# week 4.pdf - 8/09/2020 Section B (Weeks 4 – 7) One variable...

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8/09/20201Section B(Weeks 4 – 7)One variable analysis1Map of courseSectionWeeksTitle…containsA1 - 3Informal inference with 1 and 2variables; probabilityCritical thinking; variables; data display – bar charts, boxplots,histograms, scatterplots; summary statistics; probabilitydefinitions and rulesB4 - 7Formal inference: hypothesistesting with 1 variableBinomial and Normal probability distributions; Sign Test; t-testfor a mean; Z-test for a mean; Central Limit TheoremC8 - 11Formal inference: hypothesistesting using 2 variablesChi-square test; ANOVA; t-test for 2 independent means;linear regressionD12ConsolidationRevision, linking the topics together2Binomial distributionWeek 43
8/09/20202We consider the possible outcomes of a random variable, and theprobabilitiesof each of those outcomesThis gives us ourprobability distributionSo probability distributions are also either discrete or continuousThediscrete probability distributionwe will use is the Binomial Distribution(The continuous one we’ll use shortly is the Normal Distribution)Probability distributions are mathematical models of real life – they neverperfectly fit reality!(If they do, we should probably be suspicious)Probability distributions4Coin toss exercise: the random variable ‘X’ counts the number ofheads obtained (out of 3 trials)Experimental (observed) resultsTheoretical (expected) resultsܲܺ = 0=ܲܺ = 1=ܲܺ = 2=ܲܺ = 3=5ܺ = 0 :ܺ = 1 :ܺ = 2 :ܺ = 3 :read as ‘the probability that X equals 3 is …’Probability distribution for coin toss exercise0123ࡼ(ࢄ = ࢞)0.1250.3750.3750.1256
8/09/20203With this categoricalvariable, the proportionsof people within allcategories add to 1In the same way, theprobabilities within aprobability distributionwill always add to 17Binomial distribution8Example of binomial probability distributionLet’s say that Lionel Messi (brilliant Argentinian football player) is practising hispenalty shots (against an equally talented goalkeeper)Every time Messi takes a shot, we’ll say his chance of scoring (‘success’) is0.8(we know this from past experience). Each shot is a Bernoulli trial as he will eitherscore or not scoreWe want to know the probability of the event that he scores5 or fewer times outof 9 attempts(what do you guess that this probability might be?)So we want to findܲ(ܺ ≤ 5), whereXis the random variable representing thenumber of successes (goals) out of 9 shots9‘is less than or equal to’
8/09/20204The necessary conditions for a binomialdistributionIf(and only if) the following conditions are met, we can use thebinomial distribution to find this probability1)Fixed numberof trials[݊ =]2)Fixed probabilityof success[݌ =]3)Twopossible outcomes[or]4)Trials areindependent*[he doesn’t get better or worse depending onprevious results]*Discuss!

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