The$Central$Limit$Theorem$For$Sample$Proportions$$Lecture$15$$$Review$• Recall$that$probability$tells$us$how$frequently$we$should$observe$and$event$if$we$could$repeat$the$experiment$over$and$over$again$infinitely$• Last$lecture,$we$tried$to$simulate$“infinitely”$$• But$we$need$to$go$further,$we$need$to$rely$upon$a$theory$to$help$us$find$real$infinite$$àthe$Central$limit$theorem$$• Tells$us$that$our$sample$distribution$will$approximate$a$normal$distribution$mean=p$(population$proportion)$and$a$standard$deviation$which$is$equal$to$the$standard$error$$!" = !! ∗ (1 − !)!$$$$$Central$Limit$Theorem$(CLT)$conditons$for$a$normally$shaped$sampling$distribution$for$Sample$proportions$ $$• Random$and$independent:$The$sample$is$collected$randomly$and$the$observations$are$independent$of$each$other$$• Large$sample:$the$sample$has$at$least$10$successes,$this$can$be$checked$by$using$np$≥$10,$and$at$least$10$failures$using$n(1Lp)$≥ 10$• Large$population:$if$the$sample$is$collected$without$replacement,$then$the$population$size$is$at$least$10$times$larger$than$the$sample$size$$$$The$Key$Point$• The$CLT$(Central$Limit$Theorem)$for$sample$proportions$ $tells$us$that$if$we$collect$one$random$sample$from$a$population$and$the$sample$is$large(np≥ 10!and$! 1 − ! ≥ 10)$and$if$the$population$size$is$much$larger$(at$least$10$times)$than$the$sample$size,$then$the$sampling$distribution$of$ $is$NORMAL$with$mean$p$and$standard$deviation$!∗(!!!)!$$or$! !!∗(!!!)!$CLT conditions for a normally shaped sampling distribution for sample proportions Random and Independent: The sample is collected randomly and the observations are independent of each other.Large Sample: The sample has at least 10 successes, this can be checked using np ! 10, and at least 10 failures using n(1 – p) ! 10.Large Population: If the sample is collected without replacement, then the population size is at least 10 times larger than the sample size.CLT conditions for a normally shaped sampling distribution for sample proportions Random and Independent: The sample is collected randomly and the observations are independent of each other.Large Sample: The sample has at least 10 successes, this can be checked using np ! 10, and at least 10 failures using n(1 – p) ! 10.Large Population: If the sample is collected without replacement, then the population size is at least 10 times larger than the sample size.CLT conditions for a normally shaped sampling distribution for sample proportions Random and Independent: The sample is collected randomly and the observations are independent of each other.Large Sample: The sample has at least 10 successes, this can be checked using np ! 10, and at least 10 failures using n(1 – p) ! 10.Large Population: If the sample is collected without replacement, then the population size is at least 10 times larger than the sample size.$$How$to$Check$$$$$$$$$$$$How to CheckLet’s suppose that a year ago (May 2011) when William married Kate, 36% of British adults wanted Charles to be king.This more recent (May 2012) poll of 2,019 British adults reveals that only 31% want Charles to be king.Question: What is the probability of getting a result as low as 31% or lower if 36% is the correct current percentage of British support for “King Charles”?• Lets$suppose$that$a$year$ago$when$William$married$Kate,$36%$of$British$adults$wanted$Charles$to$be$king$•A$more$recent$poll$of$2,019$British$adults$reveals$that$only$31%$want$Charles$to$be$king$$$• Question:$what$is$the$probability$of$getting$a$result$as$low$as$31%$of$lower$if$36%$is$the$correct$current$percentage$of$British$support$for$King$Charles$?$$TAKE$NOTE$• There$is$a$parameter$(36%)$and$a$sample$statistic$(31%)$in$this$question.$And$a$sample$size$n$(2019).$$Checking$• Random$and$independent:$we$will$assume$it$was$a$random$sample$of$British$adults$$• Large$sample:$np$≥ 10! 2019 ∗ .36 = 726.84 $and$$!)1 − !) ! ≥ 10!(2019 ∗1 − .36 = 1292.16)$• Large$population:50$million$is$more$than10$times$larger$than$2019$$If$the$conditions$are$met$$then……$• The$sampling$conditons$will$be$normal.$The$mean$of$sampling$distribution$will$be$equal$to$p=.36$and$the$standard$distribution$will$be$equal$to$p=.36$and$the$standard$deviation$of$the$sampling$distribution$will$be$equal$to$the$Standard$Error$(SE)$or$!∗(!!!)!$=$.!"∗(!!. !")!"#$=$.0107$$o We$could$use$technology$to$figure$out$the$probability$of$getting$a$.31$in$a$N(.36,$.0107)$sampling$distribution$but$we$can$also$use$Z$scores$$$FINDING$THE$Z$SCORE$$• In$the$standard$unit,$the$poll$result$of$.31$is$z=.!"!.!".!"=!.!".!"= −5$$$$Conclusion:$there$is$virtually$no$chance$that$a$random$sample$would$result$in$a$.31$(31%)$if$the$true$parameter$is$.36$(36%).$$$Finding the Z scoreIn standard units, a poll result of .31 isA lookup in Table 2 is 0.000000287 (virtually zero)Conclusion: there is virtually no chance that a random sample would result in a .31 (31%) if the true parameter is .36 (36%)If$we$look$L5$up$on$Table$2$the$value$is$virtually$zero$or$.000000287$INTERPRETATION$• If$the$true$proportion$of$support$for$King$Charles$is$.36,$the$CLT$tells$us$that$the$sampling$distribution$(this$is$the$population$of$all$possible$samples)$should$have$a$mean$or$.36$and$a$standard$deviation$of$.0107$and$be$normally$distributed$$• Therefore$if$the$survey$had$no$serious$flaws,$it$is$extremely$unlikely$(probability$of$.000000287)$to$draw$a$single$sample$with$proportion$.31$• I$would$conclude$that$the$parameter$may$have$decreased,$Charles$is$less$popular$now$than$a$year$ago$$$Why$could$this$be$concluded$from$our$data?$à$Probabilities.$If$the$shape$of$he$sampling$distribution$looks$like$the$previous$slide,$then$the$empirical$rule$tells$us$that$68%$of$all$possible$sample$proportions$will$be$between$about$35%$and$37%$(that’s$36%$plus$of$minus$1%)$and$about$95%$of$the$possible$sample$proportions$should$fall$between$34%$and$38%$• So$either$the$news$organization$did$something$horribly$wrong$OR$Charles$is$slipping$in$popularity$$$$$$$ANOTHER$WAY$à$suppose$instead$last$year$that$Charles$popularity$was$33%.$Then$p=.33$and$$the$Standard$Error(SE)$would$have$been$.0105$so$N(.33,$.0105).$The$ZLscore$would$have$been$L1.90$and$the$corresponding$table$value$is$.0287$• !∗(!!!)!$=$.!!∗(!!. !!)!"#$=$.0105$• z=.!"!.!".!"=!.!".!"= −.2$o