pijan'othtes kai statistik'h january 18, 2011sf lmata 1 epilègoume kai kajorÐzoume to a...

Pijanìthtec kaiStatistik

S. Malef�kh

M�jhma 90

S. Malef�kh Tm ma QhmeÐac Pijanìthtec kai Statistik 18 IanouarÐou 2011

'Elegqoi Upojèsewn

Up�rqei èna f�rmako me to opoÐo an gÐnei jerapeÐa emfanÐzontai tapr¸ta apotelèsmata mèsa se 10 mèrec. AnakalÔptetai èna kainoÔriof�rmako gia thn Ðdia asjèneia kai h farmakeutik etaireÐa isqurÐzetai ìtito nèo f�rmako fèrnei apotelèsmata se ligìtero qronikì di�sthma. Anloipìn eÐnai gnwstì ìti o mèsoc qrìnoc m gia thn emf�nish twn pr¸twnapotelesm�twn eÐnai 10 mèrec (dhlad µ = 10), qrei�zetai na elegqjeÐ anpragmatik� to kainoÔrio f�rmako dra suntomìtera (dhlad µ < 10).

H diadikasÐa aut onom�zetai èlegqoc upìjeshc statistikìc èlegqoc.

H upìjesh ìti µ = 10 lègetai mhdenik upìjesh

H upìjesh ìti µ < 10 lègetai enallaktik upìjesh

H apìfash an ja apodeqtoÔme ja aporrÐyoume th mhdenik upìjesh,basÐzetai se mÐa sun�rthsh twn dedomènwn mac.

H perioq pou aporrÐptetai h mhdenik upìjesh onom�zetai perioq

apìrriyhc krÐsimh perioq .

'Elegqoi Upojèsewn

H diadikasÐa tou statistikoÔ elègqou

1 OrÐzoume th mhdenik upìjesh (H0 : µ = µ0)

2 OrÐzoume thn enallaktik upìjesh (H1 : µ 6= µ0) (H0 : µ < µ0) (H0 : µ > µ0)

3 UpologÐzoume thn tim thc statistik c sun�rthshc sto deÐgma mac

4 OrÐzoume thn perioq apìrriyhc thc H0

5 Ex�goume ta sumper�smata mac.

'Elegqoi Upojèsewn

Sf�lmata

1 Sf�lma tÔpou I (na aporrÐyoume th mhdenik upìjesh en¸ eÐnaiswst )

a = P(apìrriyh H0|H0 swst )

2 Sf�lma tÔpou II (na apodeqtoÔme th mhdenik upìjesh en¸ eÐnail�joc)

β = P(apodoq H0|H0 l�joc)

IsqÔc enìc elègqou onom�zetai h pijanìthta apìrriyhc thc H0 ìtan h H0

eÐnai pragmatik� l�joc.

γ = P(apìrriyh H0|H0 l�joc) = 1− β

Sf�lmata

1 Epilègoume kai kajorÐzoume to a kai b�sh autoÔ orÐzetai h perioq apìrriyhc thc mhdenik c upìjeshc. To a eÐnai sun jwc polÔ mikrì(p.q. 0.05, 0.01, 0.1)

2 ParathroÔmeno epÐpedo shmantikìthtac apl� shmantikìthta touelègqou onom�zetai h pijanìthta na parathrhjeÐ mÐa tim thcstatistik c sun�rths c megalÔterh apì aut pou èdwse tosugkekrimèno deÐgma

P(Z > |z ||H0 swst )

ìpou z h tim tou statistikoÔ gia to sugkekrimèno deÐgma.

H parap�nw pijanìthta anafèretai se monìpleurouc elègqouc en¸gia dÐpleurouc elègqouc diplasi�zetai

Sf�lmata

Genik idèa

'Estw ìti jèloume na elègxoume thn upìjesh

H0 : θ = θ0 kat� H1 : θ 6= θ0

O plhjusmìc me b�sh k�poion kanìna qwrÐzetai se dÔo perioqèc, sthnperioq apodoq c thc H0 kai sthn perioq apìrriyhc thc. Katìpin, gÐnetaio èlegqoc b�sh tou krithrÐou pou orÐsthke kai apofasÐzetai an h H0

gÐnetai deqt ìqi

'Estw θ ektimht c tou θ

Xèroume ìti θ ∼ N (θ, τ 2)Upì thn mhdenik upìjesh èqoume

θ ∼ N (θ0, τ2)⇒ Z =

θ − θ0

τ∼ N (0, 1)

'Ara an h H0 eÐnai swst , dhlad an to deÐgma mac proèrqetai apì ènan

plhjusmì me mèsh tim θ0 tìte h metablht Z ja paÐrnei timèc me meg�lh

pijanìthta (1− a) gÔrw apì to 0.

Genik idèa

H0 : θ = θ0 kat� H1 : θ 6= θ0

gÐnetai deqt ìqi

'Estw θ ektimht c tou θXèroume ìti θ ∼ N (θ, τ 2)

Upì thn mhdenik upìjesh èqoume

θ ∼ N (θ0, τ2)⇒ Z =

θ − θ0

τ∼ N (0, 1)

Genik idèa

H0 : θ = θ0 kat� H1 : θ 6= θ0

gÐnetai deqt ìqi

'Estw θ ektimht c tou θXèroume ìti θ ∼ N (θ, τ 2)Upì thn mhdenik upìjesh èqoume

θ ∼ N (θ0, τ2)⇒ Z =

θ − θ0

τ∼ N (0, 1)

Genik idèa

H0 : θ = θ0 kat� H1 : θ 6= θ0

gÐnetai deqt ìqi

'Estw θ ektimht c tou θXèroume ìti θ ∼ N (θ, τ 2)Upì thn mhdenik upìjesh èqoume

θ ∼ N (θ0, τ2)⇒ Z =

θ − θ0

τ∼ N (0, 1)

Genik Idèa

Poia eÐnai h perioq apìrriy c ìtan h enallaktik upìjesh

eÐnai

H1 : θ > θ0

eÐnai

H1 : θ < θ0

Genik Idèa

eÐnai

H1 : θ > θ0

eÐnai

H1 : θ < θ0

'Elegqoc upìjeshc gia to mèso m enìc plhjusmoÔ

me gnwst diaspor�

H0 : µ = µ0 kat� H1 : µ 6= µ0 µ < µ0 µ > µ0

X ∼ N (µ, σ2/n) isodÔnama X−µσ/√n∼ N (0, 1)

Z =X − µ0

σ/√n

1 an H1 : µ 6= µ0 tìte aporrÐptw th mhdenik upìjesh an

Z < −za/2 Z > za/2

2 an H1 : µ > µ0 tìte aporrÐptw th mhdenik upìjesh an

Z > za3 an H1 : µ < µ0 tìte aporrÐptw th mhdenik upìjesh an

Z < −za

me gnwst diaspor�

H0 : µ = µ0 kat� H1 : µ 6= µ0 µ < µ0 µ > µ0

Z =X − µ0

σ/√n

Z < −za/2 Z > za/2

Z < −za

me gnwst diaspor�

H0 : µ = µ0 kat� H1 : µ 6= µ0 µ < µ0 µ > µ0

Z =X − µ0

σ/√n

Z < −za/2 Z > za/2

Z > za

3 an H1 : µ < µ0 tìte aporrÐptw th mhdenik upìjesh an

Z < −za

me gnwst diaspor�

H0 : µ = µ0 kat� H1 : µ 6= µ0 µ < µ0 µ > µ0

Z =X − µ0

σ/√n

Z < −za/2 Z > za/2

Z < −za

me �gnwsth diaspor�

H0 : µ = µ0 kat� H1 : µ 6= µ0 µ < µ0 µ > µ0

X−µs/√n∼ tn−1

T =X − µ0

s/√n

T < −tn−1,a/2, T > tn−1,a/2

T > tn−1,a

T < −tn−1,a

H0 : µ = µ0 kat� H1 : µ 6= µ0 µ < µ0 µ > µ0

T =X − µ0

s/√n

T < −tn−1,a/2, T > tn−1,a/2

T > tn−1,a

T < −tn−1,a

H0 : µ = µ0 kat� H1 : µ 6= µ0 µ < µ0 µ > µ0

T =X − µ0

s/√n

T < −tn−1,a/2, T > tn−1,a/2

T > tn−1,a

T < −tn−1,a

H0 : µ = µ0 kat� H1 : µ 6= µ0 µ < µ0 µ > µ0

T =X − µ0

s/√n

T < −tn−1,a/2, T > tn−1,a/2

T > tn−1,a

T < −tn−1,a

'Elegqoc upìjeshc gia to gia th diafor� twn mèswn

µ1 − µ2 dÔo anex�rthtwn plhjusm¸n

me gnwstèc diasporèc

H0 : µ1−µ2 = δ κατά H1 : µ1−µ2 6= δ ή µ1−µ2 < δ ή µ1−µ2 > δ

Z =X − Y − δ√σ2

1/n + σ22/m

1 an H1 : µ1 − µ2 6= δ tìte aporrÐptw th mhdenik upìjesh

an Z < −za/2 Z > za/2

2 an H1 : µ1 − µ2 > δ tìte aporrÐptw th mhdenik upìjesh

an Z > za3 an H1 : µ1 − µ2 < δ tìte aporrÐptw th mhdenik upìjesh

an Z < −za

Z =X − Y − δ√σ2

1/n + σ22/m

an Z < −za/2 Z > za/2

an Z < −za

Z =X − Y − δ√σ2

1/n + σ22/m

an Z < −za/2 Z > za/2

an Z > za

3 an H1 : µ1 − µ2 < δ tìte aporrÐptw th mhdenik upìjesh

an Z < −za

Z =X − Y − δ√σ2

1/n + σ22/m

an Z < −za/2 Z > za/2

an Z < −zaS. Malef�kh Tm ma QhmeÐac Pijanìthtec kai Statistik 18 IanouarÐou 2011

me �gnwstec diasporèc kai Ðsec

T =X − Y − δ

1/n + 1/m

ìpou s2p =

(n−1)s21 +(m−1)s2

2n+m−2

an T < −tn+m−2,a/2 T > tn+m−2,a/2

an T > tn+m−2,a

an T < −tn+m−2,a

T =X − Y − δ

1/n + 1/m

ìpou s2p =

(n−1)s21 +(m−1)s2

2n+m−2

an T < −tn+m−2,a/2 T > tn+m−2,a/2

an T > tn+m−2,a

T =X − Y − δ

1/n + 1/m

ìpou s2p =

(n−1)s21 +(m−1)s2

2n+m−2

an T < −tn+m−2,a/2 T > tn+m−2,a/2

an T > tn+m−2,a

me �gnwstec diasporèc kai �nisec

T =X − Y − δ√s2

1/n + s22/m

1 αν H1 : µ1 − µ2 6= δ τότε απορρίπτω τη μηδενική υπόθεση ανT < −tv ,a/2 ή T > tv ,a/2

2 αν H1 : µ1 − µ2 > δ τότε απορρίπτω τη μηδενική υπόθεση ανT > tv ,a

3 αν H1 : µ1 − µ2 < δ τότε απορρίπτω τη μηδενική υπόθεση ανT < −tv ,a

− > n = m tìte v = 2(n − 1)

− > n 6= m tìte v =

n−1+

T =X − Y − δ√s2

1/n + s22/m

− > n = m tìte v = 2(n − 1)

− > n 6= m tìte v =

n−1+

T =X − Y − δ√s2

1/n + s22/m

− > n = m tìte v = 2(n − 1)

− > n 6= m tìte v =

n−1+

T =X − Y − δ√s2

1/n + s22/m

− > n = m tìte v = 2(n − 1)

− > n 6= m tìte v =

n−1+

m−1S. Malef�kh Tm ma QhmeÐac Pijanìthtec kai Statistik 18 IanouarÐou 2011

'Elegqoc upìjeshc gia th diafor� twn mèswn µ1 − µ2

dÔo exarthmènwn plhjusm¸n

Kataskeu�zw ta zi = xi − yi , i = 1, . . . , n.

H0 : z = δ kat� H1 : z 6= δ z < δ z > δ

T =z − δsz/√n

1 an H1 : z 6= δ tìte aporrÐptw th mhdenik upìjesh an

T > |tn−1,a/2|2 an H1 : z > δ tìte aporrÐptw th mhdenik upìjesh an

T > tn−1,a

3 an H1 : z < δ tìte aporrÐptw th mhdenik upìjesh an

T < −tn−1,a

H0 : z = δ kat� H1 : z 6= δ z < δ z > δ

T =z − δsz/√n

T > tn−1,a

T < −tn−1,a

H0 : z = δ kat� H1 : z 6= δ z < δ z > δ

T =z − δsz/√n

T > |tn−1,a/2|

2 an H1 : z > δ tìte aporrÐptw th mhdenik upìjesh an

T > tn−1,a

T < −tn−1,a

H0 : z = δ kat� H1 : z 6= δ z < δ z > δ

T =z − δsz/√n

T > tn−1,a

T < −tn−1,a

H0 : z = δ kat� H1 : z 6= δ z < δ z > δ

T =z − δsz/√n

T > tn−1,a

T < −tn−1,a

'Elegqoc upìjeshc gia thn analogÐa p enìc plhjusmoÔ

H0 : p = p0 kat� H1 : p 6= p0 p < p0 p > p0

Z =p − p0√p0(1−p0)

1 an H1 : p 6= p0 tìte aporrÐptw th mhdenik upìjesh an

Z < |za/2|2 an H1 : p > p0 tìte aporrÐptw th mhdenik upìjesh an

Z > za3 an H1 : p < p0 tìte aporrÐptw th mhdenik upìjesh an

Z < −za

H0 : p = p0 kat� H1 : p 6= p0 p < p0 p > p0

Z =p − p0√p0(1−p0)

Z < |za/2|

2 an H1 : p > p0 tìte aporrÐptw th mhdenik upìjesh an

Z < −za

H0 : p = p0 kat� H1 : p 6= p0 p < p0 p > p0

Z =p − p0√p0(1−p0)

Z > za

3 an H1 : p < p0 tìte aporrÐptw th mhdenik upìjesh an

Z < −za

H0 : p = p0 kat� H1 : p 6= p0 p < p0 p > p0

Z =p − p0√p0(1−p0)

Z < −za

'Elegqoc upìjeshc gia th diafor� twn analogÐwn

p1 − p2 dÔo anex�rthtwn plhjusm¸n

H0 : p1 − p2 = δ kat� H1 : p1 − p2 6= δ p1 − p2 < δ p1 − p2 > δ

Se aut thn perÐptwsh diakrÐnoume peript¸seican δ 6= 0

Z =p1 − p2 − δ√

p1(1− p1)/n + p2(1− p2)/m

an δ = 0

Z =p1 − p2√

p(1− p)(1/n + 1/m)

ìpou p = x+yn+m

1 an H1 : p1 − p2 6= δ tìte aporrÐptw th mhdenik upìjesh an Z > |za/2|2 an H1 : p1 − p2 > δ tìte aporrÐptw th mhdenik upìjesh an Z > za

3 an H1 : p1 − p2 < δ tìte aporrÐptw th mhdenik upìjesh an Z < −za

Se aut thn perÐptwsh diakrÐnoume peript¸seic

an δ 6= 0

Z =p1 − p2 − δ√

p1(1− p1)/n + p2(1− p2)/m

an δ = 0

Z =p1 − p2√

p(1− p)(1/n + 1/m)

ìpou p = x+yn+m

Z =p1 − p2 − δ√

p1(1− p1)/n + p2(1− p2)/m

an δ = 0

Z =p1 − p2√

p(1− p)(1/n + 1/m)

ìpou p = x+yn+m

Z =p1 − p2 − δ√

p1(1− p1)/n + p2(1− p2)/m

an δ = 0

Z =p1 − p2√

p(1− p)(1/n + 1/m)

ìpou p = x+yn+m

1 an H1 : p1 − p2 6= δ tìte aporrÐptw th mhdenik upìjesh an Z > |za/2|

2 an H1 : p1 − p2 > δ tìte aporrÐptw th mhdenik upìjesh an Z > za

Z =p1 − p2 − δ√

p1(1− p1)/n + p2(1− p2)/m

an δ = 0

Z =p1 − p2√

p(1− p)(1/n + 1/m)

ìpou p = x+yn+m

Z =p1 − p2 − δ√

p1(1− p1)/n + p2(1− p2)/m

an δ = 0

Z =p1 − p2√

p(1− p)(1/n + 1/m)

ìpou p = x+yn+m

3 an H1 : p1 − p2 < δ tìte aporrÐptw th mhdenik upìjesh an Z < −zaS. Malef�kh Tm ma QhmeÐac Pijanìthtec kai Statistik 18 IanouarÐou 2011

'Elegqoc upìjeshc gia th diaspor� σ2 enìc plhjusmoÔ

H0 : σ2 = σ20 kat� H1 : σ2 6= σ2

0 σ2 < σ20 σ2 > σ2

X 2 =(n − 1)s2

(n − 1)s2 =∑

(xi − X )2 =∑

x2i − nX 2

1 an H1 : σ2 6= σ20 tìte aporrÐptw th mhdenik upìjesh an

X 2 < X 2n−1,1−a/2, X 2 > X 2

n−1,a/2

2 an H1 : σ2 > σ20 tìte aporrÐptw th mhdenik upìjesh an

X 2 > X 2n−1,a

3 an H1 : σ2 < σ20 tìte aporrÐptw th mhdenik upìjesh an

X 2 < X 2n−1,1−a

H0 : σ2 = σ20 kat� H1 : σ2 6= σ2

0 σ2 < σ20 σ2 > σ2

X 2 =(n − 1)s2

(n − 1)s2 =∑

(xi − X )2 =∑

x2i − nX 2

X 2 < X 2n−1,1−a/2, X 2 > X 2

n−1,a/2

X 2 > X 2n−1,a

X 2 < X 2n−1,1−a

H0 : σ2 = σ20 kat� H1 : σ2 6= σ2

0 σ2 < σ20 σ2 > σ2

X 2 =(n − 1)s2

(n − 1)s2 =∑

(xi − X )2 =∑

x2i − nX 2

X 2 < X 2n−1,1−a/2, X 2 > X 2

n−1,a/2

X 2 > X 2n−1,a

X 2 < X 2n−1,1−a

H0 : σ2 = σ20 kat� H1 : σ2 6= σ2

0 σ2 < σ20 σ2 > σ2

X 2 =(n − 1)s2

(n − 1)s2 =∑

(xi − X )2 =∑

x2i − nX 2

X 2 < X 2n−1,1−a/2, X 2 > X 2

n−1,a/2

X 2 > X 2n−1,a

X 2 < X 2n−1,1−a

'Elegqoc upìjeshc gia to lìgo σ21/σ

22 twn diaspor¸n dÔo

anex�rthtwn plhjusm¸n

H0 : σ21/σ

22 = 1 κατά H1 : σ2

1/σ22 6= 1 ή σ2

1/σ22 < 1 ή σ2

1/σ22 > 0

1 an H1 : σ21/σ

22 6= 1 tìte aporrÐptw th mhdenik upìjesh an

F < Fn,m,1−a/2 F < Fn,m,a/2

2 an H1 : σ21/σ

22 > 1 tìte aporrÐptw th mhdenik upìjesh an

F < Fn,m,a3 an H1 : σ2

1/σ22 < 1 tìte aporrÐptw th mhdenik upìjesh an

F > Fn,m,1−a

H0 : σ21/σ

22 = 1 κατά H1 : σ2

1/σ22 6= 1 ή σ2

1/σ22 < 1 ή σ2

1/σ22 > 0

1 an H1 : σ21/σ

F < Fn,m,1−a/2 F < Fn,m,a/2

2 an H1 : σ21/σ

F > Fn,m,1−a

H0 : σ21/σ

22 = 1 κατά H1 : σ2

1/σ22 6= 1 ή σ2

1/σ22 < 1 ή σ2

1/σ22 > 0

1 an H1 : σ21/σ

F < Fn,m,1−a/2 F < Fn,m,a/2

2 an H1 : σ21/σ

F < Fn,m,a

3 an H1 : σ21/σ

22 < 1 tìte aporrÐptw th mhdenik upìjesh an

F > Fn,m,1−a

H0 : σ21/σ

22 = 1 κατά H1 : σ2

1/σ22 6= 1 ή σ2

1/σ22 < 1 ή σ2

1/σ22 > 0

1 an H1 : σ21/σ

F < Fn,m,1−a/2 F < Fn,m,a/2

2 an H1 : σ21/σ

F > Fn,m,1−a

Sqèsh metaxÔ elègqwn upojèsewn kai diasthm�twn

empistosÔnhc

'Estw ìti èqoume ton èlegqo

H0 : µ = µ0 kat� H1 : µ 6= µ0

Pìte apodeqìmaste th mhdenik upìjesh?

−za/2 < Z < za/2

−za/2 <X − µ0

σ/√n< za/2

X −za/2σ√

n< µ0 < X +

za/2σ√n

H mhdenik upìjesh gÐnetai apodeqt se epÐpedo shmantikìthtac a ìtan htim µ0 an kei sto 100(1− a)% di�sthma empistosÔnhc gia th mèsh tim tou plhjusmoÔ.

O trìpoc autìc elègqei pollèc mhdenikèc upojèseic tautìqrona.

empistosÔnhc

H0 : µ = µ0 kat� H1 : µ 6= µ0

−za/2 < Z < za/2

−za/2 <X − µ0

σ/√n< za/2

X −za/2σ√

n< µ0 < X +

za/2σ√n

empistosÔnhc

H0 : µ = µ0 kat� H1 : µ 6= µ0

−za/2 < Z < za/2

−za/2 <X − µ0

σ/√n< za/2

X −za/2σ√

n< µ0 < X +

za/2σ√n

empistosÔnhc

H0 : µ = µ0 kat� H1 : µ 6= µ0

−za/2 < Z < za/2

−za/2 <X − µ0

σ/√n< za/2

X −za/2σ√

n< µ0 < X +

za/2σ√n

empistosÔnhc

H0 : µ = µ0 kat� H1 : µ 6= µ0

−za/2 < Z < za/2

−za/2 <X − µ0

σ/√n< za/2

X −za/2σ√

n< µ0 < X +

za/2σ√n

Mègejìc deÐgmatoc

Pìso prèpei na eÐnai to mègejoc tou deÐgmatoc ètsi ¸ste na

èqw di�sthma empistosÔnhc pl�touc d ?(X −

za/2σ√n, X +

za/2σ√n

d =2za/2σ√

pijan'othtes kai statistik'h january 18, 2011sf lmata 1 epilègoume kai kajorÐzoume to a...

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