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Lecture

# Analysis of Variance (4).docx

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School
Department
Statistics
Course
STATS 2B03
Professor
Aaron Childs
Semester
Fall

Description
October 31 , 2012 Stats 2B03: Statistical Methods for Science Analysis of Variance (4) - In our example, to test: H 0 μ1=μ 2 H A μ 1μ 2 Xavg.1xavg.27.59-7.24=0.35 √ HSD*= q 05, 4, 23  .8828 from ANOVA table, q from q table  do not reject H 0ince |.35|<1.517 Table 2 Hypothesis HSD* Conclusion H 0μ 1μ 2 1.517 Do not reject H 0ince .35<1.517 H 0μ 1μ 3 1.385 Do not reject H 0ince 1.26<1.385 H 0μ 1μ 4 1.341 Reject H 0ince 1.58>1,341 H 0μ 2μ 3 1.517 Do not reject H 0ince 0.91<1.517 H 0μ 2μ 4 1.477 Do not reject H 0ince 1.23<1.477 H 0μ 3μ 4 1.341 Do not reject H 0ince 0.32<1.341 - smoking would have the effect of decreasing the birth weight - equivalently, we can compute Tukey confidence intervals for μ – μ andirejjct H 0μ 1μ jf and only if 0 is not in the confidence interval - in our example, a Tukey 95% C.I for μ -μ 1s 2 7.59-7.24±1.57 0.35±1.57
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