4. Axial stiffness (kips/in) for four different lengths of metal plate-connected

Question

4. Axial stiffness (kips/in) for four different lengths of metal plate-connected trusses yielded the observations in the Table below. Let , , , 4 be the unknown means for the four lengths used. ANOVA model is valid. (Answers rounded up to whole numbe Assume the usual L1 309 410 L3 392 366 2 402 347 311 361 351 405 331 349 382 4 407 442 420 411 473 441 466 327 317 350 310 357 410 367 382 368 375 21 437 26 sample means sample stdev sample variances 37 29 1339 816 434676 a test of Ho : 1 = 2-13- vs Ha: not all equal at the -0.05 ( 8 marks) Perform level. Be sure to state your conclusion. a) b) ( 6 marks) Compute a 99% lower one sided CI on - assuming for each Length is KNOWN to be 30. Relate this interval to your result in Part(a)

Explanation / Answer

(a) here we use F-test and do one way analysis of variance and using ms-excel following ANOVA is generated

since p-value of between groups of lengths is less than alpha=0.05, so we reject null hypothesis

( or critical F=3.008787 is less than calcuated F=15.8986548 , so we reject H0

(b) with 99% confidence Critical difference or LSD=sqrt(2*MSE/r)*t(alpha,error df)=sqrt(2*820.3929/7)*t(0.01/2,24)=

=sqrt(2*820.3929/7)*2.7969=42.82

lower 99% CI on mu4-mu1=(437-375)-42.82=19.18

ANOVA Source of Variation SS df MS F P-value F crit Between Groups 39129.42857 3 13043.14 15.8986548 6.63E-06 3.008787 Within Groups 19689.42857 24 820.3929 Total 58818.85714 27

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