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# STATISTICAL METHODS II STAT 516

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##### Biol 1020 - 002 (BIOL, Christine A. Sunderman, Principles of Biology (1020))

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###### Class Notes

##### Biol 1020 - 002 (BIOL, Christine A. Sunderman, Principles of Biology (1020))

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This 16 page Class Notes was uploaded by Shane Marks on Monday October 26, 2015. The Class Notes belongs to STAT 516 at University of South Carolina - Columbia taught by B. Habing in Fall. Since its upload, it has received 29 views. For similar materials see /class/229660/stat-516-university-of-south-carolina-columbia in Statistics at University of South Carolina - Columbia.

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Date Created: 10/26/15

Type I and Type III Sums of Squares Supplement to Section 83 Brian Habing 7 University of South Carolina Last Updated February 4 2003 PROC REG PROC GLM and PROC INSIGHT all calculate three types of F tests 0 The omnibus test The omnibus test is the test that is found in the ANOVA table Its F statistic is found by dividing the Sum of Squares for the Model the SSR in the case of regression by the SSE It is called the Testfor the Model by the text and is discussed on pages 355356 0 The Type III Tests The Type 111 tests are the ones that the text calls the Tests for Individual Coef cients and describes on page 357 The pvalues and F statistics for these tests are found in the box labeled Type III Sum of Squares on the output 0 The Type I Tests The Type Itests are also called the Sequential Tests They are discussed brie y on page 372374 The following code uses PROC GLM to analyze the data in Table 82 pg 346347 and to produce the output discussed on pages 353358 Notice that even though there are 59 observations in the data set seven of them have missing values so there are only 52 observations used for the multiple regression DATA fw08X02 INPUT Obs age bed bath size lot price CARDS 1 21 3 30 0951 64904 30000 2 21 3 20 1036 217800 39900 3 7 1 10 0676 54450 46500 ltsnip rest of data as found on the book s companion web sitegt 58 1 3 20 2510 189500 59 33 3 40 3627 17760 199000 1 PROC GLM DATAfw08X02 MODEL price age bed bath size lot RUN With five independent variables this procedure produces seventeen pvalues for testing hypotheses A 7 This is the test associated with the ANOVA table It always even for cases that aren t regression tests the null hypothesis that none of the independent variables linearly predict the dependent variable The alternate hypothesis is that at least one of the independent variables does linearly predict the dependent variables Because it tests all of these at once it is sometimes called the omnibus test For this example it is testing the null hypothesis H0 all of Bage0 Bbed0 5133350 Bsize0 and 51050 The alternate hypothesis is that at least one of them is not zero In this example the pvalue is less than 00001 and we would reject the null hypothesis You can check these SS MS and F values with what the text gives on page 356 The GLM Procedure Dependent Variable price Sum of Source DF Squares Mean Square F Value Pr gt F Model 5 6569579292 1313915858 4293 lt0001 4A Error 45 13774 04972 306 08999 Corrected Total 50 7946984265 ReSquare Coeff Var Root MSE price Mean 0826676 1598770 1749543 1094305 Source DF Type SS Mean Square F Value Pr gt F age 1 52613923 52613923 172 01965 413 bed 1 1071309272 1071309272 3500 lt0001 4C bath 1 20049 55143 20049 55143 65 50 lt 0001 4D size 1 3393941003 3393941003 11088 lt0001 4E lot 1 467 59951 467 59951 153 02229 4F Source DF Type I ll SS Mean Square F Value Pr gt F age 1 946 23859 946 23859 3 09 00855 4G bed 1 197260916 197260916 644 00147 4H bath 1 15683009 15683009 051 04778 41 size 1 3273975553 3273975553 10696 lt0001 4J lot 1 467 59951 467 59951 153 02229 4K Standard Parameter Estimate Error t Value Pr gt ltl Intercept 3528792164 1411712107 250 00161 4L age 70 34980453 0 19895262 71 76 00855 4M bed 71123820158 442691173 7254 00147 4N bath 7454015206 634278904 7072 04778 40 s I Ze 65 94646658 6 37644152 10 34 lt 0001 4P lot 006205081 0 05020362 124 02229 4Q B through F 7 These are the Type I or sequential tests The line with each of the pvalues always begins with which independent variable is being tested given that those listed above it are already included in the model The pvalue of less than 01965 for Fdf1745172 on line B thus tests the hypotheses H0 Bage0 given that no other independent variables are included in the model vs H A Bageio given that no other independent variables are included in the model This result is similar to but not exactly the same as what you would get if you performed a simple linear regression to predict price from age shown here Analysis of Variance Source DF Sum of Squares Mean Square F Stat Pr gt F Model 1 5261392 5261392 033 05703 Error 49 789437034 16110960 C Total 50 794698426 Notice that the Sum of Squares for the Model in this simple linear regression is identical to the Type I Sum of Squares for the variable Age Also note that the TSS and total degrees of freedom are the same The reason for the difference in the F statistics and the pvalues comes because the Type Itest has taken all of the other variables errors out of the SSE That is the Type Itest uses SSE and df from the ANOVA table for the entire multiple regression Now the pvalue of less than 00001 for FdH17453500 on line C takes into account that it appears on the line below height It thus tests the hypotheses H0 Bbed0 given that age is included in the model vs HA Bbedy O given that age is included in the model We could continue in this way down to line F This pvalue of 02229 for Fdf1745153 tests the hypotheses H0 B1010 given that age bed bath size and lot are included in the model vs HA B109 0 age bed bath size and lot are included in the model By line B we would thus conclude that age by itself is not linearly related to weight By line C we would conclude that bed is linearly related to weight after accounting for age etc If we wanted to reverse the order in which these independent variables were considered we simply could have listed water before height in the model line It is important to note that when using a Type Itest that it makes no sense to decide to keep something in the model if you don t also include everything listed above it Because of this it is best to list the variables for the model in order from what you think should be most important to what you think should be least important Also notice the relationship between the Type Itests and the omnibus test Notice that the sum of square on line BF add up to the SSR on line A It can be shown algebraically that the Type I sum of squares will always add up to the sum of squares on the model line G through K These are the Type III tests Just like the Type Itests each line always begins with which independent variable is being tested In this case however each of these tests is done given that all of the other variables are included in the model The pvalue of 00855 for FdH1745644 on line G thus tests the hypotheses H0 Bage0 given that bed bath size and lot are included in the model vs HA Bage7 0 given that bed bath size and lot are included in the model Similarly the pvalue of less than 00147 for Fdf1745644 on line H tests the hypotheses H0 Bbed0 given that age bath size and lot are included in the model vs HA Bbedy O given that age bath size and lot are included in the model By line G at 1005 we would thus conclude that age is not a significant predictor of price if bed bath size and lot are already included in the model By line H we would conclude bed still serves as a significant predictor even after age bath size and lot are already included in the model 3 It is important to note that anything found signi cant by a Type 111 test is adding to the ability of the multiple regression model to predict the independent variable Say you nd that several variables are not statistically signi cant however In this case it is important to realize that you found that they were not signi cant given that all of the others were included in the model You can thus only safely get rid of one of those variables maybe the others become signi cant then if that one is not included For a Type 111 test the sum of squares will generally not add up to the SSR it only happens if the independent variables are all orthogonal to each other However the last of the Type 111 tests will always equal the last ofthe Type Itests Reread the hypotheses being tested to see that this is so L through Q Lines L through Q are simply the ttests discussed on pages 358359 Notice for lines G and M that 309 176 and that the pvalues are equal Recalling that tdgnzFdH1yny it is no surprise that these ttests are exactly the same as the Type III Ftests The reason for having SAS output this section is because it also gives the estimates of the regression coef cients and their standard errors Compare these lines of output to page 367 for example Line L is the test of the hypothesis that the intercept is zero Contrasts and Multiple Comparisons Supplementfor Section 65 Brian Habing 7 University of South Carolina Last Updated February 26 2004 As discussed in Section 64 and at the beginning of Section 65 the F test from the ANOVA table allows us to test the null hypothesis The population means of all of the groupstreatments are equal The alternate hypothesis is simply that At least two are not equal Often this isn t what we want to know Say we are comparing 20 possible treatments for a disease The ANOVA F test sometimes called the omnibus test could only tell us that at least one of the treatments worked differently than the others We might however want to be able to rank the 20 from best to worst and say which of these differences are signi cant We might want to compare all the treatments produced by one company to those of another or maybe all the treatments based on one idea to those based on another An obvious suggestion in each of these cases would be to simply do a large number of ttests To rank the 20 from best to worst we could simply do a separate ttest for each possible comparison there are 190 of them To compare the two companies or two ideas we could simply group all of the observations from the related methods together and use ttests to see if they differ One difficulty with this is that the OLlevel probability of a Type I error may no longer be what we want it to be Sidak s Formula Stepping back from the ANOVA setting for a minute say we wish to conduct onesample ttests on twenty completely independent populations If we set 1005 for the first test that means that 005 0L Preject Hg for test one H0 is true for test one We could write the same for the other nineteen populations as well Ifwe are concerned about all twenty populations though we might be more interested in the probability that we reject a true null hypothesis at all That is 0LT Preject Hg for test one U reject Hg for test two U U reject Hg for test 20 H0 is true for all tests We call this quantity the family wise or experiment wise error rate The 0L for each individual test is called the comparison wise error rate The family or experiment in this case is made up of the twenty individual comparisons Using the rules of probability and the fact that we assumed the tests were independent for this example we can calculate what 0LT would be if we used 1005 for the comparisonwise rate onpreject Ha forl v rejectHu for 2 v v rejectHu for 20 Hu is true for all tests 17 Pfail to reject Hg for 1 n n fail to reject HD for 20 Hg is true for all tests 17 Pfail to reject Hg for 1 Hg is true for all tests Pfail to reject Hg for 2 Hg is true for all tests 11OL1OL 1rot 1r1rot 1 7 17005 1 0952 064 The chance ofmaking at least one error ctT isn t 5 it s nearly 64 r39 39 t t itnktestswe et 06 1 r 1700 When the tests are independent Ifwe know What 1139 We mnt We can solve for the needed X to get 0clrlroc1 k cam Lo 1 T individual comparison Bonferroni s Formula In the case ofANOVA the various tests will o en not be independent wae mnt to conduct the ttests to compare 20 possible medical treatments to each other then clearly the comparison ofl to 2 and 1 to 3 will not be independent they both contain treatment 1 The diagram ueiu 39 quot 39 39 L 4 due is I A 5quot V 17a 17a 141T T is as large as possible a is in between T is as small as possible assume the worst what usually happens assume the best Bunrerruni 7777 Fisher The worst possible case in terms of 0LT would be if the type I errors for the individual tests were mutually exclusive In this case 0LT Preject Hg for l U reject Hg for 2 U U reject Hg for k H0 is true for all tests Preject Hg for l H0 is true for all tests Preject Hg for k H0 is true for all tests OL0L 0L koc to amaXimumofone or equivalently OL XTk This is Bonferrom39 s formula The best possible case in terms of 0LT would be if the type I errors for the individual tests all overlapped In this case 0LT 0L So far then If we are performing a set of tests that are independent then we can use Sidak s adjustment to gure out what comparisonwise 0L we should be using If the tests are not independent then we have a choice We could be liberal and reject true null hypotheses too often use 0LT 0L or be conservative and not reject the true null hypotheses as much as we should for our desired 0LT use Bonferroni In terms of OCT we would be better being conservative then The problem with this is that if we do not reject the true null hypotheses enough we also will not reject the false ones enough In the case of comparing the means of treatments if we are liberal using 0LT 0L we will find lots of differences that are there but also lots of differences that aren t real If we are conservative we won t find lots of fake differences but we will also miss the real ones Fisher s LSD One method for dealing with the fact that using 0LT 0c is too liberal is called the Fisher Least Significant Difference LSD test pages 254256 The idea is to only check to see ifthe means of groups are different if you reject the omnibus Ftest This makes some obvious sense if you fail to reject that there are no differences why would you continue looking While this helps keep the number of false rejections down it does have two downsides The first problem can occur when you fail to reject the overall ANOVA null hypothesis Because the omnibus test from the ANOVA table is looking at all of the groups at once it will sometimes miss a difference between just two means It has to sacrifice power for each individual comparison in order to test them all at once The second problem can occur when we do reject the overall ANOVA null hypothesis and proceed to do the other comparisons of the group means The omnibus test may have rejected because of a difference between only two means but because using 0LT 0c is liberal we may find more differences than are really there Because of these two difficulties Fisher s LSD can t be highly recommended The Holm Test The Holm test is a method for dealing with the fact that the Bonferroni procedure is too conservative The main idea comes from noticing that we always used the condition H0 is true for all tests instead of using the condition that it was true only for the speci c test we were doing The procedure behind the Holm test is to rst nd all of the pvalues for all of the individual tests we were performing and then rank them from smallest to largest Compare the smallest to Powk If you fail to reject the null hypothesis for the rst step then you stop here If you do reject then compare the next smallest to 0LXTk 1 Again if you fail to reject the null hypothesis then you stop here if you do reject continue on and use XLTk2 You do not need to check the omnibus Ftest rst thus avoiding the rst problem with Fisher s LSD For example say you have ve hypotheses you are testing you wanted 0LT 005 and you observed p values of 001 1 0751 0020 0030 and 0001 respectively Test Number P value Compare To nnc39lu ion 5 0001 0055001 reject Hg for test 5 1 0011 005400125 reject Hg for test 1 3 0020 005300166 fail to reject for test 3 4 0030 no comparison made fail to reject for test 4 2 0751 no comparison made fail to reject for test 2 Notice that Bonferonni s test would only have rejected for test 5 Using 0LT 0L would have rejected for tests 5 l 3 and 4 Thus the power of the Holm test is somewhere in between that of the Bonferroni procedure and Fisher s LSD While it is more powerful than Bonferroni s method it rejects more false Ho s it still makes sure that 0LT is held to the desired level unlike Fisher s LSD Notice that if all the null hypotheses are true we make an error if we reject any of them The chance that we reject any is the same as the chance that we reject the rst which is XTk We are thus safe for the same reason that Bonferroni s formula works Now assume that we rejected the rst null hypothesis because it was false There are only k I tests left and so when we go to the second test we can start as if we were using Bonferroni s formula with k I instead of k And we continue in this way While this argument is not a proof that the Holm Test protects the family wise error rate 0LT it should make the general idea fairly clear While there are many other methods for making multiple comparisons see pages 256267 the Holm test performs fairly well compared to all of them controls 0LT at the desired level and is fairly easy to understand Because of this it will be the method that we will focus on Contrasts In order to perform any of these tests though we must be able to tell SAS what we want done The building blocks for many of the SAS procedures that we will have SAS use are called contrasts They are discussed on pages 243249 ofthe text I I The key formulas are that for the contrast L 2 a ul where 2 611 0 are i1 39 1 where i is normally distributed if the ANOVA assumptions are met Since we have the standard error for i we could make a con dence interval for L or test the null hypothesis that L0 This is exactly what we will ask SAS to do in the examples below Independence Orthogonal Contrasts and Sidak s Formula an aside note As discussed on pages 246 249 two contrasts are said to be orthogonal if the dotproduct sum of the cross products of their coef cient vectors is zero The reason to care if two contrasts are orthogonal is that the estimates that go with a set of orthogonal contrasts are independent This will play an important roll in constructing the ANOVA table for factorial designs in Chapter 9 In the present situation though the test statistics will not be independent however as they both contain the same MSE in the denominator Because of this it is not technically appropriate to use Sidak s formula even in this situation Tying it All Together When we approach an ANOVA problem there are three basic types of questions we could have in mind 1 Are there any differences between any of the group means Choose 0L and simply use the F test from the ANOVA table the omnibus test N Do the means of some palticular groups differ from the means of some other particular groups Choose 0LT and come up with the contrasts you wish to test Find the pvalues for the tests that go with these contrasts and then use the Holm test procedure to see which are signi cant F What is the order of the group means and which are signi cantly different from each other Choose 0LT Make all of the contrasts that compare two means to each other nd their pvalues and use the Holm test procedure to see which are signi cantly different Then make a simple graph to display the result It is important to note that you should decide which one of these questions you want to answer before you look at any of the output If for some reason you don t know why you are looking at the data in advance then Scheffe s method discussed on pages 259260 can be used Also you should only pick one of these three questions It doesn t make sense to look at more than one of them does it Finally in all cases remember to check the assumptions Example Shrimp Larvae Diets The follow pages contain the code and output for answering each of the questions above for Example 66 on pages 264267 The write up assumes that the desired familywise error rate is 0LT005 Enter the Data from Table 621 The seven treatment groups can be broken into four experimental diets that contain a basal compound and cafol corn and sh oil in a 11 ratio calo2 corn and linseed oil in a 11 ratio faso3 sh and sun ower oil in a 11 ratio or falcgt4 sh and linseed oil in a 11 ratio And three standard diets that were bc5 the basal compound diet lma6 live micro algae lmaa7 live micro algae and Artemia DATA shrimpweights INPUT diet weight I cafol 47 O cafol 50 9 cafol 45 2 cafol 48 9 cafol 48 2 calo2 38 l calo2 39 6 calo2 39 l calo2 33 l calo2 4O 3 faso3 57 4 faso3 55 l faso3 54 2 faso3 56 8 faso3 52 5 falcgt4 54 2 falcgt4 57 7 falcgt4 57 l falcgt4 47 9 falcgt4 53 4 bc5 385 bc5 420 bc5 38 7 bc5 38 9 bc5 44 6 lma6 48 9 lma6 47 O lma6 47 O lma6 44 4 lma6 46 9 8 lmaa7 81 7 lmaa7 73 3 lmaa7 82 7 lmaa7 74 8 lmaa7 87 l Check the Assumptions Using PROC INSIGHT and the Modi ed Levene test PROC INSIGHT OPEN shrimpweights FIT weightdiet RUN l PROC GLM DATAshrimpweights ORDERDATA CLASS diet MODEL weightdiet MEANS diet HOVTESTBF l Pweight RNweig ht The GLM Procedure Brown and Forsythe39s Test For Homogenelty of welght Varlance ANOVA of Absolute DeVIatIons From Group Medlans Sum of Mean Source DF Squares Square F Value Pr gt F dlet 6 414549 69091 132 02793 Error 28 1461 52166 From the residual vs predicted plot the means for each of the three groups seem to be near zero in fact they must always be for a oneway ANOVA However it is not clear from the residual vs predicted plot if the variances of the errors for the seven groups are the same Using the modi ed Levene test we fail to reject that they are different with a pValue of 02793 Finally from the QQ plot of the residuals it appears that the distribution of the errors is approximately normally distributed although its arguable at the ends Since it is a randomized experiment the assumption of independence is also satis ed so that all four assumptions for the ANOVA are met in this case Possible uestion 1 Are there an differences between an of the rou means PROC GLM DATAshrimpweights ORDERDATA CLASS diet MODEL weightdiet l The GLM Procedure Dependent Varlable welght Sum of Source DF Squares Mean Square F Value Pr gt F Model 6 5850774857 975129143 8814 lt000 l Error 28 309 792000 11064000 Corrected Total 34 6160566857 As the pValue of lt0001 is less than 005 we reject the null hypothesis HOZHcafol Healol 21503 Hfalo4 Hch Hlma6 Hlmaa7 and conclude that at least one of the diets has an effect that differs from the other six Say that we are interested in the seven specific comparisons discussed on page 265 1 How the experimental diets compare to the standard diets on average newold 2 How diets containing corn compare to those without corn corn 3 How diets containing fish oil compare to those without fish 4 How diets containing linseed oil compare to those without lin 5 How diets containing sun ower oil compare to those without sun 6 How diets containing micro algae compare to those without mic 7 How diets containing Artemia compare to those without art The seven corresponding null hypotheses could be written as 1 H0 HeafollrlcalozHfaso3Hfalo44 lec5Hlma6Hlmaa73 2 H0 Heafollrlcaloz2 Hfaso3Hfalo4lrlbc5Hlma6Hlmaa75 3 H0 HeafolHfaso3Hfalo43 21102Hbc5lllma6lllmaa7y4 4 H0 HealoZHfalo42 HeafolHfaso3Hbc5lrllma6lllmaa7y5 l l39faso3 l lcafoll39lcaloz l l39t 2tlo4l39l39bc5l l39l39lm216l39l39lmaa7y6 H1ma6Hlmaa72 HeafolHcaloZHfaso3Hfalo4 l l oc55 Hlmaa7 HeafolHcaloZHfaso3Hfalo4Hbc5Hlma66 5 H0 6 H0 7 H02 It is important that the data was entered in the correct order and that we use the ORDERDATA command 3 3 4 2 2 2 4 4 3 2 5 2 6 l l 2 2 2 l l l 4 4 2 2 3 3 2 2 l l 5 5 l 6 I ll ll ll ll ll ll l CLASS di et MODEL weightdiet PROC GLM DATAshrimpweights ORDERDATA ESTIMATE newold diet 3 3 3 3 4 4 4 divisor12 ESTIMATE corn diet 5 5 2 2 2 2 2 divisor10 ESTIMATE fish diet 4 3 4 4 3 3 3 divisor12 ESTIMATE lin diet 2 5 2 5 2 2 2 divisor10 ESTIMATE sun diet 1 l 6 l l l l divisor6 ESTIMATE mic diet 2 2 2 2 2 5 5 divisor10 ESTIMATE art diet 1 l l l l l 6 divisor6 RUN Standard Parameter Estlmate Error t Value Pr gt 1t1 newold 769783333 1 13613379 7614 lt0001 corn 712 3000000 124457222 79 88 lt0001 Flsh 10633333 113613379 094 03573 In 780860000 124457222 7650 lt0001 Sun 3 9366667 160673582 245 00208 mlC 162740000 124457222 1308 lt0001 art 32 9400000 160673582 2050 lt0001 Since we have seven tests that we are conducting we need to determine which pvalues to compare them to We could do this by hand as follows note we have put the tests in order of pvalue Test T P value Compare To nnc39ln ion art 2050 lt0001 005700071 reject H0 mic 1308 lt0001 005600083 reject H0 corn 988 lt0001 005500100 reject H0 lin 6 50 lt0001 005400125 reject H0 newold 6l4 lt0001 005300167 reject H0 sun 245 00208 005200250 reject H0 fish 094 03573 005 fail to reject H0 We could simultaneously make the following conclusions then N E 4 V39 0 gt1 We have signi cant evidence to conclude that the average effect of the four experimental diets is different from the average effect of the three standard diets we estimate it to be 698 lower We have significant evidence to conclude that the average effect of the two diets with corn oil is different from the average effect of the five that don t we estimate it to be 123 lower We do not have significant evidence to reject that the average effect of the three diets with fish oil are the same as the average effect of the four diets that don t We have significant evidence to conclude that the average effect of the two diets with linseed oil is different from the average effect of the five that don t we estimate it to be 809 lower We have significant evidence to conclude that the average effect of the diet with sun ower oil is different from the average effect of the siX that don t we estimate it to be 394 higher We have significant evidence to conclude that the average effect of the two diets with micro algae is different from the average effect of the five that don t we estimate it to be 1627 higher We have significant evidence to conclude that the effect of the diet withArtemz39a is different from the average effect of the siX that don t we estimate it to be 3294 higher PROC MULTTEST could also be used to produce the above ttest pvalues w automatically adjust them with the Holm procedure what SAS calls the Stepdown Bonferroni procedure This procedure will not produce the means that go with the contrasts however A sample of the code is contained below in answering the third possible question Possible uestion 3 What is the order ofthe rou means and which are signi cantly different from each other We will use PROC MULTTEST to conduct this procedure In reading the output note that SAS also calls the Holm test the Stepdown Bonferroni test To compare the means for all of the groups we rst need to enter all of the contrasts for comparing the means There will always be k choose 2 of them Because there are 62 groups in this example there will be 21 such contrasts We could also have entered these in PROC GLM but PROC MULTTEST will automatically adjust all ofthe 0L levels so that we don t have to PROC MULTTEST DATAshrimpweights ORDERDATA HOLM SS 39 C CONTRAST 1 VS 2 l l O O O O O CONTRAST 1 VS 3 l O l O O O O CONTRAST 1 VS 4 l O O l O O O CONTRAST 1 VS 5 l O O O l O O CONTRAST 1 VS 6 l O O O O l O CONTRAST 1 VS 7 l O O O O O l CONTRAST 2 VS 3 O l l O O O O CONTRAST 2 VS 4 O l O l O O O CONTRAST 2 VS 5 O l O O l O O CONTRAST 2 VS 6 O l O O O l O CONTRAST 2 VS 7 O l O O O O l CONTRAST 3 VS 4 O O l l O O O CONTRAST 3 VS 5 O O l O l O O CONTRAST 3 VS 6 O O l O O l O CONTRAST 3 VS 7 O O l O O O l CONTRAST 4 VS 5 O O O l l O O CONTRAST 4 VS 6 O O O l O l O CONTRAST 4 VS 7 O O O l O O l CONTRAST 5 VS 6 O O O O l l O CONTRAST 5 VS 7 O O O O l O l CONTRAST 6 V 7 O O O O O l 1 TEST mean weight RUN The Multtest Procedure Contlnuous Var lab e Tabu atlons Standard Varlable dlet Num0bs Mean DeVIatIon welght cafoi l 5 480400 2 1267 welght caloi2 5 38 0400 2 8754 welght Faso3 5 55 2000 l 9812 welght Faloi4 5 54 0600 3 9017 Welght bci5 5 405400 2 6857 welght ma76 5 468400 1 6009 welght maa77 5 800600 5 9777 p7Va lues Stepdown Variable Contrast Raw Bonferroni weight 1 vs 2 lt 0001 0 0006 weight 1 vs 3 0 0020 0 0132 weight 1 vs 4 0 0079 0 0315 weight 1 vs 5 0 0013 0 0106 weight 1 vs 6 0 5729 1 0000 weight 1 vs 7 lt 0001 lt 0001 weight 2 vs 3 lt 0001 lt 0001 weight 2 vs 4 lt 0001 lt 0001 weight 2 vs 5 0 2447 0 7340 weight 2 vs 6 0 0003 0 0026 weight 2 vs 7 lt 0001 lt 0001 weight 3 vs 4 0 5922 1 0000 weight 3 vs 5 lt 0001 lt 0001 weight 3 vs 6 0 0005 0 0041 weight 3 vs 7 lt 0001 lt 0001 weight 4 vs 5 lt 0001 lt 0001 weight 4 vs 6 0 0019 0 0132 weight 4 vs 7 lt 0001 lt 0001 weight 5 vs 6 0 0057 0 0285 weight 5 vs 7 lt 0001 lt 0001 weight 6 vs 7 lt 0001 lt 0001 We could use the column labeled Raw and compare the values to 0LT 21 OCT 20 etc after putting them in order Instead however the column labeled Stepdown Bonferroni has already been adjusted so that we can just compare those values directly to 0LT Thus we would reject the null hypotheses that the rst and second group have the same mean but accept fail to reject the null hypothesis that the rst and sixth group have the same mean Notice that at 1005 there are only three hypotheses we fail to reject but there are many more at 1001 What way of presenting this would be to list the groups in order of their means and have them share letters if they cannot be said to be signi cantly different Group Mean Weight Groupings at a005 Groupings at a00l lmaa7 80 0600 A A faso3 552000 B B B B falo4 54 0600 B C B C B cafol 48 0400 C C B D C C D lma6 46 8400 C C D D bC5 40 5400 D E D D E calo2 38 0400 D E Note Notice it is possible in some cases to have the groups overlap Remember if we fail to reject we are not saying the means are equal merely that we don t have enough eVidence to say which are different for sure 1001 means you need more eVidence to say they are different than 1005 12

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