Statistics for Economics & Business

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1 Statistics for Ecoomics & Busiess Cofidece Iterval Estimatio Learig Objectives I this chapter, you lear: To costruct ad iterpret cofidece iterval estimates for the mea ad the proportio How to determie the sample size ecessary to develop a cofidece iterval estimate for the mea or proportio How to use cofidece iterval estimates i auditig

2 Chapter Outlie Cofidece Itervals for the Populatio Mea, µ whe Populatio Stadard Deviatio σ is Kow whe Populatio Stadard Deviatio σ is Ukow Cofidece Itervals for the Populatio Proportio, π Determiig the Required Sample Size Poit ad Iterval Estimates A poit estimate is a sigle umber, a cofidece iterval provides additioal iformatio about the variability of the estimate Lower Cofidece Limit Poit Estimate Width of cofidece iterval Upper Cofidece Limit

3 Poit Estimates We ca estimate a Populatio Parameter Mea Proportio µ π with a Sample Statistic (a Poit Estimate) X p Cofidece Itervals How much ucertaity is associated with a poit estimate of a populatio parameter? A iterval estimate provides more iformatio about a populatio characteristic tha does a poit estimate. Such iterval estimates are called cofidece itervals.

4 Cofidece Iterval Estimate A iterval gives a rage of values: Takes ito cosideratio variatio i sample statistics from sample to sample Based o observatios from 1 sample Gives iformatio about closeess to ukow populatio parameters Stated i terms of level of cofidece e.g. 95% cofidet, 99% cofidet Ca ever be 100% cofidet Cofidece Iterval Example Cereal fill example Populatio has µ = 368 ad σ = 15. If you take a sample of size = 25 you kow 368 ± 1.96 * 15 / 25 = (362.12, ) cotais 95% of the sample meas Whe you do t kow µ, you use X to estimate µ If X = the iterval is ± 1.96 * 15 / 25 = (356.42, ) Sice µ the iterval based o this sample makes a correct statemet about µ. But what about the itervals from other possible samples of size 25?

5 Cofidece Iterval Example (cotiued) Sample # X Lower Limit Upper Limit Cotai µ? Yes Yes No Yes Yes Cofidece Iterval Example (cotiued) I practice you oly take oe sample of size I practice you do ot kow µ so you do ot kow if the iterval actually cotais µ However, you do kow that 95% of the itervals formed i this maer will cotai µ Thus, based o the oe sample you actually selected, you ca be 95% cofidet your iterval will cotai µ (this is a 95% cofidece iterval) Note: 95% cofidece is based o the fact that we used Z = 1.96.

6 Estimatio Process Populatio (mea, µ, is ukow) Radom Sample Mea X = 50 I am 95% cofidet that µ is betwee 40 & 60. Sample Geeral Formula The geeral formula for all cofidece itervals is: Poit Estimate ± (Critical Value)(Stadard Error) Where: Poit Estimate is the sample statistic estimatig the populatio parameter of iterest Critical Value is a table value based o the samplig distributio of the poit estimate ad the desired cofidece level Stadard Error is the stadard deviatio of the poit estimate

7 Cofidece Level Cofidece Level Cofidece the iterval will cotai the ukow populatio parameter A percetage (less tha 100%) Cofidece Level, (1-α) Suppose cofidece level = 95% Also writte (1 - α) = 0.95, (so α = 0.05) A relative frequecy iterpretatio: 95% of all the cofidece itervals that ca be costructed will cotai the ukow true parameter A specific iterval either will cotai or will ot cotai the true parameter No probability ivolved i a specific iterval (cotiued)

8 Cofidece Itervals Cofidece Itervals Populatio Mea Populatio Proportio σ Kow σ Ukow Cofidece Iterval for µ (σ Kow) Assumptios Populatio stadard deviatio σ is kow Populatio is ormally distributed If populatio is ot ormal, use large sample Cofidece iterval estimate: X ± Zα/2 σ where X Z α/2 σ/ is the poit estimate is the ormal distributio critical value for a probability of α/2 i each tail is the stadard error

9 Fidig the Critical Value, Z α/2 Cosider a 95% cofidece iterval: 1 α = 0.95 so α = 0.05 Z α/2 = ± 1.96 α 2 = α 2 = Z uits: X uits: Z α/2 = Z α/2 = 1.96 Lower Cofidece Limit Poit Estimate Upper Cofidece Limit Commo Levels of Cofidece Commoly used cofidece levels are 90%, 95%, ad 99% Cofidece Level 80% 90% 95% 98% 99% 99.8% 99.9% Cofidece Coefficiet, 1 α Z α/2 value

10 Itervals ad Level of Cofidece Samplig Distributio of the Mea α /2 1 α α/2 Itervals exted from X Z α / 2 to X + Z α / 2 σ σ µ x = µ x 1 x 2 Cofidece Itervals x (1-α)x100% of itervals costructed cotai µ; (α)x100% do ot. Example A sample of 11 circuits from a large ormal populatio has a mea resistace of 2.20 ohms. We kow from past testig that the populatio stadard deviatio is 0.35 ohms. Determie a 95% cofidece iterval for the true mea resistace of the populatio.

11 Example (cotiued) A sample of 11 circuits from a large ormal populatio has a mea resistace of 2.20 ohms. We kow from past testig that the populatio stadard deviatio is 0.35 ohms. Solutio: σ X ± Zα/2 = 2.20 ± 1.96 (0.35/ = 2.20 ± µ ) Iterpretatio We are 95% cofidet that the true mea resistace is betwee ad ohms Although the true mea may or may ot be i this iterval, 95% of itervals formed i this maer will cotai the true mea

12 Cofidece Itervals Cofidece Itervals Populatio Mea Populatio Proportio σ Kow σ Ukow Do You Ever Truly Kow σ? Probably ot! I virtually all real world busiess situatios, σ is ot kow. If there is a situatio where σ is kow, the µ is also kow (sice to calculate σ you eed to kow µ.) If you truly kow µ there would be o eed to gather a sample to estimate it.

13 Cofidece Iterval for µ (σ Ukow) If the populatio stadard deviatio σ is ukow, we ca substitute the sample stadard deviatio, S This itroduces extra ucertaity, sice S is variable from sample to sample So we use the t distributio istead of the ormal distributio Assumptios Cofidece Iterval for µ (σ Ukow) Populatio stadard deviatio is ukow Populatio is ormally distributed If populatio is ot ormal, use large sample Use Studet s t Distributio Cofidece Iterval Estimate: X ± tα / 2 (where t α/2 is the critical value of the t distributio with -1 degrees of freedom ad a area of α/2 i each tail) S (cotiued)

14 Studet s t Distributio The t is a family of distributios The t α/2 value depeds o degrees of freedom (d.f.) Number of observatios that are free to vary after sample mea has bee calculated d.f. = - 1 Degrees of Freedom (df) Idea: Number of observatios that are free to vary after sample mea has bee calculated Example: Suppose the mea of 3 umbers is 8.0 Let X 1 = 7 Let X 2 = 8 What is X 3? If the mea of these three values is 8.0, the X 3 must be 9 (i.e., X 3 is ot free to vary) Here, = 3, so degrees of freedom = 1 = 3 1 = 2 (2 values ca be ay umbers, but the third is ot free to vary for a give mea)

15 Studet s t Distributio Note: t Z as icreases Stadard Normal (t with df = ) t-distributios are bellshaped ad symmetric, but have fatter tails tha the ormal t (df = 13) t (df = 5) 0 t Studet s t Table Upper Tail Area df Let: = 3 df = - 1 = 2 α = 0.10 α/2 = α/2 = 0.05 The body of the table cotais t values, ot probabilities t

16 Selected t distributio values With compariso to the Z value Cofidece t t t Z Level (10 d.f.) (20 d.f.) (30 d.f.) ( d.f.) Note: t Z as icreases Example of t distributio cofidece iterval A radom sample of = 25 has X = 50 ad S = 8. Form a 95% cofidece iterval for µ d.f. = 1 = 24, so t α /2 = t = The cofidece iterval is S X ± t α /2 = 50 ± (2.0639) µ

17 Example of t distributio cofidece iterval Iterpretig this iterval requires the assumptio that the populatio you are samplig from is approximately a ormal distributio (especially sice is oly 25). This coditio ca be checked by creatig a: Normal probability plot or Boxplot (cotiued) Cofidece Itervals Cofidece Itervals Populatio Mea Populatio Proportio σ Kow σ Ukow

18 Cofidece Itervals for the Populatio Proportio, π A iterval estimate for the populatio proportio ( π ) ca be calculated by addig a allowace for ucertaity to the sample proportio ( p ) Cofidece Itervals for the Populatio Proportio, π Recall that the distributio of the sample proportio is approximately ormal if the sample size is large, with stadard deviatio σ p = π(1 π ) We will estimate this with sample data (cotiued) p(1 p)

19 Cofidece Iterval Edpoits Upper ad lower cofidece limits for the populatio proportio are calculated with the formula where p ± Z α /2 p(1 p) Z α/2 is the stadard ormal value for the level of cofidece desired p is the sample proportio is the sample size Note: must have X > 5 ad X > 5 Example A radom sample of 100 people shows that 25 are left-haded. Form a 95% cofidece iterval for the true proportio of left-haders

20 Example A radom sample of 100 people shows that 25 are left-haded. Form a 95% cofidece iterval for the true proportio of left-haders. (cotiued) p = 100 * 0.25 = 25 > 5 & (1-p) = 100 * 0.75 = 75 > 5 Make sure the sample is big eough p ± Z α /2 p(1 p)/ = 25/100 ± (0.75)/100 = 0.25 ± 1.96 (0.0433) π Iterpretatio We are 95% cofidet that the true percetage of left-haders i the populatio is betwee 16.51% ad 33.49%. Although the iterval from to may or may ot cotai the true proportio, 95% of itervals formed from samples of size 100 i this maer will cotai the true proportio.

21 Determiig Sample Size Determiig Sample Size For the Mea For the Proportio Samplig Error The required sample size ca be foud to obtai a desired margi of error (e) with a specified level of cofidece (1 - α) The margi of error is also called samplig error the amout of imprecisio i the estimate of the populatio parameter the amout added ad subtracted to the poit estimate to form the cofidece iterval

22 Determiig Sample Size Determiig Sample Size For the Mea X ± Z α / 2 σ Samplig error (margi of error) e = Z α / 2 σ Determiig Sample Size Determiig Sample Size (cotiued) For the Mea e = Z α / 2 σ Now solve for to get = Z 2 2 α / 2 σ 2 e

23 Determiig Sample Size (cotiued) To determie the required sample size for the mea, you must kow: The desired level of cofidece (1 - α), which determies the critical value, Z α/2 The acceptable samplig error, e The stadard deviatio, σ Required Sample Size Example If σ = 45, what sample size is eeded to estimate the mea withi ± 5 with 90% cofidece? 2 Z σ 2 e (1.645) = = 2 (45) 2 = So the required sample size is = 220 (Always roud up)

24 If σ is ukow If ukow, σ ca be estimated whe determiig the required sample size Use a value for σ that is expected to be at least as large as the true σ Select a pilot sample ad estimate σ with the sample stadard deviatio, S Determiig Sample Size Determiig Sample Size (cotiued) For the Proportio e = Z π(1 π) Now solve for to get = Z 2 π (1 π) 2 e

25 Determiig Sample Size (cotiued) To determie the required sample size for the proportio, you must kow: The desired level of cofidece (1 - α), which determies the critical value, Z α/2 The acceptable samplig error, e The true proportio of evets of iterest, π π ca be estimated with a pilot sample if ecessary (or coservatively use 0.5 as a estimate of π) Required Sample Size Example How large a sample would be ecessary to estimate the true proportio defective i a large populatio withi ±3%, with 95% cofidece? (Assume a pilot sample yields p = 0.12)

26 Required Sample Size Example Solutio: For 95% cofidece, use Z α/2 = 1.96 e = 0.03 p = 0.12, so use this to estimate π 2 Zα = /2 = 2 π (1 π) = 2 e 2 (1.96) (0.12)(1 0.12) (0.03) (cotiued) So use = 451 Applicatios i Auditig Six advatages of statistical samplig i auditig Samplig is less time cosumig ad less costly Samplig provides a objective way to calculate the sample size i advace Samplig provides results that are objective ad defesible Because the sample size is based o demostrable statistical priciples, the audit is defesible before oe s superiors ad i a court of law

27 Applicatios i Auditig (cotiued) Samplig provides a estimate of the samplig error Allows auditors to geeralize their fidigs to the populatio with a kow samplig error Ca provide more accurate coclusios about the populatio Samplig is ofte more accurate for drawig coclusios about large populatios Examiig every item i a large populatio is subject to sigificat o-samplig error Samplig allows auditors to combie, ad the evaluate collectively, samples collected by differet idividuals Cofidece Iterval for Populatio Total Amout Poit estimate for a populatio of size N: Populatio total = NX Cofidece iterval estimate: N X ± N ( tα / 2 ) S N N 1 (This is samplig without replacemet, so use the fiite populatio correctio factor i the cofidece iterval formula)

28 Cofidece Iterval for Populatio Total: Example A firm has a populatio of 1,000 accouts ad wishes to estimate the total populatio value. A sample of 80 accouts is selected with average balace of $87.6 ad stadard deviatio of $22.3. Costruct the 95% cofidece iterval estimate of the total balace. Example Solutio N = 1000, = 80, X = 87.6, S = 22.3 S N NX ± N( t ) α /2 N , = (1, 000)(87.6) ± (1, 000)(1.9905) 80 1, = 87, 600 ± 4, The 95% cofidece iterval for the populatio total balace is $82, to $92,362.48

29 Cofidece Iterval for Total Differece Poit estimate for a populatio of size N: Total Differece = ND Where the average differece, D, is: D = i= 1 D where D i i = audited value - origial value Cofidece Iterval for Total Differece (cotiued) Cofidece iterval estimate: N D ± N SD ( tα / 2) N N 1 where S D = i= 1 ( D i 1 D) 2

30 Oe-Sided Cofidece Itervals Applicatio: fid the upper boud for the proportio of items that do ot coform with iteral cotrols Upper boud = p + Zα p(1 p) N N 1 where Z α is the stadard ormal value for the level of cofidece desired p is the sample proportio of items that do ot coform is the sample size N is the populatio size Ethical Issues A cofidece iterval estimate (reflectig samplig error) should always be icluded whe reportig a poit estimate The level of cofidece should always be reported The sample size should be reported A iterpretatio of the cofidece iterval estimate should also be provided

31 Summary Itroduced the cocept of cofidece itervals Discussed poit estimates Developed cofidece iterval estimates Created cofidece iterval estimates for the mea (σ kow) Determied cofidece iterval estimates for the mea (σ ukow) Created cofidece iterval estimates for the proportio Determied required sample size for mea ad proportio cofidece iterval estimates with a desired margi of error Summary (cotiued) Developed applicatios of cofidece iterval estimatio i auditig Cofidece iterval estimatio for populatio total Cofidece iterval estimatio for total differece i the populatio Oe-sided cofidece itervals for the proportio ocoformig Addressed cofidece iterval estimatio ad ethical issues

32 Statistics for Ecoomics & Busiess Additioal Topic (optioal) Estimatio & Sample Size Determiatio For Fiite Populatios Topic Learig Objectives I this topic, you lear: Whe to use a fiite populatio correctio factor i calculatig a cofidece iterval for either µ or π How to use a fiite populatio correctio factor i calculatig a cofidece iterval for either µ or π How to use a fiite populatio correctio factor i calculatig a sample size for a cofidece iterval for either µ or π

33 Use A fpc Whe Samplig More Tha 5% Of The Populatio (/N > 0.05) Cofidece Iterval For µ with a fpc X ± t α/2 S N N 1 Cofidece Iterval For π with a fpc p ± z α / 2 p(1 p) N N 1 A fpc simply reduces the stadard error of either the sample mea or the sample proportio Cofidece Iterval for µ with a fpc Suppose N = 1000, = 100, X = 50, s = 10 s N 95% CI for µ : X ± tα / 2 N ± = ± 1.88 = (48.12, 51.88) =

34 Determiig Sample Size with a fpc Calculate the sample size ( 0 ) without a fpc For µ: For π: Z σ 2 2 α /2 = 0 2 e Z π (1 π ) α e 2 /2 = 0 2 Apply the fpc utilizig the followig formula to arrive at the fial sample size (). = 0 N / ( 0 + (N-1)) Topic Summary Described whe to use a fiite populatio correctio i calculatig a cofidece iterval for either µ or π Examied the formulas for calculatig a cofidece iterval for either µ or π utilizig a fiite populatio correctio Examied the formulas for calculatig a sample size i a cofidece iterval for either µ or π

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