Review. Statistics and Quantitative Analysis U4320. Review: Sampling. Review: Sampling (cont.) Population and Sample Estimates:

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1 Stattc ad Quattatve Aaly U430 Segmet 6: Cofdece Iterval Prof. Shary O Hallora URL: Revew Populato ad Sample Etmate: Populato Sample N X X Mea = = µ = X = N Varace N ( X µ ) ( X X ) = = σ = = N The mea defe cetral tedecy dtrbuto. The varace defe dpero of the dtrbuto. Revew: Samplg Whe we ample from a populato, our ample hould be repreetatve of the uderlyg populato. That, our ample hould be ubaed. A mple radom ample elected uch a way that each member of the populato ha the ame chace of beg cluded the ample. Samplg varablty the varace of ample etmate aroud populato parameter Th heret the amplg proce. Revew: Samplg (cot.) Two ource of amplg varablty: Samplg error occur by chace It mply the dfferece betwee the value of a ample tattc ad the value of the correpodg populato parameter. Samplg Error = µ x No-ample Error Error that occur the collecto, recordg, ad tabulato of that data. Ug o-radom ample pollg Over-amplg oe cla or group Uder-amplg other cla or group

2 Revew: Samplg (cot.) Example: Coder a populato of fve employee alare: Salary Thouad $ Idvdual Mea 30.8 take a radom ample Samplg error = x µ = = $.87 thouad No-Samplg error = x µ = = $ 53. thouad = $ 53-$.87. = $.66 thouad No-amplg error record ample Salary data Thouad $ Idvdual Mea Samplg Error Oop! Samplg error Salary Thouad $ Idvdual Mea Noamplg Error 3.33 Copyrght 3.67 Shary O'Hallora µ=30.80 Typed 37 tead of 35! Revew: Cetral Lmt Theorem(cot.) If a mple radom ample take from ay populato wth mea µ ad tadard devato σ, Populato µ, σ draw a ample X,σ Sample X, SE(X) make ferece about how good a etmate X of µ. A creae, the amplg dtrbuto of ted toward the true populato mea µ. X Revew: Cetral Lmt Theorem(cot.) The Cetral Lmt Theorem tate that the amplg dtrbuto of the ample mea wll be ormally dtrbuted wth: σ X ~ N µ, Sample Mea E( X) = µ; ad σ Stadard Error of the amplg proce SE( X) = = = =6 Revew: Cetral Lmt Theorem(cot.) Implcato From the Cetral Lmt Theorem we are able to how that eve f the populato ot ormally dtrbuted, but the ample ze large, the amplg dtrbuto of X ca be approxmated by a ormal dtrbuto. Th allow u to ue the tadard ormal table to make ferece about the populato from our ample etmate.

3 Revew: Iferece To make ferece about the populato from a gve ample, though, we make oe correcto: Itead of dvdg by the tadard devato σ, we dvde by the tadard error of the amplg proce: σ SE = We ca the tadardze by covertg oberved value to z-value: X µ Z = SE Ad the ue the tadard ormal table fd the probablty of evet. Revew: Iferece Thk of th proce a oe of chagg hat We frt put o our tattca hat to tudy dtrbuto the abtract We tart wth ome gve dtrbuto wth mea µ ad tadard devato σ We the dcover that the mea of a ample of ze wll be dtrbuted N(µ, σ/ ) Th lke a cotrolled expermet; we get to chooe the tal dtrbuto ourelve Revew: Iferece Ug th reult, we ow put o our practtoer' hat. A a reearcher, we have ome data, but o dea what the real paret dtrbuto. Say we have a ample of ze from a dtrbuto wth tadard devato σ Th ample happe to have mea X The our bet gue that the paret dtrbuto ha mea X a well. Cofdece Iterval Motvato We ow wat to develop tool that allow u to determe how cofdet we are of the that our ample etmate are repreetatve of the uderlyg populato. We kow that, o average, X equal to µ. We wat ome way to expre how cofdet we are that a gve X ear the actual µ of the populato. We do th by cotructg a cofdece terval, whch ome rage aroud X that mot probably cota µ. 3

4 Cofdece Iterval (cot) Defto A cofdece terval cotructed aroud a pot etmate (e.g., X), ad t tated that th terval lkely to cota the correpodg populato parameter (e.g., µ). Two compoet: The tadard error a meaure of how much error there the amplg proce. The level of cofdece attached to the terval. The cofdece level aocated wth a cofdece terval tate how much cofdece we have that the terval cota the true populato parameter. The cofdece level deoted by (-α)00% Commo value are 90%, 95% ad 99% Correpodg α-level are.0,.05, ad.0. Cofdece Iterval: σ kow(cot.) Cotructg a 95% Cofdece Iterval Graph Frt, we kow from the cetral lmt theorem that the ample mea X dtrbuted ormally, wth mea µ ad tadard error σ SE = X Cofdece Iterval (cot.) Secod, we determe how cofdet we wat to be our etmate of µ. Defg how cofdet you wat to be called the α-level. A 95% cofdece terval ha a aocated α-level of.05. We fd a rage uder the curve wth area of If we are cocered wth both hgher ad lower value, the the relevat rage wll have α/ probablty each tal. Cofdece Iterval (cot.) Thrd, we fd a terval aroud X that cota 95% of the area uder the curve The actual terval [.96*SE] o ether de of the ample mea. We the kow that 95% of the tme, th terval wll cota µ. Th terval defed by: α = ( 0.95) =.05or 5% level SE = α Area = 95% Copyrght Shary XO'Hallora α 95% cofdece terval.05 [-z.05 * SE Area = 95% [-.96 * SE ] X [z.05 * SE] [.96 * SE].05 4

5 Cofdece Iterval (cot.) Cofdece Iterval (cot.) How hould we terpret cofdece Iterval? What' the probablty that the populato mea µ wll fall wth the terval ±.96 * SE?.05 [-z.05 * SE [-.96 * SE ] Area = 95% X [z.05 * SE] [.96 * SE] Now, let' take th terval of ze [-.96 * SE,.05 Thk of t a a game of horehoe.96 * SE] ad ue t a a meaurg rod µ Say the true amplg dtrbuto ha mea µ ad tadard devato σ/ The 95% of the tme the cof. terval X ±.96*SE geerated wll cota µ The larger the terval, the le certa we are of our etmate. Cofdece Iterval (cot.) I geeral: We kow from the rule that a 95% cofdece terval wll be about tadard devato o ether de of X. To be prece, from the z-table, we fd the z-value aocated wth a.05 probablty.96. If we take a radom ample of ze from the populato, 95% of the tme the populato mea wll be wth the rage: σ σ X -(Z.05 * ) < µ < X + (Z.05 * ) µ = ± Zα *SE / Cofdece Iterval (cot.) Example: Calculatg a 95% cofdece terval Say we ample 80 people ad ee how may tme they ate at a fat-food retaurat a gve week. Sample ze =80 The ample ha a mea of 0.8, ad The populato tadard devato σ Calculate the 95% cofdece terval for thee data. 5

6 Cofdece Iterval (cot.) Awer:.48 Step: CalculateSE = = Step : Calculate Marg of Error = z Step 3: CalculateCofdece Iterval =.8±.07, or [.75< µ <.89] α level=0.05/= * SE =.96*0.036 = 0.07 SE = *.036= *.036=.89 α level=0.05/=.05 Cofdece Iterval (cot.) Example : Calculatg a 90% cofdece terval A radom ample of 6 obervato wa draw from a ormal populato wth Stadard devato, σ = 6, ad Sample mea µ = 5. Fd a 90% (α=.0) cofdece terval for the populato mea. 0% of the area le outde the cofdece terval α=.05 σ=6 90% of Area µ = 5 α=.05 Cofdece Iterval (cot.) Frt, fd Z.0/ the tadard ormal table: Z.05 =.64 Secod, calculate the 90% cofdece terval µ = X ± Z.05 * σ/ µ = 5 ±.64 * 6/ 6 SE=.5 µ = 5 ±.64 *.5 = 5 ± < µ < % of the tme, the mea le wth th rage. Cofdece Iterval (cot.) What f we wated to be 99% of the tme ure that the mea fall wth the terval? Select α-level: Z.005 =.58 Calculate marg of error:.58 *.5 = 3.87 Calculate Cofdece Iterval: 5 ± 3.87 or.3 µ 8.87 What happe whe we move from a 90% to a 99% cofdece terval? α=.005 α=.05 Area.3.53 µ = % cofdece terval The rage get larger 99% cofdece terval α=.05 α=.005 6

7 Cofdece Iterval(cot.) Why 95%? It tadard to accept that our etmate wll be wrog out of 0 tme. We could reduce the poblty of error, of coure, by makg the terval larger. Icreag the terval, however, make our etmate le prece. That, the marg of error creae [z α-level *SE] Trade off preco for the probablty that the true mea le a gve rage. Cofdece Iterval: σ Ukow Cofdece Iterval whe σ ukow We have bee calculatg cofdece terval aumg that we kow the populato tadard devato σ. Of coure, mot cae, we are ot oly ucerta of the mea µ, but alo of the uderlyg varace of the paret populato. Whe th the cae, we mut etmate σ. The bet etmate of σ the ample tadard devato : = = ( X X ) Th troduce a ew ource of error that mut be take to accout. Cofdece Iterval: σ Ukow (cot.) Charactertc of a Studet-t dtrbuto Shape the tudet t-dtrbuto The t-dtrbuto chage hape a the ample ze get larger, ad the lmt t become detcal to the ormal. Normal Dtrbuto Studet-t Cofdece Iterval: σ Ukow (cot.) Cotructg Cofdece Iterval ug t-dtrbuto 95% cofdece terval : where: = = ( X X ) X ± t. 05. Whe to ue t-dtrbuto σ ukow Sample ze mall (<30) µ t.05 Z.05 Z.05 t.05 The ze of the cofdece terval chage a ample ze chage..05 [-t d.f X *SE] [t d.f * SE]

8 Cofdece Iterval: σ Ukow (cot.) Ug t-table Gve a ample ze, what the crtcal value to get 95% of the area uder the curve? Step : Fd Degree of Freedom Degree of freedom the amout of formato ued to calculate the tadard devato,. We deote t a d.f. = - Step : Look up the t-table Now we go dow the de of the table to the degree of freedom ad acro to the approprate t-value. That' the cutoff value that gve you area of.05 each tal, leavg 95% uder the mddle of the curve. Applcato: Suppoe we have ample ze =5 ad t.05. What the crtcal value?.3 Cofdece Iterval: σ Ukow (cot.) Comparo to the ormal dtrbuto A d.f. get large the hape of the curve ted toward a ormal dtrbuto. A get larger, t. 05 get cloer ad cloer to.96 ad wth fte degree of freedom, t equal.96. A the ample ze grow, the dfferece betwee the t ad the ormal dtrbuto dappear. Look back at the tadard ormal table What table do I ue? Cofdece Iterval: σ Ukow (cot.) Cofdece Iterval: Example: σ Ukow (cot.) Four tudet had grade o a tet of 64, 66, 89, 77. Calculate a 95% cofdece terval for the cla average. Sample Varace ( ) Mea X = = 74 4 (64-74) + (66-74) + (89-74) = 3 = 3.7 Sample Stadard Devato = 3.7 =.5 + (77-74) 8

9 Cofdece Iterval: Awer: σ Ukow (cot.) Calculate Marg of Error: Not very prece wth a ample of ze SE = = = t.05 * SE = 3.8 * 5.76 = 8 d.f. = 3 t.05 = 3.8 Calculate cofdece terval: µ = ± 8 = 56< µ < /= /=.05 Cofdece Iterval: Dfferece of Mea We ca ue thee ame techque to addre a umber of dfferet queto. For example, we may wh to determe f two populato (e.g., me ad wome) have the ame mea (e.g., alary). Other example: How two ecto of the ame cla dd o a exam. The comparatve effectvee of two drug treatg the ame deae. Cofdece Iterval: Dfferece of Mea (cot.) Populato Varace Kow (σ-kow) We are tereted etmatg the value (µ - µ ) by the ample mea, ug ( X X ). Take ample of the ze ad from the two populato. Etmate the dfferece two populato mea. To tell how accurate thee etmate are, we ca cotruct the famlar cofdece terval aroud ther dfferece: σ σ ( µ - µ ) = (X - X ) ± z.05 + Th jut the tadard error Th hold f the ample ze large ad we kow both σ ad σ. Cofdece Iterval: Dfferece of Mea(cot.) Populato Varace Ukow (σ-ukow) If, a uual, we do ot kow σ ad σ, the we ue the ample tadard devato tead. Whe the varace of populato are ot equal ( ): Example: ( µ - µ ) = (X - X ) ± t.05 Tet core of two clae where oe from a er cty chool ad the other from a affluet uburb. + 9

10 Cofdece Iterval: Pooled Sample Varace(cot.) Pooled Sample Varace, = (σ ukow) If both ample come from the ame populato (e.g., tet core for two clae the ame chool), we ca aume that they have the ame populato varace,. where p p 95% Cofdece Iterval ( µ - µ ) = (X - X) ± t.05 * p + ( X X) + ( X X ) = ( ) + ( ) The degree of freedom are ( -) + ( -), or ( + -). ( µ - µ ) = (X - X ) ± t.05 * p + p Cofdece Iterval: Example: Pooled Sample Varace(cot.) Two clae from the ame chool take a tet. Calculate the 95% cofdece terval for the dfferece betwee the two cla mea. Obervato Cla Cla Sum Mea p = Cofdece Iterval: Awer Pooled Sample Varace(cot.) Step : Calculate ample etmate X = 74; X = 4 = 4; = 3 X X = 60 ( ) + ( ) = X X X X p ( ) + ( ) [ (64-74) + (66-74) + (89-74) + (77-74) ] + [ (56 60) + (7 60) + (53 60) ] p ( 4 -) + ( 3-) p = 0.8. = ( ) / (3 + ) = 7 d.f. = 5 t t Cofdece Iterval: =.57 Pooled Sample Varace(cot.) Step : Calculate tadard error SE = p + = = Step 3: Calculate 95% Cofdece Iterval *SE =.57 *8.6 = 0.05/=.05-7 ( µ - µ ) = (X - X ) ± = 4 ± - 7 ( µ µ ) 35. Eve though the frt cla look lke they are dog better, we ca t gore the poblty that 0.05/=.05 Cla dog better tha Cla. 4 µ Copyrght -µ 35 Shary O'Hallora 0

11 Cofdece Iterval: Matched Sample Matched Sample Defto Matched ample are oe where you take a gle dvdual ad meaure hm or her at two dfferet pot ad the calculate the dfferece. Advatage Oe advatage of matched ample that t reduce the varace becaue t allow the expermeter to cotrol for may other varable whch may fluece the outcome. Cofdece Iterval: Matched Sample(cot.) Calculatg a 95% Cofdece Iterval For each dvdual we ca calculate ther dfferece D from oe tme to the ext. We the ue thee D' a the data et to etmate, the populato dfferece. The ample mea of the dfferece wll be deoted. The tadard error wll jut be: Ue the t-dtrbuto to cotruct 95% cofdece terval: D = D ± t.05 D SE = D Cofdece Iterval: Example: d.f. = - = 3 t.05 = 3.8 = 3.9 D Matched Sample(cot.) Studet X (Fall) X (Sprg) D = X-X Trmble Wlde Gao Ame Sum Mea (7-) + (9-) + (6-) + (-) 46 D = = = % Cofdece Iterval Notce that the tadard error much maller tha our umatched par of equal ample ze. D 3.9 SE = = =.96 4 SD = D ± t.05 * t * SE = 3.8*.96 = 6.05 ± 6 = 5 to7 5 < < 7 Cofdece Iterval: Proporto Example: Jut before the 996 predetal electo, a Gallup poll of about 500 voter howed 840 for Clto ad 660 for Dole. Calculate the 95% cofdece terval for the populato proporto π of Clto upporter. = 500 Sample proporto P: P = 840 = That, our ample of 500 dvdual, 840 people repoded that they preferred Clto to Dole.

12 Cofdece Iterval: Proporto Create a 95% cofdece terval: where π ad P are the populato ad ample proporto, repectvely, ad the ample ze. π = P ± amplg allowace π = P ±.96 π =.56 ±.96 P( P). 56(. 56), 500 π =.56 ±.03. Varace of Bomal Dt. I geeral, the varace the expected value of (x-µ) Take a bomal wth P(x=) = π P(x=0) = - π Mea µ = π * + (- π) * 0 = π Varace = π * ( - π) + ( - π) * (0 - π) That, wth 95% cofdece, the proporto of voter for Clto the whole populato wa betwee 53% ad 59%. Prob (x-µ) Prob x= f x= x=0 (x-µ) f x=0 Varace of Bomal Dt. I geeral, the varace the expected value of (x-µ) Take a bomal wth P(x=) = π P(x=0) = - π Mea µ = π * + (- π) * 0 = π Varace = π * ( - π) + ( - π) * (0 - π) = (- π)*[π(- π)] + π*[π(- π)] = [π(- π)]

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