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1/2555 สมศักดิ์ ศิวดำรงพงศ์ somsaksi@sut.ac.th 525201 Statistics and Numerical Method Part I: Statistics Week V: Decision Making (II sample) 1/2555 สมศักดิ์ ศิวดำรงพงศ์ somsaksi@sut.ac.th

Introduction of 2 samples

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

Inference on the Mean of 2 population, Variance known

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)=0.0059 <0.05 Then reject H0

Sample size

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

Case I: Equal variances (Pooled t-test)

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 C5 8 92.26 2.39 0.84 C8 8 92.73 2.98 1.1 Difference = mu (C5) - mu (C8) Estimate for difference: -0.48 95% CI for difference: (-3.37, 2.42) T-Test of difference = 0 (vs not =): T-Value = -0.35 P-Value = 0.729 DF = 14 Both use Pooled StDev = 2.7009

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

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 10 12.50 7.63 2.4 Rural 10 27.5 15.3 4.9 Difference = mu (Metro) - mu (Rural) Estimate for difference: -15.00 95% CI for difference: (-26.71, -3.29) T-Test of difference = 0 (vs not =): T-Value = -2.77 P-Value = 0.016 DF = 13

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.

Paired t-test

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 9 1.3401 0.1460 0.0487 Lehigh method 9 1.0662 0.0494 0.0165 Difference 9 0.2739 0.1351 0.0450 95% CI for mean difference: (0.1700, 0.3777) T-Test of mean difference = 0 (vs not = 0): T-Value = 6.08 P-Value = 0.000

5-6 Inference on 2 population proportions

Inference on 2 population proportions

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 1 253 300 0.843333 196 300 0.653333 Difference = p (1) - p (2) Estimate for difference: 0.19 95% CI for difference: (0.122236, 0.257764) Test for difference = 0 (vs not = 0): Z = 5.36 P-Value = 0.000