Chapter 2 Sampling distribution

Transcription

1 [ 05 STAT] Chapte Samplig distibutio. The Paamete ad the Statistic Whe we have collected the data, we have a whole set of umbes o desciptios witte dow o a pape o stoed o a compute file. We ty to summaize impotat ifomatio i the sample ito oe o two umbes, called a statistics. Fo each statistic, thee is a coespodig summay umbe i the populatio, called a paamete.. Measues fo quatitative vaiables Thee ae two basic statistics whe we have a quatitative vaiables, the mea ad the vaiace. The mea measues the cete of data while the vaiace measues the spead out of data fom its mea. The populatio mea is deoted by The populatio vaiace is deoted by The stadad deviatio of the populatio is deoted by The sample mea is deoted by The sample vaiace is deoted by The stadad deviatio of the sample is deoted by 0

2 [ 05 STAT].3 Measues fo qualitative vaiables The measue fo the qualitative vaiables(the chaacteistic which we wat to study i the populatio) is called the popotio, thus we have : - the populatio popotio = - the sample popotio = Note that the popotio must be umbes betwee 0 ad. A pecetage may be obtaied by multiplyig the popotio by Samplig Methods Thee ae two basic types of samplig: - Pobability samplig: evey populatio elemet has a chace of beig chose fo the sample with kow pobability. - No pobability samplig: Not evey populatio elemet has a chace of beig chose fo the sample o the pobability of choosig a elemet is ukow. Fo the statistical methods, it is useful to use the fist type. Thus, we do ot coside the secod type. The simple adom samplig is the basic kid of pobability samplig. We have two cases fo samplig:

3 [ 05 STAT] - Samplig with eplacemet The umbe of all possible samples of size fom a populatio of size N with eplacemet is = - Samplig without eplacemet The umbe of all possible samples of size fom a populatio of size N without eplacemet is = = =!!! Ex() If we have a populatio of size 5 ad we wat to choose samples of size. a) With eplacemet b) Without eplacemet Solu. The umbe of all possible samples is: a) = 5 = 5 b) = 5! = = 0!!.5 Samplig distibutio of the sample mea We ca use the followig steps to obtai the samplig distibutio of the sample mea : - Fid all possible samples of size fom a populatio of size N.

4 [ 05 STAT] - Calculate fo each sample, whee =. 3- Costuct the fequecy table, fo all diffeet values of ad also the fequecy of each value( the total of fequecies =k). To study the samplig distibutio of the sample mea ad its elatioship with the populatio paametes (, ), we eed to fid its mea = ad its vaiace =, as follows: = ad = 4- Its elatioship with populatio paametes, is: a) = b) = (samplig with eplacemet) ad = (Samplig without eplacemet) Whee the factio is called coectio facto. We may igoe this coectio facto i pactice if , that is the sample size is less tha o equal to 5% of the populatio size N because that facto will appoaches to. 5- Detemie the fom o shape of the samplig distibutio. Sice it depeds o the distibutio of the vaiable i the populatio( Nomal o ot omal), thus we have two cases fo its fom: a) The populatio ad the vaiable X i that populatio is omally distibuted (I.e.,,.The, (with o without eplacemet), 0, 3

5 [ 05 STAT] b) The populatio ad the vaiable X i that populatio has ay distibutio othe tha omal distibutio with mea ad vaiace, thus by applyig the Cetal Limit Theoem : (i) With eplacemet, 0, (ii) Without eplacemet,.5. Cetal Limit Theoem If the sample size lage eough, the some statistics (such as the sample mea ad the sample popotio) have a appoximate omal distibutio with the same mea ad vaiace as that obtaied fo the samplig distibutio of that statistics. Wheeve we ca igoe the coectio facto fo the vaiace of,as becomes lage ( > 30), the has a appoximate omal distibutio, 0,.This esult is vey impotat i the ext chapte. EX() Coside a small populatio of five childe with the followig weights: = 3.6, = 4.7, = 3.4, = 5, = 4. ()Fid the samplig distibutio of the sample mea fo samples of size with ad without eplacemet. () ca we igoe the coectio facto? (3) ca we apply the cetal limit theoem? (4) what is the fom(type) of the samplig distibutio of the mea Solu. -(a)with eplacemet, the umbe of all possible samples of size is: = 5 = 5 4

6 [ 05 STAT] F Total Thus: = = 4.8 ad = =.(.) = (b)without eplacemet, the umbe of all possible samples of size is: = 5 5! =! 3! = 0 F Total Thus : =. = =. = 4.8 5

7 [ 05 STAT] ad Note that: = =.(.) = 0.46 = =. = 4.8 ad = =.(.) = we ca veify the elatioship betwee populatio paametes ad the samplig distibutios of the mea as follows: ) = 4.8 = (With o without eplacemet) ) = =. = (With eplacemet) = = = 0.46 (without eplacemet) () The coectio facto caot be igoed because = 5 = 0.4 > 0.05 (3) Also, we caot apply the cetal limit theoem, sice the populatio is ot omal ad < 30. (4) we caot kow the fom of this samplig distibutio..6 Samplig distibutio of the sample popotio Aothe statistics that will be see is the sample popotio that has oe of two possible values of a qualitative vaiable. The samplig distibutio fo the sample popotio ca be deduced by the pevious steps fo fidig the samplig distibutio fo the sample mea, i.e., Fo each 6

8 [ 05 STAT] possible sample, we will fid the value of the sample popotio, ad the pepae a fequecy table which gives the samplig distibutio fo the sample popotio. - the populatio popotio = = - the sample popotio = =.6. The samplig distibutio fo the sample popotio without eplacemet - Fid all possible samples of size fom a populatio of size N. - Calculate fo each sample, whee =. 3- Costuct the fequecy table, fo all diffeet values of ad also the fequecy of each value( the total of fequecies =k). This table is the samplig distibutio fo the sample popotio. To study the samplig distibutio of the sample popotio ad its elatioship with the populatio popotio, we eed to fid its mea ad its vaiace, as follows: = ad = 4- Its elatioship with populatio popotio is: c) = (with ad without eplacemet) d) = () (samplig with eplacemet) 7

9 [ 05 STAT] ad = () (Samplig without eplacemet) Also, We may igoe the coectio facto i pactice if Ex(3 ) A populatio cosists of five childe, we asked each child if he like milk o ot. we get the followig esults: =, =, =, =, = - Fid the populatio popotio of the childe who liked milk. - Fid the samplig distibutio fo the sample popotio of size 3 without eplacemet fo childe who liked milk. 3- Deduce the mea ad the vaiace of the samplig distibutio of the sample popotio. 4- Veifyig the elatio betwee the populatio popotio,ad the mea ad the vaiace of the sample popotio. Solu. A 3 () P = = = 0. 6 N 5 () K = = c 5 = 3 0 c N Samples Nyy, NyN, Nyy, NyN, Nyy, NNy, yyn, yyy, yny, yny f 6 (3) µ = = = 0. 6 K 0 = a 3,, ,, ,, f kµ = k ad σ = = F 4 0(0.6) 0 8

10 [ 05 STAT] P( P) N (0.6)(.4) 5 3 = N 3 5 (4) µ = P = 0.6 ad σ = = The samplig distibutio fo the sample popotio with eplacemet I this case, we ca use the Biomial distibutio to fid the values of the sample popotio ad thei fequecies by usig the followig table: a a a A = f = ca ( A) ( N A) 0 0 = 0 0 c 0 ( A) ( N A) c ( A)( N A) c ( A) ( N A) :... Ex( 4) N = c ( A) ( N A) 0 A populatio cosists of 0 pesos, two of them have iflueza vius. Fid: - The populatio popotio of pesos who have iflueza vius. - The samplig distibutio fo the sample popotio of size 4 with eplacemet fo pesos who have iflueza vius. 3- The mea ad the vaiace of the samplig distibutio fo the sample popotio. 4- Apply the elatio betwee the populatio popotio,ad the mea ad the vaiace of the sample popotio. 9

11 [ 05 STAT] Solu. N = 0 A = N-A = 8 A () P = = = 0. N 0 () K = N = 0 4 = 0000 A = a f = 4 C a ( a 4 a ) ( 8) o 4 C 0 () (8) = = 0. 5 C ()(8) = = 0. 5 C () (8) = = C 3 () (8) = ( ) 4 (8) 0 C 6 4 = 4 (3) µ = = 0. K f Total 0000 f kµ = k ad σ = P ( P ) 0. x0. 8 (4) µ = P = 0. ad σ = = = Detemiig the fom of the samplig distibutio fo the sample popotio To Detemie the fom o shape of the samplig distibutio fo the popotio we have two cases: () If the sample size is lage ( > 30), thus we ca apply the Cetal Limit Theoem as follows: a) With eplacemet, () b) Without eplacemet, () 30

12 [ 05 STAT] whee the coectio facto appoaches to ad thus ca be igoed if () If the sample size is small ( < 30), thus we ca t apply the Cetal Limit Ex(5): Theoem ad also, we ca t detemie the fom of the samplig distibutio fo the popotio. A populatio of 500 wome, 00 of them have a high blood pessue. If we Solu. made a samplig of size 5 with eplacemet, fid: - The populatio popotio of wome with high blood pessue The fom of the samplig distibutio of the popotio. 00 () P = = () µ = = σ P ( P ) 0. ( 0. 8) = = = (3) sice, < 30,we ca't apply the cetal limit theoem ad thus, we ca't detemie the distibutio fo the sample popotio. Ex(6): I a city, we made a study about youth ad maiage, we obtai that 80% of them maied. If we take a sample of size 50 with eplacemet. Aswe the followig questios: - what is the value of the populatio popotio of ot maied? 3

13 [ 05 STAT] -? 3- ca we apply the cetal limit theoem? 4- What is the type of the distibutio of the popotio? Solu. () P = 0.0 P ( P ) () µ =P= 0. ad σ = = = thus σ = (3) Yes, we ca apply the cetal limit theoem. (4) sice, > 30,we ca apply the cetal limit theoem ad thus, the distibutio fo the sample popotio is: N ( µ = 0., σ = 0.003) 3

14 [ 05 STAT] Chapte 3 Estimatio of Cofidece Itevals 3. A Poit Estimate: A poit estimate of some populatio paamete θ, is a sigle value θˆ of a statistic θˆ. Fo example, x of the statistic X computed fom a sample of size is a poit estimate of the populatio paamete µ.similaly a = is a poit estimate of the tue popotio P fo a biomial expeimet. x is a estimato of the populatio paamete µ, but the value of x is a estimate of µ. The statistic that oe uses to obtai a poit estimate is called a estimato o a decisio fuctio. Hece the decisio fuctio S = i= ( X X ) i which is a fuctio of the adom sample is a estimato of the populatio vaiace σ.the sigle umeical value that esults fom evaluatig this fomula is called a estimate of the paamete σ. A estimato is ot expected to estimate the populatio paamete without eo. 3. Iteval Estimate: A iteval estimate of a populatio paamete θ is a iteval of the fom θˆ θ< ˆ < whee θ ˆL ad θ ˆU deped o the value of the statistic θˆ fo a L θ U paticula sample ad also o the samplig distibutio of θˆ.sice diffeet samples will geeally yield diffeet values of θˆ ad theefoe, diffeet 33

15 [ 05 STAT] values of θ ˆL ad θ ˆU. Fom the samplig distibutio of θˆ we shall be able to detemie θ ˆL ad θ ˆU such that the P( θ ˆ ˆ L< θ< θu ) is equal to ay positive factioal value we cae to specify. If fo istace we fid θ ˆL ad θ ˆU such that: P( θˆ < θ < θˆ ) = fo 0 < < L U, the we have a pobability of ( ) selectig a adom sample that will poduce a iteval cotaiig θ. of The iteval θˆ θ< θˆ L < computed fom the selected sample, is the U called a ( )00% cofidece iteval, the factio ( ) is called cofidece coefficiet o the degee of cofidece ad the ed poits θ ˆL ad θ ˆU ae called the lowe ad uppe cofidece limits. Thus whe = we have a 95 % cofidece iteval ad so o, that we ae 95% cofidet that θ is betwee θ ˆL, θ ˆU 3.3 Estimatig the Mea: 3.3. Cofidece Iteval of µ whe σ is Kow: If X is the mea of a adom sample of size fom a populatio with kow vaiace σ, a ( )00% cofidece iteval fo µ is give by: σ X Z < < X + Z / µ / () whee Z / is theζ-value leavig a aea of to the ight. σ θˆ L σ ˆ σ = X Z /, θu = X + Z / () 34

16 [ 05 STAT] EX (): The mea of the quality poit aveages of a adom sample of 36 college seios is calculated to be.6. Fid the 95 % ad 99 % cofidece itevals fo the mea of the etie seio class. Assume that the populatio stadad deviatio is 0.3. Solu: = 36, X =.6, σ= % cofidece iteval fo the mea µ : at = 0.95 = 0.05 = 0.05 Z =.96, X ± Z σ ± (.96)( ).6 ± < µ <.698 P (.50 < µ <.698) = 0.95 We ae 95% cofidet that µ lies betwee.50, % cofidece iteval fo the mea µ : At = 0.99 = 0.0 = Z =.57, ± (.57)( ).6 ± X ± Z.475 < µ <.785 P (.475 < µ <.785) = 0.99 we ae 99% cofidet that µ lies betwee.473,.788 σ 35

17 [ 05 STAT] Theoem (): If X is used as a estimate of µ, we ca the be ( )00% cofidet that the eo will ot be exceed Z σ. Fo example (): d= (.96) (0.3/6)=0.098 o Theoem (): d = (.575) (0.3/6) = 0.88 If X is used as a estimate of µ, we ca be ( )00% cofidet that the eo will ot exceed a specified amout,,whe the sample size is: =. The factio of is ouded up to ext whole umbe. EX (): How lage a sample is equied i Ex. () if we wat to be 95 % cofidet that ou estimate of µ is off by less tha 0.05 (the eo is 0.05)? Solu: =. =.(.) = is ouded up to whole umbe. 36

18 [ 05 STAT] 3.3. Cofidece Iteval of µ whe σ is Ukow ( < 30) : If X ad S ae the mea ad stadad deviatio of a adom sample fom a omal populatio with ukow vaiace σ, a ( )00% cofidece iteval fo µ is give by: S X t < µ < X + t,, S (4) whee t is the value with - degees of feedom leavig a aea of to the ight. EX (3): The cotets of 7 simila cotaies of sulphuic acid ae 9.8, 0., 0.4, 9.8, 0, 0., 9.6 littes. Fid a 95 % cofidece iteval fo the mea of all such cotaies assumig a appoximate omal distibutio. Solut io: Fo 95 % cofidece iteval fo the mea µ : = 7, X = 0, S = 0.83, at : = 0.95 = 0.05 = 0.05 t = t0.05,6 = t0.975, 6 =.447 X,, S 0.83 ± t ( ) 0 ± (.447)( ) 0 ± < µ < 0.6 P (9.738 < µ < 0.6) =

19 [ 05 STAT] 3.4 Estimatig Cofidece Iteval fo popotio P (with Lage Sample): If is the popotio of successes i a adom sample of size the the ( )00% cofidece iteval fo the populatio popotio is give by: < < + Whee whee Z = () is the Z - value leavig a aea of to the ight ad is the maximum value of the eo. EX (4): A ew ocket lauchig system is beig cosideed fo deploymet of small, shot ag ockets. The existig system has P= 0.8 as the pobability of a successful lauch. A sample of 40 expeimetal lauches is made with the ew system ad 34 ae successful. Costuct a 95% cofidece iteval fo P. Solutio: 95 % cofidece iteval fo the popotio P : = 40, = = 0.85, =

20 [ 05 STAT] = 0.95 = 0.05 = =.96 = = Theoem (3): < < < < 0.96 = 0.95 If is used as a estimate of, we ca be ( )00% cofidet that the eo will ot exceed = EX (5): I Ex. 4, fid the eo of P. Solutio:. The eo will ot exceed the followig value: Theoem (4): = = = 0. If is used as a estimate of P we ca be ( )00% cofidet that the eo will be less tha a specified amout appoximately: whe the sample size is = The the factio of is ouded up. 39

21 [ 05 STAT] EX (6): How lage a sample is equied i Ex. 4 if we wat to be 95 % cofidet that ou estimate of P is withi 0.0? Solutio: = 0.0, =.96, = 0.85, = 0.5 = = Estimatig the Diffeece betwee Two populatios Meas 3.5. Idepedet samples : Cofidece Iteval fo µ µ whe σ ad σ Kow: If X ad X ae the meas of idepedet adom samples of size ad fom populatios with kow vaiaces σ ad σ espectively, a ( )00% cofidece iteval fo µ µ is give by: σ σ ( X X ) ± Z + (5) whee Z is theζ-value leavig a aea of to the ight. EX (7): A stadadized chemisty test was give to 50 gils ad 75 boys. The gils made a aveage gade of 76, while the boys made a aveage gade of 8. Fid a 96 % cofidece iteval fo the diffeece µ µ whee µ is the mea scoe of all boys ad µ is the mea scoe of all gils who might take this test. Assume that the populatio stadad deviatios ae 6 ad 8 fo gils ad boys espectively. 40

22 [ 05 STAT] Solutio: Gils boys = 50 = 75 X = 76 X = 8 σ = 6 σ = 8 96% cofidece iteval fo the mea µ µ : = 0.94 = 0.04 = 0.0 Z =.05 σ σ ( X X ) ± Z (8 76) ± (.05) + 6 ± < µ µ < 8.57 P (3.49 < µ µ < 8.57) = Cofidece Iteval fo µ µ whe σ ad σ Ukow but equal vaiaces: X ad X ae the meas of idepedet adom samples of size ad espectively fom appoximate omal populatios with ukow but equal vaiaces, a ( )00% cofidece iteval fo µ µ is give by: ( X X ) ± t S P + (6),, v Whee S P = ( ) S + ( ) S + (7) t is the pooled estimate of the populatio stadad deviatio ad, is the v t value with v = + degees of feedom leavig a aea of to the ight. 4

23 [ 05 STAT] EX (8): The idepedet samplig statios wee chose fo this study, oe located dowsteam fom the acid mie dischage poit ad the othe located upsteam. Fo mothly samples collected at the dowsteam statio the species divesity idex had a mea value X ad a stadad deviatio = 0.77 = 3. S while 0 mothly samples had a mea idex value X. 04 ad a stadad deviatio S = Fid a 90 % cofidece iteval fo the diffeece betwee the populatio meas fo the two locatios, assumig that the populatios ae appoximately omally distibuted with equal vaiaces. Solutio: Statio Statio = = 0 X = 3. X =.04 S = 0.77 S = % cofidece iteval fo the mea µ µ : = S P (0.77) + 9(0.448) = = at = 0.90 = 0. = 0.05 t t 0.95, 0 =.75, + (3..04) ± (.75)(0.646) +.07 ± < µ µ <.547 P (0.593 < µ µ <.547) =

24 [ 05 STAT] 3.5. depedet (paied) samples : Sometimes samples fom two populatios ae ot idepedet but paied by some impotat chaacteistics. Fo example, studyig the height of the fathe ad his fist so, studyig the weights of twis. So we ca say that samples fom two populatios will be paied o depedet samples wheeve the uit take fom oe populatio is elated to a uit take fom the secod populatio. i the case of paied samples, it is moe efficiet to look at the diffeeces i the values fo the paied populatios. Also, we ca say that we have paied samples whe we have oe populatio ad get a adom sample the we measue two chaacteistics o the same expeimetal uit, fo example, suppose we take a sample of 0 people ad measue the weights befoe execisig ad afte execisig. The, we have oe populatio ad two vaiables: - The weights befoe execisig - The weights afte execisig But the both values ae fo the same peso ad thus, ae paied. Suppose we have a sample of size fom the fist populatio,, ad a sample of size fom the secod populatio,,, the study depeds o the diffeece betwee the values of the two samples that we deoted by = which has omal distibutio with mea = ad ukow vaiace.thus, to begi aalyze, we eed the followig ifomatio fom samples: 43

25 [ 05 STAT] = h h = = h h = Theefoe, the cofidece iteval fo the diffeece is give by: < < + Whee Ex(9) page =, locatio Time Time = Total -9 = 9 6 = 3.83 = = 9.49 ad = 0.05 =,. = = < <

26 [ 05 STAT] 3.6 Estimatig Cofidece Iteval fo Populatio Vaiace We must assume that the populatios ae omal fo all tests o populatio vaiaces. We will also eed two ew distibutios: oe fo vaiace ad the othe fo the case of compaig two populatio vaiaces Estimatig Cofidece Iteval fo Populatio Vaiace If we have a adom sample of size fom a omal populatio with ukow vaiace σ ; thus the cofidece iteval is give by: ( ) S χ, < σ < ( ) S χ, 3.6. Estimatig Cofidece Iteval fo stadad deviatio By takig the squae oot of the pevious cofidece iteval fo, we obtai the cofidece iteval of the populatio stadad deviatio as: ( ) S χ, < σ < ( ) S χ, Note that: To fid the table value of chi-squae at a degee of feedom does't exist i the table, we use the followig ule χ df df ( χ χ ), = χ V, + V, V, v v v Whee df: is the eeded degee of feedom V:the closest degee of feedom lowe tha the eeded df. V: the closest degee of feedom geate tha the eeded df. 45

27 [ 05 STAT] Ex(0) Fid the table value χ 34,0.05 Solu. `Sice df=34 is ot i the table, we fid V = 30 ad V = 35. Fom the ule we have: χ 34,0.05 = χ 30, ( χ χ ) 35, , = ( ) 8 0 = ( 3.97) = = Ex() A adom sample of size 48 uit of a cetai type of baaa, the aveage of the legth is5.7 cm ad vaiace.. assume that the legth of baaa have a omal distibutio, the - Estimate the 90% cofidece iteval fo the vaiace of legth. - Estimate the 90% cofidece iteval fo the stadad deviatio of legth. Solu. = 48 S =., - = 0.90 ( ) S χ, < σ < ( ) S χ, 46

28 [ 05 STAT] 47 (.) χ 47,0.95 < σ < 47 (.) χ 47, χ 47,0.05 = ( ) = χ 47,0.95 = = (.) <σ < ( ) (.) 3.78 Thus the cofidece iteval fo the vaiace is: < σ < Also, the cofidece iteval fo the stadad deviatio is: ( < σ < ) 3.7 Estimatig Cofidece Iteval fo two Populatio Vaiaces If we have two methods to estimate somethig ad we fid that the two meas ae the same. Which method should we choose? We should choose the method which is less vaiable. Theefoe, fo tests ivolvig two vaiaces, it is ecessay to have idepedet samples fom two omal populatios. The, the cofidece iteval fo thei atio is :,,,, 47

29 [ 05 STAT] Note that Ex() F,, = F,, F 0.995,,5= table, so Ex(3) 4.5 F 5 but we caot fid the value of 0.005,, diectly fom the F ,,5= = = F 0.995,5, 0.86 I a study of milk i Riyadh makets i 979, idepedet samples of aw milk ad pasteuized milk wee collected. The total bacteia cout pe ml(divided by 03) was measued: Sample size Mea Stadad deviatio Raw Milk Pasteuized Milk Assumig omal distibutios with uequal vaiaces, fid a 99% cofidece iteval fo the atio of the vaiaces of aw ad pasteuized milk. Solu. Iteval fo thei atio is :,,,, 48

30 [ 05 STAT] Sice,,, =.,, =.98,, =.,, = Theefoe, the cofidece iteval will be as follows: =.,,.8 = Estimatig Cofidece Iteval fo two Populatios Popotios If we have two idepedet adom samples, the we ca obtai the cofidece iteval of the diffeece fo the two populatios popotios as follows: < < + d = Z ( ) ( + ) Ex(4) Two machies A,B. If we get a adom sample of size 40 fom machie A with defective popotio 0.8 ad a adom sample of size 60 fom machie B with defective popotio 0.4. Estimate the diffeece of defective popotio with 95% cofidece iteval. 49

31 [ 05 STAT] Solu. = 40 = = 0.95 = 60 = 0.4 thus, Z =. 96 < < + d = Z ( ) + ( ) d = 0.8 ( 0.8 ) 0.4 ( 0.4 ) (.96 ) + = Ex(5) < < < < 0.78 If take two samples A,B of childe with the followig aswes about dikig milk duig meals i the two populatios: samples Yes No Total A B Fid with 99% cofidet the diffeece of the popotio fo the childe who always dik milk duig meals i the two populatios. 50

32 [ 05 STAT] Solu = 0 a = 77 = = = 80 a = 88 = = = 0.99 = Z = Z = < < + d = Z ( ) + ( ) d = 0.35 ( 0.35 ) ( ) (.575 ) + = ( ) < P P < ( ) < P P <