Chapter 5 Anova2009


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  • Sort all the random numbers and rank them from 1-12.
  • One factor: anticoagulant; three level: A, B, C
  • The alternative hypothesis is that at least two of the group means are not the same.
  • The squared deviations are summed over all observations over all groups. Partitioning the sum of squares.
  • The results from the analysis of variance are typically displayed in an ANOVA table.
  • The ANOVA test lets us detect when at least two groups have different underlying means, but it does not let us state which of the groups have means that differ from each other. The usual practice is to perform the overall F test . If H0 is rejected, then specific groups are compared, as discussed in this section.
  • Df is the degree of freedom of MSE.
  • Gene, heredity; same breed. Same race
  • Chapter 5 Anova2009

    1. 1. Analysis of Variance Yuantao Hao 26th,Oct. 2009 Chapter5
    2. 2. Review: <ul><li>The basic steps and logic of hypothesis testing; </li></ul><ul><li>One-sample t test; </li></ul><ul><li>Two-sample t test; </li></ul><ul><li>Paired –sample t test; </li></ul><ul><li>F test for homogeneity of variances; </li></ul><ul><li>Z test for the parameters of binomial distribution and Poisson distribution when sample size is large enough. </li></ul>
    3. 3. main steps: <ul><li>(1) Set up the statistical hypotheses </li></ul><ul><li>(2) Select statistics and calculate its current value </li></ul><ul><li>(3) Determine the P- value </li></ul>
    4. 4. <ul><li>(4) Decision and conclusion </li></ul><ul><li>Comparing the P- value with the pre-assigned small probability , if P ≤ , then reject ; otherwise, not reject . Finally, issue the conclusion incorporating with the background. </li></ul>
    5. 5. P value <ul><li>P- value is defined as a probability of the event that the current situation and even more extreme situation towards appear in the population. </li></ul>
    6. 6. <ul><li>The P-value can also be thought of as the probability of obtaining a test statistic as extreme as or more extreme than the actual test statistic obtained, given that the null hypothesis is true. </li></ul>
    7. 7. Question: <ul><li>One-sample; </li></ul><ul><li>Two-sample; </li></ul><ul><li>Paired –sample; </li></ul><ul><li>Three or more samples? </li></ul>
    8. 8. An alysis o f va riance (ANOVA) : <ul><li>One-way ANOVA is used to test for differences among two or more independent groups . </li></ul><ul><li>Typically, however, the one-way ANOVA is used to test for differences among at least three groups , since the two-group case can be covered by a t-test . </li></ul>
    9. 9. 5.1 One-Way ANOVA for the Completely Random Design <ul><li>The completely random design </li></ul><ul><li>For this design, there is only one treatment factor with G (≥2) levels. The term level refers to the possible status planned for the treatment factor . </li></ul>
    10. 10. <ul><li>Example 5.1 Randomly assign 12 laboratory blood specimens (experiment units) into three groups with 4 blood specimens in each group. </li></ul><ul><li>How to assign the 12 units into three groups randomly ? </li></ul>
    11. 11. 12 11 10 9 8 7 6 5 4 3 2 1 Unit number 60 36 72 92 08 89 82 66 00 22 90 39 Random number 6 4 8 12 2 10 9 7 1 3 11 5 Rank (R) 2 1 2 3 1 3 3 2 1 1 3 2 Grouping result
    12. 13. <ul><li>Example 5.2 12 blood specimens are randomly assigned into three groups according to Table 2. Group 1 receives the treatment of anticoagulant (抗凝血剂) A; group 2 receives anticoagulant B; and group 3 receives anticoagulant C. </li></ul>
    13. 14. <ul><li>For each blood specimen, the erythrocyte sedimentation rate (ESR ,红细胞沉降率 ) after receiving the treatment is measured. The aim is to test whether the three mean ESRs are significantly different . The results are showed in Table 11.2. </li></ul>
    14. 17. <ul><li>If then reject H 0 </li></ul><ul><li>If then not reject H 0 </li></ul>
    15. 22. Two assumptions on analysis of variance: <ul><li>follows normal distribution , ; </li></ul><ul><li>homogeneity of variances, . </li></ul>
    16. 23. Bartlett’s test : are not all equal : :
    17. 24. Example 5.2 (cont.) <ul><li>Test the homogeneity of variances for the three populations in Table 2. </li></ul>P >0.10
    18. 25. Test for normality and transformations:
    19. 26. 5.2 Multiple comparisons <ul><li>To examine whether a specified two means are equal or not . LSD- t test . </li></ul><ul><li>To examine whether all the means of comparison groups are equal or not . SNK- q test. </li></ul>
    20. 27. LSD- t test <ul><li>(least significant difference t test) </li></ul>H 0 is rejected if :
    21. 28. SNK- q test <ul><li>(Student-Newman-Keuls q test) </li></ul>
    22. 29. <ul><li>All means should be sorted from the smallest to the biggest to form contrasts. </li></ul><ul><li>Each contrast may contain a means, a =2,3,…, G . </li></ul><ul><li>In Example 11.2, the means of three groups is sorted as 9.3, 11.3 and 16.0; if 9.3 and 11.3 are selected to form a contrast, a =2 ; </li></ul><ul><li>With the parameters a and, the critical value of SNK- q test can be find out from Table 11 of Appendix 2. </li></ul>
    23. 31. H 0 is rejected if :
    24. 33. 5.3 Two-Way ANOVA for the Randomized Complete-Block Design <ul><li>There are n blocks and each block contains G experimental units to receive G treatments randomly. The total number of observations is N = nG . </li></ul>
    25. 34. <ul><li>Example5.4 12 mice have been grouped into 4 blocks according to their birth litters and each block has 3 mice. Randomly assign 3 kinds of food to the 3 mice in each block. </li></ul>
    26. 36. <ul><li>The advantage of this design comparing with the completely random design is to reduce the effect of the variation among the experimental units if the difference among blocks was a main source of variation. </li></ul>
    27. 37. <ul><li>The disadvantage is that all the sizes of different blocks (or say, the numbers of experimental units in different blocks) should equal to the number of treatments, otherwise, the statistical analysis will be difficult. </li></ul>
    28. 39. <ul><li>Example 5.5 The investigator used randomized block design to carry out the experiment to compare the anti-tumor effects of three anti-tumor drugs A, B, C on mice sarcoma (肉瘤) . 15 mice of the same race were selected and three anti-tumor drugs A, B, C randomly allocated into 3 mice within the same block. </li></ul>
    29. 40. <ul><li>With the observations of sarcoma’s weight, the experiment result sees Table 5. Please test if the effects of three anti-tumor drugs are different. </li></ul>
    30. 41. 5
    31. 43. 0.532
    32. 44. Summary: <ul><li>Basic logic of ANOVA; </li></ul><ul><li>ANOVA for completely random design data; </li></ul><ul><li>Multiple comparison; </li></ul><ul><li>ANOVA for randomized complete-block design data. </li></ul>
    33. 45. THE END <ul><li>THANKS ! </li></ul>