1. 1/2555 สมศักดิ์ ศิวดำรงพงศ์ somsaksi@sut.ac.th
Statistics and Numerical Method Part I: Statistics Week V: Decision Making (II sample)
1/2555
สมศักดิ์ ศิวดำรงพงศ์

2. Introduction of 2 samples

3. 5-2 Inference on the Mean of 2 population, Variance known

4. Inference on the Mean of 2 population, Variance known

5. Ex: Drying time of primer has SD=8 min
Ex: Drying time of primer has SD=8 min. 10 Samples of Primer1has x-bar1=121 min. and another 10 samples of Primer II has x-bar2=112 min. Is the primer 2 has effect to drying time.
H0: 1-2=0 H1:1-2>0
n1=n2 1=2
Z0=2.52
P=1-(2.52)= <0.05
Then reject H0

6. Sample size

7. 5-3 Inference on the Mean of 2 population, Variance unknown

8. Case I: Equal variances (Pooled t-test)

10. H0: 1-2=0 H1:1-2≠0 n1=n2=8 1=2 T0=-0.35 P=0.729 > 0.05
Then accept H0
Two-Sample T-Test and CI: C5, C8
N Mean StDev SE Mean C C
Difference = mu (C5) - mu (C8)
Estimate for difference:
95% CI for difference: (-3.37, 2.42)
T-Test of difference = 0 (vs not =): T-Value = P-Value = DF = 14
Both use Pooled StDev =

12. 5-3 Inference on the Mean of 2 population, Variance unknown

14. H0: 1-2=0 H1:1-2≠0 n1=n2=10 1 ≠ 2 T0=-2.77 P=0.016 < 0.05
Then reject H0
Two-Sample T-Test and CI: Metro, Rural
N Mean StDev SE Mean
Metro
Rural
Difference = mu (Metro) - mu (Rural)
Estimate for difference:
95% CI for difference: (-26.71, -3.29)
T-Test of difference = 0 (vs not =): T-Value = P-Value = DF = 13

16. 5-4 Paired t-test
A special case of the two-sample t-tests of Section 5-3 occurs when the observations on the two populations of interest are collected in pairs.
Each pair of observations, say (X1j , X2j ), is taken under homogeneous conditions, but these conditions may change from one pair to another.
For example; The test procedure consists of analyzing the differences between hardness readings on each specimen.

17. Paired t-test

19. H0: D=1-2=0 H1: D= 1-2≠0 n1=n2=9 1 ≠ 2 T0=6.08
P=0.000 < 0.05
Then reject H0
Paired T-Test and CI:
Karlsruhe method, Lehigh method
N Mean StDev SE Mean
Karlsruhe method
Lehigh method
Difference
95% CI for mean difference: (0.1700, )
T-Test of mean difference = 0 (vs not = 0): T-Value = P-Value = 0.000

20. 5-6 Inference on 2 population proportions

21. Inference on 2 population proportions

22. H0: p1=p2 H1: p1≠p2 Z0=5.36 P=0.000 < 0.05 Then reject H0
Test and CI for Two Proportions
Sample X N Sample p
Difference = p (1) - p (2)
Estimate for difference: 0.19
95% CI for difference: ( , )
Test for difference = 0 (vs not = 0): Z = 5.36
P-Value = 0.000