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Lecture

RCBD III; Latin squares.pdf

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Department
Statistics
Course
STAT368
Professor
Douglas Wiens
Semester
Winter

Description
81 11. RCBD III; Latin squares Levene’s test for equal variances in each block. The blocks correspond to the 4 columns of the data, so we rst compute the medians of the columns, sub- tract ¯hen from ench column o¯d take absolute values: = ¯ 1 ¯. The do an anova to h i see if is the same in each block. data.matrix F) blocks 3 0.022500 0.007500 0.5806 0.6389 Residuals 12 0.155000 0.012917 Using treatments rather than blocks gives = 698 (you should check this). 82 The analogue of the Kruskal-Wallis test, for a RCBD, is ‘Friedman’s test’. The observations are replaced by their ranks within each block, and the usual ANOVA is run. This method does not take account of the fact that the denominator of the is not random (as before); the function friedman.test will do so. The di erences are generally slight. > friedman.test(y,treatments,blocks) Friedman rank sum test data: y, treatments and blocks Friedman chi-squared = 8.8462, df = 3, p-value = 0.03141 83 So - the assumptions seem to be met, and at least some of the di erences in the treatment means, i.e. in the mean readings = + , are signicant - the readings of the hardness testing device depend on which tip is being used. This is bad news for the engineers. Is there any one tip responsible for the di erences? We should look at all of the di erences ˆ ˆ = ¯ ¯ to see which are signicant. • Method 1: Fisher’s LSD (“Least Signicant Dif- ference”). A 100(1 )% condence interval on one di erence is s µ ¶ 1 1 ¯ ¯ ± 2 ( ) + s µ ¶ 2 = ¯ ¯ ± 2 9 00889 4 With = 05, the 95% interval is ¯ ¯ ± 151. Converting this to a hypothesis test, we see that the hypothesis of equality is rejected if ¯ ¯ ¯¯ ¯ ¯ = 151 84 Since |¯ ¯ | = 025 151 1 2 |¯1 ¯3|
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