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week5

Course: STAT 651, Fall 2008
School: Texas A&M
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Week Stat651 5 Feb 1014, 2003 Copyright (c) Li Zhu 1 This Week's Reading Chapter 6 8.1, 8.2 Copyright (c) Li Zhu 2 The essential issue is that you have two populations to compare You compare them on the basis of a single variable You can compare their means, or their variability. Week 4 Review: Comparing Two Populations Copyright (c) Li Zhu 3 Week 4 Review: Paired Comparisons We...

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Week Stat651 5 Feb 1014, 2003 Copyright (c) Li Zhu 1 This Week's Reading Chapter 6 8.1, 8.2 Copyright (c) Li Zhu 2 The essential issue is that you have two populations to compare You compare them on the basis of a single variable You can compare their means, or their variability. Week 4 Review: Comparing Two Populations Copyright (c) Li Zhu 3 Week 4 Review: Paired Comparisons We have shown how to test : population mean difference = 0; : population mean difference 0; using tstatistics (confidence intervals and tests). Copyright (c) Li Zhu 4 Week 4 Review: Paired Comparisons There are two alternative methods that are not so affected by outliers These are the Wilcoxon signed rank test and the sign test Copyright (c) Li Zhu 5 There are two populations Week 4 Review: Comparing Two Populations Take a sample from each population The sample sizes need not be the same Population 1: Population 2: n1 n2 Copyright (c) Li Zhu 6 Week 4 Review: Comparing Two Populations How do we compare the population 1 means and ???? 2 The usual way is to take their difference: = - 1 2 If the population means are equal, what is their difference? Copyright (c) Li Zhu 7 The CI can of course be used to test hypotheses This is the same as Week 4 Review: Comparing Two Populations H 0 : 1 = 2 vs H a : 1 2 0 H 0 : 1 - 2 =0 vs H a : 1 - 2 So we just need to check whether 0 is in the interval, just as we have done Copyright (c) Li Zhu 8 1. Comparing Two Population Means The data: The populations: The aim: Copyright (c) Li Zhu 9 1. Comparing Two Populations The aim: CI for 1 - 2 The estimate of the difference in population means is the This is a random variable: it has sample to sample variability Copyright (c) Li Zhu 10 1. Comparing Two Populations Difference of sample means X1 - X 2 "Population" mean from repeated sampling is - 1 2 The s.d. for the population is, which is unknown + n1 n 2 2 1 Copyright (c) Li Zhu 2 2 11 1. Comparing Two Populations If you can reasonably believe that the population sd's are nearly equal, it is customary to pick the equal variance assumption and estimate the common standard deviation by (n1 - 1)s + (n 2 - 1)s sp = n1 + n 2 - 2 2 1 Copyright (c) Li Zhu 2 2 12 1. Comparing Two Populations The standard error of is the X1 - X 2 value sp 1 1 + n1 n 2 The number of degrees of freedom is n1 + n 2 - 2 Copyright (c) Li Zhu 13 1. Comparing Two Populations 1 - 2 A (1-) 100% CI for is t /2 (n1 +n 2 -2)s p 1 1 + n1 n 2 X1 - X 2 Note how the sample sizes determine the CI length Copyright (c) Li Zhu 14 1. Comparing Two Populations Generally, you should make your sample sizes nearly equal, or at least not wildly unequal. Consider a total sample size of 100 X1 - X 2 t /2 (n1 +n 2 -2)s p 1 1 + n1 n 2 1 1 = 1 if n1 = 1, n2 = 99 + n1 n2 = 0.20 if n1 = 50, n2 = 50 Thus, in the former case, your CI would be Copyright (c) Li Zhu 15 1. Comparing Two Populations The CI can of course be used to test hypotheses This is the same as H 0 : 1 = 2 vs H a : 1 2 So we just need to check whether 0 is in the interval, just as we have done Copyright (c) Li Zhu 16 1. Comparing Two Populations: The t test H 0 : 1 - 2 =0 vs H a : 1 - 2 0 There is something called a ttest, which gives you the information as to whether 0 is in the CI. It does not tell you where the means lie however, so it is of limited use. Pvalues tell you the same thing. Copyright (c) Li Zhu 17 1. Comparing Two Populations: The t test The tstatistic is defined by X1 - X 2 t= 1 1 sp + n1 n 2 Copyright (c) Li Zhu 18 1. Comparing Two Populations: The t test You reject equality of means if |t| > t /2 (n1 +n 2 -2) In this case, is p < or is p > ? Copyright (c) Li Zhu 19 The question is whether arsenic ingestion is related to squamous cell carcinoma 2. Arsenic and Squamous Cell Skin Cancer We used the transformation X = log(0.005 + toe arsenic) Copyright (c) Li Zhu 20 2. Arsenic and Squamous Cell Skin Cancer 0. 00 15% Healthy Percent 10% 5% 0% 1. 00 15% Percent 10% Cancer 5% 0% -5.00 -3.00 -1.00 1.00 log(0.005 + Toe Arsenic Level) Copyright (c) Li Zhu 21 2. Arsenic and Squamous Cell Skin Cancer 2 1 1 0 3 5 7 12 13 14 18 20 -1 2 4 6 9 8 10 11 15 16 17 19 21 log(0 .005 + Toe Arsenic) -2 -3 804 805 808 -4 -5 N= 806 807 284 524 Cancer Healthy Squamous Cell Carcinoma Status Copyright (c) Li Zhu 22 2. Arsenic and Squamous Cell Skin Cancer, Healthy Cases Normal Q-Q Plot of log(0.005 + Toe Arsenic) 0 -1 -2 Expected Normal Value -3 -4 -5 -5 -4 -3 -2 -1 0 Observed Value Copyright (c) Li Zhu 23 2. Arsenic and Squamous Cell Skin Cancer: Cancer Cases Normal Q-Q Plot of log(0.005 + Toe Arsenic) 0 -1 -2 Expected Normal Value -3 -4 -5 -5 -4 -3 -2 -1 0 1 Observed Value Copyright (c) Li Zhu 24 Healthy, s = 0.5838, IQR = 0.6913 2. Arsenic and Squamous Cell Skin Cancer Squamous, s = 0.6629, IQR = 0.7582 These statistics and box plots indicate that the two populations do not have vastly different variabilities. Copyright (c) Li Zhu 25 Healthy: mean = 2.3057, n = 524, se = 0.02551 Squamous: mean = 2.3365, n = 284, se = 0.03934 Mean difference = 0.03077, se = 0.04515 95% CI= [0.0579, 0.1194] p = 0.496: what does this mean? Copyright (c) Li Zhu 26 2. Arsenic and Squamous Cell Skin Cancer Because sample means and standard deviations are sensitive to outliers, so to are comparisons of populations based on them Rank tests form a robust alternative, that can be used to check the results of tstatistic inferences You are looking for major discrepancies, and then trying to explain them Copyright (c) Li Zhu 27 3.Robust Inference via Rank Tests 3.1 Motivation Rank tests are very easy to compute, and SPSS provide them. The algorithm is to assign ranks to each observation in the pooled data set Then apply a ttest to these ranks Robust because ranks can never get wild Copyright (c) Li Zhu 3. Robust Inference via Rank Tests 3.2 Method Typically called the Wilcoxon rank sum test 28 Here is how data are ranked 3. Robust Inference via Rank Tests 3.3 Numerical Example 22 50 45 55 81 56 Data #1 3 7 #2 28 44 Ranks #1 1 2 3 6 10 #2 4 5 7 8 9 Now run a ttest Copyright (c) Li Zhu 29 The rank tests give the same answer no matter whether you take the raw data, their logarithms or their square roots. 3. Robust Inference via Rank Tests 3.4 Properties If you have data (raw or transformed) that pass qq plots tests, then Wilcoxon and ttest should have much the same pvalues In this case, you can use the latter to get CI's Copyright (c) Li Zhu 30 In SPSS, you can get Wilcoxon rank sum tests as follows (SPSS calls them Mann Whitney U) "Analyze", "Nonparametric Tests", "2 independent samples" 3. Robust Inference via Rank Tests 3.5 SPSS Copyright (c) Li Zhu 31 3. Robust Inference via Rank Tests 3.6 Example Illustrate with NHANES to look at p-values for saturated fat and its logarithm Saturated Fat pvalues: ttest = 0.057 , rank test = 0.014 Log(Saturated Fat): ttest = 0.012, rank test = 0.014 Note how the transform, which is more bell shaped, agrees more closely with the rank test! Copyright (c) Li Zhu 32 We now turn to making inferences when there are 3 or more This populations is classically called Analysis of Variance, or ANOVA It is somewhat more formula dense than what we have been used to. Tests for normality are also somewhat more complex Copyright (c) Li Zhu 33 4 How About More Than Two Populations? 4.1 ANOVA Introduction Suppose we form three populations on the basis of body mass index (BMI): BMI < 22, 22 <= BMI < 28, BMI > 28 This forms 3 populations We want to know whether the three populations have the same mean caloric intake, or if their food composition differs. Copyright (c) Li Zhu 34 4.1 ANOVA Introduction There is considerable controversy as to how to compare 3 or more populations. One possibility is to compare them 2 at a time In the body mass example, there are 3 such comparisons. In general, if there are t populations, there are t(t1)/2 twosample comparisons. Copyright (c) Li Zhu 35 4.1 ANOVA Introduction The controversy revolves around the concept of Type I error Specifically, if we do six 95% confidence intervals, what is the probability that at least one of them does not include the true mean? It is higher than 5%! Copyright (c) Li Zhu 36 4.1 ANOVA Introduction One school of thought is to stick to a few major hypotheses (24 say), do 95% CI on them, and label anything else as exploratory, with possibly inflated Type I errors Another school thinks that you should control the chance of making any errors at all to be 5% I worry about Type II error (power). To be 95% confident that every single one of my conclusions is correct, I will not have much power to detect meaningful changes or differences. Copyright (c) Li Zhu 37 4.1 ANOVA Introduction One procedure that is often followed is to do a preliminary test to see whether there are any differences among the populations Then, once you conclude that some differences exist, you allow somewhat more informality in deciding where those differences manifest themselves The first step is the ANOVA Ftest Copyright (c) Li Zhu 38 4.2 ANOVA Notations The Ftest is easy to compute, and provided in all statistical packages The populations are 1, 2, ..., t n1 ,L, n t The population means are 1 ,L , t The sample sizes are The hypothesis to test is H 0 : 1 = 2 L = t Copyright (c) Li Zhu 39 4.2 ANOVA Notations The data from population i are The sample mean from population i is The sample mean of all the data is Yi1 , Yi2 ,L , Yin i Y iC YC The total sample size is the total number of observations, called n T Copyright (c) Li Zhu 40 4.3 ANOVA Table The ANOVA Table (demo in SPSS in class of ACS data) "Analyze" "Compare Means" "1way ANOVA": I'll now show you what each item is ANOVA Baseline FFQ Sum of Squares 960.287 15275.639 16235.925 df 2 181 183 Mean Square 480.143 84.396 F 5.689 Sig. .004 Between Groups Within Groups Total Copyright (c) Li Zhu 41 4.3 ANOVA Table The sample mean from population i is The sample mean of all the data is Y iC The idea of the Ftest is based on distances The distance of the data to the overall mean is 2 YC TSS = (Yij - Y C ) ij TSS = Total Sum of Squares Copyright (c) Li Zhu 42 4.3 ANOVA Table nT -1 This has degrees of freedom So, the total sample size is the degrees of freedom for TSS + 1 Copyright (c) Li Zhu 43 4.3 ANOVA Table Next comes the "Between Groups" row ANOVA Baseline FFQ Sum of Squares 960.287 15275.639 16235.925 df 2 181 183 Mean Square 480.143 84.396 F 5.689 Sig. .004 Between Groups Within Groups Total Copyright (c) Li Zhu 44 4.3 ANOVA Table The sum of squares between groups is n i (Y iC - Y ) i 2 It has t1 degrees of freedom, so the number of populations is the degrees of freedom between groups + 1. Copyright (c) Li Zhu 45 4.3 ANOVA Table Next comes the "Within Groups" row ANOVA Baseline FFQ Sum of Squares 960.287 15275.639 16235.925 df 2 181 183 Mean Square 480.143 84.396 F 5.689 Sig. .004 Between Groups Within Groups Total Copyright (c) Li Zhu 46 4.3 ANOVA Table The distance of the observations to their sample means is SSW = ij (Yij - Y iC) 2 This is the Sum of Squares Within nT - t It has degrees of freedom Copyright (c) Li Zhu 47 4.3 ANOVA Table Next comes the "Mean Squares" These are the different sums of squares divided by their degrees of freedom ANOVA Baseline FFQ Sum of Squares 960.287 15275.639 16235.925 df 2 181 183 Mean Square 480.143 84.396 F 5.689 Sig. .004 Between Groups Within Groups Total Copyright (c) Li Zhu 48 4.3 ANOVA Table Next comes the Fstatistic It is the ratio of th...

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# BLASTP 2.2.17 [Aug-26-2007]# Query: T0445 # Database: /projects/compbio/data/pdb/dunbrack-pdbaa# Fields: Query id, Subject id, % identity, alignment length, mismatches, gap openings, q. start, q. end, s. start, s. end, e-value, bit scoreT04451
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T0445.t06 near-backbone-113 2 1M IK L I A T D I D GT L V KD G S LL I D PE Y MS V I D R L ID K G II F V VC S G RQ F S SE F K L F AP I K H K L L Y I T D GG T V V R T P KE I L K T Y P MD E D I W K G MC R M V R D E LP 1 10 20 30 40 50 60 70 80 90AG J
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Berkeley - CE - 130
Michigan State University - CEM - 332
Chapter 23 PotentiometryRead: pp. 659 692 Problems: 23-1,4,9,13,16 Potentiometric methods are based upon measurement of the potential of electrochemical cells in the absence of appreciable current. Equilibrium measurement, so Nernst equation is
Cal Poly - CPE - 269
Digilent, Inc. 125 SE High Street Pullman, WA 99163 (509) 334 6306 (Voice and Fax) www.digilentinc.comPRELIMINARYDigilab XCR Reference ManualRevision: April 1, 2002 Overview The Digilab XCR (DXCR) circuit board featuring the Xilinx CoolRunner XC
Michigan State University - LBS - 118
Quiz 9: LBS118 (001-002) April. 4, 2006 Problem 1 (3 pts): Find the following limit: ln(x2 ) x1 x - 1 limName: Section:Answer: Initially this is indeterminate of the form 0 . Using L'Hospital's Rule, we have: 0 ln(x2 ) x1 x - 1 lim = = Problem 2
Michigan State University - WILL - 1272
Time Friday, November 3, 2006 6:00 - 9:00 8:00 - 10:00 10:00 - Until Saturday, November 4, 2006 7:30 - 10:00 7:30 - 8:20 8:30 - 9:20 9:30 - 10:20 10:30 - 11:20 12:00 - 2:00 2:10 - 3:00 3:00 - 5:00 5:00 - 7:00 7:30 - 10:00 10:30 - 1:30 Sunday, Novembe
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Rebecca Ann Parkerparke110@msu.edu Education: Michigan State University, East Lansing, MI Bachelor Arts in Integrated Elementary Education and Language Arts, ZA Endorsement in Early Childhood Development Expected Graduation: May 2006 Catholic Centra