7/28/2019 Pertemuan 11 Comparison of Two Means

1/14

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)

7/28/2019 Pertemuan 11 Comparison of Two Means

2/14

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.

7/28/2019 Pertemuan 11 Comparison of Two Means

3/14

Cont...

1. Anova/Anava/ Analysis of Varians satu jalur

2. Anova dua jalur

Uji ini digunakan untuk menganalisis lebih daridua sampel.

7/28/2019 Pertemuan 11 Comparison of Two Means

4/14

Paired Sample T Test

Paired sample t test merupakan uji beda dua sampelberpasangan. Sampel berpasangan merupakansubjek yang sama namun mengalami perlakuan yang

berbeda.

7/28/2019 Pertemuan 11 Comparison of Two Means

5/14

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

7/28/2019 Pertemuan 11 Comparison of Two Means

6/14

Cont...

Satu arahHo : 1 2

Ho: 1-2 0

HA: 1-2 < 0 Dua arah

= 1-2

Ho: =0HA: 0

7/28/2019 Pertemuan 11 Comparison of Two Means

7/14

=

n

7/28/2019 Pertemuan 11 Comparison of Two Means

8/14

Cont...

Sd =

n-1

7/28/2019 Pertemuan 11 Comparison of Two Means

9/14

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%.

7/28/2019 Pertemuan 11 Comparison of Two Means

10/14

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%

7/28/2019 Pertemuan 11 Comparison of Two Means

11/14

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

7/28/2019 Pertemuan 11 Comparison of Two Means

12/14

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%.

7/28/2019 Pertemuan 11 Comparison of Two Means

13/14

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

7/28/2019 Pertemuan 11 Comparison of Two Means

14/14

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.