Chapter 28final.pdf

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University of Alberta
Biology (Biological Sciences)
Blaine Mullins

Chapter 28: Analysis of Variance In this chapter, we will talk about:  Hypothesis tests about the equality of two or more population means  An analysis of variance (ANOVA) table for a one-way layout Suppose we would like to compare test k populations wherek  2. Let, for i k1, 2, ,, th  i i Population mean th  i i Ppoulation Standard deviation Therefore, we would like to test: H :k      , where 2 0 1 2 3 k Note that:  1 2 3dn,a,  k 1 2k 3   are all unknown.  We are not interested about  1 2 3  k .Hence, they are called nuisance parameters.  We are testing : H 0 12 3  k(All population means are equal) versus H a H 0s not true (Not all population means are equal)  Some students think H a k1 2 3  .This is NOT true.  The true statement about alternative hypothesis is: H1: At least two population means are not equal or H a At least one population mean is different from other population means. The necessary assumptions for one-way analysis of variance are: a) Each sample is a simple random sample b) Populations are normal c) Samples are independent. d) Population standard deviations are equal  1 2 3    k Suppose there are k populations. Let, fori k1, 2, , , n i Number of observationsin samplei, Y iSample mean of sample i S i SampleStandard deviation of sample i Nnnn   n  Total number of observations 1 2 3 k Define n11n2 2 n3 3 nY kk Y  Grand mean N SS(Between Groups)=SS(Treatments)  n1 1Y ) 2 2(Y 2 Y 3 32n (Y Y )  n (YI k Y ) SS(Between Groups) MS(Between Groups)  M S(Treatments) k-1 2 2 2 2 SS(Within Groups)=SS(Error) (n1 1 2S2 33n1)S (n 1)S  (n k k1)S SS(Within Groups) MS(Within Groups)  MS(Error) = Nk Then, the test statistic for testiH :    is: 0 1 2 3 k TS  MS(BetweenGroups) ~ F- distribution, iH is true. MS(Within Groups) 0 Example 28.1: Find SS(Treatments) , SS(Error), and the value of the test statistic for testing for the following data: H 0 A  B C A B C 4 6 2 7 2 4 1 3 4 An analysis of variance table for a one-way layout is Source df SS MS F-Stat P-value Treatments k 1 SS(Between Groups) MS(B) MS(B)/MS(W) Error Nk SS(Within Groups) MS(W) Total N 1 Example 28.1 (Cont.): Find ANOVA table for data in example 28.1. Example 28.2: A researcher wishes to determine whether there is any difference in weight gains of athletes following one of three special diets. Athletes are randomly selected and assigned to one of the three groups. Each group is then placed on one of the diets for six weeks. The weight gains (in pounds) are as follows: A B C 4 6 2 7 14 8
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