# A one sample test because the same group is being

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a one-sample test because the same group is being tested twice (or sometimesrelated/dependent groups e.g. pairs of twins)25If we rejectܪwhen it is true, this is atype I errorIf we fail to rejectܪwhen it is false, this is atype II error26Types of error
4/02/201914Central Limit TheoremIf the population sampled from is Normally distributed, theܺdistribution will be Normally distributedwhatever the size of݊Mean ofܺdistribution isߤStandard deviation ofܺdistribution is(also called ‘standard error’)Variance ofܺdistribution is~ࡺ(ࣆ,)Central Limit Theorem: if we take many random samples of size݊andif݊ ≥ 30, theܺdistribution will approximate a Normal distributioneven when the population sampled from is not Normally distributedscan be substituted in place ofso thatreplacesin calculations27Distribution of the sample meanWhen݊ ≥ 30and sampling is random, sample means (ܺ) will beNormally distributed (Central Limit Theorem)test statistic isݖ =௫̅ିఓ(=௦௔௠௣௟௘ ௠௘௔௡ି௛௬௣௢௧௛௘௦௜௦௘ௗ ௠௘௔௡௦௧௔௡ௗ௔௥ௗ ௘௥௥௢௥)can substitutesinstead ofand replaceݖ =௫̅ିఓwithݖ ≈௫̅ିఓtest statistic is nowZ,hence conducting aZ-test (using Normal dist)When݊ < 30, can also substitutesinstead ofand useappropriatet-distributionbut only if sampling from a Normally distributed populationAsnincreases (and therefore݂݀increases), thet-distributionapproaches the Normal distributionsot-tests andZ-tests give increasingly similar results28
4/02/201915Using a sample mean or median to test a hypothesis aboutpopulation central location,whenis unknown and sampling is randomis samplesize large?is samplingfrom Normallydist.population?nonoyes(CLT)Z-test (can also uset-test)Usesinstead ofߪSign Testyes29t-testUsesinstead ofߪSection C(Weeks 8 – 11)Two variable analysis30
4/02/201916Chi-square test(2 cat. variables)When working with two variables, our motivating question willgenerally be“is there an association between the variables?”The chi-square)test is used to detect the presence of anassociation between twocategoricalvariablesTheχdistribution is asymmetrical and skewed to the rightExpected values are calculated by multiplying the appropriate rowtotal by the column total and dividing by the total number ofobservationsܧ =ோ×஼Hypotheses when performing a chi-square test will be:ܪ:there isnoassociation between variablesܪ:thereisan association31Chi-square testIf there were perfect agreement between theobservedandexpectedfrequencies,then we would expect the߯test statistic to equal zeroThe more the observed and expected frequencies disagree, the greater the valueof the test statisticA߯test is a one-tailed testThe degrees of freedom for the߯test are calculated as݂݀ =ܴ − 1 ×ܥ − 1,whereܴis no. of rows andܥis no. of columnsNecessary conditions for you to check when conducting a߯test are1.observations must be independent2.

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