grand Sum of Squares X^2 = 639+1374+551+679 = 3243. Correction Factor CF = (( X)^2)/(N) = (265^2)/(23) = (70225)/(23) ≈ 3053.2609. Step 4: Calculate Sum of Squares Sum of Squares Total (SST): SST = X^2 - CF = 3243 - 3053.2609 = 189.7391 Sum of Squares Between Treatments (SSTreatment): SSTreatment = _j=1^k (( x_j)^2)/(n_j) - CF SSTreatment = ( (55^2)/(5) + (90^2)/(6) + (57^2)/(6) + (63^2)/(6) ) - 3053.2609 SSTreatment = ( (3025)/(5) + (8100)/(6) + (3249)/(6) + (3969)/(6) ) - 3053.2609 SSTreatment = (605 + 1350 + 541.5 + 661.5) - 3053.2609 SSTreatment = 3158 - 3053.2609 = 104.7391 Sum of Squares Within Treatments (SSError): SSError = SST - SSTreatment = 189.7391 - 104.7391 = 85.0000 Step 5: Calculate Degrees of Freedom (df) df for Treatments (df_1 = k-1): 4-1 = 3 df for Error (df_2 = N-k): 23-4 = 19 df Total (N-1): 23-1 = 22 Step 6: Calculate Mean Squares (MS) Mean Square for Treatments (MSTreatment): MSTreatment = (SSTreatment)/(df_1) = (104.7391)/(3) ≈ 34.9130 Mean Square for Error (MSError): MSError = (SSError)/(df_2) = (85.0000)/(19) ≈ 4.4737 Step 7: Calculate the F-statistic F = (MSTreatment)/(MSError) = (34.9130)/(4.4737) ≈ 7.8041 Step 8: Determine the Critical F-value For = 0.05, df_1 = 3, and df_2 = 19, the critical F-value from the F-distribution table is approximately F_critical = 3.127. Step 9: Make a Decision Since the calculated F-statistic (7.8041) is greater than the critical F-value (3.127), we reject the null hypothesis (H_0). Step 10: State the Conclusion There is sufficient evidence to conclude that there are significant differences between the means of the treatments (samples A, B, C, D) at the 5% level of significance. The final answer is Yes, there are significant differences between the means of the treatments. Send me the next one 📸

Consider the following four simples A, B, C, D. Are there significant different between the mean of the treatment of the samples on the basis of the 5% of level of significant?

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Consider the following four simples A, B, C, D. Are there significant different between the mean of the treatment of the samples on the basis of the 5% of level of significant?

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