A study of the properties of m
A study of the properties of metal plate-connected trusses usedfor roof support yielded the following observations on axialstiffness index (kips/in.) for plate lengths 4, 6, 8, 10, and 12in:
| 4: | 340.2 | 409.5 | 311.0 | 326.5 | 316.8 | 349.8 | 309.7 |
| 6: | 432.1 | 347.2 | 361.0 | 404.5 | 331.0 | 348.9 | 381.7 |
| 8: | 390.4 | 366.2 | 351.0 | 357.1 | 409.9 | 367.3 | 382.0 |
| 10: | 359.7 | 452.9 | 461.4 | 433.1 | 410.6 | 384.2 | 362.6 |
| 12: | 413.4 | 441.8 | 419.9 | 410.7 | 473.4 | 441.2 | 465.8 |
Does variation in plate length have any effect on true averageaxial stiffness? State the relevant hypotheses using analysis ofvariance.
H0: μ1 ≠ μ2 ≠μ3 ≠ μ4 ≠ μ5Ha: at least two μi‘s areequal H0: μ1 ≠ μ2 ≠μ3 ≠ μ4 ≠ μ5Ha: all five μi‘s areequal H0: μ1= μ2 = μ3 = μ4 =μ5Ha: all five μi‘s areunequal H0: μ1 = μ2 =μ3 = μ4 = μ5Ha: at least two μi‘s areunequal
Test the relevant hypotheses using analysis of variance with α =0.01. Display your results in an ANOVA table. (Round your answersto two decimal places.)
| Source | Degrees offreedom | Sum ofSquares | MeanSquares | f |
|---|---|---|---|---|
| Treatments | 2 | 3 | 4 | 5 |
| Error | 6 | 7 | 8 | |
| Total | 9 | 10 |
Give the test statistic. (Round your answer to two decimalplaces.)f = 11What can be said about the P-value for the test?
P-value > 0.100 0.050 < P-value <0.100 0.010 < P-value <0.050 0.001 < P-value < 0.010 P-value <0.001
State the conclusion in the problem context.
Reject H0. There are no differences in thetrue average axial stiffness for the different plate lengths.Reject H0. There are differences in the trueaverage axial stiffness for the different platelengths. Fail to rejectH0. There are differences in the true averageaxial stiffness for the different plate lengths. Fail to rejectH0. There are no differences in the trueaverage axial stiffness for the different plate lengths.
Answer:
(1)Correct option:
H0: All five 
‘s areequal: 
Ha: At least two ‘s areunequal: 
(2)
From the given data, the following Table is calculated:
| 4 | 6 | 8 | 10 | 12 | Total | |
| N | 7 | 7 | 7 | 7 | 7 | 35 |
 |
2363.5 | 2606.4 | 2623.9 | 2864.5 | 3066.2 | 13524.5 |
| Mean | 2363.5/7=337.6429 | 2606.4/7=372.3429 | 2623.9/7=374.8429 | 2864.5/7=409.2143 | 3066.2/7=438.0286 | 13524.5/35=386.414 |
 |
805385.91 | 978186.6 | 986087.31 | 1182648.83 | 1346811.94 | 5299120.59 |
| Std. Dev. | 35.0405 | 35.852 | 20.5635 | 41.7423 | 24.929 | 46.3556 |
From the above Table, ANOVA Table is calculated as follows:
| Source | Degrees of freedom | Sum of Squares | Mean squares | F |
| Treatments | 4 | 41261.0086 | 41261.008/4=10315.2541 | 10315.2541/1059.9858=9.7315 |
| Error | 30 | 31799.5743 | 31799.5743/30=1059.9858 | |
| Total | 34 | 73060.5829 |
Test statistic F =10315.2541/1059.9858=9.73
(3)
Degrees of Freedom (4,30)
By Technology, p – value = 0.000036
So
Correct option:
P – Value < 0.001
(4)
Correct option:
Reject H0. There are differences in the true averageaxial stiffness for the different plate lengths.
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