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# Introduction to Biostatistics STAT 541

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This 21 page Class Notes was uploaded by Mrs. Triston Collier on Thursday September 17, 2015. The Class Notes belongs to STAT 541 at University of Wisconsin - Madison taught by Staff in Fall. Since its upload, it has received 42 views. For similar materials see /class/205085/stat-541-university-of-wisconsin-madison in Statistics at University of Wisconsin - Madison.

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Date Created: 09/17/15

Ismor Fischer 8112008 Smt54131 Probability Theory 31 Basic Ideas De nitions and Properties POPULATION Unlimited supply of five types of fruit in equal proportions 01 Macintosh apple 04 Cavendish banana Oz Golden Delicious apple 05 Plantain banana 03 Granny Smith apple 933633 caqumq a gag quot1 58139 13 i lllll in W Experiment 1 Randomly select one fruit from this population and record its type Sample Space The setS of all possible elementary outcomes of an experiment Any subset ofa sample space S Elementary outcomes simple events Event Event PEvent 3 A 5 7 06 B 0 4 trials of experiment 2 LO The probability of randomly selecting an apple is 06 The probability of randomly selecting a banana is 04 lsmor Fischer 8112008 Stat 541 32 General formulation may be facilitated with the use of a Venn diagram Experiment 2 Sample Space S 01 02 0 S k Om2 EventA 0102OmgS Am k Definition The probability of event A denoted PA is the longrun relative frequency with whichA is expected to occur as the experiment is repeated inde nitely Fundamental Properties of Probability For any eventA 01 02 Om in a sample space S l 0 S PA S l 2 PA Ema POlP02PO3POm il Special Cases 0 PQ 0 k 0 PS ZPOi l certainty il 3 If all the elementary outcomes of S are equally likely ie l POl POz POk g then A m P A S k 3 2 Example PA g 06 PB Q4 ImmFlschzr8HZEIEI8 Stow us Experiment 2 Select a card at iandam from a standard deck and replace Sample Space 3 AAK 8Q 9Q 10 IQ QQ KQ 393v 4V 539 6V 739 8V 939 1039 IV QV KV 3 4 5 6 7 B 9 10 I Q K Events A Selectal 2A2l2v2 M SelectaTu A 21KT Probabilities Since all elementary outcomes are equally likely it follows that New Eventx from Old Event 1 AC notl All outcomes that are in s but not inA Example AC Select either A 34 Example Expuimem Toss a coin once Events A Heads AC Tails Probabilities Fair coin PA 05 2 PAC 1 7 05 05 3mm com PA 07 2 PAC 1 7 07 03 lsmor Fischer 8112008 Stat 541 34 intersection 2 A n B A and B All outcomes in S thatA and B share in common All outcomes that result when events A and B occur simultaneously Example AnB Selecta2andaT0 24 gt Definition Two eventsA and B are said to be disjoint or mutually exclusive if they cannot occur simultaneously ie A n B Q hence PA n B 0 Example A Select a 2 and C Select a 3 are disjoint events Exercise Are A 24 34 44 54 and B 26 36 46 56 disjoint If not findA n B 3 A U B A or B All outcomes in S that are either inA or B inclusive 0 if A and B are disjoint Example A U B Select either a 2 or a 01 has probability 4 l3 1 l6 PAUB Z 52 52 52 52 Example A U C Select either a 2 or a 3 has probability 4i0 PAUC 52 52 ImmFischzr8HZEIEI8 5151541 as Note 393 d 4 d 4 geszzon 39 39 quot quot quot quot j Example Take 1 3 events S Then PA U B U C A B PA PB PC spwn3spt4nctsp3nct PA n B n C Exercise For s January December verify this formula for the the events A Has 31 days B Name ends in r and C c Name begins with a vowel Exercise A single tooth is to be randomly selected for a certain dental procedure Draw a v diagram to illustrate A u erjaw B le side an ca o and in 1 co ending p abilities Calculate the probability that all of ese three e ents occur culate the probabr ty that non ee events oc Calc the probability that exactly one th s three events oc 1 that exactly Mo of these thr ents occur Think carefully Assume equal likelihood in all cases l The three set operations s union intersection and cornplernent s can be uni ed via DeMWgMiI Lam Exercise Using a Venn diagxam convince yourself that these staternents are e in flu BC Acn BC general Then ver39fy thern for a speci c exarnple eg Pick a picture card and B Pick a black card AMBCACUBC Ismor Fischer 8112008 Stat 541 36 FACT I Random variables can be used to define events that involve measurement Experiment 3a Roll one fair die Discrete random variableX value obtained 7 i Sample Space S 1114576 3 u 7 Because the die is fair each of the six faces has an equally likely probability of occurring ie 16 The probability distribution for X determined by the probability mass function pmffx can be organized in a probability table and displayed via a corresponding probability histogram as shown Uniform Distribution Experiment 31 Roll two fair dice 2 Outcome Die 1 Die 2 Sample Space S 11 66 7 r 11 12 13 14 5 211 22 23 24 3 131 32 M 34 41 42 43 44 51 52 5 quot54 61 lt62 6 u at 39 e i 9 m L I O I m a Discrete random variableX Sum of the two dice 2 3 4 12 Events X2 11 X3 12 21 X4 13 2 2 3 1 X5 14 2 3 3 2 4 1 X6 15 2 4 3 3 4 2 51 X7 16 2 5 3 4 4 3 5 2 6 1 X 8 2635445362 X9 3 6 4 5 5 4 6 3 X 10 4 6 5 5 64 X 11 5 6 6 5 X 12 66 lanor Fischer 8llZEIEI8 Stat 541 377 Again the probability distribution forXcan be organized in a probability tab and displayed via a probability histogram both of which enable calculations be done easily le to Probability Histogram x fx PXx O 2 135 2 3 235 4 335 E 5 435 5 535 3 7 535 D x 535 9 435 g 10 335 11 235 8 12 135 D 2 3 4 5 6 7 8 9 10 11 12 1 X PX7 or X 11 Note that X7 and X 11 are ml PX7 PX 11 wameula 3 above 636 236 836 P5 nga PX5 or X6 or X7 or X8 PX5 PX 5 PX7 PX 8 435 35 535 536 I 2036 salami If PXlt10 1 e PX210 wameula1above 1 r PX10 PX11 PX12 1 7 335 235 135 1 7 535 3035 Exercise How could event E Roll doubles be characterized in terms of a random variable Hmt Let Y Difference between the two dice lsmor Fischer 8112008 Stat 541 38 The previous example motivates the important topic of Discrete Probability Distributions In general suppose that all of the distinct population values of a discrete random variable X are sorted in increasing order x1 lt x2 lt x3 lt with corresponding probabilities of occmrence xlfx2fx3 Formally then we have the following Definition fx is a probability distribution function for the discrete random variableX if for all x fx 2 0 AND 4 fx 1 In this case fx PX x the probability that the value x occurs in the population The cumulative distribution function cdf is defined as for all x Fx PXSx 2 fx fX1fX2fx allxl x Therefore F is piecewise constant increasing from 0 to 1 Furthermore for any two population values a lt b it follows that b Pa SXS b Z x Fb Fa7 a where a is the value just preceding a in the sorted population Total Area 1 Exercise Sketch the cdf Fx for Experiments 3a and 3b above 1 I F X3 I M R i x x2 E fxl 3 Fx1 5 fx i X X x1 x2 x3 x 0 x1 x2 x3 x lsmor Fischer 8112008 Stat 541 39 Population Parameters a and 0392 vs Sample Statistics I and s2 0 population mean the expected value of the random variableX the arithmetic average of all the population values Compare this with the relative equency definition of sample mean given in 23 Properties of Mathematical Expectation 1 For any constant c it follows that EcX cEX 2 For any two random variablesX and Y it followsthat 39 E X Y E X EY and via Property 1 39 EX Y EXJ E Yu Any operator on variables satisfying land 2 is said to be linear 0 population variance the expected value of the squared deviation of the random variable X from its mean u Compare the first with the definition of sample variance given in 23 The second is the analogue of the alternate computational formula Of course the population standard deviation 0 is defined as the square root of the variance Exercise Algebraically expand the expression X 7 a2 and use the properties of expectation given above lsmor Fischer 8112008 Stat 541 3 10 Experiment 4 At a party guests randomly select one pastry from two trays continually refilled where the distribution of calories X1 and X are indicated below Probability Histogram O s O 2 g 39 Uniform Distribution O a O 8 Q O Probability Table x o A 1 90 120 150 X1 Calories 13 13 equally likely outcomes Probability Histogram Probability Table 8 x f2x 30 60 90 30 36 60 26 90 16 Ismor Fischer 8112008 Stat 541 3 11 Summary Also refer back to 24 Summag POPULA TON Discrete random variable X Probability Table Probability Histogram x fx PXx 1 g EX2xfx g EltX m2 Z x Alfx E Oquot2 or D Em 2 Z Woe 2 l SAMPLE size n Relative Frequency Table Density Histogram lrqux n X and S 2 can be shown to be unbiased estimators of x f x x1 xl X32 xz 1 and 02 respectively That is EX y and E S2 0quot2 In fact they are MVUE xk xk Statistics 3 Ismor Fischer 8112008 Stat 541 312 it In it if Some Notes on General l Suppose that 6 is a xed population parameter eg y and 6 is a samplebased estimator eg Consider all the random samples of a given size n and the resulting sampling distribution of 6 values Formally define the following P OPULA TI ON Parameter 6 39 Bias E676 the difference between the expected v value of and the target parameter 6 39 Mean of 6 the expected value of SAMPLE Statistic 6 A A A 2 Variance of 6 E6E6j the expected value of the squared deviation of 6 from its mean or equivalently 2J7E62 39 Mean Squared Error lVISE E6762 the expected value of the squared difference between estimator 6 and the target parameter 6 Exercise Provequot that Vector interpretation c676 A Ecz at Ebz a 5 49 Comment A parameter estimator 6 is defined to be unbiased if E66 ie Bias 0 In this case MSE Variance so that if 6 minimizes MSE it then follows that it has the smallest variance of any estimator Such a highly desirable estimator is called MVUE Minimum Variance aniased Estimator It can be shown that the estimators X and S2 of y and 02 respectively are MVUE but finding such an estimator 6 for a general parameter 6 can be quite difficult in practice Often one must settle for either not having minimum variance or having a small amount of bias using the basic properties of mathematical expectation given earlier lsmor Fischer 8112008 Stat 541 313 Related but not identical to this is the idea that of all linear combinations 01x1 02x2cnxn ofthe data x1x2xn such as X with 0102 cn ln which are also unbiased the one that minimizes MSE is called BLUE Best Linear Unbiased Estimator It can be shown that in addition to being MVUE as stated above X is also BLUE To summarize MVUE gives Min Variance among all unbiased estimators 5 Min Variance among linear unbiased estimators Min MSE among linear unbiased estimators since MSE Var Biasg given by BLUE by def The Venn diagram below depicts these various relationships Minimum Minimum Variance MSE Minimum variance BLUE among all unbiased estimators Minimum variance among linear unbiased estimators Unbiased Linear Comment 393 j a then 6 is said to have mean square convergence to 6 This in turn implies convergence in probabilit via Markov s lne uality also used in proving Chebyshev s Inequality ie r Ismor Fischer 8112008 Stat 541 21 2 Exploratory Data Analysis amp Descriptive Statistics 21 Examples of Random Variables amp Associated Data Types gt NUMERICAL Quantitative measurements interval 0 Continuous X Length Area Volume Temp r J Time elapsed pH Mass of tumor X steps 0 Discrete X Shoe size weeks till death A Time displayed RX dose tumors I I I I I X gt CATEGORICAL Qualitative bins unranked 0 Nominal X Color 1 Red 2 Green 3 Blue r H 1D Zip Code Type of tumor X 352 quotecialsCase Binary 1 Success X 2 quotf dilutequot ranked 0 Ordinal X Dosage l Low 2 Med 3 High Year 2000 2001 2002 X Stage of tumor 1 II III IV 1 lt 2 lt 3 Random variables are important in experiments because they ensure objective reproducibility ie verifiability replicability of results Example 12 3 4 9091 92 93 94 95 96 97 98 99 100 In any given study the researcher must first decide What percentage of replicated experiments should in principle obtain results that correctly agree speci cally accept a true hypothesis and incorrectly agree speci cally reject a true hypothesis allowing for random variation Con dence Level 1 a 090 099 are common choices Signi cance Level or 010 001 the corresponding error rates Ismor Fischer 8122008 Stat 541 85 82 Estimation KaplanMeier ProductLimit Formula 1950s 1 Time 0 t1 t2 t3 Let t1 t2 t3 denote the actual times of death of the n individuals in the cohort Also let d1 d2 d3 denote the number of deaths that occur at each of these times and let n1 n2 n3 be the corresponding number of patients remaining in the cohort Note that n2 n1 d1 n3 n2 d2 etc Then loosely speaking S02 PT gt t2 Probability of surviving beyond time if depends conditionally on St1 PTgt t1 Probability of surviving beyond time t1 Likewise S03 PT gt t3 Probability of surviving beyond time if depends conditionally on S02 PT gt t2 Probability of surviving beyond time t2 etc By using this recursive idea we can iteratively build a numerical estimate 30 of the true survival function St Specifically V asquot we have St PT gt t Probability of surviving ecause no deaths have as yet occurred Therefore for all tin Recall see 32 1 For any two eventsA and B PA and B PA x PB A Let A survive to time t1 and B survive from time t1 to beyond some time t before t2 Having both events occur is therefore equivalent to the event A and B survive to beyond time t before t2 ie T gt t Hence the following holds For any time n we have St PTgt t Psurvive in 0 t1 x Psurvive in t1 t survive in 0 t1 J V V A d S t 1 x M or I ll ie 3i Similarly For any time m we have St PTgt t Psurvive in t1 Q X Psurvive in t2 t survive in t1 t2 V V ie 0 1 x L d2 or 1 11 1 12 etc lsmor Fischer 8122008 Stat 541 86 121 2 3 we have In general for This is known as the KaplanMeier estimator of the survival function St Note that it is not continuous but only piecewisecontinuous actually piecewiseconstant or step function hertz 3 Comment The Kaplan Meier estimator S t can be regarded as a point estimate of the survival function St at any time t In a manner similar to that discussed in 72 we can construct 95 confidence intervals around each of these estimates resulting in a pair of con dence bands that brackets the graph To compute the confidence intervals Greenwood s Formula gives an asymptotic estimate of the standard error of S t for large groups Ism or Fischer 8122008 Example gcont dz Twelvemonth cohort study of 39 Stat 541 87 Patient t1 months 1 32 10 0 2 55 10 0 10 1 3 67 10 1 9 1 4 67 9 1 8 2 5 79 8 2 6 1 6 84 6 1 5 3 7 84 5 3 2 1 8 84 2 1 1 0 9 103 10 alive 0 10 10 l 09 09 39 l 08 08 39 I 07 5 5 06 06 39 505 05 39 04 E 03 E 02 02 l 01 01 1 1 1 1 1 1 1 0 32 55 67 79 84 103 12 Time months lsm or Fischer 8122008 Exercise preceding example the Kaplan Meier estimator can be written simply as t t e rig l31 i 01 22quot Stat 541 88 Prove algebraically that assuming no censored observations as in the nil I l for Hint Use mathematical induction recall that ni1 ni di In light of this now assume that the data consists of censored observations as well so Example g cont d 1 Patient t months 1 32 2 55 3 67 4 67 5 79 6 84 7 84 8 84 9 103 10 alive censored S t A 1000 10 0900 09 39 F 08 0675 07 x 06 05 04 03 0270 02 0135 0391 quot 1 1 1 1 1 1 1 l l l l l l l V 0 32 55 67 79 84 103 12 lsmor Fischer 8122008 Stat 541 89 Hazard Functions Suppose we have a survival function St PTgt t where T survival time and some At gt 0 We wish to calculate the conditional probability of survival to the later time t At given survival to time t St A k l I 50 50 At i T t PtSTlttAt St StAt PSurvive in ttAt Survive after t PltTgt0 0 Sgt ISTltIAI Tgtt Therefore dividing by At PitSTlttAt Tgtt l StAt S tl At St At Now take the limit of both sides as Al gt 0 This is the hazard function or hazard rate failure rate and roughly characterizes the instantaneous probability of dying at time t in the above mathematical limiting sense It is always 2 0 Why What signs are St and S t respectively but can be gt 1 hence is not a true probability in a mathematically rigorous sense Exercise Suppose two hazard functions are linearly combined to form a third hazard function 01 h1 tczl12t l13t for any constants 0102 2 0 What is the relationship between their corresponding logsurvival functions lnS1t lnS2 t and lnS3t Its integral 9114st is the cumulative hazard rate denoted Ht and increases since H t ht 2 0 Note also that Ht ln St and so iii Hazard h Hazard hm lsmor Fischer 8122008 pExam les 39 If the hazard function is constant for t 2 0 ie h survival function is St e ie the a Timer Stat54l 8 lO t a gt 0 then it follows that the 5 Shown here is a l 0 t v 00 I Q 0 K cs E E 5 2e N O V O rr O M l l l l l U l 2 3 4 5 TImeI 39 More realistically perhaps suppose the hazard takes the form of a more general power function ie ht 053 If for scale parameter a gt 0 and shape parameter 3 gt 0 for t 2 0 Then St e m39g ie the extremely versatile and useful model with broad applications to many fields The case a l 3 2 is illustrated below Exercise Tlmet Suppose that for argument s Q 2 2 lt0 a o E z Z a g g x O d 7 l l I l l U l 2 3 4 5 Timel sake a population is modeled by the decreasing hazard function ht for t 2 0 where c gt 0 is some constant 0 Sketch the graph of the survival function St and find the median survival time lsmor Fischer 8122008 Stat 541 811 Kaplan Meier estimation of hm Fischer 2005 unpublished Let 91 If and tJH be three consecutive times of death in the cohort Then we have the following elementary estimate based on a weighted centered difference 1 ni approximation of S If and the relation 0 1 2 1 ti for any i 1 Other much more sophisticated methods exist for constructing smooth estimators of St ht Ht and related objects

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