2. © Buddy Freeman, 2015 H 0 : H 1 : α = Decision Rule: If then do not reject H 0, otherwise reject H 0. Test Statistic: Decision: Conclusion: We have found ________________ evidence at the _____ level of significance that

3. © Buddy Freeman, 2015 H 0 : H 1 : α = Decision Rule: If then do not reject H 0, otherwise reject H 0. Test Statistic: Decision: Conclusion: We have found ________________ evidence at the _____ level of significance that.05

4. © Buddy Freeman, 2015 H 0 : H 1 : α = Decision Rule: If then do not reject H 0, otherwise reject H 0. Test Statistic: Decision: Conclusion: We have found ________________ evidence at the _____ level of significance that the means of the number of parts produced per hour using methods: A, B, and C are not all equal..05 not all the means are equal μ 1 = μ 2 = μ 3 A B C

5. © Buddy Freeman, 2014 # of groups ? Parameter ? Can we make all f e > 5 ? Normal populations ? Hartley’s F max *(not in text) Resample and try again. yes no yes no chi-square df = (R-1)(C-1) pp. 368-374 yes no Kruskal-Wallis *pp. 621-625 1-way ANOVA pp. 386-395 ANOVA OK ? mean or median proportion variance or standard deviation more than 2 Parameter ? Related Samples ? mean or median proportion variance or standard deviation Normal populations ? yes no Levine-Brown-Forsythe F = S 1 2 /S 2 2 pp. 344-354 Z for proportions pp. 322-328 yes no unequal-variances t-test p. 307-315 pooled-variances t-test pp. 307-315 Wilcoxon Rank Sum *pp. 616-621 no yes Normal populations ? Normal populations ? yes no yes no n 1 > 30 and n 2 > 30 ? Z for means with σ 1 & σ 2 pp. 307-315 yes no σ 1 and σ 2 both known ? no Normal populations ? yes no yes at least interval level data ? yes no Sign Test *pp. 631-634. Wilcoxon Signed-Ranks *pp. 614-616 paired-difference t-test pp. 315-322 2 chi-square goodness-of-fit test pp. for the Multinomial Experiment 362-368 and the Normal Distribution 374-376 2 Groups and > 2 Groups Flowchart Spearman Rank Correlation test pp. 625-630 1 2 3 4 5 7 8 9 10 11 12 13 14 15 yes no n 1 > 30 and n 2 > 30 ? 6 σ 1 = σ 2 ? Levine-Brown-Forsythe Jaggia and Kelly (1 st edition) Default case * means coverage is different from text.

6. © Buddy Freeman, 2015 ANOVA OK ? There are three major assumptions for doing an Analysis of Variance (ANOVA) Test: 1. Normality: All the populations are normal. 2. Equality of Variances: The variances of all the populations are equal. 3. Independence of Error: The deviation of each value from the mean of the group containing that value should be independent of any other such deviation.

7. © Buddy Freeman, 2015 For this part of the course you may assume the third one (Independence of Error). Care taken when obtaining the sample results should minimize the possibility of a violation of this assumption. However, when data is collected over a lengthy interval of time, this assumption should be checked. We will discuss this further when we get into regression analysis. You are expected to check the other two assumptions: 1. Normality 2. Equality of Variances

8. © Buddy Freeman, 2015 This problem has the normality assumption as a given. Later on in the course, we will cover a procedure to test a population to see if it is normally distributed. How do we determine if all the population variances are equal ?

9. © Buddy Freeman, 2015 We can use the following hypothesis test: H0:H0: H 1 : not all variances are equal

10. © Buddy Freeman, 2015 If we do not reject the null hypothesis we can perform an ANOVA test to compare the means, but if we do reject the null hypothesis, the ANOVA test is NOT appropriate. If you have one or more of your populations that are definitely NOT normal the ANOVA test is NOT appropriate.

11. © Buddy Freeman, 2015 The flowchart can be used to determine the test statistic needed to test: H0:H0: H 1 : not all variances are equal

12. © Buddy Freeman, 2014 # of groups ? Parameter ? Can we make all f e > 5 ? Normal populations ? Hartley’s F max *(not in text) Resample and try again. yes no yes no chi-square df = (R-1)(C-1) pp. 368-374 yes no Kruskal-Wallis *pp. 621-625 1-way ANOVA pp. 386-395 ANOVA OK ? mean or median proportion variance or standard deviation more than 2 Parameter ? Related Samples ? mean or median proportion variance or standard deviation Normal populations ? yes no Levine-Brown-Forsythe F = S 1 2 /S 2 2 pp. 344-354 Z for proportions pp. 322-328 yes no unequal-variances t-test p. 307-315 pooled-variances t-test pp. 307-315 Wilcoxon Rank Sum *pp. 616-621 no yes Normal populations ? Normal populations ? yes no yes no n 1 > 30 and n 2 > 30 ? Z for means with σ 1 & σ 2 pp. 307-315 yes no σ 1 and σ 2 both known ? no Normal populations ? yes no yes at least interval level data ? yes no Sign Test *pp. 631-634. Wilcoxon Signed-Ranks *pp. 614-616 paired-difference t-test pp. 315-322 2 chi-square goodness-of-fit test pp. for the Multinomial Experiment 362-368 and the Normal Distribution 374-376 2 Groups and > 2 Groups Flowchart Spearman Rank Correlation test pp. 625-630 1 2 3 4 5 7 8 9 10 11 12 13 14 15 yes no n 1 > 30 and n 2 > 30 ? 6 σ 1 = σ 2 ? Levine-Brown-Forsythe Jaggia and Kelly (1 st edition) Default case * means coverage is different from text.

13. © Buddy Freeman, 2015 H 0 : H 1 : not all the variances are equal α = Decision Rule: If then do not reject H 0, otherwise reject H 0. Test Statistic: Decision: Conclusion: We have found ________________ evidence at the _____ level of significance that that the variances of the number of parts produced per hour using methods: A, B, and C are not all equal..05 A B C

14. © Buddy Freeman, 2015 H 0 : H 1 : not all the variances are equal α = Decision Rule: If then do not reject H 0, otherwise reject H 0. Test Statistic: Decision: Conclusion: We have found ________________ evidence at the _____ level of significance that that the variances of the number of parts produced per hour using methods: A, B, and C are not all equal..05 A B C More than 2 groups is an upper-tail test.

15. © Buddy Freeman, 2015 H 0 : H 1 : not all the variances are equal α = Decision Rule: If then do not reject H 0, otherwise reject H 0. Test Statistic: Decision: Conclusion: We have found ________________ evidence at the _____ level of significance that that the variances of the number of parts produced per hour using methods: A, B, and C are not all equal..05 A B C More than 2 groups is an upper-tail test. F max = 5.34 c = number of groups = 3

16. © Buddy Freeman, 2015 H 0 : H 1 : not all the variances are equal α = Decision Rule: If F max computed < 5.34 then do not reject H 0, otherwise reject H 0. Test Statistic: Decision: Conclusion: We have found ________________ evidence at the _____ level of significance that that the variances of the number of parts produced per hour using methods: A, B, and C are not all equal..05 A B C More than 2 groups is an upper-tail test. F max = 5.34 c = number of groups = 3

17. © Buddy Freeman, 2015 Calculation of Variations (and Variances)

18. © Buddy Freeman, 2015 Method Method Method A B C n 10 10 10  X = 900 840 810  X 2 = 81,882 72,076 67,048 Calculation of Variations (and Variances)

19. © Buddy Freeman, 2015 Method Method Method A B C n 10 10 10  X = 900 840 810  X 2 = 81,882 72,076 67,048 Calculation of Variations (and Variances)

20. © Buddy Freeman, 2015 Method Method Method A B C n 10 10 10  X = 900 840 810  X 2 = 81,882 72,076 67,048 Method A: Calculation of Variations (and Variances)

21. © Buddy Freeman, 2015 Method Method Method A B C n 10 10 10  X = 900 840 810  X 2 = 81,882 72,076 67,048 Method B: Calculation of Variations (and Variances)

22. © Buddy Freeman, 2015 Method Method Method A B C n 10 10 10  X = 900 840 810  X 2 = 81,882 72,076 67,048 Method C: Calculation of Variations (and Variances)

23. © Buddy Freeman, 2015 Method Method Method A B C n 10 10 10  X = 900 840 810  X 2 = 81,882 72,076 67,048 SS = 882 1,516 1,438 S 2 = 882/9 1,516/9 1,438/9 Calculation of Variations (and Variances) Largest variance Smallest variance

24. © Buddy Freeman, 2015 H 0 : H 1 : not all the variances are equal α = Decision Rule: If F max computed < 5.34 then do not reject H 0, otherwise reject H 0. Test Statistic: Decision: Conclusion: We have found ________________ evidence at the _____ level of significance that that the variances of the number of parts produced per hour using methods: A, B, and C are not all equal..05 A B C More than 2 groups is an upper-tail test. F max = 5.34 c = number of groups = 3 Do not reject H 0 Do not reject H 0. insufficient

25. © Buddy Freeman, 2015 We have found no evidence of any difference in the variances so the ANOVA test is appropriate to use to test the means. We now resume our interrupted hypothesis test of the means.

26. © Buddy Freeman, 2015 H 0 : H 1 : α = Decision Rule: If then do not reject H 0, otherwise reject H 0. Test Statistic: Decision: Conclusion: We have found ________________ evidence at the _____ level of significance that the means of the number of parts produced per hour using methods: A, B, and C are not all equal..05 not all the means are equal μ 1 = μ 2 = μ 3 A B C where k = # of groups and n T = n 1 + n 2 + … + n k More than 2 groups is an upper-tail test.

27. © Buddy Freeman, 2015 H 0 : H 1 : α = Decision Rule: If then do not reject H 0, otherwise reject H 0. Test Statistic: Decision: Conclusion: We have found ________________ evidence at the _____ level of significance that the means of the number of parts produced per hour using methods: A, B, and C are not all equal..05 not all the means are equal μ 1 = μ 2 = μ 3 A B C where k = # of groups and n T = n 1 + n 2 + … + n k dfd = n T - k = 30 - 3 = 27 dfn = k - 1 = 3 - 1 = 2 F = 3.354 More than 2 groups is an upper-tail test.

28. © Buddy Freeman, 2015 H 0 : H 1 : α = Decision Rule: If F computed < 3.354 then do not reject H 0, otherwise reject H 0. Test Statistic: Decision: Conclusion: We have found ________________ evidence at the _____ level of significance that the means of the number of parts produced per hour using methods: A, B, and C are not all equal..05 not all the means are equal μ 1 = μ 2 = μ 3 A B C where k = # of groups and n T = n 1 + n 2 + … + n k dfd = n T - k = 30 - 3 = 27 dfn = k - 1 = 3 - 1 = 2 F = 3.354 More than 2 groups is an upper-tail test.

29. © Buddy Freeman, 2015 Calculation of the ANOVA test statistic: The Denominator Variance The denominator is the variance within the groups, but remember variance = variation/df. Therefore, (S W ) 2 = SSW/df W. SSW is the variation WITHIN the groups = variation WITHIN group 1 + variation WITHIN group 2 + … + variation WITHIN group k = SS 1 + SS 2 + … + SS k. Here k = 3, so SSW = SS 1 + SS 2 + SS 3 = 882 + 1516 + 1438 = 3836. df W = df WITHIN group 1 + df WITHIN group 2 + … + df WITHIN group k = (n 1 - 1) + (n 2 - 1) + (n 3 - 1) = (n T - k) = 27.

30. © Buddy Freeman, 2015 Calculation of the ANOVA test statistic: The Numerator Variance The numerator is the variance between the groups, but remember variance = variation/df. Therefore, (S B ) 2 = SSB/df B. THE CONCEPT Here the group is the ‘entity.’ If you have k groups then you have (k-1) degrees of freedom, so df B = (k-1). Recall that the variation of a sample is defined as: SS =  (X-X) 2. We will expand this concept to groups. The X represents the group, so we will replace X with a value to represent the group, the mean of the group, X. In the sample variation the X represents the mean of all the X values, so we will replace it with the mean of all values in all the groups, X. The catch is that you must remember to multiply each square by the sample size of the group to account all the values of each group appropriately.

31. © Buddy Freeman, 2015 The definition formula for SSB is given by: The means of each group are: SSB = 10(90-85) 2 + 10(84-85) 2 + 10(81-85) 2 = 420 The next slide shows an alternative way to calculate SSB. Calculating SSB by Definition

32. © Buddy Freeman, 2015 SSB + SSW = SST, the total variation. Therefore, SSB = SST – SSW = SST – 3,836. To calculate SST, just add a column with the totals. Method Method Method A B C Total n 10 10 10 30  X = 900 840 810 2,550  X 2 = 81,882 72,076 67,048 221,006 Total: Hence, SSB = SST – SSW = 4,256 – 3,836 = 420. With SSB, SSW, and the dfs, the test statistic may be computed. Alternative Way to Calculate SSB

33. © Buddy Freeman, 2015 H 0 : H 1 : α = Decision Rule: If F computed < 3.354 then do not reject H 0, otherwise reject H 0. Test Statistic: Decision: Conclusion: We have found ________________ evidence at the _____ level of significance that the means of the number of parts produced per hour using methods: A, B, and C are not all equal..05 not all the means are equal μ 1 = μ 2 = μ 3 A B C dfd = n T - k = 30 - 3 = 27 dfn = k - 1 = 3 - 1 = 2 F = 3.354 More than 2 groups is an upper-tail test. Do not reject H 0 Do not reject H 0. insufficient