lecture 17 ANOVA pt.1

# lecture 17 ANOVA pt.1 - Multiple t Tests Given k groups...

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Hypothesis Testing with 3 or more Means Multiple t Tests? Given k groups, there are k(k - 1 )/ 2 possible pairwise comparisons that can be constructed Familywise error rate: the true probability of one or more Type I errors occurring on c hypothesis tests # Groups # PC 2 1 3 3 4 6 5 10 6 15 1

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Bonferroni Inequality FWER c α FWER is the familywise error rate c is the number of comparisons (hypotheses to be tested) α is the probability of a Type I error occurring on any single comparison # Groups # PC FWER 2 1 .05 3 3 .15 4 6 .3 5 10 .5 6 15 .75 2
A New Null Hypothesis testing procedures discussed so far are only appropriate for testing H 0 of the basic form: What’s needed is a procedure that is appropriate for evaluating H 0 of the form: H 0 : μ = μ 0 or H 0 : μ 1 = μ 2 H 0 : μ 1 = μ 2 = μ 3 = . .. = μ k 3

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F Distribution Formed by the ratio of two independent chi- square distributed random variables Positively skewed, and ranges over only positive numbers The central variant has two parameters, ν 1 and ν 2 , (numerator and denominator degrees of freedom, respectively) Asymptotically normal as ν 1 and ν 2 approach inﬁnity ( χ 2 ( ν 1 ) / ν 1 )/( χ 2 ( ν 2 ) / ν 2 ) ~ F ( ν 1 , ν 2 ) 4
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F as a Variance Ratio Given a normally distributed random variable: s 2 (n-1)/ σ 2 ~ χ 2 ( ν = n - 1) [ s 2 1 (n 1 -1)/ σ 2 1 (n 1 -1) ] / [ s 2 2 (n 2 -1)/ σ 2 2 (n 2 -1) ] ~ F (n 1 - 1, n 2 - 1) (s 2 1 / σ 2 1 )/(s 2 2 / σ 2 2 ) ~ F (n 1 - 1, n 2 - 1) If σ 2 1 = σ 2 2 ( i.e. , H 0 of equal variances) then: s 2 1 /s 2 2 ~ F (n 1 - 1, n 2 - 1) 6
as t 2 Given Z ~ N(0, 1) , and Y ~ χ 2 ( ν = n - 1) : Z/ (Y/ ν ) ~ t ( ν = n - 1) Õ Squaring yields: Z 2 /(Y/ ν ) ~ t 2 ( ν = n - 1) From the definition of χ 2 (1) as the square of a normal distribution with μ = 0 and σ 2 = 1: Z 2 ~ χ 2 ( ν = 1) (Z 2 /1)/(Y/ ν ) ~ ( χ 2 ( ν Z = 1) / ν Z )/( χ 2 ( ν Y = n - 1) / ν Y ) t 2 ( ν = n - 1) ~ F (1, n - 1)

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lecture 17 ANOVA pt.1 - Multiple t Tests Given k groups...

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