# 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: 340.2 409.5 311 326.5 316.8 349.8 309.7 6: 432.1 347.2 361 404.5 331 348.9 381.7 8: 390.4 366.2 351 357.1 409.9 367.3 382 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 = μ45Ha: all five μi‘s areunequal H0: μ1 = μ23 = 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.

(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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