1/2555 สมศักดิ์ ศิวดำรงพงศ์ [email protected]
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525201 Statistics and Numerical Method Part I: Statistics Week V: Decision Making (II sample). 1/2555 สมศักดิ์ ศิวดำรงพงศ์ [email protected]. Introduction of 2 samples. 5-2 Inference on the Mean of 2 population, Variance known. Inference on the Mean of 2 population, Variance known. - PowerPoint PPT PresentationTRANSCRIPT
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525201Statistics and Numerical Method
Part I: StatisticsWeek V: Decision Making
(II sample)
1/2555 สมศักดิ์ ศิวดำ�รงพงศ์
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Introduction of 2 samples
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5-2 Inference on the Mean of 2 population, Variance known
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Inference on the Mean of 2 population, Variance known
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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>0n1=n2 1=2Z0=2.52P=1-(2.52)=0.0059 <0.05Then reject H0
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Sample size
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5-3 Inference on the Mean of 2 population, Variance unknown
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Case I: Equal variances (Pooled t-test)
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H0: 1-2=0 H1:1-2≠0n1=n2=8 1=2T0=-0.35P=0.729 > 0.05Then accept H0Two-Sample T-Test and CI: C5, C8
N Mean StDev SE MeanC5 8 92.26 2.39 0.84C8 8 92.73 2.98 1.1
Difference = mu (C5) - mu (C8)Estimate for difference: -0.4895% CI for difference: (-3.37, 2.42)T-Test of difference = 0 (vs not =): T-Value = -0.35 P-Value = 0.729 DF = 14Both use Pooled StDev = 2.7009
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5-3 Inference on the Mean of 2 population, Variance unknown
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H0: 1-2=0 H1:1-2≠0n1=n2=10 1 ≠ 2T0=-2.77P=0.016 < 0.05Then reject H0
Two-Sample T-Test and CI: Metro, Rural
N Mean StDev SE MeanMetro 10 12.50 7.63 2.4Rural 10 27.5 15.3 4.9
Difference = mu (Metro) - mu (Rural)Estimate for difference: -15.0095% CI for difference: (-26.71, -3.29)T-Test of difference = 0 (vs not =): T-Value = -2.77 P-Value = 0.016 DF = 13
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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.
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Paired t-test
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H0: D=1-2=0 H1: D= 1-2≠0n1=n2=9 1 ≠ 2T0=6.08P=0.000 < 0.05Then reject H0
Paired T-Test and CI: Karlsruhe method, Lehigh method
N Mean StDev SE MeanKarlsruhe method 9 1.3401 0.1460 0.0487Lehigh method 9 1.0662 0.0494 0.0165Difference 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
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5-6 Inference on 2 population proportions
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Inference on 2 population proportions
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H0: p1=p2 H1: p1≠p2Z0=5.36P=0.000 < 0.05Then reject H0
Test and CI for Two Proportions Sample X N Sample p1 253 300 0.8433332 196 300 0.653333
Difference = p (1) - p (2)Estimate for difference: 0.1995% CI for difference: (0.122236, 0.257764)Test for difference = 0 (vs not = 0): Z = 5.36 P-Value = 0.000