Z. Recording in Class

8/13/2019 Z. Recording in Class

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Chapter 1:

INTRODUCTION TO

PROBABILITY AND STATISTICS1. Some concepts of Statistics

[*] Data => Are observations (Nationality, height, weight)1. Qualitative: Khc kiu s

2. Quantitative: Kiu s

a. Discrete: Ri rc

1. Finite number of elements: C hu hn phn t

2. Countable: C thm c sgi tr

b. Continuous: Lin

Possible values fill the range [a;b]: gi trc thc lp y 1 khong t[a;b]

[*] Collecting Data: phng php ly dliu1. Retrospective study (Using historical data): Nghin cu dliu qu kh

2. Observaional Sudy: Khng ng ln i ng nghin cu

3. Experiment: C t ng ln i ng nghin cu

[*] Population:

Tng thcomplete collection of all elements to be studied: By ca tt ccc phn

tc nghin cu, y lun, khng thiu ai Mnh nghin cu g, ci thuc phm vi nghin cu l

population

[*] Sample: Mu = Tp con ca population

[*] Parameter: Tham sMesurement describing some properties of population: L so no miu mt vi tnh

cht ca tng thRt ttng th

[*] Statistic: i lng thng kMesurement describing some properties of sample: L so no miu mt vi tnh

cht ca muRt tmu


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[*] Statistics:

Tp hp ca tt ccc phng php :

1. Collect data: Thu thp dliu

2. Present data: Biu din dliu

a.

Tableb. Graph

c. Histogram

3. Numerical summaries: Tnh ton cc dliu cn thit

[*] 1, 2, 3: Descriptive Statistics: Thng k m t

4. Conclude about parameters of population: Rt ra kt lun vcc tham stng th

[*] 4: Inferential Statistics: Thng k suy din

2. Some concepts of Probability

[*] Random Experiment: Php thngu nhinAn experiment results in different outcomes: L mt php thm dn n cc kt qukhc

nhau

[*] Sample Space: Khng gian muSet of all possible outcomes: Tp tt ccc kt quc thc

Ex 1: Tossing a coin 2 times:

S = {HH, HT, TH, TT} => |S| = 4

Ex 2: Tossing 2 dices

S = {(1,1), (1,2), , (6,6)- => |S| = 36

S = 2,3, ,12-|S| = 11

Khng gian mu khng phi duy nht! N phhuvohly dliu, m h php h

[*] Event: Bin cSub-space of Sample space: Con ca khng gian mu


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[*] Intersection of 2 events: Giao ca 2 bin c

A BAnd

[*] Union of 2 events: Hp ca 2 bin c

AUBOr

[*] Complement of an events: Phn b ca 1 tp hpA = A ngang = Ac= S\A

Eg:

A = (AB) U (AB)A B = (A U B)A U B = (A B)


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[*] Multiplication Rule: Quy tc nhn

[*] Multiplication Rule: Quy tc cng

[*] Permutation: Hon v

[*] Chnh hp:Rt ra k phn tttp n phn t, thtphn t

( )

[*] Thp:Rt ra k phn tttp n phn t, KHNG thtphn t

( )

[*] Hon vlpN phn t:

N1 phn t: Tnh cht 1

N2 phn t: Tnh cht 2

Nk phn t: Tnh cht k


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Chapter 2:

PROBABILITY OF AN EVENT

1. Axioms of probabiliy: Tin vxc sutProbability of A = P(A) = a real number [0,1]To quantify the likelihood of an event

P(A) [0,1] P(S) = 1

P(AUB) = P(A) + P(B) if AB =

P(A) = 1 P(A)

() If outcomes are equally likely: Nu cc khnng a kt qul nh nhau () ( )Tng xc sut tng kt qu

Eg: Flip a coin 10 times:

1. P(>= 1H) = ?2. P(exactly 2H) = ?

1. A = >= 10

A = 10T => |A| = 1

P(A) = () 2. B = Exaly 2H

|B| = => P(B) =


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2. Addition rule: Quy tc cng P(AUB) =

= P(A) + P(B)P(AB)

P(AUBUC) = P(A) + P(B)P(AB)P(AC)P(BC) + P(ABC)

Eg: P(A) = 0.8, P(B) = 0.6, P(A

B) = 0.5

a. P(AUB) = 0.8+0.6-0.5 = 0.9

b. P(AB)= 0.8-0.5 = 0.3

A = (AB) U (AB)

. P(AB) = 0.6-0.5 = 0.1

d. P(AUB)

e. P(AUB)

f. P(AUB)

g. P(AB)

* Khi c 2 b th a b ra

3. Conditional Probability: Xc su iu kin

a. P(smoke) =

b. P(smoke|female) = Probabiliy of smoke given ha female =

c. P(smoke|male) =

d. P(female|smoke) = 50/50+150

e. P(male|smoke) = 150/50+150

P(A|B) =

=

=()

() (Xc sut va c A va c B|Chc B)


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Eg: Toss a die:

S = {1, 2, 3, 4, 5, 6}

B = {1, 4}

A = {>3} = {4, 5, 6}

P(B|A) =()

()

SOLVE

P(S) = 1

P(1) P(6) = 1

M P(odd) = p => P(even) = 2p

p +2p+ p +2p+ p +2p = 1

p = 1/9

P(AB) = P(4) = 2/9

P(A) = P(4) + P(5) + P(6) = 5/9

P(B|A) = 2/5

[*] Multiplication RuleP(AB) = P(B).P(A|B) = P(A).P(B|A)--

* P(A|B) = 1 P(A|B)

4. Total Probability: Quy tc xc su y

P(Defects) = ?

SOLUTION


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P(1) = 30%; P(D|(1)) = 5%

P(2) = 40%; P(D|(2)) = 2%

P(3) = 30%; P(D|(3)) = 3%

P(D) = ?

D = [D(1)] U [D(2)] U [D(3)]

P(D) = P[D(1)] +P[D(2)] +P[D(3)]

= P(1).P(D|(1))+ P(2).P(D|(2))+ P(3).P(D|(3))

= 0.3*0.05 + 0.4*0.02+ 0.3*0.03

EiEj =

Ei = S

P(A) = P(AE1)+ P(AE2)P(AEn)

P(A) = ()()

Bayes TheoremP(Ek|A) =

()() =

()()() =

()() ()()


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Independence: Sc lp AB = A & B are disjoint/mutually exclusive: Xung khc

if P(A|B) = P(A)A & B are independence

or P(B|A) = P(B)

or P(AB) = P(A).P(B)

EG:P(A|B) = 0.2, P(B) = 0.9, P(A) = 0.3. Are the events B and complement of A independence?

SOLUTION:

Is P(AB) = P(A).P(B)?

P(AB) = P(B).P(A|B) =P(B).(1 - P(A|B)) = 0.72

No

If A & P are independene hen A & B are also independene!


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Chapter 3:

DISCRETE RANDOM VARIABLES

(BIN NGU NHIN RI RC)1. Random variableDefinition:Random variable = a variable that its values are determined by the outcomes of a random

experiment: Bin ngu nhin: l 1 bin m cc gi trca n x nh da trn kt quca 1 php

th.

Random variable:

Discrete: ri rc: Its range is finite/countable: Min l hu hn/m c.

Continuous: Lin tc: L 1 khong, nhn lin tip gi tr

2. Probability distribution: Phn bxs ca 1 bin ngu nhin ri rc

a. The probability mass function: M tbi hm phn bxc sutX x1 x2 xnf(x) P1 P2 Pn

f(xi) = Pi = P(X=Xi)

Eg:Tossing a die wie: X = ouomes.

Find PMF of X.

X 2 3 4 5 10 11 12 F(x) 1/36

[*] f(xi) = 1

b. The cumulative distribution function (CDF): Hm phn bh lyKhng tnh xc xut tng im, cng dn

f(a) = P(X = a) F(a) = P(X a)

Eg:F(4) = P(X 4) = P(X = 2) P(X = 3) P(X = 4) = 1/36 2/36 3/36 = 1/6


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Eg:

F(x) =

0 : x < 1

1/3 : 1 x < 2

1/2 : 2 x < 3 1 : x 3

Find the PMFf(x)

X 1 2 3

f(x) 1/30 = 1/3 1/21/3 = 1/6 11/2 = 1/2

Eg:

X -1 2 3

f(x) 1/5 3/5 1/5

Find F(x) = ?

0 (x < -1)

1/5 (-1 x < 0)

1/53/5 (0 x < 1)

4/51/5 (x 1)

3. Expected value (mean): K vng | Variane: Phng sai | Sandard

deviaion: lch tiu chunAssume:

X x1 x2 xn

f(x) f(x1) f(x2) f(xn)

[*] E(x) = = () [*] V(X) = ( )() = () ^2[*] Standard deviation

nb

[*] Properties

E(h(X)) = () () E(aX+b) = aE(x) + b

V(aX+b) = a2V(X)


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4. Some discrete distributions (1 sphn phi ri rc)

a. The uniform distribution (Phn phi u)X x1 x2 xnf(x) 1/n 1/n 1/n

E(x) = = x1/n xn/n = xi/nV(X) = x1^2/n xn^2/n - ^2 = xi^2/n (xi/n)^2

If(x1,x2,,xn) = a, a1,, B- a,b Zn = b-a/1 + 1 = ba + 1

= (a a 1 b)/(b a + 1) =()()

^2 = V(X) =

()

EX:

Le X has uniform disribuion on inegers 5,6,,120-

a. Find P27 x < 89-

b. Find E(X), E(2X+4)

c. Find V(X), V(2X+4)

SOLVE:

P(X = i) =

P27 x < 89- = 62/116

E(X) =

E(2X+4) = 2E(X) + 4 = 129

V(X) =()

V(2X+4) = 2^2 V(X)


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b. Binomial Distribution (Phn phi nhthc)[*] n independent trials

[*] each trial:

Success: P(s) = p

Failure: P(f) = 1-p

X = The number of trials that result in success.

X 0 1 k n

f(x) ( )

E(X) = ( )= npV(X) = np(1-p)

c. Geometric Distribution: Phn phi hnh hc[*] Trials are independent

Success: P(s) = p

Failure: P(f) = 1-p

X = The number of trials until 1 success

X 1 2 n f(x) p (1-p)p (1-p)^(n-1)p

E(X) = np( )= p ( ) = V(X) =


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d. Negative Binomial Distribution: Phn phi nhthc m[*] Trials are independent

[*] Trials:

S: P(s) = p

F: P(f) = 1-p

[*] X = The number of trials until r successes.

X R R1 n

f(x) ( )

kE(X) =

V(X) =()

e. Hypergeometric Distribution: Phn phi siu hnh hcN = 10 balls, k = 8 red.

Pick n = 4 balls without replacementSample

X: The number of red balls in sample

X 2 3 4

f(x) ()()()

()()()

()()

TNG QUT:

P(X = x red balls) =()()

()

E(X) = np (p = k/N)

V(X) = np(1-p).

= ()


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f. Poisson Distribution

Time interval: [a,b]

E(X) =

Chia khong thi gian sao cho mi khong khng c qu 1 xe vo

P(X = k) = C(k,n).p^k.(1-p)^(n-k)

Np = => p = /n

=> P(X = k) = C(k,n).(/n)^k.(1-/n)^k

Khi (n): (e^(-). ^k)/k! = P(X=k)

[*] Interval(Time, area, space)

[*] = mean number of events/interval

[*] X = The number of events/interval

E(X) =

V(X) =

= BT: 3-100

C 6 loi phn phi:

C tn trong :o Uniform

o Poission

Quan tm n sthnh cng:o Binomial: independent c lpo Hypergeometric: Without replacement Phc thuc

Quan tm sphp thcho n khi c 1 hoc r php ththnh cngo R = 1: Geometric

o R > 1: Negative Binomial


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Chapter 4:

CONTINUOUS RANDOM VARIABLE

1. The probability desity function (PDF): Hm m xc sut

P(


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3. Expected value (Mean) (E(X) = )Varriane (V(X) = 2)

Standard deviation ()

= ()

2= ( )() = ( () PROPERTIES:

E[h(x)] = ()() E(aX+b) = aEX +b

V(aX+b) = a2V(X)

4. Some continuous distributions: Cc phn phi lin tc

a. The continuous uniform distribution: Phn phi u lin tc

a b

X = is uniformly distributed if its PDF: f(x) = c vi mi x [a;b] () Cx = 1c = > f(x) =

o

if x [a;b]o 0 if x ![a;b]

P(c


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b. The normal distribution: Phn phi chun; The standard normal distribution: Phn

phi tiu chunX is normally distributed if the graph of its PPF is bell shaped.

f(x) =

if = 0; = 1X has standard normal distribution.

P(c < X < d) = ?

X = normal:

= 50

= 5

P(X < 70) = P(

) = P(Z < 4)EG:

X = N(20, 25):

N: Normal

20:

25: 2

P(0 < X < a) = 0.5434

Find a:

P(0-20/5 < X-20/5 < a-20/5) = 0.5434

P(-4 < Z < a-20/5) = 0.5434

P(Z


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Sumary:

f(x):

P(a < X < b) = () ; 2

F(x)

F(x):

P(a < X < b)

; 2 f(x)

Uniform:

P(a < X < b)

; 2

f(x), F(x)

[*] Normal approximation to the Binomial distribution & Poisson

distribution: Xp xchun ca phn phi Nhthc v phn phi Poisson

- Binomial Distr: B(n: Sphp th; p = P(success)) P(X = k success) = C(n,k) p^k(1-p)^(n-k)

P(X k suess) = ( )( ) [*] if n, p:

np > 6

np(1-p) > 5

=> B(n,p) N( = np; 2= np(1-p))

Ex: n = 1000 trials. P(s) =. P( 900 suess) = ?

- Poisson:

P(k events) =

P(k evens) =

[*] if > 5 => P() N( = , 2= )


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c. The exponential Distribution: Phn phi m* Poisson:

a = mean b

Interval

P(k events) =

X = The distance between 2 successive events

X = continuous random variable

P(X > x) = P( 0 event/interval = x) =()

P(X x) = 1 e-x= F(x) *x 0+

F(x) = 1e-x (x > 0)

f(x) = F(x) = e-x (x > 0)

= dx = 2= = * MEAN(Sskin) =

* MEAN( K/c 2 skin) =


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Chapter 6:

DESCRIPTIVE STATISTICS

1. Frequency Distribution Sample: x1, x2, , xn

X x1 x2 xk

f(x) n1 n2 nk

X *a1,a2) *a2,a3) *ak,ak+1)

f(x) n1 n2 nk

X x1 x2 xk

f(x) n1/n100% n2/n100% nk/n100%

X *a1,a2) *a2,a3) *ak,ak+1]

f(x) n1/n100% n2/n100% nk/n100%

=> Relative frequency distribution.

X x1 x2 xk

f(x) n1 n1n2 n

X *a1,a2) *a2,a3) *ak,ak+1]f(x) n1 n1n2 n

=> Frequency cumulative

2. Numerical Summaries

[*] Range: Skhc nhau ca max v min = max[xi]min[xi]

[*] Mode: Gi trc tn sxut hin nhiu nht = xk(nk max) Mode = nu n1 = n2 = = nk = 1

[*] 3 quartiles: Tphn v

Min Q1 Q2 Q3 Max

25% 25% 25% 25%


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Q1: First quartile

Q2: Median

Q3: Third quartile

Cch tnh Q2:

Z: Q2 = () ! Z: Q2 = ( [*+]

Cch tnh Q1:

Z: Q1 = ()

! Z: Q1 = ( [*+]

Cch tnh Q3:

Z: Q2 = ( )

! Z: Q2 = ( [* +] Q3Q1 = Inter quartile range = IQR

EXAMPLE

Sample: 2 3 4 5 2 4 3 6 7 8 7 9 3 5

Find:

Range: 92 = 7

Mode: 3

Quartiles: Sp xp trc khi lm

Sorted: 2 2 3 3 3 4 4 5 5 6 7 7 8 9

N = 14median: 14/2 = 7: Q2 = (x7+x8)/2 = (4 + 5)/2

Q1: n/4 = 14/4 = 3.5Q1 = x4 = 3

Q3: 3n/4 = 10.5Q3 = x11 = 7

[*] Sample mean ()= (Khng gp cc strng nhau) =

( bcc strng nhau thnh ni ln) =


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[*] Sample variance (s2)

S2= ( )

[*] Sample standard deviation (s) =

3. Histogram, Plot

a. Dot Plot

b. Stem-and-leaf diagram

c. Box plot

Thhin 5 rng: min, max, 3 quariles

Xi = outlier

Xi > Q3 + 1.5 IQR

Xi < Q1 - 1.5 IQR

d. Bar chart

e. Scatter Plot


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Chapter 7:

SAMPLE DISTRIBUTION.THE CENTRAL LIMIT

THEOREM

1. Poin Esimae ( lng im)Trch populationSample

= a point estimate for

s2= a point estimate for 2

(A) = a point estimate for p(A) a point estimate for 1 - 2 (2 population, 2 samples) (A) - (A) = a point estimate for p1 - p2


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EXAMPLE:

Sample

X Ni

20-30 5

30-40 340-50 2

Find a point estimate for , 2

--

A point estimate for is :=

A point estimate for 2is s2:

S2=()()()

2. Random sample. Sampling distribution (Mu ngu nhin, phn

phi mu)

Ex: Population {1, 2, 3}Ly ra 1 mu, n = 2 (with replacement)

Sample: {(1,1), (1,2), (1,3), (2,1), (2,2), (2,3), (3,1),(3,2),(3,3)}

Random sample:(X1, X2) m gi trca n nm trong tp hp 9 samples trn.

[***]Statistic = a function of a random sample. (Mt thng k l hm ca 1 mu ngu nhin)

Ex:

= a statistic.S2=

( )( ) = a statistic.

- Sampling Distribution is a probability distribution of a statistic L phn bxc sut ca 1 thng k!

Ex:

What is the sampling distribution of


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3. The enral limi Theorem (nh l gii hn trung tm)

Random Sample (n) If n 30, N(_= _pop; _ ) If the distribution is normal, Populaion = N(, 2) => N(_= _pop; _ ) If n1 30 && n2 301 - 2 N(1 - 2 ;

)

Population

Mean =

Var = 2


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Chapter 8:

CONFIDENCE INTERVAL ON PARAMETERS OFPOPULATION

1. nh ngha hungAssume that we have a random sample, Gisc mu ngu nhin (X1,X2,,Xn)

L = f(X1,,Xn)|| U = g(X1,,Xn)

If P(L | | P U) = (1 ) = onfidene level

[L;U] = confidence interval (two-sided) on on | | P = I on | | P

1 - = onfidene level

U1= g1(X1,,Xn) = upper bound (cn trn) for ||P with confidence level = 1 - if P(||P U1) = 1 -

L1= f1(X1,,Xn) = lower bound (cn trn) for ||P with confidence level = 1 - if P(||P U1) = 1 -

2. Construct CI on

P(L U) = 1 - (known), L;U = ? L = U = E = ?

P ( ) = 1 - P( - E E) = 1 - [*] CASE 1: (n 30 || population has normal distribution) && populationknown

P(()

Z

()

) = 1 -

P(

Z

) = 1

P(Z

) = 1 -

= Z

Tra bng


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E = ***CI on : [L;U] = [ ]Ex: 1 - = 90%Z(

)= ?

--

= 0.1Z()= Z0.05= 1.65Ex:

N = 36

= 2.6 ; = 0.31 - = 95%

--

[

]

L1, U1 = ?

P( U1) = 1 -

U1 = P( )= 1 - P(Z

)= 1 -

P(Z

)

= 1 -

Upper bound for :

. Lower bound for :

L1= . [*] Error = If:

Error E

Confidence level = 1 -

n = ?

E


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n ( )2Ly trn[*] CASE 2: n 40 && populationunknown

Two-sided CI on : [ ]Upper bound for : * () ]Lower bound for : * () ][*] CASE 3: n < 40 || population has normal distribution) && populationunknown

- Student Distribution

PDF: y = f(x, k) k NNgoi bin x cn tham skTHEOREM

If pop = normal distribution

= Student distribution with k = df = n1

P( ) P(- E ) = 1 - P(

)

E =t(

; n-1)

Two-sided CI: [ ( ) ]

Upper bound for : Lower bound for : ( )


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3. onsru I on (Sandard deviaion of populaion) Chi square distribution (X

2)

PDF: y = f(x,k)k: degree of freedom = df

THEOREMIf pop = normal distribution

()

= Chi square distribution with k = df = n 1

2 [?;?]P(L 2 U) = 1 -

P(()

()

()

) = 1 -

=>

L =()

U =()

()

=> 2[ () ()

()]

Upper bound for 2:()

()

Lower bound for 2:()

()


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4. Construct CI on P (Population proportionTltng th)

Size: n: ()Sample proportion

E: ?

P(E p + E) = 1 - P( pE

p E) = 1 -

= Binomial ( = np; 2= np(1-p))

Khi n ln N( = np; 2= np(1-p)) (Normal distribution)

P(npnE x np nE) = 1 -

P(

()

()

()) = 1 -

P(

()

(

()

) = 1 -

E = Z/2.()

=> P [ () ]

Upper: p * () ] Lower: P [ () ]

Population

P(A) = ?

Sample


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1. CI on (mean, average)

[

2. I on 2(variance/standard deviation)

2[() ()

]

[ , ]3. CI on P (Proportion, Percentage, probability)

P [ () ]

E =

()

= Error

n =


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Chapter 9:

TEST OF HYPOTHESYS

KIM NH GITHUYT THNG K1. The statistical hypothesisThe statistical hypothesis = a statement (a claim) about parameters of population

The statistical hypothesis = a statement () about parameters of population.

H0: = 1.65 m (p = 30%; = 0.5 m)

H1:

o 1.65m (two-sided)

o > 1.65m (one-sided)

o < 1.65m

H0, H1: i lp nhau, H0 quy c l dng du =.

2. Procedure to test of hypothesis on mean ()Ex :Test:

H0: = 1.65m

H1: 1.65m

- Choose 1 interval (a,b) cha 1.65

- If

( )Fail to reject H0 ( )Reject H0

2 types of errors:

1. Type I error: Reject H0 when H0 true.

2. Type II error: Fail to reject H0 when H0 false.

P(ype I error) =

P(type II error) = Ex :Test:

H0: = 1.65m

H1: 1.65m


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Assume that: Choose (a,b) = (1.63;1.67)

a. = ?

b. = ?

--

a. = P(Type I error) = P(Reject H0 when H0 true) = P( (1.63;1.67) when = 1.65Assume:

N = 36 > 30

= 0.02 m

Normal dis= P(Z (((1.63-1.65)/(0.02/n36);(1.67-1.65)/(0.02/n36))

= 1P(Z

(-6;6)) = 1(P(Z


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Conclude

If

(0 Z(/2). )Reject H0 (Strong conclusion)

(0

Z(/2).

)

Fail to reject H0 (Weak conclusion)

--

(0 Z(/2). )

(Z(/2))

Test statistic Critical values

(TK kim nh) (Gi trti hn)--

Step 1:

H0: ?

H1: ?

Step 2:

Test statistic: Z0 =

Step 3:

Find critical values: Z(/2)Step 4:

Conclude:

If Z0 (Z(/2))Reject H0 If Z0(Z(/2))Fail to reject H0


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* Test on mean

1. Case 1: n 30; pop = normal dist + knownTest statistic: Z0 =

Critical value:

: H1 Z : H>

-Z : H Z

Reject H0 when Z0 < -Z

2. Case 2: n 40; unknownTest statistic: Z0 =

Critical value:

: H1 Z : H>

-Z : H Z

Reject H0 when Z0 < -Z

3. Case 3: n < 40; unknown + normalTest statistic: T0 =

Critical value:

: H1 : H> - : H t, n-1

Reject H0 when t0 < -t, n-1

* Test on variance of population (2)

H0: 2=

2<

Test statistic: () Critical value:

H1:X2(/2;n-1); X2(1-/2;n-1)

Conclude:

Reject H0 when X02 [X2(/2;n-1); X2(1-/2;n-1)]H1: >X

2(;n-1)

Conclude:

Reject H0 when X02>X2(;n-1)

H1:


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Chapter 11:

SIMPLE LINEAR REGRESSION &CORRELATION

HI QUY TUYN TNH N &TNG QUAN1. Linear regression line Assume ha: (x1,y1),,(xn,yn) = sample.

Choose model: Y = Base on data => Fine estimations of , say, such that min. where ei = yi - (Method of least squares) min ( ) min ( ) min = L

2. orrelaion oeffiien [*] Sample correlation coefficient

r =

=()()

()()|r| 1

Sxy = || ( ) |r|: measure he srengh of he linear relaionship beween x and y: o m quan h uy nh

gia x v y

|r| 1StrongThe regression line is significant.

|r| 0No linear relationship im hn nKhng mqh tt

r > 0Sxy > 0 =>

> 0

=> Posiive orrelaion (Tng quan hun) (X angY tang)

r < 0Sxy < 0 => < 0=> Negaive orrelaion (Tng quan hun) (X angY giam)

r2= determination coefficient

r ng du


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3. Predict the value of Y when X = x0 Find r

If

o |r| 1 (0.5)Find the regression line

Y predicted = o |r| 0 (< 0.5)Y predicted =

4. Test of hypothesis on Sample: ; rPopulation: ; r: Correlation coefficient of populationa. Test on r

H0: r = 0Tes xem X v quan h uyn nh khng. || Tes for signifian of regression:

Tes xem ngha a ng hi quy hay khng.

Khng H Tuyn nh H1: r0

L Tuyn nh

Test statistic: T0 =

Critical values:

( )(r0) ;n-2 (r > 0) -;n-2 (r < 0)

Conclude:

Reject H0 when T0 [( )] Reject H0 when T0 >() Reject H0 when T0 < -()

b. Test on 1:

T0 =() ;

0:

T0 =() ;

Lu : Ly bng 5


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- Sxy, Sxx, Syy

-bea 1 m, b a 0 m

- r

- SSt,SSr,SSe

- Predict Y

- Test

Z. Recording in Class

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