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ANOVA: shear force of bolts at three positions

Short answer

One-way ANOVA: F=2.333, critical F(0.05;2,14)=3.739. Since 2.333 < 3.739, fail to reject H0; differences among the three position means are not significant at 5% level.

As a part of investigation of the collapse of the roof of a building, a testing laboratory is given all the available bolts that connected the steel structure at three different positions on the roof. The forces required to shear each of these bolts (coded values) are as follows:

Position 190, 82, 79, 98, 83, 91
Position 2105, 89, 93, 104, 89, 95, 86
Position 383, 89, 80, 94
Perform an ANOVA to test at the 0.05 level of significance whether the difference among the sample means at the three positions are significant.

One-way ANOVA with unequal group sizes (6, 7 and 4).

  1. H0: μ₁ = μ₂ = μ₃ (mean shear force is the same at the three positions) H1: at least one position mean differs (F-test, right-tailed)
  2. Level of significance: α = 0.05
  3. Computations:
Position 1   n=6  total=523  mean=87.1667
Position 2   n=7  total=661  mean=94.4286
Position 3   n=4  total=346  mean=86.5000

Grand total T = 1530,  N = 17
Correction factor CF = T²/N = 1530²/17 = 137700.0000
Total SS  = Σx² − CF = 138638.0000 − 137700.0000 = 938.0000
SS between = Σ(Ti²/ni) − CF = 137934.4524 − 137700.0000 = 234.4524
SS within  = Total SS − SS between = 703.5476
  1. ANOVA table:
SourceSSd.f.MSF
Between positions234.45242117.22622.333
Within (error)703.54761450.2534
Total938.000016
  1. Critical value: F(0.05; 2, 14) = 3.739. Reject H0 if F > 3.739.
  2. Decision: the calculated value falls in the acceptance region (p-value = 0.1335), so fail to reject H0.

Conclusion: the difference among the sample means is not significant at the 5% level.

✓ Answer checked by Claude Sonnet 5.5

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