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PENGUJIAN HIPOTESIS
STATISTIK: Uji-Z, Uji-t, Uji-F
SUJARWO, SP., MP
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Ingat kembali ruang ruang penerimaan dan
penolakan hipotesis
Two-tailed Test
H0: =
H1:
is divided equally between
the two tails of the criticalregion
Means less than or greater than
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INFERENSI TERHADAP RATA-RATAPOPULASI
Tujuan pengujian untuk mengetahui apakahrata-rata sampel sudah mewakili populasinya.
Ukuran Sampel :
Besar : lebih dari 30 data Uji-Z. Jika Zhit > Ztab
maka Ho ditolak dan Ha di terima, sebaliknya jika
Zhit < Ztab maka Ha ditolak dan Ho diterima
Kecil : Kurang dari 30 data (untuk keperluan praktis). Uji-t .
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Contoh 2: Uji-Z
Rata-rata hasil produksi mesin lama adalah 2200 kg/hari. Sebuah mesin baru diuji dalam 200 hari, ternyatahasil produksinya menyebar secara normal dengan rata-
rata produksi 2280 kg/ hari dan standart deviasi 520 kg/hari. Apakah produktifitas mesin baru lebih baik darimesin lama ?
Hipotesis : H0: = 2200 H1: 2200
Pengujian dilakukan dengan uji Z
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UjiZ Kasus 1:
200520
220022800
/n/
x
hitz
Menghitung Z-Hitung :
Z-hitung = 2,715
Z-tabel = 1,96 - alpha = 0,05 (two tailed)
Karena Z hitung lebih besar dari Z tabel maka Ho di tolak danditerima H1.
Kesimpulan : Produktifitas mesin yang baru lebih tinggi dari yanglama.
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Increasing sample size
S2closer to 2
tstatistic
Critical value of t Depends on sample size
Df
Significant level was chosen byresearcher
Ms
M
n
s
Mt
-5 -4 -3 -2 -1 0 1 2 3 4 5
X
0.0
0.1
0.2
0.3
0.4
Y
-5 -4 -3 -2 -1 0 1 2 3 4 5
X
0.0
0.1
0.2
0.3
0.4
Y
-5 -4 -3 -2 -1 0 1 2 3 4 5
X
0.0
0.1
0.2
0.3
0.4
Y
t-distribution with 2, 5, 10, 30 df
-5 -4 -3 -2 -1 0 1 2 3 4 5
X
0.0
0.1
0.2
0.3
0.4
Y
Uji statistik : Uji-t
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Critical Values of tProportion in one tail 0.25 0.1 0.05 0.025 0.01 0.005Proportion in two tails
df0.5 0.2 0.1 0.05 0.02 0.01
1 1.000 3.078 6.314 12.706 31.821 63.656
2 0.816 1.886 2.920 4.303 6.965 9.925
3 0.765 1.638 2.353 3.182 4.541 5.841
4 0.741 1.533 2.132 2.776 3.747 4.604
5 0.727 1.476 2.015 2.571 3.365 4.032
10 0.700 1.372 1.812 2.228 2.764 3.169
15 0.691 1.341 1.753 2.131 2.602 2.947
20 0.687 1.325 1.725 2.086 2.528 2.845
25 0.684 1.316 1.708 2.060 2.485 2.787
30 0.683 1.310 1.697 2.042 2.457 2.75050 0.679 1.299 1.676 2.009 2.403 2.678
100 0.677 1.290 1.660 1.984 2.364 2.626
1000 0.675 1.282 1.646 1.962 2.330 2.581
1000000 0.674 1.282 1.645 1.960 2.326 2.576
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Hypothesis testing with the t
statistic Basic form
Steps
State H0and H1Set level
Determine critical value of t One or twotailed hypothesis
level
df.
Calculate tvalue
Evaluate H0
errorstandardestimated
Hfrommeanpopulation-datafrommeansample 0t
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Computation
Ms
M
n
s
Mt
11
2
2
2
n
n
XX
n
SSs
1
2
2
2
nn
n
XX
M
n
s
M
s
Mt
M
X 2
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Example 1:
Diketahui produktifitas kedelai di suatu daerahrata-rata adalah 12 kwt/ha. Diambil sampel 15responden. Apakah rerata sampel sama denganpopulasi ?
H0: = 12, H1: 12 = .01, df= 15 - 1 = 14
Critical t(14) = 2.977
Diketahui : Rerata sampel = 11.2
standart deviasi = 3.255
X X2
12 14413 1696 36
11 12112 1448 64
11 121
7 4910 10016 25610 1007 49
14 19615 22516 256
168 2030
X2X
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8406
8733
25583
15
25583.
.
..
n
ssM
95166.8406. 8.8406. 122.11
Ms
Mt
Critical t(14) = 2.977-5 -4 -3 -2 -1 0 1 2 3 4 5
X
0.0
0.1
0.2
0.3
0.4
Y
Gimana kesimpulannya ????
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Example 2:Produktifitas padi rata-ratamenghasilkan 67 kwt/ ha. Dan adateknologi baru dengan data sampelberikut:
H0: = 67, H1: 67
= .05, df= 15 - 1 = 14
Critical t(14) = 2.145
8.7115
1077M
422.477,328.6751,7715
1,159,929751,77
15
107777751
22
2
N
XXSS
X X2
65 422576 577669 476171 504174 547678 608477 592968 462472 5184
75 562574 547664 409669 476163 396982 6724
1077 77751
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30.171115
422.4
1
2
n
SS
s 5.49330.1712 ss
1.418873.3
5.493
15
30.1712
n
ssM
3.38
1.418
8.4
1.418
678.71
Ms
Mt
thit = 3.3845
-5 -4 -3 -2 -1 0 1 2 3 4 5
X
0.0
0.1
0.2
0.3
0.4
Y
Gimana kesimpulannya ????
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Uji Statistik : Uji-F
For cases in which two population variances are to becompared, the F statistic is commonly used.
TestStatistic: F =s1
2/ s22
where s12
and s22
are the sample variances.
The more this ratio deviates from 1, the stronger the evidencefor unequal population variances.
H0: 1 = 2
Ha: 1 2
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The hypothesis that the two standard deviations are equal is
rejected if:
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Example:
Method Mean (ppm)Standard
Deviation (ppm)1 6.7 0.8
2 8.2 1.2
As an example, assume we want to see if a method (Method 1) for measuring
the arsenic concentration in soil is significantly more precisethan a second method (Method 2).Each method was tested ten times, with, yielding the following values:
Since s2 > s1, Fcalc= s22/s1
2 = 1.22/0.82 = 2.25. The tabulated value
for d.o.f. = 9 in each case, and a 1-tailed, 95% confidence level isF9,9 = 3.179. In this case, Fcalc< F9,9, so we accept the nullhypothesis that the two standard deviations are equal, and we are95% confident that any difference in the sample standard deviationsis due to random error. We use a 1-tailed test in this case becausethe only information we are interested in is whether Method 1 ismore precise than Method 2.
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