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# 67 Class Note for STAT 30100 with Professor Gundlach at Purdue

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This 9 page Class Notes was uploaded by an elite notetaker on Friday February 6, 2015. The Class Notes belongs to a course at Purdue University taught by a professor in Fall. Since its upload, it has received 15 views.

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Date Created: 02/06/15

Chapter 12 Inference for OneWay AN OVA and Comparing the Means Learning goals for this chapter 0 Know how oneway ANOVA and 2sample comparison of means techniques are related 0 Test the standard deviations to see if it is appropriate to pool the variances Understand why it is important to pool the variances in oneway ANOVA 0 Explain and check the assumptions for doing oneway ANOVA 0 Calculate R2 and the estimate for 039 I Write the correct hypotheses including the words population mean for one way ANOVA I Use the F test statistic and Pvalue from SPSS to perform the oneway ANOVA test 0 State the conclusion to a oneway ANOVA test in terms of the story I Know when to use a Bonferroni multiple comparisons test I Use SPSS to perform the Bonferroni multiple comparisons test and interpret the output both Pvalues and con dence intervals 0 State the conclusions to a Bonferroni multiple comparisons test in terms of the story I Interpret sidebyside boxplots and means plots in terms of the story Recognize the response variable factors number of levels for each factor and the total number of observations for a story 0 Identify from reading a story whether the scenario is oneway ANOVA Use OneWay ANOVA when you have one categorical and one quantitative variable and you want to compare the means If the categorical variable has 2 groups gender male or female for example use Ch 7 twosample comparison of means ttest If the categorical variable has more than 2 groups eye color blue brown black green hazel other then use Ch 12 oneway ANOVA ANOVA ANalysis Of Variance the method for comparing several means Oneway ANOVA F test for Ho y1p2 uI Q the population means are equal Ha not all the population means are equal at least one is different Is there at least one population mean that is statistically signi cantly different from the others When you rst approach a problem which involves comparing more than 2 groups here is what you should do 1 Find the size n sample mean and sample standard deviation of each group You can then plot the means on a graph Do histograms of each group to look for outliers and overall shape 2 Find the 5number summary Min Ql Median Q3 Max for each group and do sidebyside box plots to see how much overlap there is between the groups 3 Run ANOVA Standard Deviations 0 The standard deviation 039 is assumed to be the same for all of the groups even though the sample sizes ni may be different I If the largest s lt 2 smallest s we can use the methods based on the assumption of equal standard deviations above 0 If we assume all the standard deviations are equal each s is an estimate of 039 We combine these into a Pooled Estimator of 039 n n I n n 1 SP In the SPSS ANOVA output SP IMSE The ANOVA output see pg 764 for more information Source Sum of Degrees of Mean Square F Slg S uares Freedom Groups SSG MSG Between SSG DFG I l MSG 7 7 Pvalue Groups DFG I MSE Error 7 SSE 7 2 Within SSE DFE N 1 MSE 51 Groups SST Total SST DFTN1 MSTi W Note thatN the total number of observations the sum of the all the n R2 SSG SST coefficient of determination of variation in the data that is accounted for by the FIT part of the model T his tells you how good a job your model is doing of explaining the variation in your data The closer to 100 the better your model is Example Many studies have suggested that there is a link between exercise and healthy bones Exercise stresses the bones and this causes them to get stronger One study examined the effect of jumping on the bone density of growing ram There were 3 treatmenm a control with no jumping a lowjump condition the jump height was 30 cm and a highjump condition 60 cm After 8 weeks of 1 0 jumps per day 5 days per week the bone density of the ram expressed in mgcmj was measured Here are the data Group Bone density control 611 621 614 593 593 653 600 554 603 569 lowjump 635 605 638 594 599 632 631 588 607 596 high jump 650 622 626 626 631 622 643 674 643 650 112 gun Mew gala Dansmim Anawze Qvaphs Mums Addams vadaW new Enter the data into in 391 E law 531 M EIE EHEE W 1m mm 13 1 vertical columns with the mmquot 1 GM Gmwym 1 1 folloWing labels Group Density 1 511 mmquot 1 and Treatment All the densities 2 1321 cumm 1 3 W are listed in one long column cumm 1 4 593 mm 1 Group is where you list 2 1 control lowjump or 7 gm mm 1 highjump Treatment is a E 554 mm 1 numerical way of describing your a 513 mum 1 m 539 mm 1 group Make control be I 11 635 1mm 2 lowjump be 2 and higly39ump 2 EDS WW 2 be 3 For some reason ANOVA 13 E38 mum 2 M 5941mm 2 needs a numerical column for the 15 599 WW 2 factor box not stated in your 15 E32 1uw1ump 2 W Em WW 2 SPSS manual 18 588 mum 2 19 Em mum 2 2m 5913 mum 2 21 ESE mamump 3 22 E22 mamump 3 23 EZE mamump 3 24 EZE mamump 3 25 E31 mamump 3 213 E22 mamump 3 Tr Anaemia I a Identify the response variable I n and N for this study b Make a table giving the mean and standard deviation for each group of rats Make a graph of the means Is it reasonable to pool the variances Using SPSS Analyze Compare Means Means Move density into DependentList box Move group into Independent List box Click the Options box to get your summary statistics Click OK Report Bone Density Group Mean N Std Deviation Minimum Maximum Median Control 60110 10 27364 554 653 60150 Highjump 63870 10 16594 622 674 63700 Lowjump 61250 10 19329 588 638 60600 Total 61743 30 26270 554 674 621 50 son 7 am 7 am 7 Mean of Density sm7 sun 7 c Do sidebyside box plots for each group Using SPSS to get the boxplots Graphs Boxplot Define i 2 Treatm em Move density into the Variable box Move group into the CategoryAxis box Means plot created in ANOVA SPSS step below Control 1 Low Jump 2 High Jump 3 Click OK Ban 7 55a sou em 7 Bone Density san 56a d Run the analysis of vaIiance RepOIt the F statistic and Pvalue Write the hypotheses that go With this information What do you conclude Using SPSS Analyze 9CompareMeans 90neWayANOVA T J T a o i y 1 mm Highwmp anme Group Move density into the DependentList box Move treatment into the Factor box Ifyou want aplot ofthe means you can click options and Meansplot Click OK ANOVA Bone Density Sum of Squares df Mean Square F Siq Between Groups 7433867 2 3716933 7978 002 Within Groups 125 79500 27 465907 Total 20013367 29 e What is the pooled estimate for the standard deviation 0 What is R2 This means that of the variation in bone density is explained by membership in the groups of high jump low jump and control The other of the variation is due to rat to rat variation within each of these groups If H g is rejected in OneWay ANOVA that means that we have evidence that at least one of the means is different Which 0nes Multiple Comparisons Use multiple comparisons method ONLY AFTER you have rejected Hg with the F test To perform a multiple comparisons procedure compute t statistics for all pairs of means using the formula If Otherwise we conclude that the data do not distinguish between them t2 2 t we declare that the population means 1 and u are different The value for t depends on which multiple comparisons procedure we choose We will use Bonferroni s multiple comparisons SPSS will do this for you and give you the Pvalue Another multiple comparisons method is simultaneous con dence intervals for all the possible differences Find 2 2 twsp i i for all the different pairs 1 n1 n J If an interval contains 0 then that pair of means will not be declared significantly different Example Going back to the rat jumping data use the Bonferroni multiple comparisons procedure to determine which pairs of means differ significantly Summarize your results in a short report Using SPSS Analyze Compare Means One WayANOVA Move density into DependentList box Move treatment into Factor box Click PostHoc box Click Bonferroni Click Continue Click OK Remember Control1 Lowjump2 Highjump3 Multiple Comparisons Dependent Variable Bone Density Bonferroni Mean Difference 95 Confidence Interval ITreatment J Treatment IJ Std Error Siq Lower Bound Upper Bound 1 2 11400 9653 744 3604 1324 3 37 600 9 653 002 62 24 12 96 2 1 11400 9653 744 1324 3604 3 26200 9653 034 5084 156 3 1 37600quot 9653 002 1296 6224 2 26200quot 9653 034 156 5084 The mean difference is signi cant at the 05 level Example Recommendations regarding how long infants in developing countries should be breastfed are controversial Ifthe nutritional quality of the breast milk is inadequate because the mothers are malnourished then there is risk of inadequate nutrition for the infant On the other hand the introduction of other foods carries the risk of infection from contamination Further complicating the situation is the fact that companies that produce infant formulas and other foods benefit when these foods are consumed by large numbers of customers One question related to this controversy concerns the amount of energy intake for infants who have other foods introduced in to the diet at different ages Part of one study compared the energy intakes measured in kilocalories per day kcald for infants who were breastfed exclusively for 4 5 or 6 months Here are the data Breastfed for 4 months 5 months 6 months 499 490 585 620 395 647 469 402 477 485 177 445 660 475 485 588 617 703 675 616 528 517 587 465 Energy Intake 649 528 kcald 209 5 18 404 370 738 431 628 518 609 639 617 368 704 538 558 519 653 506 548 a Identify the response variable I 71 and N for this study b Make a table giving the sample size mean and standard deviation for each group of infants Is it reasonable to pool the variances Report Ene rqv Time Mean N Std Deviation Median Minimum Maximum BF4 57000 19 122 958 609 00 209 738 BF5 483 00 18 112 948 512 00 177 639 BF6 54188 8 93 963 506 50 445 703 Total 53020 45 118 906 528 00 177 738 c Show sidebyside boxplots for the 3 groups auu mu sun sun mu 7 sun Energy 2nu 22 mn an EFS an Time d Run the analysis of variance Report the F statistic and PValue Write the hypotheses for your test What do you conclude ANOVA Enerqv Sum of Squares df Mean Square F Siq Between Groups 71288325 2 35644163 2718 078 Within Grou ps 5508109 42 131 14 545 Total 622099 2 44 Should we now do Bonferroni multiple comparisons procedure for this example Why or why not

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