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E R R Y Y U D H Y A M U L Y A N I , M . S C
Comparison of Two Means
(Uji 2 Beda)
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Uji beda pada statistik parametrik:
1. Paired sample test (t-test)
2. Independent sample test (t-test)
Uji t digunakan untuk analisis dari hipotesis duakelompok data, baik yang berpasangan maupun
yang bebas.
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Cont...
1. Anova/Anava/ Analysis of Varians satu jalur
2. Anova dua jalur
Uji ini digunakan untuk menganalisis lebih daridua sampel.
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Paired Sample T Test
Paired sample t test merupakan uji beda dua sampelberpasangan. Sampel berpasangan merupakansubjek yang sama namun mengalami perlakuan yang
berbeda.
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Cont...
Tujuan : utk menguji perbedaan mean antara duakelompok data yang dependen.
Syarat :
1. Distribusi data normal2. Kedua kelompok dependen/pair
3. Jenis variabel adalah numerik dan kategori
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Cont...
Satu arahHo : 1 2
Ho: 1-2 0
HA: 1-2 < 0 Dua arah
= 1-2
Ho: =0HA: 0
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=
n
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Cont...
Sd =
n-1
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Contoh soal
The sample size is n = 63. The mean time tooccurrence of angina was xC = 3.35% during the
baseline (control) visit (when subjects werebreathing clean air on both the stress and secondmeasurement) and xT = 9.63%faster when subjects
were breathing air mixed with CO during the second(treatment) visit. The difference between the two
means isd = 6.63% with standard deviation sd =20.29%.
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1. The null hypothesis is H0 : = C T 0
2. The alternative hypothesis is
Ha : < 0 ( C < T) That is, when breathing air
mixed with CO angina occurs faster. 3. The level is 5%
4. The test statistic is
=2.59%
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5. Rejection rule: Reject H0 if T < t62;0.95 =1.673. Since T = 2.59 < 1.673 = t62;0.95 the nullhypothesis is rejected.
Conclussion:
Subjects when breathing air with CO experienceangina faster than when breathing air without CO
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Contoh soal 2
The sample size is n = 62. The mean time tooccurrence of angina was xC = 3.35% during the
baseline (control) visit (when subjects werebreathing clean air on both the stress and secondmeasurement) and xT = 9.63%faster when subjects
were breathing air mixed with CO during the second(treatment) visit. The difference between the two
means isd = 4.95% with standard deviation sd =19.05%.
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1. The null hypothesis is H0 : = C T 0 2. The alternative hypothesis is
Ha : < 0 ( C < T) That is, when breathing air mixedwith CO angina occurs faster.
3. The level is 5% 4. The test statistic is
= -2.05
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5. Rejection rule: Reject H0 if T < t62;0.95 =1.673. Since T = 2.05 < 1.673 = t62;0.95 the nullhypothesis is rejected.
Conclussion:
The patient is exposed to a low level of CO is largerthan the mean percent decrease when the patient is
not exposed.
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