# STAT TEST #1.docx - SAMPLING METHODS 1) Simple Random...

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SAMPLING METHODS1)Simple Random Sample2)Stratified Random Sample:take SRS of EACH group3)Cluster Sample:take SRS of ENTIRE groups4)Voluntary-Response Sample5)Convenience SampleSAMPLING BIAS:1)Selection:not rep. of pop2)Non-response:responses differ from those in sample3)Measurement Error:info given/recorded incorrectDESIGNED EXPERIMENT:Can determine causationCorrelation does not mean causationResponse Variable:Weight lossExplanatory Variable:Time spent playing PGHISTOGRAM:Center of data distributionVariabilityOutliersShape of distributionSHAPE OF DATA:Bell:mean = medianLeft:mean < medianRight:mean > medianUniform:mean = medianFORMULAS:MAD=Σ|x -mean|/nSample Variance=calculator method (sx2)Standard Deviation=calculator (sx)Z-Score=(x-x)/SD *mean=0, SD=1*Ex. Height is 173 cm, mean is 163cm, SD is 7.6Zscore=173-163/7.6=1.32 SD above meanEMPIRICAL RULE:1SD=68%2SD=95%3SD=99%BOX-PLOTS:5number summary:-min, Q1, median, Q3, maxLINEAR TRANSFORMATIONS:Adding:no effect on V or SDMultiplying:all mult. Var. is multiplied by c2RULES OF PROBABILITY:1)All probabilities lie between 0 & 12)Total probability has to be 1INTERSECTION OF EVENTS:P(AnB)=# of points in intersection/# of points in eventMutually Exclusive:events can’t occur at same time P(AnB)=0 *dep*UNION OF EVENTS:P(AUB)=P(A)+P(B)-P(AnB)Complementary Events:Accontains all points A doesn’tP(Ac)=1-P(A) *A and Acare mutually exclusive*CONDITIONAL PROBABILITY:P(A|B)=P(AnB)/P(B)INDEPENDENT EVENTS:event A has no effect on probability of other1)P(AnB)=P(A)x P(B)2)P(A|B)=P(A)*indicate independence*3)P(B|A)=P(B)*if 1 statement true ALL are*MULTIPLICATION RULE:P(AnB)= P(A|B) x P(B) *seatbelt example*P(Seatbelt & Serious Injury)=0.77(0.08)=0.0616BAYES THEOREM:P(B|A) = P(B) x P(A|B)/P(A) *disease test ex*P(disease|positive)=0.0047/0.0246=0.1911PERMUTATIONS & COMBINATIONS:Permutations:Order matters-50 participants, 3 different prizesP(503)=117600Combinations:Order doesn’t matter-20 applicants, 5 jobsC(205)=15 504DISCRETE & CONTINUOUS RANDOM VARIABLES:Discrete:countable or countably infinite # of possible valuesContinuous:infinite number of values-can’t count XDISCRETE DISTRIBUTION valid when:see probability rulesEXPECTED VALUE & VARIANCE OF RANDOM VARIABLE:E(X)=Σ|xP(x)-calculator meanVar (X)=Σ|E[(x -µ)2] *see sample variance but use σx2insteadStandard Deviation=calculator method use σx insteadBERNOULLI DISTRIBUTION:can take on 2 possible values: 0 or 11)X is an outcome of a one-trial experiment2)Two possible outcomes: Success=1, Failure=03)P(Success)=1 P(Failure)=0P(X=x)=px(1-p)1-xE(X)=pVar(X)=p(1-p)Ex. Security randomly checks 60% of travelers, what is probability ofselection?X=1: searchedX=0: not searchedP(X=1) = 0.601(1-0.6)1-1=0.60BINOMIAL DISTRIBUTION:can take on n+1 possible values 0,1,…,n1)There are a fixed number of trials2)Trials are independent of each other3)Two Possible Outcomes: Success or Failure4)P(Success)=1 P(Failure)=1-pP(X=x)=(nx) px(1-p)n-xE(X)=npVar(X)=np(1—p)Ex. Given probability of 0.8 & 4 shots what is P(X=2)P(X=2) = (42)0.82(0.2)4-2= 0.1536

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