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Uji Satu Sampel
© 2002 Prentice-Hall, Inc. Chap 7-1
Skope
� Uji sebuah rata-rata ( known)
� Uji sebuah rata-rata ( unknown)
� Uji sebuah proporsi
σσ
© 2002 Prentice-Hall, Inc. Chap 7-2
Hypothesis?
� hypothesis adalah klaim
(assumsi)tentang parameterpopulasi
Saya klaim rata-rataIPK mhs unpad 3.5!µ =
© 2002 Prentice-Hall, Inc. Chap 7-3
populasi
© 1984-1994 T/Maker Co.
Hypothesis Testing Process
Idetifikasi Populasi
Asumsi Usia populasi
adalah 50.( )0 : 50H µ =
© 2002 Prentice-Hall, Inc. Chap 7-4
( )
REJECT
Ambil Sampel
Null Hypothesis
No, not likely!X 20 likely if Is ?µ= = 50
0 : 50H µ =
( )20X =
Distribution Sampling
Alasan Rejecting H0
X
© 2002 Prentice-Hall, Inc. Chap 7-5
= 50µµµµ20If H0 is true
X
Klaim Kita
Sampel Kita
Level of Significance and the Rejection Region
H0: µµµµ ≥ ≥ ≥ ≥ 3 H1: µµµµ < 3
0
αααα
αααα
Critical Value(s)
Rejection Regions
© 2002 Prentice-Hall, Inc. Chap 7-6
0
0
H0: µµµµ ≤≤≤≤ 3 H1: µµµµ > 3
H0: µµµµ = = = = 3 H1: µµµµ ≠≠≠≠ 3
αααα
αααα/2
Regions
One-tail Z Test for Mean( Known)
� Assumsi
� Population berdistribusi normal
� Z test statistic
σ
© 2002 Prentice-Hall, Inc. Chap 7-7
�
/X
X
X XZ
n
µ µσ σ− −= =
Rejection Region
Reject H0Reject H0
H0: µ ≥ µµ ≥ µµ ≥ µµ ≥ µ0H1: µµµµ < µµµµ0
H0: µ ≤ µµ ≤ µµ ≤ µµ ≤ µ0H1: µµµµ > µµµµ0
© 2002 Prentice-Hall, Inc. Chap 7-8
Z0
0
Z0Z Must Be SignificantlyBelow 0 to reject H0
Small values of Z don’t contradict H0
Don’t Reject H0 !
αα
Example Solution: One Tail Test
αααα = 0.5n = 9 σ = 60
H0: µ ≤ µ ≤ µ ≤ µ ≤ 300 H1: µµµµ > 300
© 2002 Prentice-Hall, Inc. Chap 7-9
n = 9 σ = 60
Example Solution: One Tail Test
αααα = 0.5n = 9 σ = 60
Test Statistic: H0: µ ≤ µ ≤ µ ≤ µ ≤ 300 H1: µµµµ > 300
1.50X
Zµ
σ−= =
© 2002 Prentice-Hall, Inc. Chap 7-10
n = 9 σ = 60
Critical Value: 1.645
Decision:
Conclusion:Do Not Reject at α = .05
No evidence that true mean is more than 368
Z0 1.645
.05
Reject
nσ
1.50
p -Value Solution
P-Value =.0668
Use the alternative
p-Value is P(Z ≥ ≥ ≥ ≥ 1.50) = 0.0668
© 2002 Prentice-Hall, Inc. Chap 7-11
Z0 1.50
P-Value =.0668
Z Value of Sample Statistic
From Z Table: Lookup 1.50 to Obtain .9332
alternative hypothesis to find the direction of the rejection region.
1.0000- .9332.0668
p -Value Solution(continued)
(p-Value = 0.0668) ≥ ≥ ≥ ≥ (αααα = 0.05) Do Not Reject.
p Value = 0.0668
© 2002 Prentice-Hall, Inc. Chap 7-12
01.50
Z
Reject
αααα = 0.05
Test Statistic 1.50 is in the Do Not Reject Region
1.645
Example Solution: Two-Tail Test
αααα = 0.05n = 25 σ = 60
Test Statistic: H0: µ = µ = µ = µ = 300 H1: µ ≠µ ≠µ ≠µ ≠ 300
1.50X
Zµ
σ−= =
© 2002 Prentice-Hall, Inc. Chap 7-13
n = 25 σ = 60
Critical Value: ±1.96
Decision:
Conclusion:
Do Not Reject at α = .05
No Evidence that True Mean is Not 300Z0 1.96
.025
Reject
-1.96
.025
1.50
nσ
p-Value Solution
(p Value = 0.1336) ≥ ≥ ≥ ≥ (αααα = 0.05) Do Not Reject.
p Value = 2 x 0.0668
© 2002 Prentice-Hall, Inc. Chap 7-14
01.50 Z
Reject
αααα = 0.05
1.96
Test Statistic 1.50 is in the Do Not Reject Region
Reject
R SOLUTION
� x=c(300,300,368,340,300,340,320,352,350)
� library(TeachingDemos)
� z.test(x, mu = 0, stdev, alternative = c("two.sided", "less", "greater"), sd = stdev,
© 2002 Prentice-Hall, Inc. Chap 7-15
c("two.sided", "less", "greater"), sd = stdev, conf.level = 0.95, ...)
� z.test(x, mu = 300, sd = 60, conf.level = 0.9, "two.sided")
R Solution
Ztest=function(x,m0,alpha,sigma)
{
n=length(x)
Zhit=sqrt(n)*(mean(x)-m0)/sigma
© 2002 Prentice-Hall, Inc. Chap 7-16
Zhit=sqrt(n)*(mean(x)-m0)/sigma
Ztabel=qnorm(1-alpha/2)
pvalue=2*(1-pnorm(Zhit))
hasil=c(Zhit,Ztabel,pvalue)
names(hasil)=c("Zhit","Ztabel","Pvalue")
return(hasil)
}
t Test: Unknown
� Assumsi
� Population berdistribusi normal
� T test statistic with n-1 degrees of freedom
σ
© 2002 Prentice-Hall, Inc. Chap 7-17
�
/
Xt
S n
µ−=
Example Solution: One-Tail
αααα = 0.01n = 9, df = 8
Test Statistic: H0: µ ≤ µ ≤ µ ≤ µ ≤ 300 H1: µ >µ >µ >µ > 300
nS
Xt
/
µ−= =3.48
© 2002 Prentice-Hall, Inc. Chap 7-18
3.48
2.896
n = 9, df = 8
Critical Value:
Decision:
Conclusion:
Reject at αααα = .01
that true mean is more than 300t80
.01
Reject
nS /
2.896
p -Value Solution
(p Value is 0.00823) ≥≥≥≥ (αααα = 0.01). Reject H0.
p Value 0.00823
© 2002 Prentice-Hall, Inc. Chap 7-19
0 3.48 t8
Reject
p Value 0.00823
αααα = 0.01
Test Statistic 3.48 is in Reject Region
2.896
R Solution
t.test(x, y = NULL, alternative = c("two.sided", "less", "greater"), mu = 0, paired = FALSE, var.equal = FALSE, conf.level = 0.95, ...)
t.test(x, alternative ="two.sided”, conf.level =
© 2002 Prentice-Hall, Inc. Chap 7-20
t.test(x, alternative ="two.sided”, conf.level = 0.95)
R Solution
Ttest=function(x,m0,alpha){n=length(x)df=n-1Thit=sqrt(n)*(mean(x)-m0)/sd(x)
© 2002 Prentice-Hall, Inc. Chap 7-21
Thit=sqrt(n)*(mean(x)-m0)/sd(x)Ttabel=qt(1-alpha/2,df)pvalue=2*(1-pt(Thit,df))hasil=c(Thit,Ttabel,pvalue)names(hasil)=c("Thit","Ttabel","Pvalue")return(hasil)}
Uji sebuah proporsi
� Terdapat dua “outcomes”
� “Sukses” “Gagal”
Fraction atau proportion dari population in the
© 2002 Prentice-Hall, Inc. Chap 7-22
� Fraction atau proportion dari population in the “success” category didefinisikan sbg p
Proportion
� Proporsi untuk kategori sukses dinotasikan pS�
(continued)
Number of Successes
Sample Sizes
Xp
n= =
© 2002 Prentice-Hall, Inc. Chap 7-23
� mean dan standard deviation
�
Sample Sizen
sp pµ = (1 )sp
p p
nσ −=
( ) ( ).05 .04
1.141 .04 1 .04
500
Sp pZ
p p
n
− −≅ = =− −
Z Test for Proportion: Solution
ps= .05
n = 500
H0: p = = = = .04 H1: p ≠≠≠≠ .04
Test Statistic:
© 2002 Prentice-Hall, Inc. Chap 7-24
500nn = 500
Do not reject at α = .05Critical Values: ±±±± 1.96 Decision:
Conclusion:
Z0
Reject Reject
.025.025
1.96-1.961.14
We do not have sufficient evidence to reject the company’s claim of 4% response rate.
p -Value Solution
(p Value = 0.6815189 ) ≥ ≥ ≥ ≥ (αααα = 0.05). Do Not Reject.
p Value = 2 x .1271
© 2002 Prentice-Hall, Inc. Chap 7-25
01.14
Z
Reject
αααα = 0.05
1.96
Test Statistic 1.14 is in the Do Not Reject Region
Reject
R Solution
� prop.test(x, n, p = NULL, alternative = c("two.sided", "less", "greater"), conf.level = 0.95, correct = TRUE)
Prop.test(5,100,0.04, alternative =
© 2002 Prentice-Hall, Inc. Chap 7-26
� Prop.test(5,100,0.04, alternative = "two.sided", conf.level = 0.95)
R Solution
Ptest=function(ps,p0,alpha)
{
Zhit=(ps-p0)/sqrt(ps*(1-ps)/n)
Ztabel=qnorm(1-alpha/2)
© 2002 Prentice-Hall, Inc. Chap 7-27
Ztabel=qnorm(1-alpha/2)
pvalue=2*(1-pnorm(Zhit))
hasil=c(Zhit,Ztabel,pvalue)
names(hasil)=c("Zhit","Ztabel","Pvalue")
return(hasil)
}