Topic 8. Analysis of Variance (Ch. 12)
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1 Topc 8. Analyss of Varance (Ch. 1) 1) One-way analyss of varance (ANOVA) 1.1 General ratonale We studed the comparson of two means n the last chapter, and found that we could use the t and/or z statstcs to make statstcal nferences. Suppose however, that we have k > populatons, and would lke to test for dfferences n means. For example, consder the example of subects recruted from dfferent medcal centers. Let X represent the FEV (forced expraton volume) of a subect. Suppose one has three dfferent centers, and would lke to know f the dstrbutons of X are the same at each place. One hypothess of nterest would be H0 : μ1 = μ = μ3. (Mght there also be other hypotheses of nterest?). We could take a random sample at each ste. To be specfc, suppose one has the followng data: Ste 1 3 n x s Obvously, one could construct a confdence nterval for each μ. These are dsplayed on p. 88. Also, a common procedure n practce s to do parwse hypothess testng,.e. ste 1 vs, ste 1 vs 3, and ste vs 3. Is there any problem wth usng the parwse testng procedure? One factor to consder s the number of possble parwse tests. If one has k populatons, the number of parwse combnatons s k ( k)( k 1) () =. For example, wth k = 3, there are 3 combnatons. However, for k = 10, there are 45 parwse combnatons. If two means are equal (and s true) the probablty of makng a correct decson s 1 α. However, f we make r such decsons and assumng the decsons are ndependent (whch they are not), the probablty of beng correct on all r s r (1 α). The probablty of makng one or more ncorrect decsons f s true s then r 1 (1 α). 1
2 Ths s called the expermentwse error rate, and can be large. For example, f α =.05 and r = 3, ths error rate s 3 1 (1.05) = For k = 10 populatons and r = 45, the error rate s 45 1 (1.05) = Although the parwse tests are not ndependent and hence these results are not precse, they llustrate the general problem nvolved wth a hgh number of parwse combnatons. The analyss of varance (ANOVA) s the methodology for testng k means smultaneously, wth specfed α. Our hypotheses are: H μ = μ = = μ (1) 0 : 1 k H A : the means are not all equal. Three assumptons are necessary for ths procedure: 1) The ndvdual samples are ndependent. (e.g. one does not have pared or blocked samples). ) The populatons are normally dstrbuted (though wth larger sample szes ths assumpton s less mportant). 3) The populaton varances are equal,.e. σ = σ = = σ =σ () 1 k. (If ths s not true, data are often transformed). Lettng agan X denote the th observaton n the th sample, our assumptons symbolcally are X ~ IN ( μ, σ ) for = 1,, n and = 1,, k Under the assumpton of equal varances, each of the k sample varances,, estmates σ. The best estmate s the weghted average ( n 1) s sw = ( n 1) = n s n k) ( 1) /( where n= n s the total number of observatons. s w s called the wthngroups varance estmate and t generalzes the pooled varance,, wth samples. For example, for the prevous data from the three centers s p s (3)
3 (0)(0.496) + (15)(0.53) + ()(0.498) s w = ( ) = 0.54 wth s w = The queston now reduces to whether the varablty n the three sample means s surprsng f the populaton means are equal and the common standard devaton s s = 0.504, as llustrated n Fgure Sources of varaton The wthn-groups estmate,, s based on the ndvdual, whch n turn are based on the devatons of the s w Ths statstc always estmates σ. x observatons from ther respectve means, x. One can also estmate σ from the sample means f s true. Let the grand mean be x = nx / n. (4) Ths s ust the grand sum of all observatons dvded by the grand sample sze,.e. x = x / n. The between-group estmate,, s B s = n ( x x) /( k 1) (5) If s false, overestmates σ, because n addton to varablty wthn the populaton, Consder a test statstc also ncludes dfferences between the sample means. F = s / s. (6) B w Clearly, F must be postve. If n (1) s true, F should be approxmately 1. Why? The F statstc has two degrees of freedom, df and df 1. The numerator df are df1 = k 1 and the denomnator df are df = n k. For example, n our problem, the F statstc would have df1 = and df = 57. Any departure from ANOVA test s one-sded. (How would you nterpret The RR can be found for gven α from Table A.5. To complete the example, note, from (4) s expected to ncrease F (from about 1), hence the sb s < s w, yeldng F < 1?). 3
4 1(.63) + 16(3.03) + 3(.88) x = 60 =.83. Hence, from (5) + + = Therefore, from (6), F = / 0.54 = For α =.05, Table A.5 wth df1 = and df = 57 gves (usng 40 df) F.05 = (.63.83) 16( ) 3(.88.83) = Because our F < 3.3, we would not reect. Note that our p-value s.05 < p <.10. How would you nterpret the results n the context of our problem? Ths secton concludes the one-way ANOVA. Most books have alternatve formulas based on x sums rather than x and s. The present calculatons are very senstve to round-off error. There are numerous other ANOVA s for other desgns. For example, one could have a Randomzed Block Desgn, whch generalzes the pared t-test. There are whole courses at TAMU on ANOVA, and t s wdely used n research. ) Multple comparson procedures. In ths problem, we dd not reect, hence one would not pursue possble parwse dfferences. Had one set α =.10, one would reect, and could then nvestgate whch dfferences are sgnfcant. It s not good statstcal practce to have α =.10 (even though t s sometmes done). Consequently, we need a new example. Consder the Further Applcatons problem. A study was conducted to follow 3 groups of overweght males for a perod of one year. The three groups were: 1) new det program but no exercse program, ) new exercse program but no det program, 3) control group, no change n det or exercse. Let X denote weght loss. The results were: 4
5 Test H0: μ1 = μ = μ3. Group 1 3 n x s Do the three assumptons seem reasonable n ths case? Is there any concern of how the three groups were chosen? The analyss s : μ = μ = μ 1) H0 1 3 ) H : some 3) A TS : F = s / s μ are dfferent. B w 4) RR : df1 =, df = 18 f H0 F F0.05 α =.05, reect f > = ) from calculatons on p. 95 sw = 14.4 x = 3.55 sb = F = 646. /14.4 = 45.8 Therefore, reect H0: μ1 = μ = μ3, wth p <.001. The queston now s whch means dffer. If we want the expermentwse error rate to be α, the ndvdual comparson error rates should be * k α = α /( ) α (7) =. kk ( 1) Ths s called the Bonferron method. For example, to acheve an overall rate of α =.05 wth k = 3, the ndvdual rate s *.10 α = = It can be a bt of a problem to fnd the reecton regons for these new α * sgnfcance values. To test parwse comparsons one uses H 0 : μ = μ 5
6 t x = s x w n n Ths s smlar to the t statstc n Chapter 11, except that the df are now n k. Why the change and how s t useful? Often all sample szes are the same, whch means one has to calculate the denomnator n (8) only once. Let s examne the three parwse combnatons. For testng μ1 = μ 7. ( 4.0) t1 = = Usng the t table wth df = 10, we see that p <.0005, and because ths s a - sded test, we conclude p <.001. For testng μ 1 = μ 3 and μ = μ3, one has t 13 = 9.47 and t 3 = 5.74, both of whch also reect H 0 wth p <.001. Hence, all three populaton means are dfferent. (8) 3) Further consderatons. The ANOVA results are usually presented as an ANOVA table. For the above data, an abrdged ANOVA table s Source df Mean Square F P Between groups < Wthn groups Total 130 Wth the multtude of possble parwse comparsons, there s a general way to present dfferences. For the three medcal centers, consder orderng the means as follows: Center Mean St. Lous 3.03 A Los Amgos.88 A B John Hopkns.63 B In many computer programs, means that are not sgnfcantly dfferent are connected by a common letter. In the above case, St. Lous and John Hopkns are not connected by a common letter, and hence are sgnfcantly dfferent. Another way to dsplay these results s to use vertcal lnes nstead of alphabetc letters. 6
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