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# 9624 08 2 08 2 92 1 3 30 3 34 92 1 3 30 3 26 3 34 3

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9624 . ) 08 . 2 08 . 2 ( 92 . 1 3 . 30 3 . 34 / 92 . 1 3 . 30 3 . 26 ) 3 . 34 3 . 26 ( ) 92 . 1 , 3 . 30 ( ~ or ) / , ( ~ CLT By CLT) the in ' ' the ignore ll we' so (No, matter? it Does normal? Is of on distributi sampling requires which ) 3 . 34 3 . 26 ( determine to Need 2 2 a = < < - = - < - < - = < < < < Z P n X P X P N X n N X X X X P

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Auditing accounts… l According to our results Clare will almost always (96% chance) obtain a sample mean within \$4 of the population mean (designated task) l Problem could have been solved without knowledge of μ l Auditor has a measure of sampling error for the sample mean as an estimate for an unknown population mean 21 9624 . ) 08 . 2 08 . 2 ( 92 . 1 4 / 92 . 1 4 ) 4 4 ( )) 92 . 1 , ( ~ (or ) / , ( ~ CLT By large size sample because matter t Doesn' normal? Is of on distributi sampling requires which ) 4 4 ( determine to Need 2 2 = < < - = < - < - = + < < - + < < - Z P n X P X P N X n N X X X X P σ
Auditing accounts. .. l While don’t need to know μ , problem does require l Knowledge of n l But that is determined by the auditor l σ to be known which is more problematic l Later show how an estimate of ( s ) can be used instead 22

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23 Interval estimation l Point estimators produce a single estimate of the parameter of interest l The auditing example suggests some notion of the margin of error might be useful l Keller calls this bound on the error of estimation l Interval estimators produce a range of values & attach a degree of confidence associated with that interval l Hence the name confidence interval ? ? ? ? l In auditing example consider an estimator that is l How often would you expect the true population mean to be in this interval? 4 \$ ± X
24 Interval estimation… 95 . 96 . 1 96 . 1 yield to statement y probabilit this rearrange Now 95 . 96 . 1 / 96 . 1 tables) (from 96 . 1 475 . ) 0 ( 05 . choose Say we 1 / of value ed standardiz this around interval symmetric a Consider / thus and ) / , ( ~ Suppose 2 = + - = - - = = = - = - - - = n X n X P n X P a a Z P a n X a P X n X Z n N X σ μ α

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25 Interval estimation… l Be careful l Endpoints of the interval are rv’s l We have constructed a random interval l μ is a constant l For a particular sample (& sample mean value) is either in the confidence interval or not l If 100 samples were drawn we would expect 95 of them to include interval confidence a define , 96 . 1 , Endpoints n X σ ±
26 Confidence intervals l CI’s for means/proportions typically have a similar structure l Centred at sample statistics l Endpoints are ± some multiple of the standard error (standard deviation of the sampling distribution) l The “multiple” is determined by confidence level chosen by investigator

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27 Selecting sample size l Recall the auditor Clare l Suppose she needs to decide on sample size l She wants a sample size that has a margin of error of \$4 & she is willing to set the confidence level at 90% l Can now use the form of the CI to solve for required n 156 4 334 . 30 645 . 1 n or 4 require Thus interval confidence the defines 05 . 2 / × = = ± n n z n z X σ α
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9624 08 2 08 2 92 1 3 30 3 34 92 1 3 30 3 26 3 34 3 26 92 1...

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