Question

A study of the properties of metal plate-connected trusses used for roof support yielded the following...

A study of the properties of metal plate-connected trusses used for roof support yielded the following observations on axial stiffness index (kips/in.) for plate lengths 4, 6, 8, 10, and 12 in:

4: 308.2 409.5 311.0 326.5 316.8 349.8 309.7
6: 431.1 347.2 361.0 404.5 331.0 348.9 381.7
8: 388.4 366.2 351.0 357.1 409.9 367.3 382.0
10: 355.7 452.9 461.4 433.1 410.6 384.2 362.6
12: 415.4 441.8 419.9 410.7 473.4 441.2 465.8

Does variation in plate length have any effect on true average axial stiffness? State the relevant hypotheses using analysis of variance.

a) H0: μ1 = μ2 = μ3 = μ4 = μ5
b) Ha: at least two μi's are unequalH0: μ1μ2μ3μ4μ5
c) Ha: all five μi's are equal    H0: μ1μ2μ3μ4μ5
d) Ha: at least two μi's are equalH0: μ1 = μ2 = μ3 = μ4 = μ5
e) Ha: all five μi's are unequal


Test the relevant hypotheses using analysis of variance with α = 0.01. Display your results in an ANOVA table. (Round your answers to two decimal places.)

Source Degrees of
freedom
Sum of
Squares
Mean
Squares
f
Treatments
Error
Total

(the last column its just f)


Give the test statistic. (Round your answer to two decimal places.)
f =

What can be said about the P-value for the test?

a) P-value > 0.100

b) 0.050 < P-value < 0.100    

c) 0.010 < P-value < 0.050

d) 0.001 < P-value < 0.010

e) P-value < 0.001


State the conclusion in the problem context.

a) Reject H0. There are no differences in the true average axial stiffness for the different plate lengths.

b) Fail to reject H0. There are differences in the true average axial stiffness for the different plate lengths.   

c) Reject H0. There are differences in the true average axial stiffness for the different plate lengths.

d) Fail to reject H0. There are no differences in the true average axial stiffness for the different plate lengths.

Homework Answers

Answer #1

The statistical software output for this problem is:

Hence,

Hypotheses: Option A is correct.

ANOVA table:

Source Degrees of
freedom
Sum of
Squares
Mean
Squares
f
Treatments 4 44591.61 11147.90 10.24
Error 30 32650.92 1088.36
Total 34 77242.53

Test statistic = 10.24

P-value < 0.001

Conclusion: Option C is correct.

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