Statistik gia QhmikoÔc MhqanikoÔc EKTIMHSH PARA

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1 Statistik gia QhmikoÔc MhqanikoÔc EKTIMHSH PARAMETRWN AprilÐou 2013

2 Eisagwgikˆ sthn ektðmhsh paramètrwn t.m. X me katanom F X (x; θ) Parˆmetroc θ: ˆgnwsth θ µ, σ 2, p DeÐgma {x 1,..., x n }: gnwstì EktÐmhsh paramètrou 1 Shmeiak ektðmhsh: ˆθ 2 EktÐmhsh diast matoc: [θ 1, θ 2 ] 'Allo deðgma ˆlla dedomèna {x 1,..., x n } {x 1,..., x n } sumbolðzoun: 1. Parathr seic 2. t.m. {X 1,..., X n } me katanom F X (x; θ)

3 Krit ria kal n ektimhtri n Mèjodoc upologismoô thc shmeiak c ektðmhshc Shmeiak EktÐmhsh ˆθ: ektim tria thc θ ˆθ eðnai sunˆrthsh twn t.m. {x 1,..., x n } ˆθ eðnai t.m., E(ˆθ) µˆθ, Var(ˆθ) σ 2ˆθ EktÐmhsh mèshc tim c (deigmatik mèsh tim ) θ µ x = 1 n x i n EktÐmhsh diasporˆc (deigmatik diasporˆ) θ σ 2 s 2 = 1 n (x i x) 2 n ( s 2 = 1 n n ) (x i x) 2 = 1 xi 2 n x 2 n 1 n 1

4 Krit ria kal n ektimhtri n Mèjodoc upologismoô thc shmeiak c ektðmhshc Krit ria kal n ektimhtri n 1. AmerolhyÐa ˆθ amerìlhpth: E(ˆθ) = θ alli c h merolhyða eðnai b(ˆθ) = E(ˆθ) θ ParadeÐgmata H deigmatik mèsh tim x eðnai amerìlhpth: E( x) = µ ( ) 1 n E( x) = E x i = 1 n E(x i ) = 1 n µ = µ n n n H deigmatik diasporˆ s 2 eðnai amerìlhpth: E(s 2 ) = σ 2? H deigmatik diasporˆ s 2 eðnai merolhptik : b( s 2 ) = E( s 2 ) σ 2 = n 1 n σ2 σ 2 = σ2 n Asumptwtikˆ (n ) h s 2 eðnai amerìlhpth

5 Krit ria kal n ektimhtri n Mèjodoc upologismoô thc shmeiak c ektðmhshc Krit ria kal n ektimhtri n (sunèqeia) 2. Sunèpeia ˆθ sunep c: P( θ ˆθ ɛ) 1 ìtan n ParadeÐgmata x, s 2, s 2 eðnai sunepeðc H ektim tria thc µ: x d = (x min + x max )/2 den eðnai sunep c 3. Apotelesmatikìthta DÔo ektim triec ˆθ 1 kai ˆθ 2 thc θ: ˆθ 1 eðnai pio apotelesmatik apì ˆθ 2 ìtan σ 2ˆθ 1 < σ 2ˆθ 2 Parˆdeigma x eðnai piì apotelesmatik apì th x d giatð σ 2 x < σ 2 x d.

6 Krit ria kal n ektimhtri n Mèjodoc upologismoô thc shmeiak c ektðmhshc Krit ria kal n ektimhtri n (sunèqeia) 4. Epˆrkeia ˆθ eðnai epark c ìtan qrhsimopoieð ìlh thn plhroforða apì to deðgma pou sqetðzetai me th θ. ParadeÐgmata x, s 2, s 2 eðnai eparkeðc giatð qrhsimopoioôn ìlec tic parathr seic {x 1,..., x n } x d den eðnai epark c giatð qrhsimopoieð mìno x min kai x max. Parathr seic Jèma 4 Mia kal ektim tria prèpei na plhreð autèc tic idiìthtec. Bèltisth ektim tria: amerìlhpth kai me elˆqisth diasporˆ Exisorrìphsh merolhyðac kai diasporˆc ektðmhshc: to mèso tetragwnikì sfˆlma (mean square error).

7 Krit ria kal n ektimhtri n Mèjodoc upologismoô thc shmeiak c ektðmhshc Upologismìc shmeiak c ektðmhshc Jèloume na ektim soume th θ parˆmetro katanom c F (x; θ) miac t.m. X apì ta anexˆrthta dedomèna {x 1,..., x n }. Mèjodoc twn Rop n 1 EktimoÔme pr ta tic ropèc thc katanom c: ροπή πρώτου βαθμού: µ x ροπή δευτέρου βαθμού: σ 2 s 2 2 Apì th sqèsh thc θ me tic ropèc upologðzoume thn ektðmhsh ˆθ. ParadeÐgmata Kanonik katanom : parˆmetroi µ kai σ 2 eðnai oi Ðdiec ropèc (ˆmesh ektðmhsh). Omoiìmorfh katanom sto [a, b]: parˆmetroi a kai b upologðzontai apì µ = a+b 2 kai σ 2 = (b a)2 12 (èmmesh ektðmhsh).

8 Krit ria kal n ektimhtri n Mèjodoc upologismoô thc shmeiak c ektðmhshc Parˆdeigma: Metr seic por douc hlðou se gaiˆnjrakec A/A x i xi SÔnolo

9 Krit ria kal n ektimhtri n Mèjodoc upologismoô thc shmeiak c ektðmhshc Upojètoume kanonik katanom : parˆmetroi µ kai σ 2 EktÐmhsh mèshc tim c: x = 1 25 x i = = Gia s 2, upologðzoume pr ta to ˆjroisma tetrag nwn 25 xi 2 = EktÐmhsh diasporˆc: s 2 = ( 25 ) 1 xi 2 25 x = 1 24 ( ) = Upojètoume omoiìmorfh katanom sto [a, b]: parˆmetroi a kai b EktÐmhsh twn a kai b: x = a+b 2 s 2 = (b a)2 12 â = x 3s ˆb = x + 3s â = 4.61 ˆb = 6.73

10 Krit ria kal n ektimhtri n Mèjodoc upologismoô thc shmeiak c ektðmhshc Mèjodoc MegÐsthc Pijanofˆneiac DÐnontai anexˆrthta {x 1,..., x n }, x i F (x; θ) poia eðnai h pio pijan tim gia th θ? f (x i ; θ) P(X = x i ; θ) gia kˆpoia tim X = x i Sunˆrthsh pijanofˆneiac (πιθανότητα να παρατηρήσουμε {x 1,..., x n } σ ένα τυχαίο δείγμα) L(x 1,..., x n ; θ) = f (x 1 ; θ) f (x n ; θ) 'An L(x 1,..., x n ; θ 1 ) > L(x 1,..., x n ; θ 2 ) tìte θ 1 pio alhjofan c apì θ 2 H piì alhjofan s tim thc θ: aut pou megistopoieð th L(x 1,..., x n ; θ) log L(x 1,..., x n ; θ). Ektim tria megðsthc pijanofˆneiac ˆθ dðnetai apì thn exðswsh log L(x 1,..., x n ; θ) θ = 0.

11 Krit ria kal n ektimhtri n Mèjodoc upologismoô thc shmeiak c ektðmhshc Mèjodoc MegÐsthc Pijanofˆneiac (sunèqeia) Jèloume na ektim soume θ 1,..., θ m Sunˆrthsh pijanìfaneiac: L(x 1,..., x n ; θ 1,..., θ m ) Ektim triec megðsthc pijanofˆneiac ˆθ 1,..., ˆθ m dðnontai apì log L(x 1,..., x n ; θ 1,..., θ m ) θ j = 0 j = 1,..., m. Parathr seic H mèjodoc megðsthc pijanofˆneiac mporeð na efarmosjeð gia opoiod pote θ an xèroume thn katanom F X (x; θ). H mèjodoc twn rop n den efarmìzetai an h θ de mporeð na upologisjeð apì tic ropèc. H ektim tria megðsthc pijanofˆneiac eðnai amerìlhpth (asumptwtikˆ), sunep c, apotelesmatik ki epark c.

12 Krit ria kal n ektimhtri n Mèjodoc upologismoô thc shmeiak c ektðmhshc EktÐmhsh paramètrwn kanonik c katanom c {x 1,..., x n } apì kanonik katanom N(µ, σ 2 ) kai σ 2 gnwst f X (x; µ) f (x) = 1 2πσ e (x µ)2 2σ 2 ektðmhsh tou µ? sunˆrthsh pijanìfaneiac ( ) [ 1 n/2 L(x 1,..., x n ; µ) = 2πσ 2 exp 1 2σ 2 ] n (x i µ) 2 log L(x 1,..., x n ; µ) = n 2 log 2π n 2 log(σ2 ) 1 2σ 2 n (x i µ) 2 Ektim tria megðsthc pijanofˆneiac ˆµ log L µ = 0 1 n σ 2 (x i µ) = 0 ˆµ = 1 n x i = x n

13 Krit ria kal n ektimhtri n Mèjodoc upologismoô thc shmeiak c ektðmhshc EktÐmhsh paramètrwn kanonik c katanom c (sunèqeia) Kai h diasporˆ σ 2 ˆgnwsth log L µ = 0 1 σ 2 n (x i µ) = 0 log L σ 2 = 0 n 2σ σ 4 H epðlush dðnei gia µ, ˆµ = x, kai gia σ 2 n (x i µ) 2 = 0 ˆσ 2 = 1 n n (x i ˆµ) 2 = 1 n n (x i x) 2 = s 2

14 EktÐmhsh Diast matoc empistosônhc Melet same thn ektim tria ˆθ paramètrou θ: An gnwrðzoume thn katanom thc X kai eðnai F X (x; θ), tìte brðskoume th ˆθ me 1 Mèjodo rop n 2 Mèjodo megðsthc pijanofˆneiac Anexˆrthta apì thn katanom thc X èqoume touc ektimhtèc: θ := µ ˆθ = x θ := σ 2 ˆθ = s 2 ˆθ = s 2 H tim thc ektim triac ˆθ exartˆtai apì to deðgma {x 1,..., x n }. ˆθ eðnai t.m. me E(ˆθ) µˆθ, Var(ˆθ) σ 2ˆθ Katanom thc ˆθ? E(ˆθ)? Var(ˆθ)? Me bˆsh thn katanom thc ˆθ jèloume na orðsoume èna diˆsthma [θ 1, θ 2 ] pou ja perièqei me kˆpoia pijanìthta thn pragmatik tim thc θ.

15 Diˆsthma empistosônhc thc µ Ektim tria (shmeiak ektðmhsh) thc µ: x µ x = E(ˆµ)= µ σ 2 x = Var( x) = Var ( 1 n ) n x i = 1 n n 2 Var(x i ) = 1 n 2 n = 1 n 2 (n σ2 )= σ2 n σ x = σ/ n stajerì sfˆlma H katanom thc x exartˆtai apì 1 th diasporˆ thc X, σ 2 (gnwst / ˆgnwsth) 2 thn katanom thc X (kanonik ìqi) 3 mègejoc tou deðgmatoc n (megˆlo / mikrì) σ 2

16 Diˆsthma empistosônhc thc µ - gnwst diasporˆ σ 2 Gia thn katanom thc x èqoume dôo peript seic 1 X N(µ, σ 2 ) 2 n > 30 x N(µ, σ 2 /n) X N(µ, σ 2 ) n < 30 x? 1 An h katanom thc X eðnai kanonik κατανομή της X X n είναι κανονική h katanom thc x eðnai kanonik 2 An to deðgma eðnai megˆlo n > 30 Κεντρικό Οριακό Θεώρημα h katanom thc x eðnai kanonik

17 Jèma 5 An rðxoume pollˆ nomðsmata o sunolikìc arijmìc twn kefal n ja akoloujeð kanonik katanom

18 gnwstì σ 2 kai x akoloujeð kanonik katanom x N(µ, σ 2 /n) z x µ σ/ N(0, 1) n Gia kˆje pijanìthta α (kai 1 α) upˆrqoun oi antðstoiqec timèc thc z, z α/2 = z 1 α/2 : P(z < z α/2 ) = Φ(z α/2 ) = α/2 P(z > z 1 α/2 ) = 1 Φ(z 1 α/2 ) = 1 (1 α/2) = α/2 P(z < z α/2 z > z 1 α/2 ) = α P(z α/2 < z < z 1 α/2 ) =Φ(z 1 α/2 ) Φ(z α/2 ) = 1 α Apì ton statistikì pðnaka tupik c kanonik c katanom c DÐnetai pijanìthta 1 α krðsimh tim z 1 α/2 = Φ 1 (1 α/2) }

19 gnwstì σ 2, x akoloujeð kanonik katanom (sunèqeia) h z an kei sto diˆsthma [z α/2, z 1 α/2 ] = [ z 1 α/2, z 1 α/2 ] me pijanìthta 1 α. Apì to metasqhmatismì z x µ σ/ n diast matoc [ z 1 α/2, z 1 α/2 ] z 1 α/2 = x µ σ/ n LÔnoume wc proc µ èqoume gia ta ˆkra tou z 1 α/2 = x µ σ/ n µ = x + z 1 α/2 σ n µ = x z 1 α/2 σ n Diˆsthma empistosônhc thc µ se epðpedo empistosônhc 1 α [ x z 1 α/2 σ n, x + z 1 α/2 σ n ]

20 gnwstì σ 2, x akoloujeð kanonik katanom (sunèqeia) ErmhneÐa diast matoc empistosônhc me pijanìthta (empistosônh) 1 α h mèsh tim µ brðsketai mèsa s' autì to diˆsthma OQI an qrhsimopoioôsame pollˆ tètoia diast mata apì diaforetikˆ deðgmata, posostì (1 α)% apì autˆ ja perieðqan th µ NAI me 1 α pijanìthta (empistosônh) to diˆsthma autì ja perièqei thn pragmatik µ NAI

21 gnwstì σ 2, x akoloujeð kanonik katanom (sunèqeia) DiadikasÐa ektðmhshc tou diast matoc empistosônhc tou µ 1 Epilog tou 1 α, σ gnwstì, x apì to deðgma. 2 EÔresh krðsimhc tim c z 1 α/2 apì ton pðnaka gia tupik kanonik katanom. 3 Antikatˆstash ston tôpo [ x z 1 α/2 σ n, x + z 1 α/2 σ n ]

22 Parˆdeigma Shmeiak EktÐmhsh Diˆsthma empistosônhc se epðpedo 95% gia to mèso posostì por dec hlðou se gaiˆnjraka tou koitˆsmatoc A? DÐnetai σ 2 = Iστoγραμμα πoρωδoυς ηλιoυ για κoιτασμα A 7 Θηκoγραμμα πoρωδoυς ηλιoυ για κoιτασμα A συχνoτητα ευρoς 4.5 A summetrða, ìqi makrièc ourèc, ìqi akraða shmeða X N(µ, 0.38) µ =?

23 Parˆdeigma (sunèqeia) X N(µ, 0.38) x N(µ, 0.38/25) x = x i = = 5.67 DiadikasÐa ektðmhshc tou diast matoc empistosônhc tou µ 1 1 α = 0.95, σ = 0.38, x = KrÐsimh tim : z = Φ 1 (0.975) = x ± z 1 α/2 σ n 5.67 ± [5.43, 5.91] Sumpèrasma: Se 95% epðpedo empistosônhc perimènoume to mèso posostì por douc hlðou se gaiˆnjraka me bˆsh to deðgma tou koitˆsmatoc A na kumaðnetai metaxô 5.43 kai 5.91

24 Diˆsthma empistosônhc thc µ, ˆgnwsth diasporˆ σ 2 PerÐptwsh 1: megˆlo deðgma (n > 30) s 2 σ 2 : [ x z 1 α/2 s n, x + z 1 α/2 s n ] PerÐptwsh 2: mikrì deðgma (n < 30) kai X N(µ, σ 2 ) Tìte isqôei t x µ s/ n t n 1 katanom student me n 1 bajmoôc eleujerðac N(0,1) t 5 t 24 t 50 f (x) X x

25 'Agnwsth diasporˆ σ f X (x) t 24,0.025 = t 24,0.975 = x DiadikasÐa ektðmhshc tou diast matoc empistosônhc tou µ 1 Epilog tou 1 α, σ ˆgnwsto, x kai s apì to deðgma. 2 EÔresh krðsimhc tim c t n 1, 1 α/2 apì ton pðnaka gia katanom student. 3 Antikatˆstash ston tôpo [ x t n 1,1 α/2 s n, x + t n 1,1 α/2 s n ]

26 'Agnwsth diasporˆ σ 2 PerÐptwsh 3: mikrì deðgma (n < 30) kai X N(µ, σ 2 ) Mh-parametrik mèjodoc Jèma 6 Mh-parametrik mèjodoc: diˆsthma empistosônhc gia th diˆmeso (Wilcoxon) Jèma 7 Mh-parametrik mèjodoc: diˆsthma empistosônhc gia th mèsh tim me th mèjodo bootstrap

27 Parˆdeigma Shmeiak EktÐmhsh Diˆsthma empistosônhc se epðpedo 95% gia to mèso por dec hlðou gaiˆnjraka apì to koðtasma A? [σ 2 ˆgnwsto] Mikrì deðgma (n < 30) kai X N(µ, σ 2 ) t x µ s/ n t n 1, bajmoð eleujerðac: n 1 = 24 s 2 = 1 24 ( 25 ) xi 2 25 (5.67) 2 = 1 ( ) = DiadikasÐa ektðmhshc tou diast matoc empistosônhc tou µ 1 1 α = 0.95, x = 5.67, s 2 = KrÐsimh tim : t 24,0.975 = x ± t n 1,1 α/2 s n 5.67 ± [5.42, 5.92] An z = 1.96 antð t 24, = ± [5.52, 5.82]

28 EktÐmhsh diast matoc empistosônhc thc µ diasporˆ X -katanom n x -katanom d.e. gnwst kanonik z x µ σ/ N(0, 1) n x ± z 1 α/2 σ n gnwst mh kanonik megˆlo z x µ σ/ N(0, 1) n x ± z 1 α/2 σ n gnwst mh kanonik mikrì ˆgnwsth megˆlo z x µ s/ N(0, 1) x ± z n 1 α/2 s n ˆgnwsth kanonik mikrì t x µ s/ n tn 1 x ± t n 1,1 α/2 s n ˆgnwsth mh kanonik mikrì Genikˆ gia to d.e. thc µ brðsketai apì σ x ± z α/2 n s x ± t n 1,1 α/2 n

29 EÔroc diast matoc empistosônhc) To diˆsthma empistosônhc exartˆtai apì: thn katanom kai th σ 2 thc t.m. X to mègejoc n tou deðgmatoc to epðpedo empistosônhc 1 α Gia dedomèno eôroc diast matoc empistosônhc mporoôme na broôme to mègejoc n pou antistoiqeð apì ton antðstoiqo tôpo. Endeiktik perðptwsh: n < 30, X N(µ, σ 2 ) kai σ 2 ˆgnwsto eôroc tou d.e. w = 2t n 1,1 α/2 s n Gia eôroc w prèpei to deðgma na èqei mègejoc n = ( 2t n 1,1 α/2 s ) 2 ( n = 2z w 1 α/2 s ) 2 w anˆloga me to n pou brðskoume.

30 Parˆdeigma Shmeiak EktÐmhsh Sto prohgoômeno arijmhtikì parˆdeigma (por dec hlðou), qrhsimopoi ntac t-katanom br kame 95% d.e ± [5.42, 5.92] EÔroc d.e.: = 0.50 isodônama akrðbeia gôrw apì th x : = 0.25 An jèloume eôroc 0.20 ( akrðbeia 0.10), pìso prèpei na megal sei to deðgma? ( ) 2 (kanonik katanom ) n = = (katanom student) ( ) 2 t 24,0.975 = n = = ( ) 2 t 101,0.975 = n = = ( ) 2 t 94,0.975 = n = =

31 Diˆsthma empistosônhc thc σ 2 s 2 ektim tria thc σ DÐnetai χ 2 (n 1)s2 σ 2 X 2 n X f (x) X X 2 24 X x Gia polô megˆlo n: Xn 1 2 kanonik Xn 1 2 den eðnai summetrik katanom dôo krðsimec timèc: χ 2 n 1,α/2 P(χ 2 < χ 2 n 1,α/2 ) = α/2 χ 2 n 1,1 α/2 P(χ 2 < χ 2 n 1,1 α/2 ) = 1 α/2

32 Diˆsthma empistosônhc thc σ 2 (sunèqeia) X f (x) X X 2 24,0.025 = 12.4 X 2 24,0.975 =39.4 ( P x ) χ 2 n 1,α/2 < χ2 < χ 2 n 1,1 α/2 = 1 α ) χ 2 n 1,1 α/2 = 1 α P (χ 2 n 1,α/2 < (n 1)s2 σ 2 ( P (n 1)s 2 χ 2 n 1,1 α/2 ) < σ 2 < (n 1)s2 χ 2 n 1,α/2 = 1 α

33 Diˆsthma empistosônhc thc σ 2 (sunèqeia) DiadikasÐa ektðmhshc tou diast matoc empistosônhc tou σ 2 1 Epilog tou 1 α, s 2 apì to deðgma. 2 EÔresh krðsimwn tim n χ 2 n 1,α/2 kai χ2 n 1,1 α/2 apì ton pðnaka gia katanom Xn 1 2. [ ] (n 1)s 3 Antikatˆstash ston tôpo 2 (n 1)s, 2 χ 2 χ n 1,1 α/2 2 n 1,α/2 Diˆsthma empistosônhc thc tupik c apìklishc σ To 95% d.e. gia thn tupik apìklish σ èqei wc ˆkra tic tetragwnikèc rðzec twn antðstoiqwn ˆkrwn tou 95% d.e. gia th diasporˆ σ 2. [ ] (n 1)s 2, χ 2 n 1,1 α/2 (n 1)s 2 χ 2 n 1,α/2

34 Parˆdeigma Shmeiak EktÐmhsh Diˆsthma empistosônhc se epðpedo 95% gia th diasporˆ tou por douc hlðou gaiˆnjraka enìc koitˆsmatoc A? χ 2 (n 1)s2 σ 2 Xn 1 2, bajmoð eleujerðac: n 1 = 24 s 2 = DiadikasÐa ektðmhshc tou diast matoc empistosônhc tou σ α = 0.95, s 2 = KrÐsimec timèc: χ 2 24,0.025 = 12.4 kai χ2 24,0.975 [ ] = = [ , (n 1)s 2, χ 2 n 1,1 α/2 (n 1)s 2 χ 2 n 1,α/2 ] = [0.228, 0.726] To 95% d.e. gia thn tupik apìklish σ tou por douc hlðou gaiˆnjraka eðnai [ 0.228, 0.726] = [0.478, 0.852].

35 analogða p = M N M: stoiqeða tou plhjusmoô pou plhroôn mia idiìthta (epituqða) N: sônolo ìlwn twn stoiqeðwn tou plhjusmoô DeÐgma megèjouc n kai m epituqðec Ektim tria thc p : ˆp = m n Gia megˆlo n: ˆp N(p, p(1 p) n ) ˆp p z N(0, 1) p(1 p)/n διάστημα εμπιστοσύνης p(1 p) ˆp ± z 1 α/2 n p(1 p) ˆp ± z 1 α/2 n p ˆp Statistik ˆp(1 ˆp) gia QhmikoÔc MhqanikoÔc EKTIMHSH PARA

36 DiadikasÐa ektðmhshc tou diast matoc empistosônhc tou p 1 Epilog tou 1 α, ˆp apì to deðgma. 2 EÔresh krðsimhc tim c z 1 α/2 apì ton pðnaka tupik c kanonik c katanom c. 3 Antikatˆstash ston tôpo [ˆp z 1 α/2 ˆp(1 ˆp) n, ˆp + z 1 α/2 ] ˆp(1 ˆp) n EÔroc diast matoc empistosônhc ˆp(1 ˆp) EÔroc d.e. thc p: w = 2z 1 α/2 n ( ) 2z1 α/2 2 = n = ˆp(1 ˆp) allˆ p ˆgnwsto! w MporeÐ na deiqjeð ìti: max ˆp(1 ˆp) = 0.25 ( ) 2z1 α/2 2 ( ) z1 α/2 2 = n = w 0.25 = w

37 Parˆdeigma Shmeiak EktÐmhsh 95% diˆsthma empistosônhc gia thn analogða skouriasmènwn rabd n qˆluba miac apoj khc? DeÐgma apì n = 100 rˆbdouc kai m = 12 skouriasmènec. Shmeiak ektðmhsh: ˆp = = DiadikasÐa ektðmhshc tou diast matoc empistosônhc tou p 1 1 α = 0.95, ˆp = KrÐsimh tim : z = ˆp(1 ˆp) 3 ˆp ± z 1 α/2 n 0.12 ± 1.96 [0.056, 0.184] Mègisto n tou deðgmatoc gia w = 0.05? n = ( ) 2 = (1 0.12) 100

38 Jèma 8 Diˆsthma empistosônhc kai ektðmhsh megèjouc deðgmatoc gia analogða se peperasmèno plhjusmì

39 'Askhsh Shmeiak EktÐmhsh Εγιναν 15 μετρήσεις της συγκέντρωσης διαλυμένου οξυγόνου (Δ.Ο.) σε ένα ποτάμι (σε mg/l) Από παλιότερες μετρήσεις γνωρίζουμε ότι η διασπορά του Δ.Ο. είναι 0.1 (mg/l) 2. 1 Εκτιμείστε τη διασπορά της συγκέντρωσης Δ.Ο. από το δείγμα καθώς και τα διαστήματα εμπιστοσύνης σε επίπεδο 99% και 90%. Εξετάστε και για τα δύο επίπεδα εμπιστοσύνης αν μπορούμε να δεχτούμε την εμπειρική τιμή της διασποράς γι αυτό το δείγμα. 2 Εκτιμείστε τη μέση συγκέντρωση Δ.Ο. από το δείγμα και δώστε γι αυτήν 95% διάστημα εμπιστοσύνης υποθέτοντας πρώτα ότι η διασπορά είναι γνωστή και μετά χρησιμοποιώντας αυτήν του δείγματος. 3 Αν υποθέσουμε ότι για ένα εργοστάσιο δίπλα στο ποτάμι είναι σημαντικό η μέση συγκέντρωση Δ.Ο. να μην πέφτει κάτω από 1.8 mg/l, θα προκαλούσαν ανησυχία αυτές οι παρατηρήσεις (διασπορά από το δείγμα);

40 'Askhsh (sunèqeia) 4 Αν δε μας ικανοποιεί το εύρος του τελευταίου παραπάνω διαστήματος και θέλουμε να το μειώσουμε σε 0.2 mg/l πόσες επιπρόσθετες ημερήσιες μετρήσεις πρέπει να γίνουν; 5 Ενας άλλος τρόπος να ελέγξουμε αν η συγκέντρωση του Δ.Ο. πέφτει σε μη επιθυμητά επίπεδα είναι να δούμε αν το ποσοστό των ημερών που η τιμή της συγκέντρωσης Δ.Ο. πέφτει στο επίπεδο 1.6 mg/l και κάτω ξεπερνάει το 15%. Εκτιμείστε αυτό το ποσοστό από το δείγμα. Μπορείτε να δώσετε 95% διάστημα εμπιστοσύνης για το ποσοστό; Πόσο πρέπει να είναι το μέγεθος του δείγματος για να μπορεί να εκτιμηθεί 95% διάστημα εμπιστοσύνης για το ποσοστό με πλάτος το πολύ 10%;