Chap 8

ConfidenceInterval Estimation

USINGSTAilSTICS@ SaxonHome Improvement 8.1 CONFIDENCE INTERVALESTIMATION FORTHE MEAN (o KNOWN)

8.2 CONFIDENCE INTERVALESTIMATION t t

, tr

I

FORTHE MEAN (o UNKNOWN) Student's r Distribution Properties of theI Distribution TheConcept ofDegrees ofFreedom TheConfidence IntervalStatement 8.3 CONFIDENCE INTERVALESTIMATION FORTHE PROPORTION 8.4 DETERMINING SAMPLESIZE Sample SizeDetermination for theMean Sample SizeDetermination for theProportion 8.5 APPLICATIONS OF CONFIDENCE INTERVALESTIMATION IN AUDITING Estimating thePopulation TotalAmount Difference Estimation One-SidedConfidenceIntervalEstimationof the Rateof Noncompliance with InternalControls

8.7

G)(CD-ROMTOPiC)ESTTMATTON AND SAMPLESIZEDETERMINATION FORFINITEPOPULATIONS

EXCELCOMPANIONTO CHAPTER8 E8.l Computing theConfidence IntervalEstimate for the Mean (o Known) 88.2 Computingthe ConfidenceIntervalEstimate for the Mean (o Unknown) E8.3 Computingthe ConfidenceIntervalEstimate for the Proportion E8.4 Computingthe SampleSizeNeeded for Estimatingthe Mean E8.5 Computingthe SampleSizeNeeded for Estimatingthe Proportion E8.6 Computingthe ConfidenceIntervalEstimate for the PopulationTotal E8.7 Computingthe ConfidenceIntervalEstimate for the Total Difference E8.8 ComputingFinitePopulationCorrection Factors

8.6 CONFIDENCE INTERVALESTIMATION AND ETHICALISSUES

In this chapter,you learn: r To constructandinterpretconfidenceintervalestimates for the meanandtheproportion I How to determinethe samplesizenecessary to developa confidenceintervalfor the meanor proportion I How to useconfidenceintervalestimates in auditins

284

CHAPTEREIGHT Confidence IntervalEstimation

Using Statistics@ Saxon Home Irprovement SaxonHome Improvementdistributeshome improvementsuppliesin northeastern you are UnitedStates.As a companyaccountant, for the accuracyof the integratedinventorymanagementand salesin mation system.You could review the contentsof eachand everyrecord checkthe accuracyof this system,but sucha detailedreviewwould time-consumingand costly.A better approachwould be to usestatisti inferencetechniquesto draw conclusionsaboutthe populationof recordsfrom a relativelvsmall samolecollecteddurins an audit.At end of eachmonth,you could selecta sampleof the salesinvoices determinethe followins: r The meandollaramountlistedon the salesinvoicesfor themonth. r The totaldollaramountlistedon the salesinvoicesfor themonth. r Any differencesbetweenthe dollar amountson the salesinvoices the amountsenteredinto the salesinformationsystem. r The frequencyof occurrenceof errorsthatviolatethe internalcontrolpolicy of the wa Sucherrorsincludemakinga shipmentwhenthereis no authorizedwarehouse removalslip,fai ureto includethe correctaccountnumber,andshippingtheincorrecthomeimprovement item. How accurateare the resultsfrom the samplesand how do you usethis information?Are the samplesizeslargeenoughto giveyou the informationyou need?

processofusing sampleresultsto drawconclusionsaboutthecharac. Q tatisticalinferenceis the population. Inferential tJteristics of a statisticsenablesyouto estimaleunknownpopulationcharacteristicssuchasa populationmeanor a populationproportion.Twotypesof estimatesareused point estimatesand intervalestimates. to estimatepopulationparameters: A point estimateis the valueof a singlesamplestatistic.A confidenceinterval estimateis a rangeof numbers,calledan interval,constructed Theconfidenceintervalis constructed aroundthepoint estimate. suchthatthe probabilitythatthepopulationparameteris locatedsomewhere within the intervalis known. Supposeyou would like to estimatethe meanGPA of all the studentsat your university. The meanGPAfor all the studentsis an unknownpopulationmean,denotedby p. Youselecta sampleof studentsand find that the samplemeanis 2.80.The samplemean,X = 2.80,is a point estimateof the populationmean,p. How accurateis 2.80?To answerthis question,you mustconstructa confidenceintervalestimate. In this chapter,you will learnhow to constructandinterpretconfidenceintervalestimates. Recallthat the samplemean, X, is a point estimateof the populationmean,p. However,the samplemeanvariesfrom sampleto samplebecause it depends on theitemsselected in thesample.By takinginto accounttheknownvariabilityfrom sampleto sample(seeSection7.4 onthe samplingdistributionof the mean),you can developthe intervalestimatefor the population mean.The intervalconstructed shouldhavea specifiedconfidenceof correctlyestimatingthe valueof the populationparameterp. In otherwords,thereis a specifiedconfidencethat p is somewherein the rangeof numbersdefinedby the interval. Supposethat after studyingthis chapter,you find that a 95ohconfidenceinterval for the meanGPA at your universityis(2.75 < p < 2.85).Youcan interpretthis intervalestimateby statingthat you are95o/oconfidentthat the meanGPA at your universityis between2.75 and 2.85.Thereis a 5o/ochancethatthe meanGPAis below2.75or above2.85. After learningabouttheconfidenceintervalfor themean,you will learnhow to developan interval estimatefor the populationproportion.Then you will learn how large a sampleto selectwhenconstructingconfidenceintervalsandhow to perform severalimportantestimation procedures accountants usewhenperformingaudits.

8.1: ConfidenceIntervalEstimationfor the Mean (o Known)

8.1

285

CONFIDENCEINTERVALESTIMATION FORTHE MEAN (o KNOWN) In Section 7 .4, you used the Central Limit Theorem and knowledge of the population distribution to determine the percentageof sample means that fall within certain distancesof the population mean. For instance,in the cereal-fill example used throughout Chapter 7 (seeExample 7 '6 on page 268), 95o/oof all samplemeansare between362.12and 373.88grams.This statement is basedon deductivereasoning.However, incluctivereasoning is what you need here. You need inductivereasoningbecause,in statisticalinference,you use the resultsofa single sample to draw conclusionsabout the population, not vice versa.Supposethat in the cerealfill example, you wish to estimate the unknown population mean, using the information from o n l y a s a m p l e .T h u s , r a t h e rt h a n t a k e p + ( 1 . 9 6 l ( o l 1 r n f ) t o f i n d t h e u p p e ra n d l o w e r l i m i t s around p, as in Section 7 .4, you substitute the sample mean, X- for the unknown and , use I X t ( 1 . 9 6 ) ( o l ( . ' l n ) a s a n i n t e r v a l t o e s t i m a t et h e u n k n o w n p . A l t h o u g h i n p r a c t i c e y o u select a single sample of size n and compute the mean, X, in order to understandthe full meaning of the interval estimate,you need to examine a hypotheticalset of all possible samplesofr values. Supposethat a sample of n : 25 boxes has a mean of 362.3 grams. The interval developed t o e s t i m a t ep i s 3 6 2 . 3I ( l . 9 6 X l 5 ) l ( ^ 1 2 5 ) ,o r 3 6 2 . 3+ 5 . g 9 .T h e e s t i r n a t e of p is

356.42(u(368.18 Becausethe populationmean,p (equalto 368),is includedwithin the interval,this sample resultsin a correctstatement aboutp (seeFigureg.l).

FIGURE 8.1 Confidenceinterval estimatesfor five differentsamples of n = 25 taken from a p o p u l a t i o nw h e r e p=368ando:15 362.12

368 I

362.3

368.18

Xt = 362.s 356.42

373.88

I =sos.s 375.38

JOJ.bZ

x. =:oo 360 Xa= 362.12 356.24

Xu= eZa.eA

362.12

398

To continuethis hypotheticalexample,supposethat for a differentsampleof n:25 the mean is 369.5.The interval developedfrom this samoleis

boxes,

+ (1.s6)(t 36s.s sybl2s) or 369.5+ 5.88.Theestimate is 363.62SuS375.38 Becausethe populationmean,p (equalto 368), is also includedwithin this interval.this statement about u is correct.

286

IntervalEstimation EIGHTConfidence CHAPTER Now, beforeyou begin to think that correctstatementsaboutp are alwaysmadeby developing a confidenceinterval estimate,supposea third hypotheticalsampleof n :25 boxes is selectedand the_ffrmplemean is equal to 360 grams.The interval developedhere is 360 t (1.96)(15)l(a|25), or 360+ 5.88.In thiscase,theestimate of p is 354.12(p(365.88 thepopulationmean,p, is not includedin theinterThis estimateis not aconectstatement because val developedfrom this sample(seeFigure8.1).Thus,for somesamples,the intervalestimateof p is correct,but for othersit is incorrect.In practice,only one sampleis selected,andbecausethe populationmeanis unknown,you cannotdeterminewhetherthe intervalestimateis correct. To resolvethis dilemmaof sometimeshavingan intervalthatprovidesa correctestimateand sometimeshaving an interval that providesan incorrect estimate,you needto determinethe proportionof samplesproducingintervalsthatresultin correctstatements aboutthepopulationmean, p. To do this, considertwo otherhypotheticalsamples:the casein which X = 362.12 gramsrnd X =373.88grams.IfX =362.12,theintervalis362.121(1.96X15)l(425), thecaseinwhich or362.12+ 5.88.Thisleadsto thefollowinsinterval: 356.24(uS368.00 Becausethe populationmeanof 368 is at the upperlimit of the interval,the statementis a correctone(seeFigure8.1). or 373.88+ 5.88.The When X = 373.88,the intervalis 373.88t (1.96X15)l(425), intervalfor the samplemeanis 368.00ap!319.76 In this case,becausethe populationmeanof 368 is includedat the lowerlimit of the interval, the statement is correct. In Figure 8.1, you seethat when the samplemeanfalls anywherebetween362.12and 373.88grams,the populationmeanis includedsomauherewithin the interval.In Example7.6 of the samplemeansfall between362.12and373.88grams. on page268,you foundthat 95o/o Therefore,95%oof al| samplesof n : 25 boxeshavesamplemeansthat includethe population meanwithin the interval developed. Because,in practice,you selectonly one sampleand p is unknown,you neverknow for surewhetheryour specificintervalincludesthe populationmean.However,if you takeall possible samplesof n and computetheir samplemeans,95o/oof the intervalswill includethe population mean,and only 5o/oof them will not. In otherwords,you have95% confidencethat the populationmeanis somewherein your interval. in this section.A sampleof n : 25 boxes Consideronceagain,the first samplediscussed hada samplemeanof 362.3grams.The intervalconstructed to estimatep is:

362.3 t( 1.e6Xls) /( J25) + 5.88 362.3 356.42S u( 368.18 The intervalfrom356.42to 368.18is referredto asa 95% confidenceinterval. that the meanamountof cerealin the populationof boxesis some"I am9ilo/oconfident and368.18grams." wherebetween356.42

Interval E,stirnation 8.1: Confidencc for theMean(o Known) 287 In somesituations,you might want a higher degreeof confidence(such as99{t/o) of including the populationmean within the interval. In other cases,you rnight acceptlessconfidence (such as 90%) of correctlyestirnatingthe populationmean. In general,the level of confidence is symbolizedby ( I - o) x 100%o, where u is the proportion in the tails of the distributionthat is outsidethe confidenceinterval.The proportionin the uppertail of the distributionis ul2, and, the proportion in the lower tail of the distributionis ul2.You use Equation(8.1) to constructa ( I - ct) x 100% confidenceintervalestimateof the mean with o known.

CONFIDENCE INTERVAL FORTHEMEAN(o KNOWN) o x t z --r= \l n

F-zf .u
(8.1)

where Z: the value correspondingto a cumulative area of 1 - al2 from the standardized normal distribution (that is, an upper-tail probability of ul2).

The value ofZ neededfor constructinga confidenceinterval is calledthe critical value for the distribution.95o/o confidencecorrespondsto an o, value of 0.05.The critical Zvalue correspondingto a cumulativeareaof 0.9750 is L96 becausethere is 0.025 in the uppertail of the distributionand the cumulativearealessthanZ: 1.96is 0.975. There is a differentcritical value for eachlevel ofconfidence. I - cr.A level ofconfidence of 95ohleadsto a Z value of | .96 (seeFigure 8.2). 99% confidencecorrespondsto an cr value of 0.01.The Z value is approximately2.58 becausethe upper-tailareais 0.005 and the cumulat i v e a r e al e s st h a nZ : 2 . 5 8 i s 0 . 9 9 5( s e eF i c u r e8 . 3 ) .

FIGURE 8.2 N o r m a cl u r v e f o r d e t e r m i n i n gt h e Z v a l u en e e d e d for 95% confidence 'p, -1.96

0

+1.96

X Z

FIGURE 8.3 N o r m a cl u r v ef o r d e t e r m i n r ntgh e Z v a l u en e e d e d for99% confidence

x + 2 . 5 8Z

288

CHAPTEREIGHT ConfidencelntervalEstimation

Now that variouslevelsof confidencehavebeenconsideredwhv not makethe confi level ascloseto I 00% aspossible?Beforedoing so,you needto realizethat any increase in level of confidenceis achievedonly by widening (and making lessprecise)the confi interval.There is no "free lunch" here.You would havemore confidencethat the meanis within a broaderrangeof values;however,this might make the interpretationof confidenceinterval lessuseful.The trade-offbetweenthe width of the confidenceinterval the level ofconfidence is discussedin greaterdepthin the contextofdetermining the sizein Section8.4. Example8.1 illustratesthe applicationof the confidenceinterval

EXAMPLE 8.1

ESTIMATING THEMEANPAPER LENGTHWITH95% CONFIDENCE process A papermanufacturer hasa production thatoperates continuously throughout an productionshift. The paper is expectedto havea meanlength of l1 inches,and the deviationof the length is 0.02 inch. At periodic intervals,a sampleis selectedto whetherthe meanpaperlengthis still equalto 1l inchesor whethersomethinghasgone in the productionprocessto changethe lengthofthe paperproduced.You selecta random ple of 100 sheets,and the meanpaperlengthis 10.998inches.Constructa95%o interval estimatefor the populationmeanpaperlength. SOLUTION UsingEquation(8.1)on page287, with Z: |.96 for 95o/o confidence,

Xxz*=ro.essr(1.e6)g 4n

{100 = 10.998 + 0.00392 10.99408 Sp ( 11.00192

Thus, with 95% confidence,you concludethat the populationmeanis between10.99408and 11.00192inches.Becausethe intervalincludes11, the value indicatingthat the production processis working properly,you haveno reasonto believethat anythingis wrong with the ductionprocess.

To seethe effectof using a99o/oconfidenceinterval,examineExample8.2.

EXAMPLE 8.2

ESTIMATING THE MEAN PAPERLENGTHWITH99% CONFIDENCE meanpaperlength. intervalestimate for thepopulation Construct a99%o confidence (8.I ) onpage287,withZ : 2.58for 99o/o confidence, SOLUTIONUsingEquation

Xtz*=lo.eest(2.58)g 4n

{100 = 1 0 . 9 91 80 . 0 0 5 1 6

(p ( 11.00316 10.99284 Once again,becauseI I is includedwithin this wider interval,you haveno reasonto beli that anythingis wrong with the productionprocess.

8.1: Confidence Interval Estimation for the Mean (o Known)

289

As discussed in section7.4,the samplingdistributionof t is normallydistributedif the populationofXis a normal distribution.And" if thepopulationofXis not a normaldistribution, the CentralLimit Theoremensuresthat X is normally distributedwhen r is large.However, when dealingwith a small samplesizeand a populationof X that is not a normal distribution, the samplingdistributionof X is not normally distributedand thereforethe confidenceinterval discussedin this sectionis inappropriate.In practice,however,aslong asthe samplesizeis large enoughand the populationis not very skewed,you can use the confidenceinterval defined in Equation8.1 to estimatethe populationmeanwhen o is known. To assessthe assumptionof normality,you can evaluatethe shapeof the sampledataby using a histogram, stem-and-leafdisplay,box-and-whiskerplot, or normal probabilityplot.

Learningthe Basics 8.1 If X = 85, o : 8, andn:64, construct a 95% confidenceinterval estimateof the population mean,p.

c. Must you assumethat the populationamountof paint per can is normally distributedhere?Explain. d. Constructa 95ohconfidenceinterval estimate.How doesthis changeyour answerto (b)?

x =t25, o: 24. and n:ru,.:""::T.:,: M ?:,r, ffi intervalestimateof -- the ---- populalAsslsTl99% confidence r-r"'uon mean'p' 8.3 A marketresearcherstatesthat shehas 95oh confidencethat the mean monthly sales of a product are between$170,000and $200,000.Explainthe meaningof thisstatement. 8.4 Why is it not possiblein Example8.1 on page288 to have100%confidence? Explain. 8.5 Fromthe resultsof Example8.1 on page288 regarding paperproduction,is it true that 95%oof the sample means will fall between10.99408and 11.00192inches? Explain. 8.6 Is it true in Example8.1 on page288 that you do not knowfor sure whetherthe populationmean is between 10.99408 and 11.00192 inches? Explain.

Applyingthe Concepts 8.7 The managerof a paint supply store wants to estimatethe actualamountof paint contained in l-gallon cans purchasedfrom a nationally knownmanufacturer.The manufacturer'sspecifications statethat the standarddeviationof the amountof paint is to 0.02 gallon. A random sample of 50 cans is selected, andthe samplemeanamountof paint per l-gallon canis 0.995sallon. a. Constructa99o/oconfidenceintervalestimateof thepopulationmeanamountof paint includedin a l-gallon can. On the basisof theseresults,do you think the manager hasa right to complainto the manufacturer?Why?

i;1"111.1H",'J.Til:11,[Tll"jJl

deviationis | z seul shipmentof lieht bulbs.The standard EU 100 hours.A-random sampleof 64 light bulbs indicateda samplemeanlife of 350 hours. a. Constructa 95o/oconfidenceinterval estimateof the populationmeanlife of light bulbs in this shipment. b. Do you think that the manufacturerhasthe right to state that the light bulbs last an averageof 400 hours? Explain. c. Must you assumethat the populationof light bulb life is normally distributed?Explain. d. Supposethat the standarddeviationchangesto 80 hours. What areyour answersin (a) and (b)?

8.9 The inspectiondivision of the Lee County Weightsand MeasuresDepartmentwantsto estimate the actual amount of soft drink in 2-Iiter bottles at the local bottling plant of a large nationally known soft-drink company. The bottling plant has informed the inspectiondivision that the populationstandard deviationfor 2-liter bottles is 0.05 liter. A random sampleof 1002-literbottlesat this bottlingplantindicates a samplemeanof 1.99liters. a. Constructa 95ohconfidenceinterval estimateof the populationmeanamountof soft drink in eachbottle. b. Must you assumethat the populationof soft-drinkfill is normally distributed?Explain. c. Explainwhy a valueof 2.02liters for a singlebottleis not unusual,eventhough it is outsidethe confidence interval you calculated. d. Supposethat the samplemean is 1.97 liters. What is your answerto (a)?

290

CHAPTER EIGHT Confidencelnterval Estimarron

8.2

CONFIDENCEINTERVALESTIMATION FORTHE MEAN (o UNKNOWN) Just as the mean of the population, p, is usually unknown, you rarely know the actual standard deviation of the population,o. Therefore,you often need to constructa confidenceinterval estimateof p, using only the sample statistics X and S.

Student'st Distribution At the beginning of the twentieth century, William S. Gosset, a statistician for Guinness Breweries in Ireland (see reference3) wanted to make inferencesabout the mean when o was unknown. BecauseGuinnessemployeeswere not permitted to publish researchwork under their own names,Gossetadoptedthe pseudonym"Student." The distribution that he developed is known as Student's r distribution and is commonly referred to as the t distribution. If the random variableX is normally distributed"then the following statistic has a t distribution with n - I degrees of freedom: Y

-tt

s ,FN This expressionhas the same form as the Z statistic in Equatron(7 .4) on page 266, except that is discussedfurther on S is used to estimatethe unknown o. The concept of degreeso.f'.freedom pages 291-292.

Properties of the t Distribution In appearance,the / distribution is very similar to the standardizednormal distribution. Both distribrrtion.s are bell shapecl. Floweve-r, the t tlistribLrtiorr hits l77orcarea irt the tirils arrd less ln rfie ccnfcr lh:rn clocs thc sfandardized norntal distribufion (see Figtrre 8.4). Because Jis used to estirnate the unknown o. the values of 1 are more variable than those for Z.

FIGURE 8.4 Standardized n o r m a ld i s t r i b u t i o n a n d t d i s t r i b u t i o nf o r 5 degrees of freedom

-

S t a n d a r d i z endo r m a ld i s t r i b u t i o n

-

f distribution for 5 degrees of freedom

The degreesof freedom, n - 1, aredirectly related to the sample size,n. As the samplesize and degreesof freedom increase,S becomesa better estimateof o, and the r distribution gradually approachesthe standardizednormal distribution, until the two are virtually identical. With a sample size of about 120 or more, S estimateso precisely enough that there is little differencebetweenthe t and Z distributions. As statedearlier, the t distribution assumesthat the random variableX is normally distributed. In practice, howeveq as long as the sample size is large enough and the population is not very skewed, you can use the I distribution to estimate the population mean when o is unknown. When dealing with a small sample size and a skewedpopulation distribution,the validity of the confidence interval is a concern.To assessthe assumptionof normality, you can

8.2: Confidence IntervalE,stimation for theMean(o Unknown) 291 evaluate the shape of the sample data by using a histogram, stem-and-leafdisplay, box-andwhisker plot, or normal probabilityplot. You find the critical values of r for the appropriatedegreesof freedom from the table of the r distribution (seeTable E.3). The columns of the table representthe area in the upper tail of the r distribution. The rows of the table representthe degreesof freedom.The cells of the table representthe particular / value for each specific degreeof freedom. For example, with 99 degrees of freedom, if you want 95% confidence, you find the appropriatevalue of l, as shown in Table 8. L The 95% confidencelevel meansthat2.5ohof the values(an areaof 0.025)are in eachtail of the distribution. Looking in the column for an upper-tail area of 0.025 and in the row corresponding to 99 degreesof freedom gives you a critical value for t of 1.9842.Becauser is a symmetrical distribution with a mean of 0, if the upper-tail value is +1.9842, the value for the lower-tail area (lower 0.025) is - l .9842.A I value of -1.9842 means that the probability that t is lessthan -1.9842 is 0.025,or2.5oh(seeFigure 8.5).Note that for a 95% confidenceinterval, you will always use an upper-tail area of 0.025. Similarly, for a99o/oconfidence interval, use 0 . 0 0 5 ,f o r 9 8 % ou s e0 . 0 1 ,9 0 % u s e0 . 0 5 ,a n d 8 0 % u s e0 . 1 0 .

Upper-TailAreas

T A B L E8 . 1 Determining the Critical Degreesof Freedom Valuefrom the t Table I foran Areaof 0.025 2 in EachTailwith 99 J Degrees of Freedom 4 5

?{

.05

.01

1.0000 3.0777 6 . 3 1 3 8 1 2 . 7 0 . 8 1 6 5 1 . 8 8 s 6 2.9200 4 0.1649 1 . 6 3 1 7 2.3s34 3 . 824 2 . 164 0.7407 1.s332 2 . 1 l38 0.1267 1 . 4 1 s 9 2 . 0 1 s 0 2 706

96 97 98

0.6771 0.6770 0.6770

100

0.6770

Source; E.ytroctel ]inn

.10

1.2904 t.2903 t.2902 1.9840

.005

31.8207 6.9646 4.s407 3.1469 3.3.649

63.6574 9.9248 5.8409 4.604r 4.0.322

2.3658 2.3654 2.36s0 2.3646 2.3642

2.6280 2.6275 2.6269 2.6264 2.6259

Tuhle E.3

FIGURE 8.5 t d i s t r i b u t i o nw i t h 9 9 degreesof freedom

+1.9842

The Concept of Degrees of Freedom rt

In Chapter3 you learnedthat the numeratorof the samplevariance,52 [seeEquation(3.9) on page 107],requiresthe computationof

S

e n

S

-.)

L(x,-xr

292

CHAPTER EIGHT ConfidenceIntervalEstimation

In order to compute 52, you first need to know X . Therefore, only r - I of the samplevalues a are free to vary. This means that you have n - 1 degreesof freedom. For example, suppose sample of five values has a mean of 20. How many values do you need to know beforeyoucan determine the remainder of the values?The fact that n: 5 and X = 20 also tells you that il

=too

Tr

.L"t

because n

\v i=l

;

n Thus, when you know four of the values, the fifth one is not free to vary becausethe sum must add to 100. For example, if four of the values are 18, 24, 19, and 16, the fifth value must be 23 so that the sum equals100.

The Confidence lnterval Statement Equation (8.2) defines the (l - a) x 100% confidence interval estimate for the mean with o unknown.

INTERVAL FORTHEMEAN(o UNKNOWN) CONFIDENCE V +t..,

s

G

x-tn-r#.u <X+,,.#

(8.2)

where /,_, is the critical value of the r distribution, with n - I degreesof freedom for an areaof al2 in the upper tail. To illustrate the application of the confidence interval estimate for the mean when the standarddeviation, o, is unknown, recall the Saxon Home Improvement Company Using Statisticsscenariopresentedon page 284.You wanted to estimatethe mean dollar amount listed on the salesinvoices for the month. You select a sample of 100 sales invoices from the population of salesinvoices during the month, and the sample mean of the 100 salesinvoicesrs $110.27,with a sample standarddeviationof $28.95.For95o/oconfidence,the critical value from the / distribution(as shown in Table 8.1) is L9842. Using Equation(8.2), it. ^ lln

'S I

T 1n

28.95 = | t0.2+ 7 ( 1 . s 8 4) 2 {100 = 1 1 0 . 2+7 5 . 7 4 $ 1 0 4 . 5( 3u < $ 1 1 6 . 0 1 A Microsoft Excel worksheet for these data is presentedin Figure 8.6.

8.2: ConfidenceIntervalEstimationfor the Mean (o Unknown)

FIGURE8.6 MicrosoftExcel worksheet to compute a confidence interval estimate for the mean s a l e si n v o i c ea m o u n t for the SaxonHome lmprovement Company

293

f-A'*--q

j 1 rEstimate for the MeanSalesInvoiceI

m1--_

j- Q

Data

;4ffi 5ft 6 t S a m n l eS i z e 7 ConfidenceLevel B 9 l n t e r m e d i aCt ae l c u l a t i o n s 1 0 S t a n d a rEdn o ro f t h eM e a n s of Freedrlm

12 | Value SeeSectlon EB.2to create thrs.

:84/SaRT{86} =85-1 =Trilv(l - 87,811) 57443 = 8 1 2" 8 1 0

ConfidenceInterval :85 - 813 :85 * 813

Thus, with 95% confidence,you concludethat the mean amount of all the salesinvoicesis between$ 104.53and $ I 16.01. The 95% confidencelevel indicatesthat if you selectedall possible samplesof 100 (something that is never done in practice), 95o/oof the intervals developed would include the populationmean somewherewithin the interval.The validity of this confidenceinterval estimatedependson the assumptionof normality for the distributionof the amount of the salesinvoices.With a sampleof 100,the normality assumptionis not overly restrictive,and the useof the t distribution is likely appropriate.Example 8.3 further illustrateshow you construct the confidenceinterval for a mean when the population standarddeviation is unklown.

E X A M P L E8 . 3

ESTIMATING THEMEANFORCEREOUIRED TO BREAKELECTRIC INSULATORS A manufacturingcompanyproduceselectric insulators.If the insulatorsbreak when in use,a short circuit is likely. To test the strength ofthe insulators,you carry out destructivetesting to determine how rnuch,lbrzc is required to break the insulators.You measureforce by observing how many poundsare appliedto the insulatorbefore it breaks.Table 8.2 lists 30 valuesfrom this experiment, which are located in the file [![!fQ Construct a 95o/oconfidence interval estimatefor the population mean force required to break the insulator.

T A B L E8 . 2 F o r c e( i n P o u n d s ) R e q u i r e dt o B r e a k theInsulator

1,870 1,128 1,656 I,610 1,634 t,784 1,522 1,696 1,592 t,662 1 , 8 6 6 1 , 7 6 4 1 , 7 3 4 1 . 6 6 2 1 , 7 3 4 1 , 7 7 4 1 , 5 5 0 | . t 5 6 t . 76 2 r , 8 6 6 1 , 8 2 0 t , 7 4 4 1 , 7 8 8 1 , 6 8 8 1 , 8 1 0 1 , 7 5 2 1 , 6 8 0 I. 8 1 0 t . 6 5 2 t , 7 3 6 SOLUTION Figure8.7 showsthatthe samplemeanis F = 1,723.4poundsandthe sample standard deviationis S: 89.55pounds.UsingEquation(8.2)on page292to construct theconfidenceinterval,you needto determinethe criticalvaluefrom the / tablefor an areaof 0.025in

FIGURE8.7 MicrosoftExcel confidence interval estimate for the mean amountof force requiredto break electricinsulators

q_ I i

___-_,,___ L

Intermediale Calculations

llnlervalHalfWidlh

SeeSection EB.2to create this.

=BIISORT{85) -86-l -Tlilv(l - 87, 811) -812 " 810 -85 - 813 -85 . 813

294

CHAPTEREIGHT Confidence Intcrval Estin.ration each tail, with 29 degreesof freedom.From Table 8.3, you seethat t),\:2.0452. Thus, usi X = 1 , 1 2 3 . 4S, : 8 9 . 5 5 n , - 3 0 ,a n dt , o : 2 . 0 4 5 2 ,

x + t, t = |

s

G

89'55 -t)7 4 -+ 't-) "o r-s ") r'

'60

= 1 . 1 2 3 . 4+ 3 3 . 4 4 1,689.96{uS1.756.84

You concludewith95% confidencethat the meanbreakingforce requiredfor the populationr insulatorsis between1,689.96and 1,756.84pounds.The validity of this confidenceintervales mate dependson the assumptionthat the fbrce requiredis norrnally distributed.Remember,hou eveq that you can slightly relax this assumptionfor large samplesizes.Thus, with a sampleof 3( you can usethe r distributionevenif the amountof force requiredis only slightly left skewed.Fror the notmal probabilityplot displayedin Figure8.8 or the box-and-whiskerplot displayedin Figur 8.9,the amountof force requiredappearsslightly left skewed.Thus,the t distributionis appropria for thesedata. Force Requiredto Break ElectricalInsulators

FIGURE 8.8

2000 T

M i c r o s o f tE x c e ln o r m a l p r o b a b i l i t yp l o t f o r the amount of force

1800

ranr riraA tn

1400

hrorlz

e l e c t r i ci n s u l a t o r s

1600

1200

! rooo 800

See Sectlon E6.2 to create this.

600 400 200 ---#

-?.5

-1.5

.1

.0.5

0

0.5

1

Z Value

FIGURE8.9

Force Required to Break Electrical Insulators

MicrosoftExcelboxand-whisker plot for the amountof force requiredto break electricinsulators

See Section E3.4 to create this.

1700

1750

1800

2

2.5

8.2: ConfidenceIntervalEstimationfor theMean(o Unknown) 295

learningthe Basics 8.10 Determinethe criticalvalueof / in eachof the following circumstances: o " I - 0 : 0 . 9 5 .n : l 0 b . I - o : 0 . 9 9n. : l 0 c.l-0:0.95,n:32 d . I - c r : 0 . 9 5n, : 6 5 e.l-0=0.90. n:16 8.11 If X = 75,S :24, andn:36. andassuming that the populationis normally distributed, constructa 95% confidenceinterval estimateof thepopulationmean,p. 8.',2 lf X = 50,,S:15,andn:16, andassuming that the populationis normally distributed, constructa 99Yoconfidenceinterval estimateof populationmean,p. 8.13 Constructa95%oconfidenceintervalestimatefor the iopulationmean,basedon eachof the followine setsof assumingthat the populationis normally distributed: S e t1

l, l, l, 1,8,8,8,8

S e t2

1,2,3,4,5,6,7,8

lain why these data sets have different confidence eventhoughthey havethe samemeanandrange. 14 Constructa 95o/oconfidenceintervalfor the popumean,basedon the numbersl, 2, 3, 4, 5, 6, and20. the number20 to 7 and recalculatethe confiinterval.Using theseresults,describethe effectof outlier(that is, an extremevalue) on the confidence

the Concepts 15 A stationerystorewantsto estimatethe meanretail of greetingcardsthat it hasin its inventory.A random of 100 greetingcardsindicatesa meanvalue of 55anda standarddeviationof $0.44. Assuminga normal distribution,constructa95o/oconfidenceintervalestimateof the meanvalueof all greeting cardsin the store'sinventory. Suppose therewere2,500greetingcardsin the store's inventory.How arethe resultsin (a) usefulin assisting the store owner to estimate the total value of her inventory? 8.16 SouthsideHospital in Bay Shore,New York,commonlyconductsstressteststo studythe heartmuscleafter a personhas a heart attack. of the diagnosticimaging departmentconducted lity improvementproject to try to reduce the turn-

aroundtime for stresstests.Turnaroundtime is defined as the time from when the test is orderedto when the radiologist signsoff on the test results.Initially, the meanturnaroundtime for a strbsstestwas68 hours.After incorporating changes into the stress-testprocess, the quality improvementteam collecteda sampleof 50 turnaround times. In this sample,the meanturnaroundtime was 32 hours,with a standarddeviationof 9 hours(Extractedfrom E. Godin, D. Raven,C. Sweetapple,and F, R. Del Guidice, "Faster Test Results," Quality Progress, January 2004, 37(l), pp. 33-39). a. Constructa95o/oconfidenceintervalfor the population meanturnaroundtime. b. Interpretthe intervalconstructedin (a). c. Do you think the quality improvementprojectwasa success?Explain. 8.17 The U.S. Departmentof Transportation requirestire manufacturersto providetire performanceinformation on the sidewallof the tire to betterinform prospectivecustomerswhenmakingpurchasing decisions.One very important measureof tire performanceis the treadwear index, which indicatesthe tire,s resistanceto treadwearcomparedwith a tire gradedwith a baseof 100.This meansthat a tire with a grade of 200 shouldlast twice as long, on average,as a tire gradedwith a baseof 100.A consumerorganizationwantsto estimate the actualtreadwearindex of a brandnameof tires graded 200 that areproducedby a certainmanufacturer. A random sampleof n : l8 indicatesa samplemeantreadwearindex of 195.3and a samplestandarddeviationof 21.4. a. Assumingthat the populationof treadwear indicesis normallydistributed,constructa95% confidenceinterval estimateof the populationmeantreadwear index for tires produced by this manufacturerunder this brand name. b. Do you think that the consumerorganizationshould accusethe manufacturerof producingtires that do not meetthe performanceinformationprovidedon the sidewall of the tire? Explain. c. Explain why an observedtreadwear index of 2 l0 for a particular tire is not unusual,eventhough it is outside the confidenceinterval developedin (a). 8.18 The following data (storedin the file !!@@ representthe bouncedcheck fee, in dollars, chargedby a sampleof 23 banksfor direct-depositcustomerswho maintain a $100balance: 26 28 20 20 21 22 2s 2s 18 25 15 20 18 20 25 25 22 30 30 30 15 20 29 Source: Extractedfrom "The New Face of Banking," Consumer Reports,June 2000.

296

EIGHTConfidence CHAPTER Interval Estimation

a. Constructa 95ohconfidenceinterval for the population meanbouncedcheckfee. b. Interpretthe intervalconstructedin (a) 8.19 The datain the file @ft[! representthe total fat, in gramsper serving,for a sampleof 20 chickensandwichesfrom fast-foodchains.The dataareasfollows: 7 8 4 5 1620202419 30 23 30 25 t9 29 29 30 30 40 s6 "FastFood:AddingHealthto theMenu," Extractedfrom Source: Consumer Reports, September 2004,pp.28-31. a. Constructa95o/oconfidenceintervalfor the population meantotal fat, in gramsper serving. b. Interpretthe intervalconstructed in (a). 8.20 Oneof the majormeasures of the qualityof service providedby any organizationis the speedwith which it responds to customercomplaints.A largefamily-helddepartment storeselling furniture and flooring, including carpet, hadundergone a majorexpansionin thepastseveralyears.In particular,the flooring departmenthad expandedfrom 2 installationcrewsto an installationsupervisoqa measurer, and l5 installationcrews.Lastyear,therewere50 complaints concemingcarpetinstallation.The following data,in the file represent thenumberof daysbetweenthereceipt @EE, of a complaintandtheresolutionof the complaint: 54 ll t2

5 35r37 3l

27 152

2 t23 81 74 21

t9 t26 ll0 110 2961359431265 4 16s 32 29 2 8 2 9 2 6 2 s

13105274

114

13

52 30 22 36 26 20 23

33 68 a. Constructa 95o/oconfidenceinterval estimateof the meannumberof daysbetweenthe receiptof a complaint andtheresolutionof the complaint. b. What assumptionmustyou makeaboutthe population distributionin (a)? c. Do you think that the assumptionmadein (b) is seriouslyviolated?Explain. d. What effect might your conclusionin (c) haveon the validityof theresultsin (a)? 8.21 In New York State,savingsbanksare permittedto sell a form of life insurancecalledsavingsbank life insurance(SBLI). The approvalprocessconsistsof underwriting, which includesa reviewof the application,a medical informationbureaucheck,possiblerequestsfor additional medicalinformationandmedicalexams,anda policy com-

8.3

pilation stagein which the policy pagesare generatedand sentto thebank for delivery.The ability to deliverapproved policiesto customersin a timely manneris criticalto the profitabilityof this serviceto the bank.During a periodof one month,a randomsampleof 27 approvedpolicieswas selected,and the total processingtime, in days,wasas shownbelowandstoredin the file@, t3 19 t6 64 28 28 3t 90 60 56 3l 45 48 t7 t7 lt

s6 22 l8

91 92 63 50 sl 69 16 17

a. Constructa 95ohconfidenceinterval estimateof the meanprocessing time. b. What assumptionmust you make about the population distributionin (a)? c. Do you think that the assumptionmadein (b) is seriouslyviolated?Explain. 8.22 The data in the file [l$$@f!represent the battery life (in shots)for three-pixeldigitalcameras: 300 180 85 170 380 460 260 35 380 t20 rr0 240 Source: Extractedfrom "Cameras: More Featuresin the Mix," ConsumerReports,July 2005,pp. 14 18.

a. Constructa 95o/oconfidenceinterval for the population meanbatterylife (in shots). b. What assumptiondo you needto makeaboutthe populationofinterestto constructthe intervalin (a)? c. Giventhe datapresented, do you think the assumption neededin (a) is valid?Explain. 8.23 One operationof a mill is to cut piecesof steelinto partsthat are usedlater in the frame for front seatsin an automobile.The steelis cut with a diamondsawandrequires the resultingpartsto be within +0.005inch of the length specifiedby the automobilecompany.The measurement reportedfrom a sampleof 100steelparts(andstoredin the file @ is the difference,in inches,betweenthe actual lengthof the steelpart, asmeasuredby a lasermeasurement device,andthe specifiedlengthofthe steelpart. For example, the first observation,-0.002, represents a steelpartthat is 0.002inch shorterthanthe specifiedlength. a. Constructa 95% confidenceintervalestimateof the meandifferencebetweenthe actuallengthof the steel part andthe specifiedlengthofthe steelpart. b. What assumptionmust you make aboutthe population distributionin (a)? c. Do you think that the assumptionmadein (b) is seriouslyviolated?Explain. d. Comparethe conclusionsreachedin (a) with thoseof Problem2.23on page53.

CONFIDENCEINTERVALESTIMATIONFORTHE PROPORTION This sectionextendsthe conceptofthe confidenceintervalto categoricaldata.Hereyou are with estimatingtheproportionof itemsin a populationhavinga certaincharacterisconcerned tic of interest.The unknownpopulationproportionis represented by the Greeklettern. The

Interval Estirration fbr thePronortion 297 8.3: Confidence point estirnatefor n is the sampleproportion,p - Xln, where ir is the sarrple size and X is the nurnberof iter.t.rs in the sarnplehaving the characteristicof interest.E,quation(8.3)defines the confidenceinterval estir.nate for the populationproportion.

coNFTDENCE TNTERVAL ESTTMATE FOXTXSjROPORTTON ,+ 7.lP(t- Pt IT

or p-Z

p:

p(t - p)
Sampleproportion:

X

=

p(I - p) n

(8.3)

Number of items havins the characteristic

; n : populationproportion

Sr_pl. ,t^

Z : critical value from the standardizednorrnal distribution r = s a m p l es i z e assumingthat both X and n - X are greater than 5 You can usethe confidenceintervalestimateof the proportiondefinedin Equation(8.3) to estimatethe proportion of salesinvoicesthat contain errors (seethe Using Statisticsscenario o n p a g e2 8 4 ) .S u p p o s et h a t i n a s a m p l eo f 1 0 0s a l e si n v o i c e s .l 0 c o n t a i ne r r o r s T . h u s .f o r t h e s e d a t a . p: X l n : 1 0 / 1 0 0 : 0 . 1 0 . U s i n gE q u a t i o n( 8 . 3 )a n d Z - 1 . 9 6f o r 9 5 ' %c o n f i d c n c e ,

rto an

n*7

il

CS

=0.10t(1.96)

ith )nt he ral 3nt mhat the ,eel

eri:of

N

p(t- p)

( 0 r. 0 x 0 . e 0 ) 100

= 0 . 1 0t ( 1 . 9 6 ) ( 0 . 0 3 ) : 0 .l 0 + 0 . 0 5 8 8 0.0412
FIGURE 8.10 Microsoft Excel worksheet to construct ac o n f i d e n cien t e r v a l estimate for the proportion of sales invoices that contain errors

1

2 3 , 5 5 7 I I 10 11 S l a n d a r dE n o r o f t h e lntewalHalfWidth

=85/84 =NORMSTNV(I - 86)2) =sQRr(Be'(1- Bey84l =ABS(810'811)

IJ

, are srisThe

1i 15 15

*89 - 812 =89'812

298

IntervalEstimation CHAPTEREIGHT Confidence

Example8.4 illustratesanotherapplicationof a confidenceintervalestimatefor the proportion.

E X A M P L E8 . 4

PRINTED ESTIMATINGTHE PROPORTIONOF NONCONFORMINGNEWSPAPERS printedthathavea nonconwantsto estimatethe proportionof newspapers A largenewspaper pages,or duplicate page missing setup. improper ruboff, forming attribute,suchas excessive printedduring newspapers the from all is selected pages.A randomsampleof 200 newspapers Constructand nonconformance. a singleday.For this sampleof 200,35 containsometype of printed the day during interfret a 90o/oconfidenceinterval for the proportion of newspapers thathavea nonconformingattribute. SOLUTION UsingEquation(8.3),

'D

?5 - ::- = 0.175. and with a90oh level of confidenceZ =

200

8.

VC:

far (E of

Jot a.

b.

8.2 p!z

p(t - p) n

= 0 . 1 7 t5 ( .r ..6.4- s. [)o { -.;r: 7 s ) ( 0 . 8 2 s )

sm the Ap use a.i

I

= 0.175r (1.645X0.0269)

l

t75 !0.0442 0 . 1 3 0 83 n 3 0 . 2 1 9 2

I

b.i

I

You concludewith 90% confidencethat between13.08%and2l.92ohof the newspapers printedon that dayhavesometype of nonconformance. you canusethe normaldistributionto approx' Equation(8.3)containsa Z statisticbecause sizeis sufficientlylarge.In Example8.4,the the sample when distribution imatethe binomial for the populationproportion approximation provides excellent an confidenceintervalusingZ you not havea sufficiently large if do However, becauseboth X andn X are greaterthan 5. (8.3)(seereferences Equation than samplesize,you shouldusethebinomialdistributionrather 1.2, and6). The exactconfidenceintervalsfor varioussamplesizesand proportionsof suc' 2). cesseshavebeentabulatedby FisherandYates(reference

I I

c..

8.2 is: nol

wi hei

ph sai ag CI a.

b.

Learningthe Basics

Applying the Concepts

8.24 lf n : 200 andX: 50, constrwt a 95o/o @E 8.26 The telephonecompanywantsto estlmate @ thatwouldpurchase l A s s r sITtheproportionof households A S S T S Tconfidenceinterval estimate of the population I

I

proportion.

ffi

8.25 lf n = 400 andX : 25, constructa 99oh confidenceintervalestimateof the population proportion.

ffi

an additionaltelephoneline if it weremadeavailreducedinstallationcost.A ableat a substantially The results is selected. households 500 of sample random purchase the would households the indicatethat 135 of cost. installation additionaltelephoneline at a reduced

8. M

8.4:Determining Sample Size 299 a. Constructa 99o/oconfidenceinterval estimateof the populationproportionof households that would purchasethe additionaltelephoneline. b. How would the managerin chargeof promotionalprogramsconcerningresidentialcustomersusethe results in (a)?

theInternetat work.423 saidtheyusedit within limits,and 183saidthattheydid not usethe Internetat work. a. Constructa95ohconfidenceintervalfor the proportion of all workerswho usedthe Internetwithin limits. b. Constructa 95o/oconfidenceintervalfor the proportion of all workerswho did not usethe Internetat work.

8.27 Accordingto the Centerfor Work-LifePolicy,a survey of 500 highly educatedwomen who left careersfor family reasonsfound that 660/owantedto return to work (Extractedfrom A. M. Chakerand H. Stout,'After Years Off, WomenStruggleto Revive Careers,"The Wall Street Journal,May 6, 2004,p. Al). a. Constructa95ohconfidenceintervalfor the population proportionof highly educatedwomenwho left careers for family reasonswho want to returnto work. b. Interpretthe intervalin (a).

8.31 When do Americansdecidewhat to make for dinner?An online survey(N. Hellmich, 'AmericansGo for the Quick Fix for Dinner,"USAToday,February14,2005, p. 1B) indicatedthatT4o/o of Americansdecidedeitherat the last minute or that day.Supposethat the surveywas basedon 500 respondents. a. Constructa 95o/oconfidenceintervalfor the proportion of Americanswho decidedwhat to make for dinner eitherat the lastminuteor that day. b. Constructa 99o/oconfidenceintervalfor the proportion of Americanswho decidedwhat to make for dinner eitherat the last minuteor that day. c. Which intervalis wider?Explainwhy this is true.

8.28 In a surveyconductedfor AmericanExpress,21o/oof smallbusinessownersindicatedthattheynevercheckin with theoffice when on vacation("Snapshots," usatoday.com, April 18,2006.).The articledid not disclosethe samplesize usedin the study. a. Suppose thatthesurveywasbasedon 500smallbusiness owners.Constructa 95o/oconfidenceintervalestimate for the populationproportionof smallbusinessowners who nevercheckin with the office whenon vacation. that the surveywasbasedon 1,000smallbusib. Suppose nessowners.Constructa 95o/oconfidenceintervalestimatefor thepopulationproportionof smallbusinessownerswho nevercheckin with the office whenon vacation. c. Discussthe effect of samplesize on the confidence intervalestimate.

n e 's

I

in the UnitedStates 8.29 The numberof olderconsumers is growing, and they are becomingan even bigger economic force. Many feel overwhelmedwhen confronted bankingservices, with the task of selectinginvestments, healthcareproviders,or phoneserviceproviders.A telephonesurveyof 1,900older consumersfound that 27oh saidtheydidn't haveenoughtime to be goodmoneymanagers(Extractedfrom "SeniorsConfusedby Financial Choices-Study,"msnbc.com,May 6, 2004). a. Constructa95o/oconfidenceintervalfor the population proportionof older consumerswho don't think they haveenoughtime to be goodmoneymanagers. b. Interpretthe intervalin (a). 8.30 A survey of 705 workers(USA TodaySnapshots, March21, 2006,p. lB) wereaskedhow muchthey used

8.32 In a surveyof 894 respondents with salariesbelow $100,000per year,367 indicatedthat the primary reason job responsibilities for stayingon theirjob was interesting ("What Is the PrimaryReasonfor Stayingon Your Job?" USATodaySnapshots, October5,2005,p. lB). 95o/o confidence intervalfor the proportion a. Constructa of all workerswhoseprimaryreasonfor stayingon their job wasinteresting job responsibilities. in (a). b. Interpretthe intervalconstructed 8.33 A largenumberof companiesare trying to reduce the cost of prescriptiondrug benefitsby requiring employeesto purchasedrugsthrougha mandatorymailorder program.In a survey of 600 employers,126 indicated that they either had a mandatorymail-order program in place or were adoptingone by the end of 2004 (B. Martinez,"Forcing Employeesto Buy Drugs via Mail," The Wall StreetJournal, February 18,2004, p. 1B). a. Constructa95ohconfidenceintervalfor the population proportionof employerswho hada mandatorymail-order programin placeor wereadoptingoneby theendof 2004. b. Constructa99ohconfidenceintervalfor the population proportionof employerswho hada mandatorymail-order programin placeor wereadoptingoneby theend of2004. c. Interpretthe intervalsin (a) and(b). d. Discussthe effect on the confidenceintervalestimate whenyou changethe levelofconfidence.

JC

se ilA Its he

8.4

DETERMININGSAMPLESIZE In each example of confidence interval estimation so far in this chapter, the sample size was reported along with the results with little discussion with regard to the width of the resulting confidence interval. In the businessworld, sample sizes are determined prior to data

300

CHAPTER EIGHT ConfidenceIntervalEstimatron

collection to ensure that the confidence interval is narrow enough to be useful in making decisions. Determining the proper sample size is a complicated procedure,subject to the constraints of budget, time, and the amount of acceptable sampling error. In the Saxon Home Improvement example, you want to estimate the mean dollar amount of the sales invoices,you must determine in advancehow large a sampling error to allow in estimating the population mean. You must also determine in advancethe level of confidence (that is, 90%,95%, or 99oh)to use in estimatingthe population pa'rameter.

Sample Size Determination for the Mean To develop an equation for determining the appropriatesample size neededwhen constructing a confidenceintervalestimateof the mean,recall Equation(8.1) on page287:

X

o + z _T "\ln

l ln this context, some statisticiansrefer to e as the "margin of error."

The amount added to or subtractedfrom X is equal to half the width of the interval. This quantity representsthe amountof imprecisionin the estimatethat results from sampling error. r.l T h e s a m p l i n ge r r o r . ' e . i s d e f i n e da s

e=Z

o _T 1n

Solving for r gives the sample size needed to construct the appropriate confidence interval 'Appropriate" means that the resulting interval will have an acceptable estimatefor the mean. amount of sampling error.

SAMPLESIZEDETERMINATION FOR THE MEAN The samplesize,n, is equalto the productof the Z valuesquaredandthe variance, o, squared,dividedby the squareof the samplingerroq e, 22o2 n-""}:-

e

I

(8.4)

FIG 2YouuseZ insteadof t because,to determine the criticalvalue of t, you need to know the samplesize,but you do not know it yet. For most studies, the sample size neededis large enough that the standardizednormal distributionis a good approximation of the t distribution.

To determine the sample size, you must know three factors: 1. The desired confidence level, which determinesthe value of Z, the critical value from the s t a n d a r d i z endo r m a ld i s t r i b u t i o n 2 2. The acceptablesampling error, e 3. The standarddeviation,o In some business-to-businessrelationships that require estimation of important parameters, legal contracts specify acceptablelevels of sampling error and the confidence level required. For companiesin the food or drug sectors,governmentregulationsoften specify sampling errors and confidence levels. In general, however, it is usually not easy to specify

Micr won dete for e

saleJ

for t lmpr l= lG h

See I this.

8 . 4 :D c t c r n r i n i S n sa r r o lS c izc 301 the two factorsneededto determinethe samplesize.How can you determinethe level of corrfidence and sarnplingerror?Typically,thesequestionsare answeredonly by the subjectmatter expert(that is, the individual most farniliar with the variablesunder study).Although 95% is the most common confidencelevel used,if rnore confidenceis desired"then 99%,rnight be more appropriate;if less confidenceis deemedacceptable,then 90%,uright be used.For the samplingerror, you shouldthink not of how much samplingerror you would like to have(you really do not want any error) but of how much you'can toleratewhen drawing conclusions from the data. In additionto specifyingthe confidencelevel and the sarnplingerror,you needan estimate of the standarddeviation.Unfortunately,you rarely know the populationstandarddeviation,o. In some instances,you can estimatethe standarddeviationfrom past data.In other situations, you can make an educatedguessby taking into accountthe rangeand distributionofthe variable. For example,if you assumea normal distribution,the rangeis approximatelyequalto 6o (that is, +3o aroundthe mean) so that you estimateo as the rangedivided by 6. If you cannot estintateo in this way,you can conducta small-scalestudy and estimatethe standarddeviatiorr f r o r nt h e r e s u l t i u gd a l a . To explore how to determinethe samplesize neededfor estirnatingthe populationmean, consideragainthe audit at SaxortHome lmprovement.In Section8.2, yoLrselecteda sampleof 100 salesinvoicesand constructeda 95o/oconfidenceintervalestimateof the populationrnearr salesinvoice amount.How was this samplesize detennined?Should you have selecteda differentsamplesize'/ Supposethat, after consultationwith company officials, you determinethat a sarnpling e r r o r o f n o m o r e t h a n + $ 5 i s d e s i r e d ,a l o n g w i t h 9 5 o l c o n f i d e n c e .P a s td a t a i n d i c a t et h a t t h e s t a n d a r dd e v i a t i o no f t h e s a l e sa r n o u n ti s a p p r o x i m a t e l y$ 2 5 . T h u s , e : $ 5 , o - $ 2 5 , a n d Z : | . 9 6 ( f o r 9 5 o hc o n f i d e n c e ) .U s i n g E q u a t i o n( 8 . 4 ) ,

22o2

( I . 9 6) 2( 2 5 ) 2

e-

(5)"

= 96.04

Becausethe generalrule is to slightly oversatisfythe criteriaby roundingthe samplesize up to the next whole integer,you should selecta sampleof size 97. Thus, the sampleof 100 usedon page292 is closeto what is necessaryto satisfythe needsof the company,basedon the estimatedstandarddeviation,desiredconfidencelevel, and samplingerror. Becausethe calculated samplestandarddeviationis slightly higher than expecte4 $28.95 comparedto $25.00, the corrfidenceinterval is slightly wider tharrdesired.Figure 8. I I illustratesa Microsoft Excel worksheetto determinethe samnlesize.

FIGUR8 E. 1 1

Le-

/el fv

ifv

Microsoft Excel worksheet for determining samplesize torestimating the mean sales invoiceamount forthe SaxonHome improvement Company

6 7

t-r

I

I 2 3 4

q ct

ul

10 11

SeeSectionEB.4to create mls.

;;

t--t

-NORTS[{V(l- 86}21 -((89'B4)/85r2

tz IJ

-ROUr{DUP(810, 0}

302

CHAPTER EIGHT Confidence Interval Estimatron

Example 8.5 illustratesanotherapplicationof determiningthe samole size neededto develooa confidenceinterval estimatefor the mean.

E X A M P L E8 . 5

DETERMININGTHE SAMPLESIZEFOR THE MEAN want to estimatethe populationmean Returningto Example8.3 on page293, suppose.you force requiredto breakthe insulatorto within +25 poundswith 95% confidence.On thebasis ofa studytakenthe previousyear,you believethat the standarddeviationis 100pounds.Find the samplesizeneeded. SOLUTION Using Equation(8.4) on page300 and e : 25, o : 100, andZ : confidence, 'n- = -

z2o2

for 95%

(1.e6)2(loo)2

e2

Q'2

Therefore,you shouldselecta samplesizeof 62 insulatorsbecausethe generalrule for determining samplesizeis to alwaysround up to the next integervalue in orderto slightly oversatisfy thecriteriadesired. An actualsamplingerror slightly largerthan25 will resultif the samplestandarddeviationcaldeviculatedin this sampleof 62 is greaterthan 100andslightlysmallerif the samplestandard ationis lessthan 100.

Sample Size Determination for the Proportion So far in this section,you havelearnedhow to determinethe samplesizeneededfor est for the populationmean.Now supposethat you want to determinethe samplesize necessary estimatingthe proportionof salesinvoicesat SaxonHome Improvementthat containerrors. To determinethe samplesize neededto estimatea population proportion, fi, you use methodsimilar to the methodfor a populationmean.Recallthat in developingthe sample for a confidenceinterval for the mean,the samplingerror is definedby e=Z

o -T "ln

Whenestimatinga proportion,you replaceo with 1G(1-

"

) Thus,the samplingerroris

n(l- n) n Solving for n, you havethe samplesizenecessaryto developa confidenceintervalestimate a proportion. SAMPLESIZE DETERMINATIONFOR THE PROPORTION The samplesizer is equalto theZvalne squaredtimesthe populationproportion,r, times minusthe populationproportion,n, dividedby the squareof the samplingerror,e, ,=

Z2n(l - n) ",

(8.s)

8.4:Determining Sample Size 303 To determinethe samplesize,you must know threefactors: 1. The desiredconfidencelevel, which determinesthe value of Z, the critical value from the standardizednormal distribution 2. The acceptablesamplingerror,e 3. The populationproportion,n In practice,selectingthesequantitiesrequiressomeplanning.Onceyou determinethe desiredlevel of confidence,you can find the appropriateZ valuefrom the standardized normal distribution.The samplingerror,e, indicatesthe amountof error that you arewilling to tolerate in estimatingthe populationproportion.The third quantity,n, is actuallythe populationparumeterthat you want to estimate!Thus,how do you statea valuefor what you aretaking a sample in orderto determine? Hereyou havetwo alternatives.In many situations,you may havepastinformationor relevant experiencethat providean educatedestimateof n. Or, if you do not havepastinformation or relevantexperience,you cantry to providea value for a that would neverunderestimatethe samplesize needed.Referring to Equation (8.5), you can seethat the quantity n(l - ru) appearsin the numerator.Thus,you needto determinethe value of n that will makethe quantity n(l - n) as largeas possible.When n : 0.5, the productn(l - n) achievesits maximum result.To showthis, severalvaluesof n, alongwith the accompanyingproductsof a(l - n), are as follows Whenn : Whenn: Whenn : Whenn :

0.9,thenn(l 0.7,thenn(l 0.5,thenn(l 0.3,thenn(l W h e nn : 0 . 1 , t h e nn ( 1-

n): (0.9X0.1): 0.09 :0.21 n) : (0.7X0.3) :0.25 n): (0.5X0.5) :0.21 (0.3X0.7) n): n): (0.1)(0.9):0.09

Therefore,whenyou haveno prior knowledgeor estimateof the populationproportion,n, you shoulduse7r: 0.5 for determiningthe samplesize.This producesthe largestpossiblesample sizeandresultsin the highestpossiblecostof sampling.Using ru: 0.5 may overestimate the samplesizeneededbecauseyou usethe actualsampleproportionin developingthe confidence interval.You will get a confidenceintervalnarrowerthan originally intendedif the actualsample proportionis differentfrom 0.5. The increasedprecisioncomesat the cost of spending more time and moneyfor an increasedsamplesize. Returning to the SaxonHome ImprovementUsing Statisticsscenario,supposethat the auditingproceduresrequireyou to have 95%o confidencein estimatingthe populationproportion of salesinvoiceswith errorsto within +0.07.The resultsfrom pastmonthsindicatethat the largestproportionhasbeenno morethan0.15.Thus,usingEquation(8.5)on page302 and e : 0 . 0 7 , n : 0 . 1 5 , a n d Z : 1 . 9 6 f o r 9 5 %c o n f i d e n c e . Z2n(l - n) "2

( 1.e6) 2( 0.1sxO.s5) (0.07)" = 99.96 Becausethe generalrule is to round the samplesize up to the next whole integerto slightly oversatisfiithe criteria,a samplesizeof 100is needed.Thus,the samplesizeneededto satisf,i the requirementsof the company,basedon the estimatedproportion,desiredconfidencelevel, and samplingerror,is equalto the samplesizetakenon page297.The actualconfidenceinterval is narrowerthan requiredbecausethe sampleproportionis 0.10,while 0.l5 wasusedfor n in Equation(8.5).Figure8.12showsa Microsoft Excel worksheetfor determiningsamplesize.

304

CHAPTEREIGHT ConfidenceIntervalEstimation

FIGURE8.12 MicrosoftExcel worksheetfor determining sample sizefor estimatingthe proportionof sales invoiceswith errors for the SaxonHome lmprovementCompany

t-l

A B or thc Proportlon of InInor Salcr Invrlecl

4

6 7 I

0.rl

unlllno Enor funcr Lrvrl lnlermedietoCalculalions

I

! Value

t0

CelculatedSamolaSize

1

-lloRxsfirv(l - Bs]n] -(89^2. Bf. (1 - B4D/85^2

tl

t3

I\N7T

Dd ofTru. ProDofdon

I|md.

-ROUI{DUP(810,0)

r Slr.

I-ll E

See Section E8.5 to create this.

Example8.6 providesa secondapplicationof determiningthe samplesize the populationproportion.

EXAMPLE 8.6

DETERMINING THESAMPLESIZEFORTHEPOPULATION PROPORTION Youwantto have90%confidence of estimating theproportionof officeworkerswho to email within an hour to within +0.05. Becauseyou havenot previouslyundertakensuch study,thereis no informationavailablefrom pastdata.Determinethe samplesizeneeded. SOLUTION Becauseno informationis availablefrom pastdata,assumethatn: 0.50.Usi Equation(8.5)on page302 ande :0.05, n : 0.50,andZ : 1.645for 90% confidence,

(1.645)',(0.50X0.s0) (0.05)2 = 270.6 Therefore,you needa sampleof 27| office workersto estimatethe populationproportion within +0.05with 90% confidence.

Learningthe Basics 8.34 If you want to be 95%oconfidentof esti@ lAsslsr I matingthepopulationmeanto within a sampling error of +5 andthe standarddeviationis assumed to be 15,whatsamplesizeis required? 8.35 If you want to be 99o/oconfidentof estimatingthe populationmeanto within a sampling error of +20 and the standard deviation is assumedto be 100,what samplesizeis required?

ffi ffi

8.36 If you want to be 99Yoconfidentof estimating the populationproportion to within an errorof t0.04,what samplesizeis needed? 8.37 If you want to be 95o/oconfidentof estimating the populationproportion to within an error of +0.02 and thereis historicalevidence

that the populationproportionis approximately0.40,what samplesizeis needed?

Applying the Concepts 8.38 A surveyis plannedto determinethe mean annualfamily medicalexpensesof employees of a large company.The managementof the company wishesto be 95ohconfident that the samplemeanis correctto within +$50 of the populationmeanannualfamily medicalexpenses. A previousstudy indicatesthat the standarddeviationis approximately$400. a. How largea samplesizeis necessary? b. If managementwantsto be correctto within +$25,what samplesizeis necessary? 8.39 If the managerof a paint supply storewantsto estimatethe meanamountof paint in a l-gallon can to within

8.4:Determining Sample Size 305 with95% confidenceand also assumesthat deviationis 0.02 gallon.what samplesize is a quality control managerwants to estimatethe of light bulbsto within +20 hourswith 95% conalsoassumes thatthepopulationstandarddevi100hours,what samplesizeis needed? If the inspectiondivision of a county weights and departmentwants to estimatethe mean amount soft-drinkfill in 2-liter bottlesto within +0.01 liter with confidenceand also assumesthat the standarddeviais 0.05liter,whatsamplesizeis needed? 8.42 A consumergroup wantsto estimatethe meanelectric bill for the month of July for single-familyhomesin a largecity. Basedon studies in othercities,the standarddeviationis assumed be $25.The group wants to estimatethe meanbill for to within+$5 with 99% confidence. Whatsamplesizeis needed? lf 95% confidenceis desired,what samplesize is necessarv? An advertisingagencythat servesa major radio stawantsto estimatethe meanamountof time that the staaudiencespendslisteningto the radio daily.Frompast ies,the standarddeviationis estimatedas45 minutes. Whatsamplesize is neededif the agencywantsto be 90%confidentof being correctto within +5 minutes? If 99% confidenceis desired,what samplesize is necessary? A growing niche in the restaurantbusinessis breakfast,lunch,andbrunch.Chainsin this includeLe Peep,Good Egg, Eggs& I, First Watch, EggsUp Grill. The meanper-personcheckfor First is approximately$7, and the meanper-personcheck Eggs Up Grill is $6.50. (Extractedfrom J. Hayes, ition HeatsUp asBreakfastConceptsEyeGrowth," b Restaurant News,April 24,2006,pp. 8, 66.) Assuminga standarddeviationof $2.00,what sample sizeis neededto estimatethe meanper-personcheckfor GoodEggto within $0.25with 95% confidence? Assuminga standarddeviationof $2.50,what sample sizeis neededto estimatethe meanper-personcheckfor GoodEggto within $0.25with 95% confidence? Assuminga standarddeviationof $3.00,what sample sizeis neededto estimatethe meanper-personcheckfor GoodEggto within $0.25with 95% confidence? Discussthe effect of variation on selectingthe sample sizeneeded. The U.S. Departmentof Transportationdefinesan ineflight asbeing"on time" if it landslessthan 15minafterthe scheduledtime shownin the carrier'scomput-

eized reservationsystem.Cancelledand diverted flights are countedas late.A studyof the l0 largestU.S.domestic airlines found SouthwestAirlines to havethe lowest proportion of late arrivals,at 0.1577 (Extractedfrom N. Tsikriktsisand J. Heineke,"The Impact of Process Variationon CustomerDissatisfaction:Evidencefrom the U S. Domestic Airline lndustry,"Decision Sciences,Winter you wereaskedto per2004,35(l),pp.129-142).Suppose form a follow-up study for SouthwestAirlines in order to updatethe estimatedproportionof late arrivals.What sample size would you use in order to estimatethe population proportionto within an error of a. +0.06with95ohconfidence? b. +0.04 with95o/oconfidence? c. +0.02 with95o/oconfidence? 8.46 In 2005, 34ohof workersreportedthat their jobs were more difficult, with more stress,and37%oreported that they worry aboutretiring comfortably.(Extractedfrom S. Armour, " Money WorriesHinder Job Performance," USAToday,October5,2005,p. Dl). Considera follow-up studyto be conductedin the nearfuture. a. What samplesize is neededto estimatethe population proportionof workerswho reportedthat theirjobs were more difficult, with more stress,to within +0.02 with 95% confidence? b. What samplesize is neededto estimatethe population proportion of workerswho worried aboutretiring comfortably to within +0.02 with 95% confidence? c. Comparethe resultsof (a) and (b). Explain why these resultsdiffer. d. If you wereto designthe follow-up study,would you use one sampleand ask the respondentsboth questions,or would you selecttwo separatesamples?Explain the rationalebehindyour decision. 8.47 What proportion of people hit snagswith online transactions?According to a poll conductedby Harris Interactive,89% hit snagswith online transactions("Top Online TransactionTrouble," USA TodaySnapshots,April 4,2006,p.lD). a. To conducta follow-up study that would provide 95% confidencethat the point estimateis correctto within +0.04 of the populationproportion,how large a sample sizeis required? b. To conducta follow-up study that would provide 99% confidencethat the point estimateis correctto within +0.04 of the populationproportion,how largea sample sizeis required? c. To conduct a follow-up study that would provide 95% confidencethat the point estimateis correctto within +0.02 of the populationproportion,how large a sample sizeis required? d. To conduct a follow-up study that would provide 99oh confidencethat the point estimateis correct to within

306

CHAPTER EIGHT ConfidenceIntervalEstimatron

+0.02of the populationproportion,how largea sample sizeis required? e. Discussthe effectsof changingthe desiredconfidence level and the acceptable samplingerror on samplesize requirements. 8.48 A poll of 1,286youngadult cell phoneuserswas conductedin March 2006.Thesecell phoneusers,aged 78-29,wereactivelyengagedin multipleusesof their cell phones.The datasuggestthat 707 took still pictureswith their phones,604 playedgames,and 360 usedthe Internet (Extracted from "Poll: CellphonesAre Annoying but Invaluable,"usatoday.com, April 3,2006). Constructa 95%confidenceintervalestimateof thepopulationproportion of youngadultsthatusedtheir cell phoneto a. takestill pictures. b. play games. c. usethe Internet.

8.5

d. You havebeenaskedto updatethe resultsofthis study, Determinethe samplesize necessary to estimatethe populationproportionsin (a)through(c) to within +0.02 with 95o/o confidence. 8.49 A studyof 658 CEOsconductedby the Conference Board reportedthat 250 statedthat their company'sgreatest concern was sustainedand steadytop-line growth ("CEOs' GreatestConcerns,"USA TodaySnapshots, May 8 , 2 0 0 6p, . 1 D ) . a. Constructa 95ohconfidenceintervalfor the proportion of CEOs whosegreatestconcernwas sustained and steadytop-linegrowth. b. Interpretthe intervalconstructed in (a). c. To conducta follow-upstudyto estimatethe population proportionof CEOs whosegreatestconcernwassustainedand steadytop-linegrowthto within +0.01with 95% confidence,how manyCEOswouldyou survey?

APPLICATIONS OF CONFIDENCEINTERVAL ESTIMATIONIN AUDITING This chapterhasfocusedon estimatingeitherthe populationmeanor the populationproportion. In previouschapters,you have studiedapplicationto differentbusinessscenarios. Auditing is one of the areasin businessthat makeswidespreaduse of probabilitysampling methodsin orderto constructconfidenceintervalestimates. AUDITING Auditing is the collectionand evaluationof evidenceaboutinformationrelatingto an economicentity,suchas a solebusinessproprietor,a partnership,a corporation,or a governmentagency,in orderto determineand report on how well the information iorrespondsto establishedcriteria.

t,

Auditorsrarelyexaminea completepopulationof information.Instead,they rely on estimationtechniques basedon theprobabilitysamplingmethodsyou havestudiedin thistext.The followinglist containssomeof the reasonssamplingis advantageous to examiningthe entire population. . ' .

. ' '

. r

is lesstimeconsuming. Sampling Samplingis lesscostly. Samplingprovidesresultsthat are objectiveand defensible.Becausethe samplesizeis basedon demonstrable statisticalprinciples,the audit is defensiblebeforeone'ssuperion and in a court of law. Samplingprovidesan objectiveway of estimatingthe samplesizein advance. Samplingprovidesan estimateof the samplingerror. Samplingis often more accuratefor drawingconclusionsaboutlargepopulationsthan othermethods.Examininglargepopulationsis time-consuming andthereforeoftensubject to morenonsamplingerrorthanstatisticalsampling. Samplingallowsauditorsto combine,andthenevaluatecollectively,samplescollected by differentindividuals. Samplingallowsauditorsto generalize their findingsto the populationwith a knownsampling error.

8.5: Applicationsof ConfidenceIntervalEstimationinAuditing

307

Estimatingthe PopulationTotalAmount In auditing applications,you are often more interestedin developing estimatesof the population total amount than the population mean. Equation (8.6) shows how to estimate a population total amount.

ESTIMATING THEPOPULATION TOTAL Thepointestimate for thepopulation totalis equalto thepopulationsize,N, timesthe samplemean. Total = NX

(8.6)

Equation(8.7)definesthe confidenceintervalestimatefor the populationtotal. CONFIDENCEINTERVALESTIMATEFOR THE TOTAL

rvx+uu ) ': .8: ' " " --,

(8.7)

^ l nY N - l

To demonstratethe applicationof the confidenceinterval estimatefor the populationtotal amount,returnto the SaxonHomeImprovement Using Statisticsscenarioon page284.Oneof the auditingtasksis to estimatethe total dollar amountof all salesinvoicesfor the month.If thereare5.000invoicesfor thatmonthand x = $ | 10.27.thenusingEquation(8.6). ryX = (5,000X$l 10.27)= $551,350 If n : 100andS: $28.95,thenusingEquation(8.7)with tnr: 7.9842for 95o/o confidence,

,c

NX + N(tntl:"ln

N-n N-1

too =55r,350 t (s,000x1.e 84D?J-/t'999 ' ./tOO1 5,000- I

= 551,350+ 28,721.295(0.99005) = 551,350 + 28,436 total < $579,786 $522,914< Population Therefore,with95o/o you estimatethatthetotalamountof salesinvoicesis between confidence, and Figure $522,914 $579,786. 8.13showsa MicrosoftExcelworksheetfor thesedata.

8.13 Excel for the interva I ofthetotal of all invoices

Home Saxon Company

TolalAmounl of All SalosInvoicet a

3 4

5

Data PopulodonShe amph teln

6 Stmple Slze StmDlG Standard Devla$on

B fonffd**a

s

Lsval

fio^2l

ilx 2SJl 95t{

1B IntermediatsCalcularioffi t l rooulalionTotal 551358.tX*84 " 85 t2 'PCFactor 0.9s[ -sQRr(Fi-B6)r{81-l}t

1 3 itandard Enor of ths Toial E8.6to create

t d eqrees of Freedom 1 5 Value l6

r*srYalHafWidih

-tB{. 87" 812ySORTtB6l

9g - 8 6 - t I 54:

-illlvtt -m,814)

m1x.7t -Bl5 * 813

17 18 Confidencelntcrval 1 S $tsffal Lowcr Llmlt 5,7,311Jf, -811- 816 nlsrvrl Uoosr Llmlt 5797&t.72 -Sl1 + 316

308

CHAPTER EIGHT ConfidenceIntervalEstimation

Example8.7 furtherillustratesthe populationtotal.

E X A M P L E8 . 7

DEVELOPINGA CONFIDENCEINTERVALESTIMATE FOR THE POPULATIONTOTAL An auditoris facedwith a populationof 1,000vouchersandwantsto estimatethe total value the populationof vouchers. A sampleof 50 vouchersis selected, with the followingresults: Meanvoucheramount(X ) = $1.076.39 Standarddeviation(S) =$2'73.62 Constructa95o/oconfidenceintervalestimateof the total amountfor thepopulationof SOLUTION UsingEquation(8.6)on page307,thepoint estimateof the populationtotalis 7try = (1,000)(1,07 6.39)= $1,076,390 From Equation(8.7) on page307,a 95% confidenceintervalestimateof the population amountis

( 1.000x 1,076.39) I ( 1.000x2.0 ogol2J3fr = 1,076,390 + 77,762.878 (0.97517) = 1,076,390 + 75,832 < Population total < $1,152,222 $1,000,558 Therefore,with95o/oconfidence,you estimatethat the total amountof the vouchersis and$ l,l 52,222. $ 1,000,558

DifferenceEstimation An auditorusesdifferenceestimationwhenhe or shebelievesthaterrorsexistin a setof and he or shewantsto estimatethe magnitudeof the errorsbasedonly on a sample.The lowing stepsareusedin differenceestimation: l . Determinethe samplesizerequired. , Calculatethe differencesbetweenthe valuesreachedduring the audit and the original

uesrecorded.The differencein valuei, denotedD,, is equalto 0 if the auditorfindsthat original value is correct,is a positivevaluewhen the auditedvalueis largerthanthe nal value,and is negativewhenthe auditedvalue is smallerthanthe original value. 3. Computethe meandifferencein the sample,D,by dividing the total differenceby samplesize,as shownin Equation(8.8).

MEAN DIFFERENCE

2o, D-t=t whereD, : Audited value- Original value

n

(8.8)

8.5: Applicationsof ConfidenceInterval Estimationin Auditing

309

4. Computethe standarddeviationof the differences,Sr, as shownin Equation(8.9). Rememberthat any item that is not in error has a dilferencevalueof 0.

STANDARD DEVIATION OF THE DIFFERENCE

Sr-^

:.,

L(ui-Dr i=l

SD=

n-l

(8.e)

5. Use Equation(8.10)to constructa confidenceintervalestimate of the total difference in thepopulation.

INTERVAL CONFIDENCE E STIMATE FORTHETOTALDIFFERENCE

.s^ffi;

N D + N ( t 'n ) l # " , l 4 n \ lN - l

(s.10)

The auditingprocedures for SaxonHomeImprovement requirea 95o/o confidenceinterval estimateof the differencebetweenthe actualdollar amountson the salesinvoicesand the amountsenteredinto the integratedinventoryand salesinformationsystem.Supposethat in a sampleof 100 salesinvoices,you have 12 invoicesin which the actualamounton the sales invoiceandthe amountenteredinto the integratedinventorymanagement and salesinformation systemis different.Thesel2 differences(storedin the file[[@fift!) are $9.03 $1.41 $17.32 $8.30 $s.21 $ r 0.80 $6.22 $5.63 $4.97 $7.43 52.99 54.63 The other88 invoicesarenot in error.Theirdifferencesareeach0. Thus,

i,, -eo

D-t=t numerator,there differences.Each

n

='" =0.90 100

and3

last88 areequal

n

0.eF

\tn, SD=

-D)'

;-l

,-l (9.03- 0.9)2+ (7.47- 0.9)2+ ... + (0 - 0.9)2

Sn = 2'752 Using Equation(8.10),constructthe95Voconfidenceinterval estimatefor the total difference in the population of 5,000 salesinvoices as follows:

(s,000)(0.90) r (s,000x1 .s84D2+ = 4 , 5 0 0+ 2 , 7 0 2 . 9 1 51,197.09< Total difference < $'7,202.91

3 10

CHAPTER EIGHT ConfidenceIntervalEstimation

Thus,the auditorestimateswith 95o/oconfidencethat the total differencebetweenthe sales invoices,as determinedduringthe audit andthe amountoriginally enteredinto the accounting systemis between$1,797.09and$7,202.91. Figure8.14showsan Excelworksheet for thesedata. FIGURE8.14 otal DlfforcnceIn Acfual lnd Enter.d

MicrosoftExcel worksheetfor the tota differencebetweenthe invoiceamountsfound duringauditand the amountsenteredinto the accountingsystem for the SaxonHome lmprovement Company

-SUl{olficrencod).t!!AAl -80/85 -Ba'B!0 -SARTGl6) -saRT{(Bl -85}/(8r - ll} -F4 - 812'8lrySQRT(Fq -B!t - | -Tll{\t(l - 86, 815} -815'Bll

See Section E8.7 to create this. - 8 1 1. 8 1 7 -B1l + 817

In the previousexample,all 12 differencesarepositivebecausethe actualamounton the salesinvoiceis more than the amountenteredinto the accountingsystem.In somecircumyou couldhavenegativeerrors.Example8.8 illustratessucha situation. stances,

E X A M P L E8 . 8

DIFFERENCE EST]MATION Returningto Example8.7 on page308,supposethat 14vouchersin the sampleof 50 vouchers containerrors.The valuesof the 14 errorsare listedbelow and storedin the file !@@. Observethat two differencesarenesative: $75.41 $38.97 $108.54-$37.18 $62.7s $l18.32 -$88.84 $r27.74 $s5.42 $39.03 $29.4r $47.99 $28.73 $84.05 Constructa 950/oconfidenceinterval estimateof the total differencein the populationof 1,000 vouchers. SOLUTION For thesedata,

tr,

: ?i D-'='

SD=

= n50

6e0.34 =13.8068

)j = to, D)' I

n-l

(75.4r- 13.8068+ ) 2( 3 8 . 9 7- 1 3 . 8 0 6 8 + ) 2' . ' + ( 0 - 1 3 . 8 0 6 8 ) 2 = 37.427

-]

iI

I j

8.5: Applications Interval of Confidence Estinration in Auditing 3 I 1 Using Equation(8. l0) on page309, constructthe confidenceintervalestirnatefor the total differencein the population as follows:

+(1,000)(2.0096 ( 1,000x l 3.8068) ) lg = 13,806.8 + 10,372.4 < Totaldifference< 524,179.20 $3,434.40 Therefore, with 95o/oconfidence, you estimate that the vouchersis between 53.434.40 and524.179.20.

d i f f e r e n c ei n t h e p o p u l a t i o no f

One-SidedConfidence Interval Estimation of the Rate of Noncompliancewith Internal Controls Organizationsuse internal control mechanismsto ensurethat individuals act in accordance with company guidelines. For example, Saxon Home Improvement requiresthat an authorized warehouse-removalslip be completed before goods are removed from the warehouse.During the monthly audit of the company,the auditing team is chargedwith the task of estimating the proportion of times goods were removedwithout proper authorization.This is referred to as the rate of nonc'ompliancewith the internal c'ontrol.To estimate the rate of noncompliance,auditors take a random sample of salesinvoices and determine how often merchandisewas shipped without an authorized warehouse-removalslip. The auditors then compare their results with a previously establishedtolerable exception rate, which is the maximum allowable proportion of items in the populationnot in compliance.When estimatingthe rate of noncompliance,it is commonplace to use a one-sided confidence interval. That is, the auditors estimate an upper bound on the rate ofnoncompliance.Equation(8.1I ) definesa one-sidedconfidenceinterval for a proportion.

ONE-SIDED CONFIDENCE INTERVAL FORA PROPORTION - p+Z Upperbound

p(t - p) n

(8.11)

where Z : the value corresponding to a cumulative area of ( I - cr) from the standardized normal distribution (that is, a right-hand tailprobability of a).

Ifthe tolerableexceptionrate is higher than the upperbound"the auditorconcludesthat the company is in compliance with the internal control. If the upper bound is higher than the tolerable exception rate, the auditor concludesthat the control noncompliancerate is too high. The auditor may then requesta larger sample. Supposethat in the monthly audit, you select 400 sales invoices from a population of 10,000invoices.In the sampleof 400 salesinvoices,20are in violation of the internalcontrol. If the tolerableexceptionrate for this internalcontrol is 6%, what shouldyou conclude?Use a 95o/olevel of confidence. The one-sidedconfidenceinterval is cornputedusingp :201400 - 0.05 and Z: 1.645. U s i n g E q u a t i o n( 8 . 1I ) ,

= 0 . 0 5+ 1 . 6 4 5 ( 0 . 0 1 0 9 X 0=. 908. )0 5+ 0 . 0 1 7 6= 0 . 0 6 7 6

312

CHAPTEREIGHT ConfidenceIntervalEstimation

Thus,you have95o/o confidencethatthe rateof noncompliance is lessthan6.76%.Because tolerableexceptionrateis 6Vo,the rateof noncompliancemay be too high for this internal trol. In otherwords,it is possiblethat the noncompliancerate for the populationis higher the rate deemedtolerable.Therefore,you shouldrequestalarger sample. In manycases,the auditoris ableto concludethat the rateof noncompliancewith the pany'sinternalcontrolsis acceptable.Example8.9 illustratessuchan occurrence.

E X A M P L E8 . 9

ESTIMATING THE RATEOF NONCOMPLIANCE A largeelectronics firm writesI millionchecksayear.An internalcontrolpolicyfor the pany is that the authorizationto sign eachcheckis grantedonly after an invoicehasbeen tialed by an accountspayablesupervisor.The company'stolerableexceptionrate for this trol is 4%. If control deviationsare found in 8 of the 400 invoicessampled,what should auditordo?To solvethis. usea95Yolevelofconfidence. SOLUTION Theauditorconstructs a95Yoone-sidedconfidenceintervalfor the invoicesin noncomplianceand comparesthis to the tolerableexceptionrate. Using (8.11) on page3 I l, p : 81400: 0.02,andZ : 1.645for 95oloconfidence,

= 0.02+ 1.6450.02(l 0.02) = 0.02+ 1.645(0.007X0.9998) = 0.02+ 0.01l5 = 0.0315 Theauditorconcludes with95%confidence thattherateof noncompliance is lessthan3. Becausethis is less than the tolerableexceptionrate, the auditor concludesthat the i controlcomplianceis adequate.In otherwords,the auditoris more thang5ohconfident rateof noncompliance is lessthan4%.

Learningthe Basics 8.50 A sampleof 25 is selectedfrom a populationof 500 items.The samplemean is 25.7, andthe samplestandard deviationis 7.8. Constructa99ohconfidenceintervalestimateof the populationtotal. 8.51 Suppose that a sample of 200 (see the file @ft@Q is selectedfrom a populationof 10,000items. Of these,l0 items are found to haveerrors of the following amounts: 13.76 42.87 34.65 I 1.09 14.54 22.87 25.52 9.81 10.03 15.49 Constructa 95o/oconfidenceinterval estimateof the total differencein the population. 8.52 If p: 0.04,n : 300,and N: 5,000,calculatethe upper bound for a one-sidedconfidenceinterval estimate of the populationproportion,7t,using the following levels ofconfidence: a.90o/o b. 95% c. 99o/o

Applying the Concepts 8.53 A stationery store wants to estimate the total

value of the 1,000greetingcardsit has in its inventory. Constructa95%oconfidenceinterval estimateof the population total value ofall greetingcardsthat are in inventory if a randomsampleof 100 greetingcardsindicatesa mean valueof $2.55and a standarddeviationof $0.44. 8.54 The personneldepartmentof a largecorporation employing3,000 workerswantsto estimatethe family dentalexpensesof its employees to determinethe feasibility of providing a dentalinsurance plan.A randomsampleof 10 employeesrevealsthe following family dental expenses(in dollars) for the preceding year(seethe!!S@ file): ll0

362 246 85 510 208 173 425 316 r79

Constructa 90ohconfidenceinterval estimateof the total family dental expensesfor all employeesin the preceding yeat.

8.6: ConfidenceIntervalEstimationand Ethical Issues

A branchofa chainoflaree electronics storesis coning an end-of-monthinventory of the merchandisein Therewere 1,546items in inventoryat that time. A of 50 itemswas randomlvselectedand an audit with the following results: conducted, Valueof Merchandise

X = 5252.28 ,S: $93.67 a 95% confidenceintervalestimateof the total in inventoryattheendof themonth. ofthemerchandise A customerin the wholesalesarmenttrade is often to a discountfor a cashpaymentfor goods.The of discountvariesby vendor.A sampleof 150items fromapopulationof4,000 invoicesat the endofa revealedthat in 13 of time (seethe!@@file) thecustomerfailed to take the discountto which he shewasentitled.The amounts(in dollars)of the 13 disthatwerenot takenwereasfollows: t5.32 97.36 230.63 r04.18 84.92 132.76 t2 26.55 129.43 88.32 47.81 89.01 a 99ohconfidenceinterval estimateof the poputotalamountof discountsnot taken. EconeDressesis a smallcompanythat manufactures 's dresses for saleto specialtystores.It has 1,200 items,and the historicalcost is recordedon a in, first out (FIFO) basis.In the past,approximately of the inventory items were incorrectly priced. were usually not significant. ; any misstatements of 120itemswas selected(seethe@$ file), thehistoricalcost of eachitem was comparedwith the value.The resultsindicatedthat l5 items differed ir historicalcostsand auditedvalues.Thesevalues asfollows:

8.6

3 l3

Sample Historical Audited Sample Historical Audited Number Cost ($) Value ($) Number Cost ($) Value ($)

5 987 17 t8 28 35 43 51

261 201 t2l 315 4tl 249 216

240 105 276 ll0 298 356 2tl 305

60 73 86 95 96 107 ll9

21 140 t29 340 341 135 228

2r0 r52 tlz 2t6 402 97 220

Constructa 95ohconfidenceinterval estimateof the total populationdifferencein the historical cost and audited value. 8.58 Tom and Brent'sAlpine Outfittersconductan annual audit ofits financial records.An internalcontrolpolicy for the companyis that a checkcan be issuedonly after the accountspayablemanagerinitials the invoice.The tolerable exceptionrate for this internalcontrolis 0.04.During an audit,a sampleof 300 invoicesis examinedfrom a population of 10,000invoices,and 1l invoicesare found to violatethe internalcontrol. a. Calculatethe upperbound for a 95ohone-sidedconfidenceinterval estimatefor the rateof noncompliance. b. Basedon (a),what shouldthe auditorconclude? 8.59 An internal control policy for Rhonda'sOnline FashionAccessoriesrequiresa quality assurancecheck before a shipmentis made.The tolerableexceptionrate for this internalcontrolis 0.05.During an audit,500 shipping recordswere sampledfrom a populationof 5,000 shippingrecords,and 12 were found that violatedthe internalcontrol. a. Calculatethe upperbound for a 95o/oone-sidedconfidenceintervalestimatefor the rateof noncompliance. b. Basedon (a),what shouldthe auditorconclude?

ESTIMATION AND ETHICAL ISSUES INTERVAL CONFIDENCE that accompany themcan Ethicalissuesrelatingto the selectionof samplesandthe inferences occur in severalways.The major ethical issuerelatesto whetherconfidenceinterval estimates areprovidedalongwith the samplestatistics.To providea samplestatisticwithout alsoincluding the confidenceinterval limits (typically set at 95o/o),thesamplesizeused,and an interpretation of the meaningof the confidenceintervalin termsthat a laypersoncanunderstandraises ethicalissues.Failureto includea confidenceintervalestimatemight misleadthe userof the resultsinto thinking that the point estimateis all that is neededto predictthe populationcharacteristicwith certainty.Thus,it is importantthat you indicatethe intervalestimatein a prominent place in any written communication,along with a simple explanationof the meaningof the confidenceinterval.In addition,you shouldhighlightthe samplesize. Oneof the mostcommonareaswhereethicalissuesconcerningestimationoccursis in the publicationof the resultsof politicalpolls.Often,theresultsof the polls arehighlightedon the

314

IntervalEstimatron CHAPTEREIGHT Confidence front page of the newspapeq and the sampling error involved along with the methodology used is printed on the page where the article is typically continued,often in the middle of the newspaper.To ensurean ethical presentationofstatistical results,the confidence levels,samplesize, and confidence limits should be made availablefor all surveysand other statisticalstudies.

AND SAMPLESlzE o (CD-ROMTopiclESTIMATION

8.7

DETERMINATIONFOR FINITEPOPULATIONS In this section,confidence intervals are developedand the sample size is determinedfor situations in which sampling is done without replacement from a finite population. For further discussion,seeE@EEEEUon the StudentCD-ROM that accompaniesthis book.

This chapter discussesconfidence intervals for estimating the characteristicsof a population, along with how you can determine the necessarysample size. You learned how an accountantat Saxon Home Improvement can use the sample

datafrom an auditto estimateimportantpopulationparameters suchas the total dollar amounton invoicesandtheproportionof shipmentsmadewithout theproperauthorization. Table8.3providesa list oftopicscoveredin thischapter. Type ofData

TABLE 8.3 Summary of Topics in ChapterB

Tlpe ofAnalysis

Numerical

Categorical

Confidenceinterval for a populationparameter

Confidenceinterval estimate for the mean(Sections8.1 and8.2)

Confidenceinterval estimate for theproportion(Section8.3).

Confidenceinterval estimatefor the total and the meandifference (Section8.5) To determinewhat equationto usefor a particularsituation,you needto askseveralquestions: . Are you developinga confidenceinterval or are you determiningsamplesize? . Do you havea numericalvariableor do you havea categorical variable?

One-sidedconfidenceinterval estimatefor theproportion (Section8.5)

. Ifyou havea numericalvariable,do you know thepopulationstandarddeviation?If you do, usethe normaldistribution.If you do not,usethe t distribution. The next four chaptersdevelopa hypothesis-testing approachto makingdecisionsaboutpopulationparameten.

Confidence Interval for the Mean (o Unknown)

Confidence Interval for the Mean (o Known)

,s

o X + Z -T=

T,

\ln

X -'*

, \ln

<X+z*

(8.1)

s.s

X-tn-t--r
.ln

Chapter ReviewProblems 315

Confidence Interval Estimatefor the Proportion

SampleSizeDeterminationfor the Mean

L.

P = zl!!!:-!) \n

ll

--

22o2

----=-

e'

(8.4)

Sample SizeDetermination for the Proportion

p-Z

p(r- p) < 1 r < p+z n

auditing 306 intervalestimate 284 confidence criticalvalue 287 degrees offreedom 290

p(r- p) n

Z2n(l - n)

(8.3)

differenceestimation308 levelofconfidence 287 one-sidedconfidenceinterval 3l I point estimate 284

CheckingYour Understanding 8.60 Why can you neverreally have 100%confidenceof of interest? conectlyestimatingthepopulationcharacteristic 1 When do you use the / distributionto developthe intervalestimatefor the mean? Why is it true that for a given samplesize,n. an in confidenceis achievedby widening(andmakprecise) less the confidenceinterval? Underwhat circumstances do you use a one-sided idenceinterval insteadof a two-sidedconfidence When would you want to estimatethe population insteadof thepopulationmean? How doesdifferenceestimationdiffer from estimaof themean?

the Concepts Youwork in the corporateoffice for a nationwide store franchisethat operatesnearly 10,000 The per-storedaily customercount has beensteady 900for sometime (thatis, the meannumberof customers astorein onedayis 900).To increasethe customercount, franchiseis consideringcuttingcoffeepricesby approxhalf.The l2-ouncesizewill now be $.59insteadof sizewill be $.69insteadof $1.19. , andthe 16-ounce with this reductionin price, the franchisewill havea grossmargin on coffee.To test the new initiative,the isehasreducedcoffeepricesin a sampleof 34 stores, customercountshavebeenrunningalmostexactlyat

(8.s)

e-

samplingerror 300 Student'sI distribution 290 total amount 307

the nationalaverageof 900. After four weeks,the sample storesstabilizeat a meancustomercountof 974 and a standard deviationof 96. This increaseseemslike a substantial amountto you, but it also seemslike a pretty small sample. Is theresomeway to get a feel for what the meanper-store countin all the storeswill be if you cut coffeepricesnationwide? Do you think reducingcoffeepricesis a good strategy for increasingthe meancustomercount? 8.67 Companiesarespendingmoretime screeningapplicantsthan in the past.A study of 102 recruitersconducttld by ExecuNetfound that 77 did Internetresearchon candidates.(Extractedfrom P.Kitchen, "Don't Let Any 'Digital Dirt'BuryYour JobProspects," Nousday,August21,2005, p. A5e). a. Constructa 95Voconfidenceinterval estimateof the population proportion of recruiterswho do Internet researchon candidates. b. Basedon (a), is it correctto concludethat more than 70o/oofrecruitersdo Internetresearchon candidates? c. Supposethat the study uses a sample size of 400 recruiters and 302 did Internet researchon candidates. Constructa 95o/oconfidenceinterval estimateof the population proportion of recruiterswho do Internet researchon candidates. d. Basedon (c), is it correctto concludethat more than 700/oof recrtiters do Internetresearchon candidates? e. Discussthe effect of samplesize on your answersto (a) through(d). 8.68 High-fructosecorn syrup(HFCS)wascreatedin the 1970sand is usedtodav in a wide varietvof foodsand

316

CHAPTER EIGHT ConfidenceIntervalEstimation

HFCS is cheaperthan sugarand is about75% beverages. think that HFCSis sweeterthansucrose.Someresearchers linkedto the growingobesityproblemin the UnitedStates (Extracted from P. Lempert, "War of the Sugars," Progressive Grocer,April 15, 2006,p.20). The following consumerviews are from a nationwidesurvevof 1.1l4 responses: Viewson HFCS Yes No Are you concernedaboutconsumingHFCS? 80o/o20o/o Do you think HFCSshouldbe banned in foodsoldto schools? 88% 12% Do you think HFCSshouldbe banned inallfoods? 56% 44% Constructa95o/oconfidenceintervalestimateof the populationproportionof peoplewho a. areconcerned aboutconsumingHFCS. b. think HFCSshouldbe bannedin food soldto schools. c. think HFCSshouldbe bannedin all foods. d. You are in chargeof a follow-up survey.Determinethe samplesizenecessary to estimatethe proportionsin (a) +0.02 (c) through to within with 95oloconfidence. 8.69 StarwoodHotels conducteda surveyof 401 top executives who play golf (Extractedfrom D. Jones,"Many CEOs Bend the Rules (of Golf)," USA Today,June26, 2002).Among the resultswerethe following: . 329cheatat golf. . 329 hateotherswho cheatat golf. . 289believebusiness andgolfbehaviorparallel. . 80 wouldlet a clientwin to getbusiness. . 40 would call in sick to play golf. Constructa 95o/oconfidenceintervalestimatefor eachof thesequestions.Basedon theseresults,what conclusions canyou reachaboutCEOs'attitudestowardgolfl 8.70 A marketresearcherfor a consumerelectronics companywantsto studythe televisionviewing habitsof residentsof a particular area.A random sampleof 40 respondents is selected,and eachrespondentis instructed to keepa detailedrecordof all televisionviewingin a particularweek.The resultsareasfollows: . Viewingtime per week: X = 15.3hours,.t: 3.8 hours. . 27 respondents watchthe eveningnewson at least3 weeknights. a. Constructa 95o/oconfidenceinterval estimatefor the meanamountof televisionwatchedper week in this city. b. Constructa 95ohconfidenceinterval estimatefor the populationproportionwho watchthe eveningnewson per week. at least3 weeknights

Supposethat the market researcherwants to take anothersurveyin a differentcity.Answerthesequestions: c. What samplesize is requiredto be 95o/oconfidentof estimatingthe populationmeanto within *2 hoursand assumes that the populationstandarddeviationis equal to 5 hours? d. What samplesize is neededto be 95o/oconfidentof being within +0.035of the populationproportionwho watchthe eveningnewson at least3 weeknightsif no previousestimateis available? e. Basedon (c) and (d), what samplesizeshouldthemarket researcher selectif a singlesurveyis beingconducted? 8.71 The real estateassessorfor a county government wants to study various characteristicsof single-family housesin the county. A random sample of 70 houses reveals the following . Heatedareaof the houses(in squarefeet): X = 1,759,

^s:380. . 42 houseshavecentralair-conditioning. a. Constructa 99o/oconfidenceinterval estimateof the populationmeanheatedareaof thehouses. b. Constructa 95o/oconfidenceinterval estimateof the populationproportionof housesthat havecentralairconditioning. 8.72 Thepersonneldirectorof a largecorporationwishes to study absenteeism amongclerical workersat the corpo ration'scentraloffice during the year.A randomsampleof 25 clericalworkersrevealsthe following: . Absenteeism: X = 9.7 days,S: 4.0days. . l2 clericalworkerswereabsentmorethan l0 days. a. Constructa 95o/oconfidenceinterval estimateof t meannumberof absencesfor clerical workersduri the year. b. Construct a 95o/oconfidence interval estimate of the

populationproportionof clericalworkersabsent than 10 daysduring the year. thatthepersonneldirectoralsowishesto take Suppose in survey a branchoffice.Answerthesequestions: c. What sample size is neededto have 95% confidencei

estimatingthe populationmeanto within +1.5days the populationstandarddeviationis 4.5 days? d. What samplesizeis neededto have90% confidence estimatingthepopulationproportionto within +0.075 no previousestimateis available? e. Basedon (c) and(d), whatsamplesizeis neededif a gle surveyis beingconducted? 8.73 The marketresearchdirectorfor Dotfy's Store wants to study women's spendingon A surveyof thestore'screditcardholdersis designed in to estimatethe proportionof womenwho purchase

Chapter ReviewProblems 317 primarily from Dotty's DepartmentStore and the yearlyamountthatwomenspendon cosmetics.A presurvevfoundthat the standarddeviationof the amount spendon cosmeticsin a yearis approximately$ 18. samplesizeis neededto have99o/oconfidenceof imatingthe populationmeanto within +$5? samplesize is neededto have 90% confidence ,-What estimatingthe population proportion to within

.045? on the resultsin (a) and (b), how many of the 's credit card holders should be sampled?

The branch managerof a nationwide bookstore wantsto study characteristicsof her store'scusShedecidesto focus on two variables:the amount spentby customersand whetherthe customers considerpurchasingeducationalDVDs relating to preparationexams,such as the GMAT, GRE, or . Theresultsfrom a sampleof 70 customersare as Amountspent:X = $28.52,,S:$11.39. 28 customersstatedthat they would considerpurchasinsthe educationalDVDs. a 95%oconfidenceinterval estimateof the meanamountspentin the bookstore. a 90% confidenceinterval estimateof the Lonproportionof customerswho would consider ine educationalDVDs. that the branchmanaserof anotherstorein the wantsto conduct a similar survey in his store. thefollowing questions: samplesize is neededto have95% confidenceof ing the populationmean amount spent in his to within +$2 if the standarddeviationis assumed $10? samplesizeis neededto have90% confidenceof ing the populationproportionwho would conpurchasingthe educational DVDs to within ,| on your answersto (c) and (d), how large a samshouldthe managertake? The branchmanagerof an outlet (Store l) of a chainof pet supply storeswantsto study charicsof her customers.In particular,shedecidesto twovariables:the amountof moneyspentby cusandwhetherthe customersown only one dog, only ormorethanonedog and/orcat.The resultsfrom a of 70 customersare as follows: t of moneyspent:X = $21.34,S:59.22. own only a dog. customers customersown only a cat. mersown more than one dos and/or cat.

a. Constructa 95ohconfidenceinterval estimateof the populationmeanamountspentin the pet supplystore. b. Constructa 90ohconfidenceinterval estimateof the population proportion of customerswho own only a cat. The branchmanagerof anotheroutlet (Store2) wishes to conducta similar surveyin his store.The managerdoes not have accessto the information generatedby the managerof Storel. Answerthe followingquestions: c. What samplesizeis neededto have95% confidenceof estimatingthe populationmean amount spentin his storeto within +$1.50if the standarddeviationis $10? d. What samplesize is neededto have90% confidenceof estimatingthe populationproportion of customerswho own only a cat to within +0.045? e. Basedon your answersto (c) and (d), how large a sample shouldthe managertake? 8.76 The owner of a restaurantthat servescontinental food wantsto study characteristics of his customers.He decidesto focus on two variables:the amountof money spentby customersand whethercustomersorderdessert. The results from a sample of 60 customersare as follows: . Amountspent:X = $38.54,S=$7.26. . l8 customerspurchaseddessert. a. Constructa 95o/oconfidenceinterval estimateof the population mean amount spent per customerin the restaurant. b. Constructa 90ohconfidenceinterval estimateof the population proportion of customerswho purchase dessert. The ownerof a competingrestaurantwantsto conducta similar surveyin her restaurant.This owner doesnot have accessto the information of the owner of the first restaurant.Answerthe following questions: c. What samplesizeis neededto have95% confidenceof estimatingthe populationmean amount spentin her restaurantto within +$1.50,assumingthat the standard deviationis $8? d. What samplesize is neededto have90% confidenceof estimatingthe populationproportion of customerswho purchasedessertto within +0.04? e. Basedon your answersto (c) and (d), how large a sample shouldthe ownertake? 8.77 The manufacturerof "Ice Melt" claimsits product will melt snow and ice at temperaturesas low as 0o Fahrenheit.A representativefor a large chain of hardware storesis interestedin testingthis claim. The chain purchasesa largeshipmentof 5-poundbagsfor distribution. The representativewants to know with 95% confidence, within +0.05,what proportionof bagsof Ice Melt perform thejob as claimedby the manufacturer.

318

EIGHTConfidence CHAPTER Interval Estimation

a. How many bagsdoesthe representative needto test? What assumptionshouldbe madeconcerningthe population proportion?(This is called destructivetesting; that is, the productbeing testedis destroyedby the test andis thenunavailable to be sold.) b. The representative tests50 bags,and42 ofthem do the job as claimed.Constructa 95o/oconfidenceinterval estimatefor the populationproportion that will do the job asclaimed. c. How can the representativeuse the results of (b) to determinewhetherto sell the Ice Melt product? 8.78 An auditorneedsto estimatethe percentage of timesa companyfails to follow an internalcontrolprocedure. A sample of 50 from a populationof 1,000itemsis selecte4andin 7 instances, the internalcontrolprocedurewasnot followed. a. Constructa90% one-sided confidenceintervalestimate of thepopulationproportionof itemsin which the internal controlprocedurewasnot followed. b. If the tolerableexceptionrate is 0.15, what shouldthe auditorconclude? 8.79 An auditorfor a governmentagencyneedsto evaluatepaymentsfor doctors'officevisitspaid by Medicarein a particularzip code during the month of June.A total of 25,056visitsoccurredduringJunein this area.The auditor wantsto estimatethe total amountpaid by Medicareto within +$5 with 95% confidence.On the basisof past experience,she believesthat the standarddeviation is approximately$30. a. What samplesizeshouldsheselect? Using the samplesizeselectedin (a), an audit is conducted with the following results. Amountof Reimbursement

X = $93.70

,l = $34.55

In l2 of the office visits, an incorrectamountof reimbursement was provided.For the 12 office visits in which there was an incorrect reimbursement,the differencesbetween the amountreimbursedand the amountthat the auditor determinedshouldhavebeen reimbursedwere as follows (andare storedin the file@!@

amountthat the auditor determinedshouldhavebeen reimbursed. 8.80 A home furnishingsstore that sells bedroomfurniture is conductingan end-of-monthinventory of the beds (mattress,bed spring, and frame) in stock.An auditorfor the store wants to estimatethe meanvalue of the bedsin stock at that time. Shewantsto have99% confidencethat her estimateof the meanvalueis correctto within +$100. On the basisof pastexperience,sheestimatesthat the standarddeviationofthe valueofa bed is $200. a. What samplesizeshouldsheselect? b. Using the samplesize selectedin (a), an audit wasconducte4 with the following results:

X = $1,654.27

S = $184.62

Constructa 99o/oconfidenceinterval estimateof the total value of the bedsin stock at the end of the month if there were258 bedsin stock. 8.81 A quality characteristicof interestfor a tea-bagfilling processis the weight of the tea in the individual bags.In this example,the labelweight on the packageindicatesthat the meanamountis 5.5 gramsof tea in a bag.If the bags are underfille4 two problemsarise.First, customersmay not be able to brew the tea to be as strongas they wish. Second,the companymay be in violation of the truth-in-labelinglaws. On the other hand, if the mean amountof tea in a bag exceedsthe label weight, the company is giving awayproduct. Getting an exact amountof tea in a bag is problematicbecauseof variationin the temperatureand humidity insidethe factory,differencesin the densityof the tea, and the extremelyfast filling operation of the machine(approximately170bagsper minute).The following dataarethe weights,in grams,of a sampleof 50 tea bagsproducedin one hour by a singlemachine(the dataarestoredin the file S$!!@: Weight of TeaBags, in Grams

5.65 s.44 5.42 5.40 s.53 5.34 5.54 s.4s 5.52 5.41 5.57 5.40 5.53 5.54 5.s5 5.62 5.56 5.46 5.44 5.5Ii 5.47 5.40 s.47 5.61 5.53 5.32 5.67 5.29 5.49 5.55

$17 $2s $14 -$10 $20 $40 $3s $30 $28 $22 $15 $5

5.77 5.57 5.42 5.58 5.58 5.50 5.32 5.50 5.53

b. Constructa 90% confidenceintervalestimateof the popthat containerrors. ulationproportionof reimbursements c. Constructa 95o/oconfidenceinterval estimateof the populationmeanreimbursementper office visit. d. Constructa 95%oconfidenceinterval estimateof the for this geopopulationtotal amountof reimbursements graphicareain June. e. Constructa 95%oconfidenceinterval estimateof the total differencebetweenthe amountreimbursedand the

5.61 5.45 5.44 5.25 5.56 s.63 5.50 5.57 5.67 a. Constructa 99o/oconfidenceinterval estimateof populationmeanweight of the teabags. b. Is the companymeetingthe requirementsetforth on labelthatthe meanamountof teain a bag is 5.5 8.82 A manufacturingcompanyproducessteel for electricalequipment.The main componentpart of housingis a steeltroughthat is madeout of a l4-gauge

Chapter ReviewProblems 3I9 It is producedusinga 250-tonprogressivepunchpress a wipe-downoperationthat puts two 90-degreeforms flat steelto makethe troush. The distancefrom one of theform to the otheris critical becauseof weatherin outdoorapplications.The data(storedin the from a sampleof 49 troughsfollows:

8.4138.489 8.414 8.481 8.4t5 8.479 8.429 8.458 8.462

warranty period. The data file !@ containsa sample madeon the company'sBostonshinof 170 measurements glesand 140measurements madeon Vermontshingles. a. For the Boston shingles,constructa 95ohconfidence interval estimateof the meangranuleloss. b. For the Vermont shingles,constructa 95o/oconfidence interval estimateof the meangranuleloss. c. Evaluatewhetherthe assumptionneededfor (a) and (b) hasbeenseriouslyviolated. d. Basedon theresultsof (a) and(b), whatconclusions can you reach concerningthe mean granule loss of the BostonandVermontshingles?

8.M 8.4298.460 8.412 8.420 8.410 8.40s 8.323 8.420

Report Writing Exercises

8.4478.4058.439 8.411 8.427 8.420 8.498 8.409

8.85 Referringto the resultsin Problem8.82on page318 concerningthe width of a steeltrough, write a report that summarizesyour conclusions.

Width of Trough, in Inches 8.3438.3178.383 8.348 8.410 8.351 8.313 8.481 8.422 8.3828.4848.403 8.414 8.419 8.385 8.465 8.498 8.447

Constructa 95o/oconfidenceinterval estimateof the meanwidth of the troughs. Interpretthe intervaldevelopedin (a). The manufacturerof Boston and Vermontasphalt know that productweight is a major factor in the 's perceptionof quality. The last stageof the line packagesthe shinglesbeforethey areplaced woodenpallets.Once a pallet is full (a pallet for most holdsl6 squaresof shingles),it is weighed,andthe t is recorded.The data file [@ contains weight(in pounds)from a sampleof 368 pallets of shinglesand330palletsof Vermontshingles. Forthe Boston shingles,constructa 95ohconfidence intervalestimateof the meanweight. Forthe Vermont shingles,constructa 95Voconfidence intervalestimateof the meanweight. Evaluatewhetherthe assumptionneededfor (a) and (b) hasbeenseriouslyviolated. can Basedon theresultsof(a) and(b), whatconclusions you reachconcerningthe meanweight of the Boston andVermontshingles? I ll

5

fs

r' )6 h" I hr

br gs he rcl

The manufacturerof Boston and Vermont asphalt providesits customerswith a 20-yearwarranty on of its products.To determinewhethera shinglewill last longasthewarrantyperiod"acceleratedlifetestingis conat the manufacturingplant. Accelerated-lifetesting it would be subjectto in a the shingleto the stresses of normal usevia a laboratoryexperimentthat takes a few minutesto conduct.In this test, a shingleis scrapedwith a brush for a short period oftime, andthe shingle granules removed by the brushing are weighed(in grams).Shinglesthat experiencelow amounts ofgranulelossareexpectedto lastlongerin normalusethan thatexperience high amountsof granuleloss.In this shingles situation,a shingleshouldexperienceno more than 0.8 gramsof granulelossif it is expectedto lastthe lengh of the

Team Proiect 8.86 Refer to the team project on page 73 (see the file). Constructall appropriateconfidence Eil@@@ interval estimatesof the populationcharacteristicsof lowrisk, average-risk,and high-risk mutual funds. Include theseestimatesin a reportto the vice presidentfor research at the financial investmentservice.

Student Survey Database 8.87 Problem1.27on page 15 describesa surveyof 50 undergraduate students(seethe file GEEEEffiE!E$. a. For thesedata,for eachvariable,constructa95%oconfidenceinterval estimateof the populationcharacteristic. b. Write a reportthat summarizesyour conclusions. 8.88 Problem1.27on page15 describes a surveyof 50 undergraduate students(seethe fileEEE@). studentsat your a. Selecta sampleof 50 undergraduate schooland conducta similar surveyfor thosestudents. b. For the data collected in (a), repeat (a) and (b) of Problem8.87. c. Comparethe resultsof (b) to thoseof Problem8.87. 8.89 Problem1.28on page 15 describesa surveyof 50 MBA students(seethe fileEEElffiElldH). a. For thesedata,for eachvariable,constructa95o/oconfidenceinterval estimateof the populationcharacteristic. b. Write a reportthat summarizesyour conclusions. 8.90 Problem1.28on page 15 describesa surveyof 50 MBA students(seethe file!@@l!@). a. Selecta sampleof 50 graduatestudentsin your MBA programandconducta similar surveyfor thosestudents. b. For the data collected in (a), repeat (a) and (b) of Problem8.89. c. Comparethe resultsof (b) to thoseof Problem8.89.

320

CHAPTER EIGHT ConfidenceIntervalEstimatron

Managingthe SpringvilleHerald The marketing departmenthas been consideringways to increasethe number of new subscriptionsand increasethe rate of retention among customerswho agreed to a trial subscription. Following the suggestion of Assistant Manager Lauren Alfonso, the department staff designed a survey to help determine various characteristicsof readers of the newspaperwho were not home-delivery subscribers. The surveyconsistsof the following 10 questions: 1. Do you or a memberofyour householdeverpurchase the Springville Hera ld? (l)Yes (2)No [ f t h e r e s p o n d e n ta n s w e r s n o , t h e i n t e r v i e w i s terminated.] 2. Do you receive the Springville Herald via home delivery? (l)Yes (2)No

4.] [f no,skipto question 3. Do you receivetheSpringvilleHerald: ( I ) Monday-Saturday(2) Sundayonly (3) Everyday [f everyday,skip to question9.] 4. How often during the Monday-Saturdayperiod do you purchasethe SpringvilleHerald? (l) Everyday (2) Mostdays (3) Occasionally or never 5 . How often do you purchasetheSpringvilleHerald on Sundays? ( l) EverySunday (2) 2-3 Sundaysper month (3) No morethanoncea month 6 . Whereareyou mostlikely to purchasetheSpringville Herald? (l) Convenience store (2) Newsstand/candy store (3) Vendingmachine (4) Supermarket (5) Other 7 . Woufdyou considersubscribingto the Springville Heraldfor a trial periodif a discountwereoffered? (l) Yes (2)No 9.] [f no, skipto question 8 . The Springville Herald currently costs $0.50 Monday-Saturday and $1.50on Sunday, for a totalof $4.50perweek.Howmuchwouldyoubewillingto pay perweekto gethomedeliveryfor a 90-daytnal period? 9 . Do you read a daily newspaperother than the SpringvilleHerald? (l)Yes (2)No t 0 .As an incentivefor long-termsubscribers, the newspaperis considering the possibilityofoflering a card thatwouldprovidediscounts at certainrestaurants in the Springvilleareato all subscribers who pay in advancefor six monthsof homedelivery.Wouldyou wantto get sucha cardunderthe termsof this offer? ( l ) Y e s ( 2 )N o

The group agreed to use a random-digit dialing method to poll 500 local householdsby telephone.Using this approach,the last four digits of a telephonenumberare randomly selected to go with an area code and exchange (the first 6 digits of a l0-digit telephonenumber). Only those pairs of area codes and exchangesthat were for the Springville city areawere used for this survey. Of the 500 households selected"94 householdseither refusedto participate,could not be contactedafter repeated attempts, or representedtelephone numbers that were not in service.The summary results are as follows: Households That PurchasetheSpringvilleHerald

Yes No Householdswith Home Delivery

Yes No Type of Home Delivery Subscription

Monday-Saturday Sundayonly 7 daysa week PurchaseBehavior of Nonsubscribers for Monday-Saturday Editions

Every day Most days Occasionally or never PurchaseBehavior of Nonsubscribers for Sunday Editions

Every Sunday 2-3 Sundaysa month No more than oncea month Nonsubscribers'PurchaseLocation

Convenience store Newsstand/candy store Vendingmachine Supermarket Otherlocations Would Consider Trial Subscription if Offered a Discount

Yes No

Frequency

352 JA

Frequency

136 216 Frequency

l8 25 93 Frequency

78 95 43

Frequency

138 54 )4 Frequency

74 95 21 13 IJ

Frequency +0

170

Rcf-crenccs 321

RateWilling to Payper Week(in Dollars) Data file EIjEEE Trial Subscription for a 90-DayHome-Delivery

4.153.604.103.603.603.604.403.ts 4.003.754.00 3.253.753.303.753.654.004.r0 3.903.503.753.00 3.404.003.803.504.104.253.503.903.954.304.20 3.503.753.303.853.204.403.803.403.502.853.75 3.803.90 Reada Daily NewspaperOther Than the Springville Herald

Yes No Would Prepal' Six Months to Receive a RestaurantDiscountCard

Yes No

Frequency

r38 214 Frequency

EXERCISES SH8.1 Some members of the marketing departrrentare concernedabout the random-digitdialing method Preparea memousedto collect surveyresponses. randumthat examinesthe following issues: . The advantagesand disadvantages of using the git rnethod. random-di dialing ' P o s s i b l ea l t e r n a t i v ea p p r o a c h e sf o r c o n d u c t ing the survey and their advantagesand disadvantages, S H 8 . 2 A n a l y z e t h e r e s u l t so f t h e s u r v e y o f S p r i n g v i l l e households.Write a report that discussesthe urark e t i n g i m p l i c a t i o n so f t h e survey results for the Springville Heruld.

66 286

Web Case Applt' your knov'ledge about confidenc'eintervctl estimation in this Weh Case. whic'h extends the OurCamptrs! Web Case.from Chapter 6. Among its other features,the OurCampus! Web site allowscustomersto purchaseOurCampus!LifeStylesmerchandiseonline. To handle paymentprocessing,the managementof OurCampus!has contractedwith the following firrns: . P a y A F r i e n d( P A F ) : a n o n l i n e p a y m e n t s y s t e m w i t h w h i c h c u s t o m e r sa n d b u s i n e s s e s u c h a s O u r C a m p u s ! registerin order to exchangepaymentsin a secureand convenientlranner without the need for a credit card. . ContinentalBanking Company (Conbanco):a processi n g s e r v i c e sp r o v i d e r t h a t a l l o w s O u r C a m p u s ! c u s tomers to pay for merchandiseusing nationally recognizedcredit cardsissuedby a financial institution. To reducecosts,the managementis consideringeliminating one of these two payment systems. However, Virginia Duffy of the salesdepartmentsuspectsthat customersuse the two forms of payment in unequal numbers

1. Cochran,W. G., Sumplingkc'hnique.r,3rd ed. (NewYork: Wiley, 1977). 2. Fisher, R. A., and F. Yates, Sraristical Tables fiir Biological, Agricultural and Medic'al Researc'h,5thed. (Edinburgh:Oliver & Boyd, 1957). 3. Kirk, R. E., ed., Statistical Issues:A Reader /br the (Beln-ront,CA: Wadsworth, 1972). BehavinraI Sc'ienc'es

a n d t h a t c u s t o m e r sd i s p l a y d i f f e r e n t b u y i n g b e h a v i o r s when using the two forrns of payment. Therefore, she would like to first determrne a. the proportion of customersusing PAF and the proportion of customersusing a credit card to pay for their purchases. b. the mean purchaseamount when using PAF and the meanpurchaseamountwhen using a credit card. A s s i s t M s . D u f f y b y p r e p a r i r r ga n a p p r o p r i a t ea n a l y s i s basedon a random sampleof 50 transactionsthat she has preparedand placedin an internalfile on the OurCampus! Web site, www.prenhall.com/Springville/OurCampus_ P y m t S a m p l e . h t m . S u m m a r i z ey o u r f i n d i n g s a n d d e t e r mine whether Ms. Duffy's conjecturesabout OurCampus! customerpurchasingbehaviorsare correct.If you want the samplingerror to be no more than $3 when estimatingthe mean purchase amount, is Ms. Duffy's sanrple large enoughto perform a valid analysis?

4. Larsen,R. L., and M. L. Marx, An Introductionto Mqthematical Statistic'sand lts Applic'ations,4th ed. (Upper SaddleRiver,NJ: PrenticeHall, 2006). 5. Microsoft Excel 2007 (Redmond,WA: Microsoft Corp., 2007). 6. Snedecor,G. W., and W G. Cochran"StatisticulMethods, 7th ed. (Arrres,IA: Iowa StateUniversityPress,1980).

322

EXCELcoMpANIoN to chaDter 8

E8.1 COMPUTINGTHE CONFIDENCE INTERVALESTIMATEFOR THE MEAN (o KNOWN) You compute the confidence interval estimatefor the mean (o known) either by using the PHStat2 Estimate for the Mean, sigma known procedureor by making entries in the

g@@EEworkbook.

Usingthe CIE_SKWorksheet

Usinq PHStat2 Estimate for tFe Mean, Sigma Known SelectPHStat ) Confidence Intervals ) Estimate for the Mean, sigma known. ln the Estimate for the Mean, sigma known dialog box (shown below), enter values for the Population Standard Deviation and the Confidence Level. Click one of the input optionsand make the required entries.Enter a title as the Title and click OK.

D*A

ffi*r

Cmfdencs Lwd:

You compute the confidence interval estimatefor the mean (o unknown) either by using the PHStat2 Estimate for the Mean, sigma unknown procedure or by making entriesin

Inpt O$ions (? funpbstdisticJ Kno$"n Siae:

theEIEEEEEEEEIEworkbook.

Sarndefthanl (^ San& Sf*isticE Ur*nwun

Usinq PHStat2 Estimate for tFe Mean, Sigma Unknown

I

t':t I

Ttle: I l* fir*e PafxJdbn Correction : '.

r-bb

{

I

Open to the CIE_SK worksheet of the workbook. This worksheet uses the NORMSINV(P<E and CONFIDENCE(l-conJidence level,population standard deviution, sample siee) functions to compute the Z value and interval half-width for the Example 8.1 mean paper length problem on page 288. To adapt this worksheet to other problems, enter the appropriatepopulation standard deviation, sample mean, sample size, and confidence level values in the tinted cells 84 throush 87 and entera n e w t i t l e i n c e l lA l .

E8.2 COMPUTINGTHE CONFIDENCE INTERVALESTIMATEFORTHE MEAN (o UNKNOWN)

Psplatim *ardsd Daviatbnl

5ar*

If you know the sarnplesize and sample mean of your sample, click Sample Statistics Known and enter those values.Otherwise,click Sample Statistics Unknown and enter the cell range of your sample as the Sample Cell Range.

| r.'*.,-..."---.a|

ti -

oK

tl

Csrcd

I

I

SelectPHStat ) Confidence Intervals ) Estimate for the Mean, sigma unknown. In the Estimate for the Mean, sigma unknown dialog box (shown on page 323), entera Confidence Level value, click one of the input options, and make the required entries.Enter a title as the Title and click OK. If you know the sample size, sample mean, and sample standarddeviation of your sample,click Sample Statistics Known and enter those values.Otherwise,click Sample Statistics Unknown and enter the cell range of your sample as the Sample Cell Range.

E8.4: Computingthe SanrpleSizeNeededfor Estimatingthe Mean

box (shownbelow),entervaluesfor the SampleSize,the Number of Successes, andthe ConfidenceLevel.Entera title astheTitle andclick OK.

Data

lG-v"

CmfklencsLevdl Inrut O$bns {3 5end6 Stati*ics Known

D€tE

5ampbSiea:

Sarpla Siza:

SanphMeanr

Nunberaf $iccmcesr

SarnphStd. Devi*ion:

CmfidenceLevd:

ls-"a

f Sstple StatistksUr{qrown

J l.

er$rt OSions

Ttle: I f

**pt

323

finite PoprJationCorrectbn

O$ions

[* fir*e Popr.dation Csrrection

Heb

I

| ---....-.,,....,-*.,*,.1

I

li€

ol<

il

I Using the CIE_Pworksheet Heb

II

I

lit-:::! oK

il

Usingthe CIE_SUWorksheet Opento the clE_su worksheetof th.EIEEEE@IE workbook.The worksheet(seeFigure 8.6 on page 293) uses theTINV( 1-conJid en ce I eveI, d egrees of fre ed om) function to determinethe critical value from the r distribution and computethe interval half-width for the Section 8.2 Saxon HomeImprovementCompany example.To adaptthis worksheetto other problems,changethe sample statisticsand level values in the tinted cells 84 throush 87 confidence andentera new title in cell Al.

E8.3 COMPUTINGTHE CONFIDENCE INTERVALESTIMATEFORTHE PROPORTION Youcompute the confidence interval estimate for the proportion either by using the PHStat2 Estimate for the Proportion procedure or by making entries in the

q[f[[fis[[!workbook.

Open to the CIE_P worksheet of the Q@EEE workbook. The worksheet(see Figure 8. I 0 on page 297) uses the NORMSINV(P<,\') function to determine the Z value and uses the square root function to compute the standarderror ofthe proportion for the Section8.3 Saxon H o m e I m p r o v e m e n t C o m p a n y e x a r n p l e .T o a d a p t t h i s worksheetto other problerns.enterthe appropriatesample s i z e , n u m b e r o f s u c c e s s e sa, n d c o n f i d e n c e l e v e l v a l u e s i n t h e t i n t e d c e l l s 8 4 , 8 5 , a n d 8 6 a n d e n t e ra n e w t i t l e i n cellAl.

E8.4 COMPUTINGTHE SAMPLES/,ZE NEEDEDFOR ESTIMATING THE MEAN You computethe samplesize neededfor estimatingthe mean either by using the PHStat2 Determination for the Mean

procedure or by makingentriesin theEEEEEEEIEEEEE workbook. Using PHStat2 Determination for the Mean SelectPHStat) SampleSize) Determination for the

UsinqPHStat2 Estimate for tf,e Proportion SelectPHStat ) Confidence Intervals ) Estimate for theProportion. In the Estimatefor the Proportiondialog

M e a n . I n t h e S a m p l e S i z e D e t e r m i n a t i o nf o r t h e M e a n dialog box (shown on page 324), enter values for the Population Standard Deviation, the Sampling Error, and the Confidence Level. Enter a title as the Title and click OK.

324

EXCELcoMPANIoNro chanter8

Dd6 Poprlatim $ardard DcviaHon: Sanphg Errsn CmfirJenceLenrd:

Data

t--

Estimatc of Trrr ProporHml

r-

5anphrgError: Cmf*Jence Lwd:

lgs*qe

O-tp.t O$iors

G*pnt Ofrims

Using the SampleSize*MWorksheet Open to

the

SampleSize_M worksheet of

Using the SampleSize_PWorksheet the

(seeFigure workbook.Theworksheet f,!ffi!f@fl!![[! 8.1I onpage301)usestheNORMSINV@<$function to compute the Z value and usesthe ROUNDUP(value) function to round up the sample size neededto the next higher integer for the Section 8.4 Saxon Home Improvement Company example.To adapt this worksheet to other problems, enter the appropriate population standarddeviation, sampling error, and confidence level values in the tinted cells 84" 85. and 86 and entera new title in cell A l .

Open

to

the

SampleSize_P worksheet of the The worksheet(see Figure @file. 8.12 on page304) usesthe NORMSINV and ROUNDUP functions,discussedin SectionE8.3, for the SaxonHome ImprovementCompany example in Section 8.3. To adapt this worksheetto other problems,enter the appropriateestimate of true proportion, sampling error, and confidence level valuesin the tinted cells 84 throush 86 and entera new title in cell A l.

I

E8.5 COMPUTINGTHE SAMPLESIZE NEEDEDFOR ESTIMATING THE PROPORTION You compute the sample size needed for estimating the proportion either by using the PHStat2 Determinationfor the Proportion procedure or by making entries in the @workbook.

E8.6 COMPUTINGTHE CONFIDENCE INTERVALESTIMATEFORTHE POPULATIONTOTAL

(

Y o u c o r n p u t et h e c o n f i d e n c e i n t e r v a l e s t i m a t ef o r t h e population total either by using the PHStat2 Estimatefor the Population Total procedure or by making entries in the

EIEEIE[IE*orkbook.

Using PHStat2 Determination for the Proportion

Using PHStat2 Estimate for the Population Total

SelectPHStat ) $n6p1e Size ) Determination for the P r o p o r t i o n . I n t h e S a m p l e S i z e D e t e r m i n a t i o nf o r t h e Proportion dialog box (shown at right), enter values for the Estimate of True Proportion, the Sampling Error, and the Confidence Level. Enter a title as the Title and click OK.

SelectPHStat ) ConfidenceIntervals) Estimatefor the PopulationTotal. In the Estimatefor the Population Totaldialogbox (shownon page325),entervaluesforthe PopulationSizeandthe ConfidenceLevel. Click one of the inputoptionsandmakethe required entries.

t

I

!

E8.7: Computingthe ConfidenceIntervalEstimatefor theTotal Difference

325

Usinq PHStat2 Estimate for tEe Total Difference SelectPHStat ) ConfidenceIntervals) Estimatefor the Total Difference. In the Estimatefor the Total Differencedialogbox (shownbelow),entervaluesfor the SampleSize,the PopulationSize,and the Confidence Level. Enter the cell range of the differencesas the DifferencesCell Range.If the first cell in the columnof differences contains a label,click First cell containslabel, entera title astheTitle. andclick OK.

Foptd*lmSa:

l*

CmffdcnccLcnd: - brtrlt O$iont

ab

{i Sarpte5t6tirtks Knonrn Sarpla Size: Sarplafvban: sefiph sttrtdtd Dwi*hnr

(^ Sarpte$atlsticsurfoowr

J

" Dda ri

,

Sarn$ Sal PotrJationSza:

lE-s.

Corfiderre Levd:

,

li

or

il

Of,fsroncesCdlRanga: I fiZ rtstcdcor*ahsl#

J

cercd I O"tput@tims

If youknowthesamplesize,samplemean,andsample deviationof your sample,click SampleStatistics andenterthosevalues.Otherwise,click Sample Unknown and enterthe cell rangeof your sam-

TllEr I lf.*.-."'el

li

OK

ll

I

Cancd I

asthe Sample Cell Range. Enter a title as the Title and

oK.

the CIE_TWorksheet workbook. totheCIE-T worksheet of theEIEIEEIEIE (seeFigure8.13on page307)usesthe worksheet l-conjidence level, degrees offreedom) function to

inethe criticalvaluefrom the I distributionand the half-width for the Section 8.5 Saxon Home Company population total example.To adapt

to otherproblems,enterthe appropriate worksheet ationsize, sample mean, sample size, sample standeviation.and confidence level values in the tinted 84 through 88 and enter a new title in cell A I .

THECONFIDENCE 7 COMPUTING FORTHE ESTIMATE INTERVAL TOTALDIFFERENCE comoutethe confidence interval estimate for the total

nceeitherby usingthe PHStat2Estimatefor the Differenceorocedureor bv makins entriesin the workbook.

Usingthe CIE_TDWorksheet the CIE_TD worksheet of This worksheet(seeFigure 8.14 on page 310) uses the TINV(I-conJidence level, degrees of freedom) function to determine the critical value from the I distribution and the interval half-width for the Section 8.5 Saxon Home Improvement Company total difference example.The worksheet also contains a calculation area in c e l l r a n g eD 9 : E 1 6 ,a s s h o w n i n F i g u r e E 8 . 1 , t h a t c o u n t s and sums the differences listed on a DifferencesData worksheet.Figure EB.2 illustratesthe first 6 of the 13 rows in the Difference Data worksheet).

to Open gtr@file.

the

or standarddeviationof difierences of 0i6erencesNol = 0 ofDifersnces=B for DiferencesNol = 0 678

forDifisrences=O ofSouares of Differences

12 -COUtlT(DlFerencoellalalAAl 88 -85 - El1 -SUt(0lfieronc.doata!B:B)

71.n - E r 2 ' ( 4 1 0 1 2 719,6664- E l 3 + E l f 7.

-El5/815

E8.1 FIGURE Calculationsarea in the Difference Data worksheet

326

ExcEL coMPANIoNto chapter8 A

B

Dlficrcncer

2 3 4 5 5

-IlBail^i 66.C

-{A2 - gE_TDltB$10}^2 7.11 €.14rc-(/[3 . gE_riDltBtl0fz 17.fr mt.6161-0[4 - CIE_Tl]lfB$10]^ Z 8. 51.78-!45 - qE_T,-DltBtl0l^2 5"21 18.5751-(A5 - CtE-T:DttB$10r 2

9.tr

FIGUREE8.2 (partia DifferencesData worksheet l)

To adaptthis worksheetto otherproblems,you needto changeboth the CIE_TD and DifferencesDataworksheet. In the CIE_TD worksheet,enterthe appropriatepopulation size,samplesize,and confidencelevel valuesin the tinted cells 84 through86 and entera new title in cell Al. In the DiffererencesData worksheet,enterthe differencesin column A. Then adjustcolumn B by either copying down the

formula in cell Bl3 to all rows with differencedata.if havemore than 12 differences;or by deleting columnB formulas,if you havefewerthan 12 di

E8.8 COMPUTING FINITE POPULATION CORRECTION FACTORS The workbooksfor confidenceinterval estimationsof mean and proportion and for computing the sample neededfor estimatingthe meanor proportion includea worksheetthat calculatesthe confidenceinterval estimate samplesize, using a finite population conection factor Section8.7 on the StudentCD-ROM).(Opento those sheetsfor further information.) If you use PHStat2,you includethesecomputationsby clicking the Finite Correction output option and enteringthe Population beforeclicking OK in the appropriatedialogboxes.