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University of Alberta
Paul Cartledge

Ch. 28 - Comparing Several Means Use t-tools? NO! Æ Reason? Compound uncertainty - In any test, there is uncertainty such that we reject H 0 when it’s true, or Type I error. By comparing multiple means and using ONE t-test for each pair, the “overall” Type I error will compound. - For example, consider 3 means that are equal and each t-test uses α = 0.05. Thus, there’s a 5% chance to show a difference when there isn’t (recall H 0 assumes no diff.). The chance of de3ecting at least one difference among the three means is roughly 1 – 0.95 = 0.143 or 14.3% when the means are EQUAL! (Note: 14.3% is the “overall” Type I error.) - For 5 means, the “overall” Type I error becomes approximately 40%. Def’n: ANalysis Of VAriance (ANOVA) is a procedure to test the equality of three or more population means. NOTE: the name of the test refers to comparing different sources of variability; it WILL test differences among means. Test requires the following assumptions: 1. The populations are all normally distributed. 2. The populations all have the same standard deviation. 3. The samples from different populations are random and independent. Checking Assumptions: - Assumption #1 is checked with histograms/boxplots for each group. - Assumption #2 is more critical but harder to assess. Still, we can use side-by-side boxplots (or the informal rule from Ch. 24). - Assumption #3 by analyzing the experiment design. Notation: - yij observation for i subject in j group - j = 1, …, k indexes groups - i = 1, …, n jindexes subjects within groups - nj= # of observations in j group; N = ∑ nj = total # of observations - y jnd sjare sample mean and variance for the j grouph - y = grand mean = mean for combined sample: y = 1 y = 1 n y N ∑j i ij N ∑ j j j Statistical model, parameters, hypotheses: Each observation can be represented by Yij µ + τ +jε ij where Y aij independentthandom observations,th is the overall mean, τ is a paramjter associated with the j group called the j treatment effect, and ε is a rijdom error. H 0 µ 1 … = µ k H : the µ are not all equal A j (OR, at least 2 µ jre different from each other; OR at least 1 µ is dijferent) ANOVA F test statistic: For sources of variability in the model above, the ANOVA identity is k nj k nj k nj ∑∑ (yij− y) 2 ∑∑ (y+j− y) 2 ∑∑ (yij y )j 2 j=1=1 j=1=1 j=1=1 SS Y= SS T SS E where SS Y is the total sum of squares, SSTis the treatment sum of squares and SS iE the error sum of squares . Also, the presence of “squares” suggest a ratio test. Thus, for the above H 0, we have SS = n (y − y()variabbiwteenles) T ∑ j j
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