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IntroductiontoEstimation Basicdefinitionsandconcepts Theassignmentofvalue(s)toapopulationparameter basedonavalueofthecorrespondingsamplestatisticis called estimation calledestimation. POINT&INTERVALESTIMATION AND (INTRODUCTIONTOTESTING) Thevalue(s)assignedtoapopulationparameterbasedon The value(s) assigned to a population parameter based on thevalueofasamplestatisticiscalledanestimate. Thesamplestatisticusedtoestimateapopulation parameteriscalledanestimator. 2 Estimationsteps PointEstimators y APointEstimation Theestimationprocedureinvolvesthefollowingsteps: 1 1. 2. 3. 4. Thevalueofasamplestatisticthatisusedtoestimatea The value of a sample statistic that is used to estimate a populationparameteriscalledapointestimate.Usually, wheneverweusepointestimation,wecalculatethemarginof error associated with that point estimation which is errorassociatedwiththatpointestimation,whichis calculatedasfollows: Margin of error r1.96V x or r 1.96s x Pointestimateisbasedonjustonesample,wecannot expectittobeequaltothecorrespondingpopulation parameter.Indeed,eachsamplewillhaveadifferent, non of them is equal to P. But they are all unbiased nonofthemisequaltoP.Buttheyareallunbiased estimatesofP.(Recallthatunbiased meanstheir expectedvalueisequaltoP.) Selectasample. Select a sample Collecttherequiredinformationfromthemembersof the sample. thesample. Calculatethevalueofthesamplestatistic. Assignvalue(s)tothecorrespondingpopulation g () p gp p parameter. 3 4 Statistics PointEstimates y Statistics y Parameters ¾ Astatisticisapropertyofasamplefromthe ¾ Instatisticalinference,thetermparameter isusedtodenote aquantity,say,thatisapropertyofanunknownprobability T distribution. ¾ Forexample,themean,variance,oraparticularquantile p , , , p q of theprobabilitydistribution ¾ Parametersareunknown,andoneofthegoalsofstatistical inference is to estimate them inferenceistoestimatethem. y Estimation ¾ Aprocedureof“guessing”propertiesofthepopulationfrom which data are collected whichdataarecollected. ¾ Apointestimateofanunknownparameterisastatisticthat representsa“guess”oftheparameterofinterest. ¾ Theremaybemorethanonesensiblepointestimateofa There may be more than one sensible point estimate of a parameter. population population. ¾ Astatisticisdefinedtobeanyfunctionofrandom variables.So,itisalsoarandomvariable.For example,thesamplemean,samplevariance,ora particularsamplequantile. ¾ Theobservedvalueofthestatisticcanbecalculated The observed value of the statistic can be calculated fromtheobserveddatavaluesofrandomvariables. Examples of statistics: X1 X 2 X n sample mean X n sample variance S 2 5 ¦ n i 1 ( X i X )2 n 1 6 Therelationshipbetweenanunknown parameteranditspointestimator PropertiesofEstimatorsthatWeDesire y Unbiasedness: U bi d ˆ E( T) T Inotherwordswewouldwishthattheexpectedvalueof theestimatoristhesameasitstruevalue. Wedefinebiasofanestimator asthedifferencebetween theexpectedvalueoftheestimatorandthetruevaluein p p thepopulation: y Efficiency:wewishtominimizethemeansquareerror aroundthetruevalue.Theefficiencytellsushowwellthe estimatorperformsinpredicting.Amongunbiased estimatorstherefore,wewanttheonewiththesmallest variance. y 7 Consistency.Assamplesizeincreases,variationofthe estimatorfromthetruepopulationvaluedecreases. 8 Unbiasedness Efficiency P(X) Unbiased P(X) Sampling Biased Distribution of Median P Sampling Distribution of Mean P 9 10 Consistency Intervalestimation: General approach Generalapproach Larger sample size P(X) B Smaller sample size A P 11 12 IntervalEstimation ConfidenceIntervalEstimation Definition In interval estimation an interval is constructed around Inintervalestimation,anintervalisconstructedaround thepointestimate,anditisstatedthatthisintervalislikely tocontainthecorrespondingpopulationparameter. y Outline: y Procedure: y 1. Sample o point estimator ( X or p ) y 2. 2 Confidence level and Table o Z or tn-11 y 3. Formulas o compute UCL and LCL: y Px 13 P $1130 x point estimator r margin p g of error $1370 $1610 IntervalEstimationofthePopulationMean Intervalestimationofapopulation mean: The case of known mean:Thecaseofknown Eachintervalisconstructedwithregardtoagivenconfidencelevel andiscalledaconfidenceinterval. The confidence level associated with a confidence interval states how Theconfidencelevelassociatedwithaconfidenceintervalstateshow muchconfidencewehavethatthisintervalcontainsthetrue populationparameter.Theconfidencelevelisdenotedby(1– )100%. ) % The(1– )100%confidenceintervalfor (populationmean)is: x r zV x if V is known and x r zs x if V is not known, where Vx V / n and sx s / n Thevalueofz usedherecanbefoundfromthestandardnormal distributiontable,forthegivenconfidencelevel. Themaximumerrorofestimatefor,denotedbyE,isthequantity that is subtracted from and added to the value of x toobtaina thatissubtractedfromandaddedtothevalueofx to obtain a confidenceintervalfor.Thus, E zV x or zs x 17 16 IntervalEstimationofthePopulationMean when isknown:Example when is known: Example Apublishingcompanyhasjustpublishedanewcollegetextbook. p y p Beforethecompanydecidesthepriceatwhichtosellthis textbook,itwantstoknowtheaveragepriceofallsuch textbooksinthemarket.Theresearchdepartmentatthe company took a sample of 36 comparable textbooks and companytookasampleof36comparabletextbooksand collectedinformationontheirprices.Thisinformationproduces ameanpriceof$70.50forthissample.Itisknownthatthe standarddeviationofthepricesofallsuchtextbooksis$4.50. d dd i i f h i f ll h b k i $4 50 (a) Whatisthepointestimateofthemeanpriceofallsuch textbooks? What is the margin of error for the estimate? textbooks?Whatisthemarginoferrorfortheestimate? (b) Constructa90%confidenceintervalforthemeanpriceofall suchcollegetextbooks. 18 IntervalEstimationofthePopulationMeanwhen is known:AnswerstotheExample x Herewetakeadvantageofourknowledgeondistributionofto developaconfidenceintervalforP. a) n=36,x =$70.50,and =$4.5,thus: V x V 4.50 $.75 n 36 Pointestimateof =x =$70.50 ( 75) r$1.47 Margin of error = r 1.96V x r1.96(. Marginoferror= b) Confidencelevelis90%or.90;andz =1.65. x r zV x 70.50 r 1.65(. ( 75) 70.50 r 1.24 (70.50 - 1.24) to (70.50 1.24) $69.26 to $71.74 Basedonourresults,wecansaythatweare90%confidentthat the mean price of all such college textbooks is between themeanpriceofallsuchcollegetextbooksisbetween $69.26and$71.74. 19 1. Interval Estimation for Population p Mean Example 1: ( known case) In an effort to estimate the mean amount spent per customer for dinner at a major Atlanta restaurant. Data were collected for a sample of 49 customers over a three-week period. Assume a population standard deviation of $5. $5 a. At the 95% confidence, what is the margin error? b. If the sample mean is $24.80, What is the 95% confidence interval for the population mean? Example 1: Answer: • n = 49 X $24.8 V=$5 V $5 • Z: (1-D)/2 = 0.95/2 = 0.475 o Table 1: Z = 1.96 V 5 • 1. ZD / 2 x (1.96) 1.4 2. UCL n 49 X ZD /2 X ZD /2 LCL P: [23.4, 26.2] V x n V x n 24 . 8 1 . 4 26 . 2 24 . 8 1 . 4 23 . 4 IntervalEstimationofthePopulationMean when isunknown Intervalestimationofapopulation mean: The case of unknown mean:Thecaseofunknown y InsteadofpopulationstandarddeviationVwehave sample standard deviation s. samplestandarddeviations. y Insteadofnormaldistribution,wehavetdistribution , Thetdistribution isusedtoconstructaconfidence intervalabout if: 1. Thepopulationfromwhichthesampleisdrawnis Th l ti f hi h th l i d i (approximately)normallydistributed; 2. Thesamplesizeissmall(thatis,n<30); p ( , ); 3. Thepopulationstandarddeviation,,isnotknown. 22 23 Thet Distribution Thetdistribution isaspecifictypeofbellshapeddistributionwith alowerheightandawiderspreadthanthestandardnormal p g , distribution.Asthesamplesizebecomeslarger,thet distributionapproachesthestandardnormaldistribution.A specifictdistributiondependsononlyoneparameter,calledthe degrees of freedom (df) The mean of the t distribution is equal degreesoffreedom(df).Themeanofthetdistributionisequal to0anditsstandarddeviationisfoundby.The df /(df 2) graphbelowdepictsthecaseofdf=3. Thestandarddeviationofthe standardnormaldistributionis1.0 24 Thet Distribution:Example Findthevalueoft for16degreesoffreedomand.05areain th i ht t il f t distributioncurve. therighttailofa di t ib ti Area in the Right Tail Under the t Distribution Curve Thestandarddeviationofthet distributionis 9 /(9 2) 1.134 =0 25 df .10 10 .05 05 .025 025 … .001 001 1 2 3 . 16 . 3.078 1.886 1.638 … 1 337 1.337 … 6.314 2.920 2.353 … 1 746 1.746 … 12.706 4.303 3.182 … 2 120 2.120 … … … … … … … 318.309 22.327 10.215 … 3 686 3.686 … Therequiredvalueoft for16 df and.05areaintherighttail. ConfidenceIntervalforPopulationmean Using the t Distribution thet Thet Distribution(continued) Thet distributionwith16degreesoffreedom,areasunderthe right and the left tails rightandthelefttails. The (1 – )100%confidenceinterval The(1 )100% confidence interval for for is x r ts x where s x s n Thevalueoft isobtainedfromthet distributiontableforn – 1degreesoffreedomandthegivenconfidencelevel. d ff d d h fd l l .05 26 -1.746 1.746 0 27 ConfidenceIntervalforPopulationmean Using thet Distribution:Example Dr.Moorewantedtoestimatethemeancholesterollevelfor all adult men living in Hartford. He took a sample of 25 adult alladultmenlivinginHartford.Hetookasampleof25adult menfromHartfordandfoundthatthemeancholesterol levelforthissampleis186withastandarddeviationof12. Assumethatthecholesterollevelsforalladultmenin h h h l ll l f ll d l Hartfordare(approximately)normallydistributed.Construct a 95% confidence interval for the population mean . a95%confidenceintervalforthepopulationmean. ConfidenceIntervalforPopulationmean Using thet Distribution:ExampleAnswered y Confidencelevelis95%or.95,withdf =n – 1=25– 1=24 y Areaineachtail=.5– (.95/2)=.5 .4750=.025 s 12 y Thevalueoft intherighttailis2.064,and s x n 25 2.40 df = 24 .025 .025 .4750 .4750 x r tsx 186 r 2.064(2.40) 186 r 4.95 181.05 to 190.95 y Thus,wecanstatewith95%confidencethatthemean 28 Example 2: ( known case) The mean flying time for pilots at Continental Airlines is 49 hours p per month. This mean was based on a sample of 100 pilots and the sample standard deviation was 8.5 hours. a. At 95% confidence, what is the margin of error? b What is the 95% confidence interval estimate of b. the population mean flying time? c. The mean flying time for pilots at United Airlines is 36 hours per month. Discuss difference between the flying times at two airlines. 29 cholesterollevelforalladultmenlivinginHarfordliesbetween h l t ll lf ll d lt li i i H f d li b t 181.05and190.95. Example 2: Given: n = 100,,X = 49,, S = 8.5,, 1-D = .95 Think: What to estimate? Use Z or t? Answer: • Sample info (given): n = 100,X = 49, S = 8.5 • t: 1- D=0.95, so D/2=0.025, d.f.=n-1=99 o Table 2: d.f.=100, d f =100 D/2=0.025 /2=0 025 o t=1.984 t=1 984 d.f.=80, D/2=0.025 o t=1.990 Interpolation: t 1.984 (1.990 1.984) 100 99 1.9843 *Interpolation: 100 80 S 8 .5 a. m.o.e.: m .o .e . t 1 . 9843 1 . 69 D /2 n b. UCL = 49+1.69 =50.69 P: [47.31, [47 31 50 50.69] 69] 100 LCL = 49 – 1.69 = 47.31 c. 36 < LCL. The mean flying time is lower at United. Example Solution Twentyfiverandomlyselectedadultswhobuybooks forgeneralreadingwereaskedhowmuchthey usually spend on books per year. usuallyspendonbooksperyear. y Confidence level is 99% or .99 y sx s 300 n $60 25 y df = n – 1 = 25 – 1 = 24 Thesampleproducedameanof$1450anda standarddeviationof$300forsuchannualexpenses. d dd i i f $300 f h l y Area in each tail = .5 5 – (.99/2) ( 99/2) = .5 5 - .4950 4950 = .005 005 y The values of t are 2.797 and -2.797 Assume that such expenses for all adults who buy Assumethatsuchexpensesforalladultswhobuy booksforgeneralreadinghaveanapproximate normaldistribution. y The 99% confidence interval for is x r ts x Determinea99%confidenceintervalforthe correspondingpopulationmean. p gp p 32 $1450 r 2.797(60) $1450 r $167.82 $1282.18 to $1617.82 33 Intervalestimationofapopulationproportion: g p Thecaseoflargesamples Intervalestimationofapopulation proportion: The case of large samples proportion:Thecaseoflargesamples EstimatoroftheStandardDeviationof p̂ Thevalueof,whichgivesapointestimateof,is The value of s pˆ which gives a point estimate of V p̂ is calculatedas pˆ qˆ s pˆ n The(1– )100%confidenceintervalforthepopulation proportion,p,is pˆ r zs pˆ Thevalueofz usedhereisobtainedfromthestandard normaldistributiontableforthegivenconfidencelevel,and pˆ qˆ/n s pˆ 34 35 Intervalestimationofapopulationproportionin caseoflargesamples:Example Intervalestimationofapopulationproportionin caseoflargesamples:Examplesolved Accordingtoa2002surveybyFindLaw.com,20%of Americansneededlegaladviceduringthepastyearto resolvesuchthornyissuesasfamilytrustsandlandlord disputes(CBS.MarketWach.com,August6,2002).Suppose a recent sample of 1000 adult Americans showed that 20% arecentsampleof1000adultAmericansshowedthat20% ofthemneededlegaladviceduringthepastyeartoresolve suchfamilyrelatedissues. a) Whatisthepointestimateofthepopulation proportion?Whatisthemarginoferrorofthis estimate? b) Find,witha99%confidencelevel,thepercentageofall adultAmericanswhoneededlegaladviceduringthe pastyeartoresolvesuchfamilyrelatedissues. p̂ y n =1000,=.20,and,=.80 q̂ 36 y Notethatandarebothgreaterthan5. Note that npˆ and nqˆ are both greater than 5 s pˆ pˆ qˆ n (.20)(.80) 1000 .01264911 a)Pointestimateofp ==.20 p̂ s pˆ M i f Marginoferror=±1.96=±1.96(.01264911)=± ±1 96 ±1 96( 01264911) ± .025or±2.5% 025 ±2 5% b)Theconfidencelevelis99%,or.99. Thez valuefor.4950isapproximately2.58. pˆ r zss pˆ .20 20 r 2.58(.0126 2 58( 01264911) .20 20 r .033 033 37 .167 to .233 or 16.7% to 23.3% Determiningthesampleforestimationofthe mean y Giventheconfidencelevelandthestandarddeviationofthe population,thesamplesizethatwillproducea population, the sample size that will produce a predeterminedmaximumerrorE oftheconfidenceinterval estimateof is: V z 2V 2 E zV x z. n 2 ,with n E y Example: Analumniassociationwantstoestimatethemeandebtofthisyear’s collegegraduates.Itisknownthatthepopulationstandard deviation of the debts of this year’sscollegegraduatesis$11,800. deviationofthedebtsofthisyear college graduates is $11 800 Howlargeasampleshouldbeselectedsothattheestimatewitha 99%confidenceleveliswithin$800ofthepopulationmean? z 2V 2 ( 2 .58 ) 2 (11,800 ) 2 n 1448 .18 | 1449 38 E2 (800 ) 2 Determiningthesamplesizeforestimatingthe p p proportion y Given the confidence level and the values of p and q, the sample size that will produce a predetermined maximum error E of the confidence interval estimate of p is: n makesapartthatisusedinclocks.Thecompanywantstoestimatethe proportionofthesepartsproducedbythismachinethataredefective. Thecompanymanagerwantsthisestimatetobewithin.02ofthe populationproportionfora95%confidencelevel.Whatisthemost conservativeestimateofthesamplesizethatwilllimitthemaximum errortowithin.02ofthepopulationproportion? y Answer: y Thevalueofzfora95%confidencelevelis1.96; p =.5andq =.5 y n 2 (1.96) (.50)(.50) (.02) 2 and calculate to find n. p̂p and qq̂ from this sample sample. Then use them 39 Determiningthesamplesizeforestimatingthe p p proportion:Example(continued) p ( ) y ConsiderthepreviousExampleagain.Supposeapreliminary sampleof200partsproducedbythismachineshowedthat sample of 200 parts produced by this machine showed that 7%ofthemaredefective.Howlargeasampleshouldthe p y p is companyselectsothatthe95%confidenceintervalfor within.02ofthepopulationproportion? y Answer: p̂ n y Thus,ifthecompanytakesasampleof2401parts,thereis95%chance thattheestimateofp willbewithin.02ofthepopulationproportion. 41 q̂ =.07and=.93 ,thus: z 2 pˆ qˆ E2 (1.96) 2 (.07)(.93) (.02) 2 (3.8416)(. )( 07)(. )( 93) .0004 625.22 | 626 INTERVALESTIMATIONOFAPOPULATIONMEAN: LARGESAMPLES AnIntervalEstimationcont. ConfidenceIntervalfor forLargeSamples The(1– )100%confidenceintervalfor is y Definition y Eachintervalisconstructedwithregardtoagiven g g x r zV x if V is known confidencelevel andiscalledaconfidenceinterval.The confidencelevelassociatedwithaconfidenceinterval stateshowmuchconfidencewehavethatthisinterval containsthetruepopulationparameter.Theconfidence level is denoted by (1 )100%. levelisdenotedby(1– )100% 42 pq n y Take a preliminary sample of arbitrarily determined size 2401 40 zu by using p = .5 and q = .5. For a given E, these values of p and q will give the largest sample size in comparison to any other pair of values of p = .5 and q = .5 since their product is greater than the product of any other pair pair. 2 z pq E2 zV pˆ y In case the values of p and q are not known: y Take the most conservative estimate of the sample size n Determiningthesamplesizeforestimatingthe proportion:Example y LombardElectronicsCompanyhasjustinstalledanewmachinethat z 2 pq ,with E E2 where x r zs x if V is not known V x V / n and s x s / n Thevalueofz usedhereisreadfromthestandard normaldistributiontableforthegivenconfidence level. 43 Repeat, again: Finding z for a 95% confidence level. Area in the tails – what this means? Total shaded area is .9500 or 95% D D 2 .4750 .4750 x -z -1.96 2 (1 – ) 0 1.96 44 45 Interval Estimation for Population Proportion Example 3: Answer: a. Point estimate of p: Example E l 3 3: A survey asked 346 job seekers. The answer selected most (152 times) was “higher compensation.” a. What is the point estimate of the proportion of job seekers who would select “higher higher compensation compensation” as the reason of changing jobs? a What is the 95% confidence inter a. interval al estimate of the population proportion? z 0 z z p x n 152 346 .4393 b. Confidence interval: • Z: Z (1-D)/2=0.475, (1 )/2 0 475 T Table bl 1 o Z=1.96 Z 1 96 • Margin of error = Z p (1 p ) n (0.4393)(1 0.4393) = 0.0523 346 (1.96) • Confidence interval UCL p m .o .e . . 4393 . 0523 . 4916 LCL p m .o .e . . 4393 . 0523 . 3870 p: [.3870, [ 3870 .4916] 4916] Sample Size and Precision Intervalestimation: Sample size and precision Samplesizeandprecision Quality of estimation: • Confidence level: 1 - D • Precision: margin of error Confidence level: 1 - D is guaranteed by procedure. sampling distribution for sample mean. Probability that X falls between P ZD V /2 x n and P Z D V /2 x n is 1 - D. In general, any sample mean that is within this range will provide an interval that contains the population mean P. Margin of error: Given n, then (1 - D)np o margin of error np. Given 1 - D, then n np o margin of error pn. pn 48 Determine sample size to meet requirements for both g of error: confidence level and margin 1. Determine sample size for estimation of P n ª Z D2 / 2V 2 º « » 2 »¼ ¬« E E: desired margin of error [ ]: round up (“Be conservative”) p size for estimation of p 2. Determine sample n ª Z D2 / 2 p (1 p ) º » « E2 ¬ ¼ p = .5 (“Be conservative”) desired ed margin ag o of e error o E: des [ ]: round up (“Be conservative”) Example 4: Bride’s magazine reported that the mean cost of a wedding is $19,000. Assume that the population standard deviation is $9,400. Use 95% confidence, a. What a at iss the t e recommended eco e ded sa sample p e ssize e if tthe e des desired ed margin ag of error is $1,000? b. What is the recommended sample size if the desired margin of error is $500? Answer: 1D a. Z D / 2 : . 475 o Z D / 2 1 . 96 2 ª (1 . 96 ) 2 ( 9400 ) 2 º n « » >339 . 44 @ | 340 1000 2 ¬ ¼ b. n Example 6: >1357 .78 @ | 1358 Example The League of American Theatres and Producers uses an ongoing audience tracking survey that provides up-to-date i f information ti about b t Broadway B d th theater t audiences. di E Every week, k th the League distributes a one-page survey on random theater seats at a rotation roster of Broadway shows. a. How large a sample should be taken if the desired margin of error on any proportion is 0.04? Use 95% confidence. Answer: 1D . 475 , Table 1 o Z D / 2 1 . 96 2 ª (1 . 96 ) 2 . 5 (1 . 5 ) º « » >600 . 25 @ | 601 (. 04 ) 2 ¬ ¼ According to the analysis of a CNN–USA TODAY–Gallup poll conducted in October 2002, “Stress has become a common part of everyday life in the United States States. The demands of work, family, and home place an increasing burden on the average American.” According to this poll, 40% of Americans included in the survey indicated that they had a limited amount of time to relax (Gallup (Gallup.com, com November 8 8, 2002) 2002). ZD /2 : The poll was based on a randomly selected national sample of 1502 adults aged 18 and older older. n Construct a 95% confidence interval for the corresponding population l ti proportion. ti 53 Solution Example y Confidence level = 95% or .95 According to a report by the Consumer Federation of America,, National Credit Union Foundation,, and the Credit Union National Association, households with negative assets carried g of $15,528 in debt in 2002 an average (CBS.MarketWatch.com, May 14, 2002). y The value of z for .95 / 2 = .4750 is 1.96. s pˆ pˆ r zs pˆ pˆ qˆ n (.40)(.60) 1502 .01264069 Assume that this mean was based on a random sample of 400 households and that the standard deviation of debts for households in this sample was $4200. .40 r 1.96(.01264069) .40 40 r .025 025 Make a 99% confidence interval for the 2002 mean debt for all such households. .375 to .425 or 37.5% to 42.5% 54 ª (1.96 ) 2 (9400 ) 2 º « » 500 2 ¬ ¼ 55 Solution Example y Confidence level 99% or .99 y sx s 4200 n 400 Lombard Electronics Company has just installed a new machine that makes a part that is used in clocks. The company wants to estimate the proportion of these parts produced by this machine that are defective. $210 y The sample is large (n > 30) ¾ Therefore, f x r zs x we use the normal distribution z = 2.58 2 8 The company manager wants this estimate to be within .02 of the population proportion for a 95% confidence level. 15,528 r 2.58(210) 15,528 r 541.80 $14,986.20 to $16,069.80 What is the most conservative estimate of the sample size that will limit the maximum error to within .02 02 of the population proportion? Thus, we can state with 99% confidence that the 2002 mean debt for all households with negative assets was between $14,986.20 $14 986 20 and $16 $16,069.80. 069 80 56 57 Example Solution y The value of z for a 95% confidence level is 1.96. Consider previous example again. y p = .50 and q = .50 n y z 2 pq E2 (1.96) 2 (.50)(.50) ((.02) 2 Suppose a preliminary sample of 200 parts produced by this machine showed that 7% of them are defective. 2401 y Thus, if the company takes a sample of 2401 parts, there is How large a sample should the company select so that the % confidence interval for p is within .02 of the p population p 95% proportion? 95% chance that the estimate of p will be within .02 of the population proportion. 58 59 Solution pp̂ n 60 = .07 z 2 pˆ qˆ E2 and q̂ q = .93 (1.96) 2 (. ( 07)(. )( 93) 2 (.02) (3.8416)(.07)(.93) .0004 625.22 | 626