Chapter 8b 8b-1. n A confidence interval estimate is a range. n that is based on observations from 1 sample

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1 Chapter 8b 8b-1 North Seattle Commuity College Poit Estimates BUS210 Busiess Statistics Chapter 8b Iterval Estimatio We ca estimate a Populatio Parameter with a Sample Statistic. Sice the sample statistic is a sigle umber or poit o a lie, it is kow as a Poit Estimate of the parameter. The statistic x is a poit estimate of µ. The statistic p is a poit estimate of! BUS210: Busiess Statistics Itervals - 2 Itervals There is some ucertaity associated with a poit estimate of a populatio parameter. A poit estimate is a sigle umber. We dot kow how accurate it is. Because of this, we use a iterval estimate It provides more iformatio tha a poit estimate We get a idea of the variability withi the estimate. The iterval is called a cofidece iterval Iterval Estimate Lower Limit Poit Estimate Rage of cofidece iterval Upper Limit A cofidece iterval estimate is a rage that is based o observatios from 1 sample cosiders variatio from sample to sample idicates how close to ukow populatio parameter is stated i terms of % level of cofidece (cot) BUS210: Busiess Statistics Itervals - 3 BUS210: Busiess Statistics Itervals - 4 Iterval Estimate Give a populatio with µ = 368 ad σ = 15. If you take a sample of size = 25 you kow 95% of sample meas lie withi µ x ± z. 95σ x which is (!1.96)15 25 ad (1.96)15 25 so, 95% of sample meas lie betwee ad Which works great if you already kow µ But, what if you dot? (cot) Iterval Estimate Whe you dot kow µ, you use x to estimate µ You would estimate that µ is somewhere withi x ± z.95! x So, if you had a sample where x = 362.3, the the iterval estimate is ± (1.96)(15/ 25) = to Sice µ , the iterval based o this sample makes a correct statemet about µ. Limits But what about the itervals from ay other possible sample of size 25? (cot) Note: For our example, we will use 95% cofidece (Z = 1.96). BUS210: Busiess Statistics Itervals - 5 BUS210: Busiess Statistics Itervals - 6 NSCC BUS210 Itervals

2 Chapter 8b 8b-2 Iterval Estimate Samplig distributio of meas: 1-! of all α/2 x values α/2 Iterval Iterval z! / 2 " x z! / 2 " x does ot icludes µ Sample iclude µ #1 [ [ x ] ] Sample [ [ x ] ] #2 Sample #3 [ [ x ] ] BUS210: Busiess Statistics Itervals - 7 x (cot) Iterval Estimate Typically, you oly take oe sample of size you do ot kow µ you do ot kow if the iterval actually cotais µ However, you do kow that 95% of the itervals will cotai µ So, based o just oe sample, you ca be 95% cofidet your iterval will cotai µ (We say there is a 95% cofidece iterval) Note: For 95% cofidece, we use a Critical Value of Z= (cot) BUS210: Busiess Statistics Itervals - 8 Populatio (µ is ukow) Estimatio Process Radom Sample Sample Critical Value Mea X = 50 I am 95% cofidet that µ is betwee 40 & 60. For all cofidece itervals, the geeral formula is: Where: Margi Of Error = (Critical Value)(Stadard Error) Thus: Iterval Poit Estimate ± Margi Of Error CI = x ± z cv σ x BUS210: Busiess Statistics Itervals - 9 BUS210: Busiess Statistics Itervals - 10 Level Itervals The Level is The cofidece that the iterval will cotai the ukow populatio parameter Expressed as a percetage (<100%) Also writte as (1 - α) If cofidece level = 95%, the (1- α) = 0.95 & α = 0.05 Ay specific iterval will either cotai or ot cotai the true parameter No probability ivolved i a specific iterval σ Kow Populatio Mea Itervals σ Ukow Populatio Proportio BUS210: Busiess Statistics Itervals - 11 BUS210: Busiess Statistics Itervals - 12 NSCC BUS210 Itervals

3 Chapter 8b 8b-3 Iterval of Mea σ Kow Assumptios Populatio stadard deviatio σ is kow Populatio is ormally distributed If populatio is ot ormal, use large sample iterval estimate: x ± z!/ 2 " where z α/2 is the critical value for a probability of /2 i each tail Cosider a 95% cofidece iterval:! 2 = Z uits: X uits: σ Kow Fidig the Critical Value 1!" = 0.95 so " = 0.05 Z α/2 = Z α/2 = 1.96 Lower Limit Poit Estimate Upper Limit! 2 = BUS210: Busiess Statistics Itervals - 13 BUS210: Busiess Statistics Itervals - 14 Levels of Itervals Samplig Distributio of the Mea Most commo values Level 80% 90% 95% 98% 99% 99.8% 99.9% Coefficiet, Critical Value 1!" z! Each iterval rages " X ± Z! / 2 ( ) of possible itervals do ot cotai µ; α /2 1 α α/2 µ x = µ x 2 x 1 x (1- ) of possible itervals cotai µ; BUS210: Busiess Statistics Itervals - 15 Itervals BUS210: Busiess Statistics Itervals - 16 Iterval of Mea σ Kow 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. X ± Z!/ 2 " = 2.20 ±1.96 (0.35/ 11) = 2.20 ± so, # µ # Iterval of Mea σ Kow Iterpretatio: Although the true mea may or may ot be i this iterval, 95% of itervals formed i this maer will cotai the true mea So, we ca safely state, We are 95% cofidet that the true mea resistace for the populatio is betwee ad ohms BUS210: Busiess Statistics Itervals - 17 BUS210: Busiess Statistics Itervals - 18 NSCC BUS210 Itervals

4 Chapter 8b 8b-4 Itervals Iterval of Mea σ Ukow σ Kow Populatio Mea Itervals σ Ukow Populatio Proportio Do You Ever Truly Kow σ? Probably ot! I most real world busiess situatios, σ is ot kow. If σ is kow, the µ is also kow Because you eed to kow µ to calculate σ! If you kow µ, you dot eed a sample to estimate it. BUS210: Busiess Statistics Itervals - 19 BUS210: Busiess Statistics Itervals - 20 Iterval of Mea σ Ukow If the σ is ukow, we ca substitute the sample stadard deviatio, s But s varies from sample to sample which itroduces extra ucertaity. So we use the t distributio istead of the z (ormal) distributio Iterval of Mea σ Ukow Assumptios: Populatio stadard deviatio, µ, is ukow Populatio is ormally distributed If populatio is ot ormal, use large sample Iterval Estimate: x ± t! / 2 s where: t α/2 is the critical value with -1 degrees of freedom BUS210: Busiess Statistics Itervals - 21 BUS210: Busiess Statistics Itervals - 22 Studet s t Distributio The t distributio is a family of distributios The t α/2 value depeds o degrees of freedom (d.f.) which is the umber of observatios that are free to vary after sample mea has bee calculated for a sample of oe variable: d.f. = - 1 BUS210: Busiess Statistics Itervals - 23 Degrees of Freedom (d.f.) Number of observatios that are free to vary after sample mea has bee calculated Suppose the mea of 3 umbers is 8.0 Let X 1 = 7 Let X 2 = 8 What is X 3? Sice 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 of the values ca be ay umber, but the third value is ot free to vary for a give mea) BUS210: Busiess Statistics Itervals - 24 NSCC BUS210 Itervals

5 Chapter 8b 8b-5 Studet s t Distributio Studet s t Distributio t-distributios are bell-shaped ad symmetric, but fatter tha the ormal. Stadard Normal (t with df = ) 0 t (df = 13) t (df = 5) t Upper Tail Area df The body of the table cotais t values, ot probabilities Let: = 3 df = - 1 = 2 α = 0.10 α/2 = α/2 = 0.05 t BUS210: Busiess Statistics Itervals - 25 BUS210: Busiess Statistics Itervals - 26 Studet s t Distributio t t t Z Level (10 d.f.) (20 d.f.) (30 d.f.) ( d.f.) Note: t approaches Z as icreases Iterval of Mea σ Ukow A radom sample of = 25 has X = 50 ad S = 8. Form a 95% cofidece iterval for µ d.f. = - 1 = 24, so t!/2 = t = The cofidece iterval is: X ± t!/2 S = 50 ± (2.0639) µ BUS210: Busiess Statistics Itervals - 27 BUS210: Busiess Statistics Itervals - 28 Itervals Iterval for Populatio Proportio σ Kow Populatio Mea Itervals σ Ukow Populatio Proportio A iterval estimate for π ca be obtaied from the sample proportio, p. But, must add a allowace for ucertaity. We will estimate this usig sample data, so the stadard deviatio ow becomes:! p = p(1" p) We use p from the sample istead of π BUS210: Busiess Statistics Itervals - 29 BUS210: Busiess Statistics Itervals - 30 NSCC BUS210 Itervals

6 Chapter 8b 8b-6 Therefore: The upper ad lower cofidece limits for the populatio are calculated with the formula Iterval for Populatio Proportio p ± z!/2 p(1" p) where Z α/2 is the stadard ormal value for the level of cofidece desired p is the sample proportio is the sample size Note: To do this, we must have p > 5 ad (1-p) > 5 BUS210: Busiess Statistics Itervals - 31 Iterval for Populatio Proportio A radom sample of 100 people shows that 25 are left-haded. Form a 95% cofidece iterval for the true proportio of left-haders. p ± Z!/ 2 p(1" p)/ = 25/100 ± (0.75)/100 = 0.25 ±1.96(0.0433) = to BUS210: Busiess Statistics Itervals - 32 Iterval for Populatio Proportio 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 all itervals formed from samples of size 100 will cotai the true proportio. For the Mea For the Proportio BUS210: Busiess Statistics Itervals - 33 BUS210: Busiess Statistics Itervals - 34 The required sample size ca be foud give a desired margi of error (e), ad a specified level of cofidece (1 - α) The margi of error is also called samplig error. It is the amout of imprecisio i estimatig the populatio parameter. added ad subtracted to the poit estimate to form the cofidece iterval. For the Mea: Sice CI = x ± z! / 2 ", the e = z! / 2 Rearragig ad solvig for, we get = z 2! / 2 " 2 e 2 Samplig error (margi of error) " BUS210: Busiess Statistics Itervals - 35 BUS210: Busiess Statistics Itervals - 36 NSCC BUS210 Itervals

7 Chapter 8b 8b-7 Therefore: To determie the required sample size for the mea, you must kow 3 thigs: The desired level of cofidece (1 - α), which determies the critical value, z α/2 The acceptable samplig error, e The stadard deviatio, σ If σ = 45, what sample size is eeded to estimate the mea withi ± 5 with 90% cofidece? = z2! 2 = (1.645)2 (45) 2 = " 220 e (Always roud up) So, the required sample size is = 220 BUS210: Busiess Statistics Itervals - 37 BUS210: Busiess Statistics Itervals - 38 If σ is ukow, it ca be estimated by Select a pilot sample ad use the sample stadard deviatio, s, as a estimate of σ For the Proportio: Sice! p = "(1# " ), the e = z $ / 2! p = z $ / 2 "(1# " ) Samplig error (margi of error) Solvig for = z2! (1"! ) e 2 BUS210: Busiess Statistics Itervals - 39 BUS210: Busiess Statistics Itervals - 40 Required Example Therefore: To determie the required sample size for the proportio, you must kow 3 thigs: 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 If ecessary, π ca be estimated with a pilot sample (or coservatively use 0.5 as a estimate of π) (Assume a pilot sample yields p = 0.12) How large a sample would be ecessary to estimate the true proportio defective i a large populatio withi ±3%, with 95% cofidece? = z!/ 2 2 " (1# " ) e 2 = (1.96)2 (0.12)(1# 0.12) (0.03) 2 = So, the required sample size is = 451 BUS210: Busiess Statistics Itervals - 41 BUS210: Busiess Statistics Itervals - 42 NSCC BUS210 Itervals

8 Chapter 8b 8b-8 Ethical Issues Whe reportig a poit estimate, you should always iclude a cofidece iterval estimate i order to reflect samplig error. report the level of cofidece. report the sample size. provide a iterpretatio of the cofidece iterval estimate BUS210: Busiess Statistics Itervals - 43 NSCC BUS210 Itervals