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Ch08(2)

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Ch08(2)

2. 2. Chap 8-2Copyright ©2012 Pearson Education Chap 8-2Learning ObjectivesIn this chapter, you learn: To construct and interpret confidence interval estimatesfor the mean and the proportion How to determine the sample size necessary todevelop a confidence interval estimate for the mean orproportion How to use confidence interval estimates in auditing
3. 3. Chap 8-3Copyright ©2012 Pearson Education Chap 8-3Chapter Outline Confidence Intervals for the PopulationMean, μ when Population Standard Deviation σ is Known when Population Standard Deviation σ is Unknown Confidence Intervals for the PopulationProportion, π Determining the Required Sample Size
4. 4. Chap 8-4Copyright ©2012 Pearson Education Chap 8-4Point and Interval Estimates A point estimate is a single number, a confidence interval provides additionalinformation about the variability of the estimatePoint EstimateLowerConfidenceLimitUpperConfidenceLimitWidth ofconfidence intervalDCOVA
5. 5. Chap 8-5Copyright ©2012 Pearson Education Chap 8-5We can estimate aPopulation Parameter …Point Estimateswith a SampleStatistic(a Point Estimate)MeanProportion pπXμDCOVA
6. 6. Chap 8-6Copyright ©2012 Pearson Education Chap 8-6Confidence Intervals How much uncertainty is associated with apoint estimate of a population parameter? An interval estimate provides moreinformation about a population characteristicthan does a point estimate. Such interval estimates are called confidenceintervals.DCOVA
7. 7. Chap 8-7Copyright ©2012 Pearson Education Chap 8-7Confidence Interval Estimate An interval gives a range of values: Takes into consideration variation in samplestatistics from sample to sample Based on observations from 1 sample Gives information about closeness tounknown population parameters Stated in terms of level of confidence e.g. 95% confident, 99% confident Can never be 100% confidentDCOVA
8. 8. Chap 8-8Copyright ©2012 Pearson Education Chap 8-8Confidence Interval ExampleCereal fill example Population has µ = 368 and σ = 15. If you take a sample of size n = 25 you know 368 1.96 * 15 / = (362.12, 373.88) contains 95% ofthe sample means When you don’t know µ, you use X to estimate µ If X = 362.3 the interval is 362.3 1.96 * 15 / = (356.42,368.18) Since 356.42 ≤ µ ≤ 368.18 the interval based on this samplemakes a correct statement about µ.But what about the intervals from other possible samplesof size 25?2525DCOVA
9. 9. Chap 8-9Copyright ©2012 Pearson Education Chap 8-9Confidence Interval Example(continued)Sample # XLowerLimitUpperLimitContainµ?1 362.30 356.42 368.18 Yes2 369.50 363.62 375.38 Yes3 360.00 354.12 365.88 No4 362.12 356.24 368.00 Yes5 373.88 368.00 379.76 YesDCOVA
10. 10. Chap 8-10Copyright ©2012 Pearson Education Chap 8-10Confidence Interval Example In practice you only take one sample of size n In practice you do not know µ so you do notknow if the interval actually contains µ However, you do know that 95% of the intervalsformed in this manner will contain µ Thus, based on the one sample you actuallyselected, you can be 95% confident yourinterval will contain µ (this is a 95% confidenceinterval)(continued)Note: 95% confidence is based on the fact that we used Z = 1.96.DCOVA
11. 11. Chap 8-11Copyright ©2012 Pearson Education Chap 8-11Estimation Process(mean, μ, isunknown)PopulationRandom SampleMeanX = 50SampleI am 95%confident thatμ is between40 & 60.DCOVA
12. 12. Chap 8-12Copyright ©2012 Pearson Education Chap 8-12General Formula The general formula for all confidenceintervals is:Point Estimate (Critical Value)(Standard Error)Where:•Point Estimate is the sample statistic estimating the populationparameter of interest•Critical Value is a table value based on the samplingdistribution of the point estimate and the desired confidencelevel•Standard Error is the standard deviation of the point estimateDCOVA
13. 13. Chap 8-13Copyright ©2012 Pearson Education Chap 8-13Confidence Level Confidence Level Confidence the interval will containthe unknown population parameter A percentage (less than 100%)DCOVA
14. 14. Chap 8-14Copyright ©2012 Pearson Education Chap 8-14Confidence Level, (1- ) Suppose confidence level = 95% Also written (1 - ) = 0.95, (so = 0.05) A relative frequency interpretation: 95% of all the confidence intervals that can beconstructed will contain the unknown trueparameter A specific interval either will contain or willnot contain the true parameter No probability involved in a specific interval(continued)DCOVA
15. 15. Chap 8-15Copyright ©2012 Pearson Education Chap 8-15Confidence IntervalsPopulationMeanσ UnknownConfidenceIntervalsPopulationProportionσ KnownDCOVA
16. 16. Chap 8-16Copyright ©2012 Pearson Education Chap 8-16Confidence Interval for μ(σ Known) Assumptions Population standard deviation σ is known Population is normally distributed If population is not normal, use large sample Confidence interval estimate:where is the point estimateZα/2 is the normal distribution critical value for a probability of /2 in each tailis the standard errornσ/2ZX αXnσ/DCOVA
17. 17. Chap 8-17Copyright ©2012 Pearson Education Chap 8-17Finding the Critical Value, Zα/2 Consider a 95% confidence interval:Zα/2 = -1.96 Zα/2 = 1.960.05so0.951 αα0.0252α0.0252αPoint EstimateLowerConfidenceLimitUpperConfidenceLimitZ units:X units: Point Estimate01.96/2ZαDCOVA
18. 18. Chap 8-18Copyright ©2012 Pearson Education Chap 8-18Common Levels of Confidence Commonly used confidence levels are90%, 95%, and 99%ConfidenceLevelConfidenceCoefficient, Zα/2 value1.281.6451.962.332.583.083.270.800.900.950.980.990.9980.99980%90%95%98%99%99.8%99.9%1DCOVA
19. 19. Chap 8-19Copyright ©2012 Pearson Education Chap 8-19μμxIntervals and Level of ConfidenceConfidence IntervalsIntervalsextend fromto(1- )x100%of intervalsconstructedcontain μ;( )x100% donot.Sampling Distribution of the Meannσ2/αZXnσ2/αZXxx1x2/2 /21DCOVA
20. 20. Chap 8-20Copyright ©2012 Pearson Education Chap 8-20Example A sample of 11 circuits from a large normalpopulation has a mean resistance of 2.20ohms. We know from past testing that thepopulation standard deviation is 0.35 ohms. Determine a 95% confidence interval for thetrue mean resistance of the population.DCOVA
21. 21. Chap 8-21Copyright ©2012 Pearson Education Chap 8-212.40681.99320.20682.20)11(0.35/1.962.20nσ/2ZXμαExample A sample of 11 circuits from a large normalpopulation has a mean resistance of 2.20ohms. We know from past testing that thepopulation standard deviation is 0.35 ohms. Solution:(continued)DCOVA
22. 22. Chap 8-22Copyright ©2012 Pearson Education Chap 8-22Interpretation We are 95% confident that the true meanresistance is between 1.9932 and 2.4068ohms Although the true mean may or may not bein this interval, 95% of intervals formed inthis manner will contain the true meanDCOVA
23. 23. Chap 8-23Copyright ©2012 Pearson Education Chap 8-23Confidence IntervalsPopulationMeanσ UnknownConfidenceIntervalsPopulationProportionσ KnownDCOVA
24. 24. Chap 8-24Copyright ©2012 Pearson Education Chap 8-24Do You Ever Truly Know σ? Probably not! In virtually all real world business situations, σ is notknown. If there is a situation where σ is known, then µ is alsoknown (since to calculate σ you need to know µ.) If you truly know µ there would be no need to gather asample to estimate it.
25. 25. Chap 8-25Copyright ©2012 Pearson Education Chap 8-25 If the population standard deviation σ isunknown, we can substitute the samplestandard deviation, S This introduces extra uncertainty, sinceS is variable from sample to sample So we use the t distribution instead of thenormal distributionConfidence Interval for μ(σ Unknown)DCOVA
26. 26. Chap 8-26Copyright ©2012 Pearson Education Chap 8-26 Assumptions Population standard deviation is unknown Population is normally distributed If population is not normal, use large sample Use Student’s t Distribution Confidence Interval Estimate:(where tα/2 is the critical value of the t distribution with n -1 degreesof freedom and an area of α/2 in each tail)Confidence Interval for μ(σ Unknown)nStX 2/α(continued)DCOVA
27. 27. Chap 8-27Copyright ©2012 Pearson Education Chap 8-27Student’s t Distribution The t is a family of distributions The tα/2 value depends on degrees offreedom (d.f.) Number of observations that are free to vary aftersample mean has been calculatedd.f. = n - 1DCOVA
28. 28. Chap 8-28Copyright ©2012 Pearson Education Chap 8-28If the mean of these threevalues is 8.0,then X3 must be 9(i.e., X3 is not free to vary)Degrees of Freedom (df)Here, n = 3, so degrees of freedom = n – 1 = 3 – 1 = 2(2 values can be any numbers, but the third is not free to varyfor a given mean)Idea: Number of observations that are free to varyafter sample mean has been calculatedExample: Suppose the mean of 3 numbers is 8.0Let X1 = 7Let X2 = 8What is X3?DCOVA
29. 29. Chap 8-29Copyright ©2012 Pearson Education Chap 8-29Student’s t Distributiont0t (df = 5)t (df = 13)t-distributions are bell-shaped and symmetric, buthave ‘fatter’ tails than thenormalStandardNormal(t with df = ∞)Note: t Z as n increasesDCOVA
30. 30. Chap 8-30Copyright ©2012 Pearson Education Chap 8-30Student’s t TableUpper Tail Areadf .10 .05 .0251 3.078 6.314 12.7062 1.8863 1.638 2.353 3.182t0 2.920The body of the tablecontains t values, notprobabilitiesLet: n = 3df = n - 1 = 2= 0.10/2 = 0.05/2 = 0.05DCOVA4.3032.920
31. 31. Chap 8-31Copyright ©2012 Pearson Education Chap 8-31Selected t distribution valuesWith comparison to the Z valueConfidence t t t ZLevel (10 d.f.) (20 d.f.) (30 d.f.) (∞ d.f.)0.80 1.372 1.325 1.310 1.280.90 1.812 1.725 1.697 1.6450.95 2.228 2.086 2.042 1.960.99 3.169 2.845 2.750 2.58Note: t Z as n increasesDCOVA
32. 32. Chap 8-32Copyright ©2012 Pearson Education Chap 8-32Example of t distributionconfidence intervalA random sample of n = 25 has X = 50 andS = 8. Form a 95% confidence interval for μ d.f. = n – 1 = 24, soThe confidence interval is2.06390.025t/2αt258(2.0639)50nS/2αtX46.698 ≤ μ ≤ 53.302DCOVA
33. 33. Chap 8-33Copyright ©2012 Pearson Education Chap 8-33Example of t distributionconfidence interval Interpreting this interval requires theassumption that the population you aresampling from is approximately a normaldistribution (especially since n is only 25). This condition can be checked by creating a: Normal probability plot or Boxplot(continued)DCOVA
34. 34. Chap 8-34Copyright ©2012 Pearson Education Chap 8-34Confidence IntervalsPopulationMeanσ UnknownConfidenceIntervalsPopulationProportionσ KnownDCOVA
35. 35. Chap 8-35Copyright ©2012 Pearson Education Chap 8-35Confidence Intervals for thePopulation Proportion, π An interval estimate for the populationproportion ( π ) can be calculated byadding an allowance for uncertainty tothe sample proportion ( p )DCOVA
36. 36. Chap 8-36Copyright ©2012 Pearson Education Chap 8-36Confidence Intervals for thePopulation Proportion, π Recall that the distribution of the sampleproportion is approximately normal if thesample size is large, with standard deviation We will estimate this with sample data(continued)np)p(1n)(1σpDCOVA
37. 37. Chap 8-37Copyright ©2012 Pearson Education Chap 8-37Confidence Interval Endpoints Upper and lower confidence limits for thepopulation proportion are calculated with theformula where Zα/2 is the standard normal value for the level of confidence desired p is the sample proportion n is the sample size Note: must have X > 5 and n – X > 5np)p(1/2Zp αDCOVA
38. 38. Chap 8-38Copyright ©2012 Pearson Education Chap 8-38Example A random sample of 100 peopleshows that 25 are left-handed. Form a 95% confidence interval forthe true proportion of left-handersDCOVA
39. 39. Chap 8-39Copyright ©2012 Pearson Education Chap 8-39Example A random sample of 100 people showsthat 25 are left-handed. Form a 95%confidence interval for the true proportionof left-handers./1000.25(0.75)1.9625/100p)/np(1Zp /20.33490.1651(0.0433)1.960.25(continued)DCOVAnp = 100 * 0.25 = 25 > 5 & n(1-p) = 100 * 0.75 = 75 > 5Make surethe sampleis big enough
40. 40. Chap 8-40Copyright ©2012 Pearson Education Chap 8-40Interpretation We are 95% confident that the truepercentage of left-handers in the populationis between16.51% and 33.49%. Although the interval from 0.1651 to 0.3349may or may not contain the true proportion,95% of intervals formed from samples ofsize 100 in this manner will contain the trueproportion.DCOVA
41. 41. Chap 8-41Copyright ©2012 Pearson Education Chap 8-41Determining Sample SizeFor theMeanDeterminingSample SizeFor theProportionDCOVA
42. 42. Chap 8-42Copyright ©2012 Pearson Education Chap 8-42Sampling Error The required sample size can be found to obtaina desired margin of error (e) with a specifiedlevel of confidence (1 - ) The margin of error is also called sampling error the amount of imprecision in the estimate of thepopulation parameter the amount added and subtracted to the pointestimate to form the confidence intervalDCOVA
43. 43. Chap 8-43Copyright ©2012 Pearson Education Chap 8-43Determining Sample SizeFor theMeanDeterminingSample Sizenσ2/αZXnσ2/αZeSampling error(margin of error)DCOVA
44. 44. Chap 8-44Copyright ©2012 Pearson Education Chap 8-44Determining Sample SizeFor theMeanDeterminingSample Sizenσ2/Ze(continued)2222/ σeZnNow solvefor n to getDCOVA
45. 45. Chap 8-45Copyright ©2012 Pearson Education Chap 8-45Determining Sample Size To determine the required sample size for themean, you must know: The desired level of confidence (1 - ), whichdetermines the critical value, Zα/2 The acceptable sampling error, e The standard deviation, σ(continued)DCOVA
46. 46. Chap 8-46Copyright ©2012 Pearson Education Chap 8-46Required Sample Size ExampleIf = 45, what sample size is needed toestimate the mean within ± 5 with 90%confidence?(Always round up)219.195(45)(1.645)eσZn 222222So the required sample size is n = 220DCOVA
47. 47. Chap 8-47Copyright ©2012 Pearson Education Chap 8-47If σ is unknown If unknown, σ can be estimated whendetermining the required sample size Use a value for σ that is expected to beat least as large as the true σ Select a pilot sample and estimate σ withthe sample standard deviation, SDCOVA
48. 48. Chap 8-48Copyright ©2012 Pearson Education Chap 8-48Determining Sample SizeDeterminingSample SizeFor theProportion22e)(1ZnππNow solvefor n to getn)(1Zeππ(continued)DCOVA
49. 49. Chap 8-49Copyright ©2012 Pearson Education Chap 8-49Determining Sample Size To determine the required sample size for theproportion, you must know: The desired level of confidence (1 - ), which determines thecritical value, Zα/2 The acceptable sampling error, e The true proportion of events of interest, π π can be estimated with a pilot sample if necessary (orconservatively use 0.5 as an estimate of π)(continued)DCOVA
50. 50. Chap 8-50Copyright ©2012 Pearson Education Chap 8-50Required Sample Size ExampleHow large a sample would be necessaryto estimate the true proportion defective ina large population within 3%, with 95%confidence?(Assume a pilot sample yields p = 0.12)DCOVA
51. 51. Chap 8-51Copyright ©2012 Pearson Education Chap 8-51Required Sample Size ExampleSolution:For 95% confidence, use Zα/2 = 1.96e = 0.03p = 0.12, so use this to estimate πSo use n = 451450.74(0.03)0.12)(0.12)(1(1.96)e)(1Zn 2222/2 ππ(continued)DCOVA
52. 52. Chap 8-52Copyright ©2012 Pearson Education Chap 8-52Applications in Auditing Six advantages of statistical sampling inauditing Sampling is less time consuming and less costly Sampling provides an objective way to calculate thesample size in advance Sampling provides results that are objective anddefensible Because the sample size is based on demonstrablestatistical principles, the audit is defensible before one’ssuperiors and in a court of lawDCOVA
53. 53. Chap 8-53Copyright ©2012 Pearson Education Chap 8-53Applications in Auditing Sampling provides an estimate of the sampling error Allows auditors to generalize their findings to the populationwith a known sampling error Can provide more accurate conclusions about the population Sampling is often more accurate for drawingconclusions about large populations Examining every item in a large population is subject tosignificant non-sampling error Sampling allows auditors to combine, and thenevaluate collectively, samples collected by differentindividuals(continued)DCOVA
54. 54. Chap 8-54Copyright ©2012 Pearson Education Chap 8-54Confidence Interval forPopulation Total Amount Point estimate for a population of size N: Confidence interval estimate:XNtotalPopulation1nS)2/(NnNtNXN α(This is sampling without replacement, so use the finite populationcorrection factor in the confidence interval formula)DCOVA
55. 55. Chap 8-55Copyright ©2012 Pearson Education Chap 8-55Confidence Interval forPopulation Total: ExampleA firm has a population of 1,000 accounts and wishesto estimate the total population value.A sample of 80 accounts is selected with averagebalance of \$87.6 and standard deviation of \$22.3.Construct the 95% confidence interval estimate of thetotal balance.DCOVA
56. 56. Chap 8-56Copyright ©2012 Pearson Education Chap 8-56Example SolutionThe 95% confidence interval for the population totalbalance is \$82,837.52 to \$92,362.48/2S( )1n22.3 1,000 80(1,000)(87.6) (1,000)(1.9905)1,000 18087,600 4,762.48N nN X N tN22.3S,6.87X80,n,1000NDCOVA
57. 57. Chap 8-57Copyright ©2012 Pearson Education Chap 8-57 Point estimate for a population of size N: Where the average difference, D, is:DNDifferenceTotalConfidence Interval forTotal Differencevalueoriginal-valueauditedDwherenDDin1iiDCOVA
58. 58. Chap 8-58Copyright ©2012 Pearson Education Chap 8-58 Confidence interval estimate:where1nS)( D2/NnNtNDNConfidence Interval forTotal Difference(continued)1)D(12nDSniiDDCOVA
59. 59. Chap 8-59Copyright ©2012 Pearson Education Chap 8-59One-Sided Confidence Intervals Application: find the upper bound for theproportion of items that do not conform withinternal controls where Zα is the standard normal value for the level of confidence desired p is the sample proportion of items that do not conform n is the sample size N is the population size1NnNnp)p(1ZpboundUpper αDCOVA
60. 60. Chap 8-60Copyright ©2012 Pearson Education Chap 8-60Ethical Issues A confidence interval estimate (reflectingsampling error) should always be includedwhen reporting a point estimate The level of confidence should always bereported The sample size should be reported An interpretation of the confidence intervalestimate should also be provided
61. 61. Chap 8-61Copyright ©2012 Pearson Education Chap 8-61Chapter Summary Introduced the concept of confidence intervals Discussed point estimates Developed confidence interval estimates Created confidence interval estimates for the mean(σ known) Determined confidence interval estimates for themean (σ unknown) Created confidence interval estimates for theproportion Determined required sample size for mean andproportion confidence interval estimates with adesired margin of error
62. 62. Chap 8-62Copyright ©2012 Pearson Education Chap 8-62Chapter Summary Developed applications of confidence intervalestimation in auditing Confidence interval estimation for population total Confidence interval estimation for total differencein the population One-sided confidence intervals for the proportionnonconforming Addressed confidence interval estimation and ethicalissues(continued)
63. 63. Chap 8-63Copyright ©2012 Pearson EducationOnline TopicEstimation & Sample SizeDetermination For FinitePopulationsBasic Business Statistics12th EditionGlobal Edition
64. 64. Chap 8-64Copyright ©2012 Pearson EducationTopic Learning ObjectivesIn this topic, you learn: When to use a finite population correction factor incalculating a confidence interval for either µ or π How to use a finite population correction factor incalculating a confidence interval for either µ or π How to use a finite population correction factor incalculating a sample size for a confidence interval foreither µ or π
65. 65. Chap 8-65Copyright ©2012 Pearson EducationUse A fpc When Sampling More Than5% Of The Population (n/N > 0.05)DCOVAConfidence Interval For µ with a fpc/21S N nX tNnConfidence Interval For π with a fpc1)1(2/NnNnppzpA fpc simplyreduces thestandard errorof either thesample mean orthe sample proportion
66. 66. Chap 8-66Copyright ©2012 Pearson EducationConfidence Interval for µ with a fpcDCOVA10s50,X100,n1000,NSuppose)88.51,12.48(88.15011000100100010010984.1501:forCI95% 2/NnNnstX
67. 67. Chap 8-67Copyright ©2012 Pearson EducationDetermining Sample Size with a fpc Calculate the sample size (n0) without a fpc For µ: For π: Apply the fpc utilizing the following formula toarrive at the final sample size (n). n = n0N / (n0 + (N-1))DCOVA2 2/ 20 2Zne2/ 20 2(1 )Zne
68. 68. Chap 8-68Copyright ©2012 Pearson EducationTopic Summary Described when to use a finite population correction incalculating a confidence interval for either µ or π Examined the formulas for calculating a confidenceinterval for either µ or π utilizing a finite populationcorrection Examined the formulas for calculating a sample size ina confidence interval for either µ or π