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Minitab notes for unit 1 of stat 3503 at csu hayward, focusing on one-factor anova as a generalization of the two-sample t test. It covers data preparation, descriptive methods, and statistical tests. Students are guided on how to enter data, create subscripts, and perform descriptive statistics and one-way anova using minitab.
Typology: Study notes
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These data are from Table 6.2 (page 269) Ott and Longnecker: An Introduction to Statistical Methods and Data Analysis (^) 5th ed., Duxbury, 2001.
MTB > name c1 'Potency' MTB > set c DATA> 10.2, 10.5, 10.3, 10.8, 9.8, 10.6, 10.7, 10.2, 10.0, 10. DATA> 9.8, 9.6, 10.1, 10.2, 10.1, 9.7, 9.5, 9.6, 9.8, 9. DATA> end
DATA (MANIP in Releases 13 and earlier) ➯➯➯➯ Display Data MTB > print c
Type Group atop column 2 in the Worksheet CALC ➯➯➯➯ Patterned Data, Simple, values from 1 to 2, each individual value repeated 10 times MTB > name c2 'Group' MTB > set c DATA> (1:2) DATA> end
MTB > name c11 'Fresh' c12 'Stored' MTB > set c DATA> 10.2, 10.5, 10.3, 10.8, 9.8, 10.6, 10.7, 10.2, 10.0, 10. DATA> end MTB > set c DATA> 9.8, 9.6, 10.1, 10.2, 10.1, 9.7, 9.5, 9.6, 9.8, 9. DATA> end MTB > stack c11 c12 c13; SUBC> subs c14.
MTB > gstd MTB > boxp c1; SUBC> by c2.
MTB > gpro MTB > boxp c1 * c MTB > dotp c1 * c
1.3. t Test and One-Factor ANOVA
STAT ➯➯➯➯ Basic (^) ➯➯➯➯ 2-sample t, one column, assume equal variances MTB > twot c1 c2; SUBC> pool.
Two-sample T for Potency
Group N Mean StDev SE Mean 1 10 10.370 0.323 0. 2 10 9.830 0.241 0.
Difference = mu (1) - mu (2) Estimate for difference: 0. 95% CI for difference: (0.272230, 0.807770) T-Test of difference = 0 (vs not =): T-Value = 4.24 P-Value = 0.000 DF = 18 Both use Pooled StDev = 0.
STAT ➯➯➯➯ ANOVA (^) ➯➯➯➯ Oneway MTB > oneway c1 c2 (Alternatively: MTB > onew 'Potency' 'Group')
One-way ANOVA: Potency versus Group
Source DF SS MS F P Group 1 1.4580 1.4580 17.95 0. Error 18 1.4620 0. Total 19 2.
S = 0.2850 R-Sq = 49.93% R-Sq(adj) = 47.15%
Individual 95% CIs For Mean Based on Pooled StDev Level N Mean StDev ----+---------+---------+---------+----- 1 10 10.370 0.323 (-------------) 2 10 9.830 0.241 (-------------) ----+---------+---------+---------+----- 9.75 10.00 10.25 10.
Pooled StDev = 0.
STAT ➯➯➯➯ ANOVA (^) ➯➯➯➯ Balanced, select 'Potency' as Response, 'Group' as Model MTB > anova Potency = Group
Factor Type Levels Values Group fixed 2 1, 2
Analysis of Variance for Potency
Source DF SS MS F P Group 1 1.4580 1.4580 17.95 0. Error 18 1.4620 0. Total 19 2.
S = 0.284995 R-Sq = 49.93% R-Sq(adj) = 47.15%
STAT ➯➯➯➯ ANOVA ➯➯➯➯ General linear model MTB > glm Potency = Group
Factor Levels Values Group 2 1 2
Analysis of Variance for Potency
Source DF Seq SS Adj SS Adj MS F P Group 1 1.4580 1.4580 1.4580 17.95 0. Error 18 1.4620 1.4620 0. Total 19 2.
S = 0.284995 R-Sq = 49.93% R-Sq(adj) = 47.15%
Unusual Observations for Potency
Obs. Potency Fit Stdev.Fit Residual St.Resid 5 9.8000 10.3700 0.0901 -0.5700 -2.11R
R denotes an obs. with a large st. resid.
1.5. Nonparametric Alternatives
STAT > Nonparametric.