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TMA 4255 Applied statistics

TMA 4255 Applied statistics. Spring 2010 Part 2. One-Way ANOVA. General setup – One-Way ANOVA. MINITAB analysis of Table 3.1. (MINITAB analysis of Table 13.1). (Used in MINITAB). 13.9 Randomized Complete Block Designs. Model with interactions between treatment and block:.

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TMA 4255 Applied statistics

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  1. TMA 4255Applied statistics Spring 2010 Part 2

  2. One-Way ANOVA

  3. General setup – One-Way ANOVA

  4. MINITAB analysis of Table 3.1

  5. (MINITAB analysis of Table 13.1)

  6. (Used in MINITAB)

  7. 13.9 Randomized Complete Block Designs

  8. Model with interactions between treatment and block:

  9. Degrees of freedom:

  10. Example 9 It is assumed that the maximal voltage of a storing battery depends on the material used in the plates of the battery, and of the temperature in the room where the battery is stored. To investigate this one takes four replicates for each of three materials and three temperatures.

  11. 13.12 Random Effects Model

  12. s

  13. MINITAB (Example 13.8)

  14. Control Charts

  15. Control Charts for Defects (Poisson)

  16. Chi-square testing of model H0: Data are from N(75,10^2) n=60 Expected 4.0 5.5 9.0 11.5 11.5 9.0 9.5 Observed 7 11 16 7 8 5 6 Chi-square statistic: 19.1 Distribution under H0: Chi-Sq with df=7-1=6. Crit.value 5%: 12.59 P-value: P(Chi-Sq6 > 19.1) = 0.004

  17. H0: Data are normally distributed Estimated mean: 71.4, estimated st.dev. 10.7 Expected 8.6 7.9 10.4 11.0 9.4 6.6 6.1 Observed 7 11 16 7 8 5 6 Chi-square statistic: 6.6 Distribution under H0: Chi-Sq with df=7-1-2=4. Crit.value 5%: 9.49 P-value: P(Chi-Sq4 > 6.6) = 0.16

  18. 10.15 Test for independenceEye color: 1=blue, 2=brownHair color: light blond, dark blond, darkMINITAB: Chi-Square Test: l.blond; d.blond; dark Expected counts are printed below observed counts Chi-Square contributions are printed below expected counts l.blond d.blond dark Total 1 29 12 7 48 23,40 14,40 10,20 1,340 0,400 1,004 2 10 12 10 32 15,60 9,60 6,80 2,010 0,600 1,506 Total 39 24 17 80 Chi-Sq = 6,860; DF = 2; P-Value = 0,032

  19. 10.16 Test for homogeneityBlood types: A, B, AB, OPopulations: 1, 2, 3MINITAB: Chi-Square Test: A; B; AB; O Expected counts are printed below observed counts Chi-Square contributions are printed below expected counts A B AB O Total 1 176 41 19 164 400 192,00 37,14 16,57 154,29 1,333 0,401 0,356 0,612 2 112 16 6 66 200 96,00 18,57 8,29 77,14 2,667 0,356 0,631 1,610 3 48 8 4 40 100 48,00 9,29 4,14 38,57 0,000 0,178 0,005 0,053 Total 336 65 29 270 700 Chi-Sq = 8,200; DF = 6; P-Value = 0,224 1 cells with expected counts less than 5.

  20. From Gunnar Løvås: Statistikk for universiteter og høgskoler. Universitetsforlaget. Simpson’s paradox Sometimes the results of a statistical investigation may be misleading if an important variable is overlooked or ignored. This is called Simpson’s Paradox. Here is an example: Example 258: A questionnaire about car damage has been sent to car Drivers. The results could have been reported like this: Damage No damage Total Man 233 323 556 Woman 87 194 281 Total 320 517 837 Thus: A proportion 233/556 = 0.42 had car damage, while for women this was 87/281 = 0.31. Does this mean that women are better drivers than men?

  21. Solution: A further study of the questionnaire reveal that a further categorization can be made, namely the size of the cars. Then we could present the results as: Big cars Small cars Damage No damage Total Damage No damage Total Man 150 35 185 83 288 371 Woman 16 2 18 71 192 263 Total 166 37 203 154 480 634 Now things have changed: Big cars: 100*16/18 = 88% of women have damage, and 100*150/185=81% for men. Small cars: 100*71/263= 27% for women; 100*83/371 = 22% for men So women have more damage for both small and big cars! How does this relate to the result on the previous slide?

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