Upload
duc-filan
View
217
Download
0
Embed Size (px)
Citation preview
8/13/2019 Z. Recording in Class
1/40
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
8/13/2019 Z. Recording in Class
2/40
[*] 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
8/13/2019 Z. Recording in Class
3/40
[*] 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)
8/13/2019 Z. Recording in Class
4/40
[*] 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
8/13/2019 Z. Recording in Class
5/40
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) =
8/13/2019 Z. Recording in Class
6/40
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)
8/13/2019 Z. Recording in Class
7/40
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
8/13/2019 Z. Recording in Class
8/40
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) =
()() =
()()() =
()() ()()
8/13/2019 Z. Recording in Class
9/40
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!
8/13/2019 Z. Recording in Class
10/40
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
8/13/2019 Z. Recording in Class
11/40
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)
8/13/2019 Z. Recording in Class
12/40
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)
8/13/2019 Z. Recording in Class
13/40
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) =
8/13/2019 Z. Recording in Class
14/40
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).
= ()
8/13/2019 Z. Recording in Class
15/40
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
8/13/2019 Z. Recording in Class
16/40
Chapter 4:
CONTINUOUS RANDOM VARIABLE
1. The probability desity function (PDF): Hm m xc sut
P(
8/13/2019 Z. Recording in Class
17/40
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
8/13/2019 Z. Recording in Class
18/40
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
8/13/2019 Z. Recording in Class
19/40
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= )
8/13/2019 Z. Recording in Class
20/40
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) =
8/13/2019 Z. Recording in Class
21/40
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%
8/13/2019 Z. Recording in Class
22/40
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) =
8/13/2019 Z. Recording in Class
23/40
[*] 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
8/13/2019 Z. Recording in Class
24/40
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
8/13/2019 Z. Recording in Class
25/40
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
8/13/2019 Z. Recording in Class
26/40
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
8/13/2019 Z. Recording in Class
27/40
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
8/13/2019 Z. Recording in Class
28/40
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
8/13/2019 Z. Recording in Class
29/40
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 : ( )
8/13/2019 Z. Recording in Class
30/40
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:()
()
8/13/2019 Z. Recording in Class
31/40
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
8/13/2019 Z. Recording in Class
32/40
1. CI on (mean, average)
[
2. I on 2(variance/standard deviation)
2[() ()
]
[ , ]3. CI on P (Proportion, Percentage, probability)
P [ () ]
E =
()
= Error
n =
8/13/2019 Z. Recording in Class
33/40
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
8/13/2019 Z. Recording in Class
34/40
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
8/13/2019 Z. Recording in Class
35/40
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
8/13/2019 Z. Recording in Class
36/40
* 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:
8/13/2019 Z. Recording in Class
38/40
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
8/13/2019 Z. Recording in Class
39/40
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
8/13/2019 Z. Recording in Class
40/40
- Sxy, Sxx, Syy
-bea 1 m, b a 0 m
- r
- SSt,SSr,SSe
- Predict Y
- Test