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# Selected Topics in Archaeology ANTH 5589

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This 19 page Class Notes was uploaded by Laurie Lind on Monday September 21, 2015. The Class Notes belongs to ANTH 5589 at University of Virginia taught by Fraser Neiman in Fall. Since its upload, it has received 46 views. For similar materials see /class/209707/anth-5589-university-of-virginia in anthropology, evolution, sphr at University of Virginia.

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Date Created: 09/21/15

R vary The inquot Placequot H 6B Nanhm The Munnayernse E Pronequot 3 Quadatdametel10 n In on an an 1m Nnnhmg 39 1n Samvletlle m s u Prunequot s 5 n n 393 393 u u n quadaks scmg 10 n Zn on an an 1m Nam Analysis E e 3 D u u m Placequot n 2 u u D a ram the quotWe u DAD 5 Interpolat on D U m SD M an Interpolation IDW a gt Many methods 1 3 n 1n 3 2 E 39 7 2 Z w z 395 1 lnverse d1539tance We1ght1ng 1va g 5 7 394 J 1 I I i 5 I K 4 s o z 2 Krigmg 1 5 1U 15 3 Others Ensquotum d n 5 1D 15 r o1ynom1a1 regression local global 391 E f 7W5 tnangmated 1rregularnetwork5 5 39quot9 lxl spines radial bests tuncuons Pom l I 1 mm M W I 1 and 2 both make esttmates of value ofme 2 variable at an unsampled 1 ll 5 4 4 47 EB N 2 6 lEl 3 3 16 El 32 El 95 Where Z values are RHOWH 3 E lEl 2 316 El 32 El 63 4 4 4 7 4 24 El 24 l 65 50 Sun 119 413 37a 5 7 7 Pesky Questions about IDW what value for p maximum value for d beyond which d0 p2 magtltd6 p2 maxd2 3 p4 maxmm More Pesky Question about IDW Why base weights on Inverse Distance at am rWhy not make try to make the weights Optimal m a stattsucal sense theyminimize sum afsquaredpmdlclwn struts mey are unbiased Kriging rAn nterpolaton method n which the weights depend on the spatial autocorrelzmon structure of the data as re ected in the vartogram AND that produces estimates of Z that are designed to mnmze the sum ofsquared predcton errors ue Kr ing Approach Predlct 1 at polnt 5 as awelghted average A o V known values at polnts s Zs 2ampZs lzl A and welghis 0 such that asquot Zia el 1o do this we need a model Regiona ed Variable iean N ms e s 6 25 the value ofthe RV at place s ms the detexmlnlshc component or trend l e the mean or expected val 111 E a E e Fquot e s e ependent component a spatlally mtg mtem noise Ifl and 2 are true then the Semlval lal lce can be eStlmated rrorn the sample data 1 a h Zzltsgte zlts 11 e m 3 2 The panda distance and unleash P a my amihe xkh pamtlnspnce n n 2 Z 2er z i z 1 Wehnveseenihls velslan anhe h w mommacw c 222 W 11 11 2 compute an dmanoe and lquare mnerence pom z Cumhme aver clinicquot laser 4 ma madd varlOgram 139 I3939 ll L123 325 7 If M run ll ordinary Kr g g Assumptions 1 ms ls constant ncrossthe E 25 7 Zs h 2 vanance otdltreences at dltreent p md dlrechon only not absolute locat E 25 study nenrno mend or dun o laces ls a tundlon of dlstance lon ZG h E 539 ETS 1W1 zv h semlvanance Question How limiting ls the nortxend assumpaonr l l l n 2 a s x The Problem Predlct tlne value at 14 The Krlglng solution 1 Model spaoal dependence uslng tlne sennnvanognam 2 Use 1mm Pmdicl ian variogram Lingo snl rorlargervaluesorhthe varl glam levels out lnulca lngthat eq al lo the Variance 51 arms 2 values Range ls the value or hwhere the slll occurs or 95 or the value orthe slll Thls ls the ulstance hevonu whlch palrs orvalues are no longer autocorrelateu Nugget variance a nunezeru value for vwhen h u Produced hv varlous sources or unexplalneu error e 9 measurement error algebra to gmemte an 3951 Linear Unbiased ELIP slll c 5 an anger r l c r model 1 Common Model Variograms 5 39 39 39 39 based on their Euclidean distances using lab 0 15 51 5mm A Linear a if th H i mi oi i Sill mp 21 U in i r i 39 i mm mm am 1439 39 39 i ii in uni i mm M ii am i 7 I Mi it in iii 1 mi ii i g g ysm s39 4L C M 93 39 1 am i W a 5 39 92 a mum in 39 39 the knowns based on their Euclidean distances usin V e ned by Ii 1 9 J g 01551i different equations Exixmsimai c Gaussian D new qVOGiw MI ISAg e a a 34 2 270 7 um 35 gt g 45 EIEZ 1269 7 I It 5 675 I We i 1 5 39 v 39 v 39 I 7 Use the wei its lambda no make the prediction based on their Euclidean distances in in im w m 2 3 23 M i Liiimis i emu ms 103st 45 Wquot 4933 3 i 1 F m m quot N 39 HUIquot H5 4679 i e in iiquot m i ii i SEQ m an Kyiglng mm lt ii 7h in u m i i i i I t i 5 39 s 92 39 39 the knowns based on theirEuclidean distances lt vain 757 002ll3 m Poi Dianne VDGIM39 1 l J 239 10250 009834 dle i quot2 i is g 1 A It use 2 2 046982 77 quot 3 5 E75 2 ijltm l I l he I H 7 Solve the kriging equations forlambda In matrix algebra form 0Ol46 39 39 I I I i I i i i i i 1 g 7Fquotg I 2 i 4 lt 7 Use the weightsianibda to estimate the prediction error Kr39glng VS IDW G Vector Weights A g Veclor Times Weights 135 045757 6312195 c i i i u 270 0 09534 255515 model that tailors weights to the for p and maxd 1 5 0 45952 6 34257 the data 42 59 0 02113 0 90204 67 5 410146 390 9855 I 018231 0 i923 Kriging Variance 13 2396 Kriging Sid Error 3 me Correspondence Analysis Kinds of data counm or types or taxa in assemblages the sort ofcounm that it makes sense to compute percentages assemblages from strati ed deposits spatially scaner quadrats both presenc absence or occurrence data Typical applications frequency seriation examples to follow occurrence seriation e g medieval deeds ordination seeing similarity among observations in a low dimensional space seeing similarity among variables in lowdimensional space seriation plus time is not the only dimensio CA as a form of PCA 1 Transform the counts X 5 10 20 10 40 5 Some data 3 types and 4 assemblages 40 5 0 50 1 0 x 186 Sumothecounts P X quot Counts sum of the counts 0 0268817 0 0537634 0 1075269 0 0537634 0 2150538 0 0268817 0 2150538 0 0268817 0 2688172 0 0053763 0 1 Correspondence Anal 1 Eigenanalysis of a funky covariance matrix a form of PCA Inertia varianc vs e Chisquared distances vs Euclidean distances 2 Reciprocal averaging or indirect gradient analysis A connection to M D s and seria ion Approximation to ML estimation under the Gaussian response mo e The arch effect vs PCA horseshoe 3 Dretrended CA a problem not a solution 4 Examples Woodland assemblages from Georgia 0 San Marcos Pue tl uznkr to Karen Smith 1 Transform the counts continued p J 0 5645161 0 3010753 0 1344086 The column masses marginals 0 188172 F l 0 2956989 The row masses 0 2419355 0 2741935 0 0268817 0 0537634 0 1075269 0 0537634 0 2150538 0 0268817 0 2150538 0 0268817 0 2688172 0 0053763 0 p 1 Look lamilia Qeq p y 1 17 1 Transform the counts continued p 7pm p F J 0 5645161 0 3010753 0 1344086 0 0268817 0 0537634 0 1075269 P p y 0 0537634 0 2150538 0 0268817 R 0 2956989 0 2150538 0 0268817 19355 0 2688172 0 0053763 0 0 2741935 Q 0 243445 0 012144 0 517089 0 276976 0 422375 0 06452 0 2123518 0 170288 0 180328 0 2898373 0 268608 0 191974 1 Transform the counts continued CA transformation 7 PHIL q 417 17 Chisquared table entries 0y 7 E y Z 1 Residuals from expected E y und r hy s independence chi squared residuals Still looks familiar 11ransrorrn the counts continued 4 transformation p 7pm qy lam ChrrsquaredtaDe entries Residuals from expected unde null Wpomesrs of morassocrarron IV 7 0o 7 EV pv 7 pep XH PPz zscores 7 m FHFH Q 9V 7 How much vana omsmelemme FHA transformed data7 Y 5 2 ToriVana onw quotInertiaquot I e Z qu e 0 8856929 How much variation Is there IN the cmaquare residuals Tota Variation W quotTotal CmSgualequot Inertia 2 Measuring the amount of variation continued F FF meme of the 7 m rm 3 Get the sum of Squares and cross Products Mat o a m A W I 2123518 r17288 r18328 transpose I 2898373 9 26868 79191974 Q 124344 9 276976 I 2123518 I 2898373 M22375 r17288 VD 26868 9 96452 r18328 79191974 C Q39Q n 255nm 1228M in 2mm VA SSCP matrix in w n 2795954 I M7422 Ivariancesv on the magena 7n 2mm n M7422 n 34mm 15096 77 wcovanancesv esewhere 3 Get the SSCP Matrix continued c on UZ SUBUAQOZBUA I in in 228u4 M74 sn2 in 2mm n M7422 n 34mm O 3 vquot 4 mu 1 a t one no z compute the eigenvalues and eigenvectors of Q rFmd the line muugh the pmnls that maximizes the mantra Dr maua along the line arms perpendicu ar pmjecnuns 39um ach mW pmnl tn the line a k a Dimmsm 1 Axis n rFmda second line unhugunal in he rst that maximizes mane Dimension 2 The eigenvalues one for each axis 1 n 52 755 1 rWhy only 2 that are hohazemv rWhat Is ther sum equal 07 The eigenvecm he vector for each axis u us t1 n 6492391182U83U7513429 coemcehts de hhg the rst Ihe Dmehsoh 1 on the hass of the mw pohts the mws of the q matrx 4 Use the eigenvectors of C to get the scores of the row points on each dimension QUs t1 t2 13 Hz u3 junk A a 43445 0 012144 0 5170 0 1182083 0 7513429 B 0 276976 06452 0 658476 0 5487033 C 0 2123518 0 170288 0 180328 0 7432609 0 3666178 D 0 2898373 0 268608 0 191974 Dim 2 DB 13 A n o 0 3635513 5 836E17 g 04 3 B 86 3 823216E1 E M II C 0 326495 0 0032014 4 7E g k D 0 433964 0 0684464 2 126E17 9 5 39 2 5 5 an 4 I 5 I ll Dlrrensnn1 Check the sum ofsquares 5 Rescale the eigenvectors so they AND the scores are cooler p I 0 5645161 0 3010753 0 1344086 Start with the column masses J 1 3309503 1 8224787 2 7276363 Get the reciprocals 0 the square roots 1 Put them in a diagonal matrix and D7 U V rescale the eige vectors 39 u 1 u3 junk 1 3309503 0 t1 0 649239 0 1182083 0 7513429 0 1 8224787 2 0 5151065 0 658476 0 5487033 0 3 0 0 2 7276363 5596022 0 7432609 0 3666178 v1 v2 v3 junk 1 1 0 864104 0 1573294 1 2 0 9387707 1 200058 1 3 1 5263913 2 0273455 1 Check the results What has happened 11 1 5 Rescale the eigenvectors so they AND the scores are cooler continued D137 P V F Multiply the row pro les relative frequencies by the rescaled coefficients to get the scores 1 1 f2 1 3 v1 v2 v3 junk 0 1428571 0 2857143 0 5714286 U 0 864104 0 1573294 1 0 1818182 0 7272727 0 0909091 x 2 0 9387707 1 200058 1 0 t3 1 5263913 2 0273455 1 Uhmb o m m m m m m o o 0 9803922 0 0196078 0 f1 f2 f3 1 0170003 0 8380852 chums a a on a m o o o o a o o m a 0 828754 0 130714 6 Why is rescaling cool 39 rwnmmw their direction 39 re quot 39areraeof r quot fore 1 in I I r i 1 i i The eponymous correspondence us 3 2 13 up 15 E N a 392 a 1 n2 a 1 a E 39 0A E 1 5 V to a M m M a n5 n5 1 15 n on US b 5 12 391 12 Dlrrenslnn 1 Dlrrenslnn 1 Before rescaling After rescaling 6 Why is rescaling cool continued 3 Euclidean distances on the CA axes among observations equal chisquare distances among the observations Squared differences among proportions 0 each type at the two sites Column marginals weighted mean proportion of each type 7 CA diagnostics 2 IIS 1 N 1 5 9a quot 1 5 1 d i g 0 s a 5 1 1e 0 n on 1 In Dimension1 a partial contribution to inertia of dimension 1 from row a 7 large Values mean the row is important in determining the orientation of the dimension C0SH2 correlation between row b and dimension 1 show well the dimension 1 score capture row b s true position 7 GA Biplot interpretation 2 ID is u w 5 1 as E A 5 v l e en M as l is O 391 utz DImensIon1 How should we interpret me positions ome points in Lhejomt spaee distances amen rows7 e elaaonslup between rows and eolsv distances among c0157 5 me forms of scaling 2 Switchrows and columns in me analysis as above 7 e 1 mn scores now become weighted averages oftherow scores 7 s anees among columns are Ghlrsquaredddstances arow poaaons are 3 Combination ofl and 2 keep me row scores from 1 keep me columns scores from 2 distances betweenrowsAND columns are elusquared stances relationships among them are onlymterpretedmterms of recnons Defaullfoemosl so ware including SAS But 1 usually makes more sensem archaeology andecology ewhyl CA as Reclpmcal Avetayinu x1 The Gaussian response m ode I An enwanmemal gament a g elmhun mean rum Drpmba liw all men laxan dela39mmed by paauan un me gament rewunse la the gadmm xv exp h1aj75 t2 J me eaunlfarmeilhlaxanalmel msamplelmaaneamom mammum abundance DtheJ Lh laxan mlaxanemelmuanamsmaae x 1 me apumum gamentwlue furthej if melalaanee manna Dfihe j ih laxan g thalaman Dthe ample alangme mmml 1 Direct Gradient Analysis continued Look famlllat 1 Direct Gradient Analysis continued Ecological gradient 1 Direct Gradient Analysis continued Eng Nalve esllmae of location on me gradlem Xl vau Naive estimate of 1 me sample 2n locallononme m m a w n 6n 7n Sn a in amen g Naive estimate of e samp e location on the gradlem Look familiar 2 Reciprocal Averaging after Mo Hill 1973 1 Start with random site scores 0 2 Compute new taxon scores 1 as weighted averages of site scores 3 Compute new taxon scores as weighted averages of site scores x 4 For the rst axis go to Step 5 For second and higher axes make the site scores x uncorrelated with the orthogonalization w 5 Keep going until convergence 6 Standardize the site scores x See below for the standardization procedure 4 1 Orthogonalization regress site scores for the i th axis 39 for the previous axis Compute residuals Use these above 5 1 Standardization Compute mean and standard deviation the the site scores Note that the standard deviation equals the eigenvalue Convert the site scores to zscores Tax39q scores g2 11 Site scores 3 The Arch x1 l c ounts relative frequency The Arch continued Taxon space coun ts T xon space relative frequencies amp I o 1 cubic l o l 2 2 o l 2 9 9 9 etc etc etc 5 occurs with abundance and occurrence data with real data random variation swamp higherorder polynomial relationships gt2 rendering them invisible amount of arch is other things being equal a function or the amount oftaxonomic turnover along the gradient A Dimension 1 recovers the correct gradient order The Arch continued PCA makes it a horseshoe ma etc etc etc hlg function or the amoun along the gradien coccurs w tn abundance and occurrence data with real data random variation swamps 39 herorder polynomial relationships gt2 rendering them invisible amount of arch is oth er things being equal a t of taxonomic turnover t PCA Dimension 1 does NOT recovers the Pm correct gradient orderll Why is CA bettet than PCA with Gaussian Iesponses Atch vs Hotseshoe 6 Why is tescaling cool r widiiiiuic 39 quot their direction Under Ga ends of gradients ussian response rare types occur at the re quot aiemeofL r quot for inthem 39 1 39 us 3 2 3 n6 5 392 Equot oa 392 g H 2 g oa E 39 0A E t 5 V A 5 7 as u 5 5 v in b rue n5b1 ta 39 5 at 39 tz mnensum 1 mnensum 1 Belote tescaling Altet tescaling 6 Why is tescaling cool continued 3 Euclidean distances on the CA axes among observations equal chisquare distances among the observations proportions 0 each type at the two sites Column marginals weighted mean proportion of each type Rare types get greater weight Squared differences among Multiple Gtadienls So far a single gradient with Gaussian responses What happens if there are 2 or more underlying gradients that detenninet axonomic responses simultaneousl 9 AND the two gradients are to some extent independent Archaeological examples tr39me and social status CST tr39me and activity variation CA shouldrecover BOTH gradients if the rst gradient is short CA dimension 2 will a the second vari an if there is an arch the second gradient will o en em e 39 ension 3 although it may be partially visible on dimension 1 as well CA the Big Question tes ltom CA appto imate what we ML estimation of site locations along one ot mote gtadients based o known taxon patametets ltom the Gaussian Iesponse model When will estima would get ltom Four conditionsquot equal or unifome distributedtolerances along each gradient equal or uniformly distributed maxima along each gradient qually spaced or uniformly distributed optima along each gradient equally spaced or uniformly distributed sample points along each gradi ent L we 394 mi me Caussianmodels n u r w inn mmnaaalienmenmael momma Muslim 4 Dretrended CA solution or new problem CA ptoblems earc compression ofscores along axis ends DCA solution Chop 2nd axis up into N subsegments default 25 Then rescale points in each segment to have common mean Rescale 15 axis scores so withinsample varianceis constant across axis 1 Assumes species pacldng model ts the data a aquot usx Spatial Data Analysis in Archaeology Anthropology 589b Fraser D Nei man University of Virginia 21807 Spring 2007 Evaluation Goodness of Fit in TrendSurface Analysis rsquare sums of squares mean squares and F statistics As we have seen it can be helpful to think about spatial variation as being comprised of three additive components a global trend spatially dependent local variation and independently and identically distributed noise lfwe think this model might be useful in a particular case how can we go about identifying the trend There are many ways of doing this The oldest and most widespread is trendisurface analysis Trendisurface analysis attemps to extract the global trend underlying a spatial distributed variable by predicting that variable39s values as a function ofits Cartesian coordinates in space The function in question is a polynomial equation of some degree The first degree is the simplest case Here the predictions are based the X and Y coordinates of the location in question 63 the northings and eastings In the secondidegree case the predictions are a function of the X and Y coordinate and their squares and cross products Third forth fifth and higheridegree equations are possible as well In deciding what order of trend surface is the quotrightquot one it would be helpful to know how well the predictions account to the actual values ofz These notes describe how this is done Trend surface analysis is one avor what39s known in statistics as quotthe general linear modelquot This phrase denotes a single framework that can accommodate a variety of statistical methods including titests regression analysis ofvariance analysis of covariance What unites these different methods under the GLM banner is that can all be seen as attempts to predict the values ofa dependent variable as a function of one or more independent variables They all assume that errors in the model predictions of the value of the dependent variable follow a Gaussian quotnormalquot distribution They also assume that predictions are linear functions of the independent variables So GLMs all look like this 213 le1 bZX2 b3X3bPXP Here the Fs are the coefficients estimated as part of the model Z is the predicted value of the dependent variable The X39s are thep independent variables Geek note ifwe relax the Gaussianidistribution assumption we end up with generalized linear models GLIM39s The math behind GLMs is designed so that the estimates of the values of the F that result produce predictions of Z that minimize the sum of squared deviations from the predicted values to the actual ones I Two issues are critical to understanding GLMs The first is how the goodness of fit of the model to the data is judged just how good are the predictions This is usually measured using risquare the proportion of the variation in the dependent variable accounted for by the GLM The second issue is the statistical significance of the fitted model does the model account for more variation than we could expect as a result chance alone An odd question It simply means this Imagine that we had a very very large population of observations on the variables in our model Imagine too that in the population as a whole the independent variables included in our model had no predictive relevance to the dependent variable at all Suppose we drew multiple independent samples from the population of the size we actually have Suppose that we then estimated the parameters of our model based on each sample The question of statistical signi cance of the model boils down to an estimate of the proportion of times under repeated random sampling from such a population we would get a model that t as well as ours actually does has as high an r7 square simply as a result of saInpling error R sq ua re How can we measure variation A simple way is the sum of squared deviations SS of the value where is the dependent variable But deviations have to the deviations FROM some quantity What the deviations are from depend on what kind ofvariation we want to measure When we are trying to build a GLM to predict the value of some variable based on the value or one or more independent variables there are three kinds ofvariation that are of interest 1 Total SS the sum of squared deviations of the values from their mean This is the total variation in the variable 2 Model SS the sum of squared deviations of the values of predicted by the model from the mean of This is the amount ofvariation accounted for by the model 3 Residual or Error SS the sum of squared deviations of the values from the values for predicted by the model This is the quantity that is minimized in coming up with the estimates of the k s for the model In GLMs it turns out that Total SS Model SS Error SS Cool huh So it makes sense to summarize how well the values predicted by the model match the actual values by computing SS Model SS Total This is risquared W the proportion of the total variation accounted for by the model You can also compute risquare like this 17Error SS Total SS Can you see why Statistical Significance Now that we know how to measure variation in general and the proportion ofvariation accounted for by a GLM in particular the neXt question is is the amount ofvariation accounted for by the model greater than what I could expect to result from saInpling error 7 or from chance alone In other words is the model doing any real work or just coasting on chance effects The answer to that question depends on four quantities 1 How much variation does the model account for The more variation accounted for the less likely chance alone is responsible for the t of the model summarized by risquared 2 How much variation was not accounted for the more variation NOT accounted for the more likely chance alone is responsible for the t of the model 3 The sample size number of observations on The greater the sample size the less likely chance effects account for the t of the model Bigger samples mean less sampling error right 4 The number ofparaIneters contained in the model the number ofindependent variables for which the model estimates k coef cients The greater the number of paraIneters we estimate to get the t the more likely those paraIneters are capitalizing of chance effects To think about this realize you can always get a perfect t r square 1 by including as many paraIneters in the model as observations 1 and 2 are measured using SS as described above 3 and 4 are measured using the concept of degrees of freedom Just as there is a SS associated with the total variation in 239 the model and the residuals There are degrees of freedom associated with each 1 Total DF The number of observations 71 We loose one DF because in order to estimate the Total SS we had to estimate a paraIneter the mean of z Say there are 41 observations Ifwe have our estimate of the mean we can perfectly predict the value of the 4139st observation given knowledge of only 40 of them So there are 40 degrees of freedom associated with the deviations from the mean 2 Model DF The number ofindependent variables in the model for which you have k coef cient estimates For a linear trend surface this is 2 For a quadratic it39s 5 for a cubic it39s 9 3 Error DF or Residual DE This is the number of observations minus the number ofparaIneters you had to estimate to compute the residuals To get the residuals we had to estimate the k coefficients and the mean of Q So Error DFNumber of observations 7 Model DF 7 1 The error DF is the number of observations that are free to vary once we have made the predictions Say there are 41 observations and 2 paraIneters estimated by out model The error DF is 39 417271 This means that ifwe know the mean and the 2 paraIneters we can perfectly predict the value of Z for each of the 3 remaining observations once we know the value the other 39 of them Now it turns out that Total DF Model DF Error DF Now we have two sets of quantities the three sums of squares SS and their corresponding degrees of freedom Here is the final step So far we have just measured variation in terms of sums of squared deviations This is helpful But it does not take into account the number of degrees of freedom we had to burn to account for that variation It39s kind oflike saying you bought 4 bananas 7 good to know But it is also good to know how much each banana cost In the banana case you divide the total number ofbananas by the cost In the sum of squares case we divide the amount ofvariation accounted for by the model by the number of degrees of freedom we had to spend to build the model the Model DF In doing this division we have converted the SS into a variance recall that a variance is just a MEAN of squared deviations This variance is special so it has a special name quotthe mean squarequot 7 or the mean of the sum of squares accounted for by the model Actually it39s the mean amount ofvariation accounted for by each parameter estimated in the model Now we are almost there The question is Is the model mean square variance bigger than what we would expect as a result of sampling error Answering this question requires estimating the size of the variance that we could expect to account for as a result of sampling error It turns out that we already have the ingredients to build an estimate of the variance that is likely to arise as a result of saInpling error in a sample like the one we have given the model we are trying to evaluate One ingredient of the residual sum of squares 7 the variation NOT accounted for by the model The other ingredient is the residual degrees of freedom By dividing the Error SS by the Error DE we get an error variance 7 the mean amount of variation from the predictions of the model associated with those observations of Z whose values are free to vary once we have made the predictions So we have two variances the Model variance and the Error variance and the question is is the Model variance significantly greater that the Error variance To answer this we take the ratio of the two F Model variance Error variance or F Model mean square Error mean square This new quantity is a ratio ofvariances usually abbreviated with the letter F in honor of RA Fisher the guy who figured all this out in the 192039s Our question now becomes is the observed value of F greater than what we should expect to arise as a result of sampling error To answer this question we need to check out the sampling distribution of the F statistic This distribution gives the probability ofgetting a variance ratio as big or bigger than we one we actually have as a result of chance The F distribution just like any other distribution with which you may be more familiar For eXaInple the binomial distribution gives the probability ofgetting K success in N trails given an overall probability p ofa success Now the shape of the binomial distribution is governed by two parameters N and p The shape of the F distribution is also governed by two parameters the DF association with the numerator variance and the DF associated with the numerator variance Given an estimate ofF the ratio of the model to error mean squares and the degrees of freedom associated with its numerator and denominator the F distribution tells us the probability ofgetting an F value that big or bigger by chance That quantity is the thing we have been looking for the probability that the fit achieved by the model summarized by r7 square could have arisen by chance Here is how SAS summarizes all this information in the results from Proc GLM General Linear Model The model is a lidegree trend surface for the house diameter data from site AR 5 So the model looks like this Diameter neeMing existing Tm mMPmowum Dependent Variable diam Sum of Source DF Squares Mean Square F Value Pr gt F Model 2 331132446 165566223 285 00611 Error 140 81 29448253 058067488 Corrected Total 142 8460580699 RSquare Coeff Var Root MSE diam Mean 0039138 2479722 0762020 3073007 The F value is the ratio of the two mean squares or variances 16654 285 Given 2 and 140 degrees of freedom the probability ofgetting an F value this big or bigger is 06 So there39s only about a l in 20 chance that goodness of fit of the model to the data as summarized in the risquare value of 039 could have arisen by chance alone So we have a marginally significant result 7 in the statistical sense But very little predictive power The GLM framework also allows us to partition the variation that is accounted for by the model into components one associated with each of the independent variables Here again the variation is measured in SS This makes is possible to estimate how much of the variation in is accounted for by each of the independent variables And it is possible to test the hypothesis that we could have gotten a SS this big by chance alone We proceed exactly as we did before we compute a corresponding mean square or variance by dividing the SS by the appropriate DF We then form an Feratio of the independentevariable mean square with the Error mean square 570 33158 And we compute the corresponding probability from the F distribution with in this case 1 and 140 DF Source DF Type I SS Mean Square F Value Pr gt F northing 1 331090958 331090958 570 00183 Easting 1 000041488 000041488 000 09787 Here we encounter an ambiguity of sorts The amount a variation SS that each independent variable accounts for dependx on the order in whiih it oppeorx in the Inodel equation This makes intuitive sense ifyou think about each successive independent variable in the equation only being able to account for that variation which the previously entered variables have not accounted for SAS calls the SS that are computed for each independent variable in the order pen ed in the GLM equation speci ed in the Proc GLM Model statement the Typeel SS These quantities are shown above But it is also almost always helpful to know how much variation each variable would account for ifit were entered LAST after all the other variables have had a shot SAS calls this the Typeelll SS Again the corresponding mean squares DF Feratios and pevalues are given as well Source DF Type III SS Mean Square F Value Pr gt F northing 1 195012383 195012383 336 00690 Easting 1 000041488 000041488 000 09787 So far we have talked glibly about variation accounted for 7 and not accounted for7 without actually looking under the hood to check out the paraIneter estimates that are doing the predicting and accounting for Here they are Standard Parameter Estimate Error t Value Pr gt t Intercept 02252949403 176269471 013 08985 northing 00035544790 000193960 183 00690 Easting 0000246172 000092096 003 09787 The tstatistics and correspondingp values are tests of the hypothesis that the actual values of the h coefficients in the underlying population are 0 In other words that the independent variable has no effect onie the gym ofoll the other independent mriohle hove heen toeen into orionnt Note these probability values are identical to those associated with the Typeelll SS Now the problem with the significance tests in the traditional GLM framework is that they all depend on the assumption that the errors are identically and indqjendenth distributed This assumption is built into the use of error mean square in the denominator of the Feratio Spatial autocorrelation among the residuals from the model39s predictions is a violation of the independence assumption It means that the Error DF overstates the actual number of degrees of freedom in the data 7 if the residuals are correlated with one another the number ofvalues that are free to vary is smaller than it would be if the residuals were independent Table l V quotquotquotquot A33 Dates nflszevsnn 5 mum Mmmtelln 1 WIN m mum u 7 u n nu um n nun n nu u m ux1 um u on l Innmun nunu o u would nunu M u u m um uvzsll39mvmln meow mull nu n w mm nun nuno M u u n u mun nunu a z u mu m u M l n NW n W lumluunn quotquot quot quot quot uuuuu nu u an n u lw w nanu mum n kuuu unxlv u m n u u W Table II amhdays and Assnnated mums fur Hemlnqs39s Ehlldren Model 1 rEmh dale Df r mm and enneepunn date xed I pom rRandumly pmko mls U524 1545 Mm rUseMC m l39s tn cunstmctMC enneepunn dates m m emu n 5N rHuwmanyMC enneepunn daleswas TJpresml fm 7 um ma um mum magnum Ice Ice Model 2 Hum eel21135 yup um n 51 11 753mm may Ann1mm sm mum 4 55 15 rsun MC IEI39s befure eunsnuenng MC enneepnnns my nu 1193 615 mu 1 51 n 1193 Model 3 nun mylnm m nupzum so is u rChuuse slxMC bmh dales Um 5 1795Mdy 21 1303 mum 1mm ms Wynn n 1 11 em IEI39s gt525 nu mynm ma Aug um 11 is 15 USEMC m WWW r tMC enneepunn dates rHuw many Mc enneepunn dates Was TJ presEnl row Model 4 Same as 3 rsun Mc IEI39s befure ennsnuenng Mc enneepnnns Table m Relatlve Frenuenq msmhunnus furthe Number nf cuunenunus am Fall during or three Days heme 3 Jefferson us for me four Mantercarln Models Number of Io7060mm Model 1 Model 2 Model 3 Model 4 u um um uw mm 1 nu 61 u z m m m m 1 m w m m o m m m ms 5 s 1 m m m s n 1 L5 M u Bayes39 Tllentem How should we use new evldence to modlfy pnor bellefs about the world7 FUND PHxPDlH 39 PHgtltPD H P xPDl PtJ l6 Palm PW 6 PyxlPt x 04 Lel39ssay Pm 5 39 PO l 6 5 02 Two kinds of probability questions 1 Whats the probablhty ofgetbng a 6 and a lgt 2 Whats the probablllty that T was the father of all Hemmgs s chll gt Bayes39 Theorem Derivation For any two events H and D PH amp D PHXPD l H PDampH PDxPHlD PDampH PHampD 39 PHXPD H PDXPH D PP D PHWlDlH PD PD PHx PD l H P x PD l I7 SO PHx PD l H PHx PDl H P x PD l I7 PP D Ta Da Bayes39 Theorem Why the derivation matters I t 80 ofwomen wth breast cancer wlll get posltwe mammogaphles What is the probablllty that she actually has breast cance PP gtlt PDl P Bayes39 Theorem Why the derivation matters 80 ofwomen wlth breast cancer wlll get posltwe mammographles What is the probablllty that she actually has breast cancerv 39 D PPXPDlPPPxPDlP PPlD PPXPD PPXPDU I Pclm PcXPmlcPcxPmlc Bayes39 Theo Bayes39 Theorem rem Why the derivation matters I t 80 ofwomen wth breast cancer wlll get posltwe mammogaphles What is the probablllty that she actually has breast cance PXPD P PPlpP PPxPDl PPFxPDl F I PC m P gtltPmc Pcgtlt Pm c PEgtlt Pm E Pclm o7s 01gtlt899gtlt096 0008095 PH D PHgtltPDlH PHgtlt PD l H P x PD l E I A simpler version PH D 0c PHgtltPD H Posterior probabilit PHypothesls Data Prior probability P th Likelihood function Hypo esls P Data Hypethesrs Or PosteIior oc Likelihood gtlt Prior Bayesian Methods for Proportions Beihca ens ZUShetds fmm 1551a qundmt o afihem he Ctumwne N 2n xo Assumz me am came am i binamul distnbuhan wiih pammetem x n andI Fe 2 v in me IiJmIihind mc nnis me himmhl distrihmiun tmtiun mm H Ham 7 Lyme n KKK Ky 39 ie ukelihood Function mam E quot E a m a m i i Meiy 39 ie Like hood Function We eeumiee me ueumu unemqu empixu v ues meu n2 3 Whmuuem mmee me ueumu eggemg me abeeweuvuus m min p I x m lThisisLhc MmmumLikclihaadEsBmm min I I n M I To camguze equznhst can dence iemis min me nam Gaussian app mximman 7 pilg fun at equzntms we at dune Ianyesmns we at na Why le like hood Function We zstimne me hkelihaad funchan using zmpitczlvzlues mes n2 Whimiue afrz mzximtzes ihelikelihaadafgemng me absexved v ues E and n nu Ip7xn lThisisiheMndmnmLikdihdeslimzm m n To campute fuquenhst can dence iemis using me namml Gmsim appxaxtmnhan siplap pm fun at eequenueswe ne dune IfBzyesims we mum th ammo PHXPD H Bayesian Methods for Proportions We used the hinumizl dislrihill39mn mzlinnta madelthe Iihclihuud PDH Ham 7 mi 71 mm x We use me hm demily n39clinnta madcl me prinr WWW mi 7 eei PHP7reail bil rr l r Pmlnbmy u39 in II II vi vi Bayesian Methods for Proportions N My 4 7 a 39 RH P aeiywew 0 a5 aln h5 b m i39 m 3 e hs H E39quot 1quot 15quot Bayesian Methods fol Plopoltions Bayes39aquot Methnds nquot Pmpnmnns Computing lhe posterior dishibulion I PH DOCPHXPD H PHDocPDIIgtltPH m Binomial m Beta 39P7quotXgtquot PUJIVOXPUO x6 n20 a5 b5 n n n ab 1 I Xe Pltnlxn nlrnW M arn Em Em nnmx ailbml 14 a 12 1 m ai 70an r ri r ri n i n H n M H namlnmxbml n Beta E n 311 b19 I The beta is the comugzlte pilot for the binomial 39e H siThe posteliol distlibution E Bayesian Methods fol Plopoltions Bayesian Methods fol Plopoltions Comp 39 uhng the moments mean and variance of the likelihood prior and posterior Back to Beth and her surface collection he knows X6 an n o Thellkellhood p x hi S paw How can she gure out a and b for the priorgt 1 onfess total ignoranoe Let a1 b1 e a uniform prior a 2 Choosevalues thatloolt cool on the ap The prior 7r 7 a 17 3 Base the prior on prior apalencer say the samples she collected at adlacent sites we bl Slle Direamware Total D 1 6 In El 6D I The posterior x a 2 B I 89 7 V 3 ID 15 l 67 V a 17 a s 6 n 83 1 5 14 Zn l 7U y i a m a 7 1 Eshmate the mean and varianoe of the prior from the data I The Bayesian posterior 95 pmbabililylnterval RObmSO 1999 555 a 2 2 5 Emma 3911 96o39 2395 2 a2 Farr gt 5D A crude nttemptto getnd afeffect atsarnpling errar eteets aria Bayesian Methods fol Plopoltions BayeSiaquot Methnds nquot PmPMtinns I Site ra creamware Tmal 1 a 1D I Recall that 7r a 2 9 a b 3 1D 15 a 5 5 Hz Kai 5 14 2D ab1 niiai 43 on Weights Mam n 72 n EIEIBB I So after some heroic Beegraoe algebra I Esamate the moments mean and wuznce ottheprior 2X 77 2 anz 53r52pr 0088 Z I Esamate the parameters otthe prior 401 7 M4 1ng 0088 17 7271 5 25 Question What are the effects of variauon in the sample size and variance ofthe prior on our posterior estimates

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