1.  List the characteristics of the F distribution.  Conduct a test of hypothesis to determine whether the variances of two populations are equal.  Discuss the general idea of analysis of variance.  Organize data into an ANOVA table.  Conduct a test of hypothesis among three or more treatment means. Our Objectives

2.  The F Distribution.  Comparing Two Population Variances.  The ANOVA Test-One Way.  Inferences about Pairs of Treatment Means. Analysis of Variance

3.  A probability distribution used to test whether two samples are from populations having equal variances.  Also used when comparing several population means simultaneously using a technique called analysis of variance (ANOVA).  In both above situations, the populations must follow a normal distribution. The F Distribution

4.  There is a family of F distributions.  Each determined by 2 parameters: the degrees of freedom (df) in the numerator and the df in the denominator  Ex: df=(19,6)  Its shape changes as the df change Characteristics of the F Distribution

5.  The distribution is continuous.  It cannot be negative.  It is positively skewed, or skewed to the right  The long tail of the distribution is to the right-hand side  As the df in both numerator and denominator increases, the F- distribution tends to a Normal distribution  It is asymptotic.  Appendix B.4 calculates the Critical Values for a 0.05 and 0.01 level of significance (α). Characteristics of the F Distribution

6. α = 0.05 Critical Value = 3.87 Accept H 0 For an F(6,7), the area after 3.87 is 0.05.

7.  We assume that the ratio of the sample SDs follow an F distribution with n 1 -1 and n 2 -1 degrees of freedom. Comparing Two Population Variances. H 0 : σ 1 2 = σ 2 2 H 1 : σ 1 2 ≠ σ 2 2

8. Hypothesis Testing (Two-Tailed) H 0 : σ 1 2 = σ 2 2 H 1 : σ 1 2 ≠ σ 2 2 α/2 Critical Value Accept H 0  This is a Two-tailed test. So divide the significance level in half (α/2).  Select samples: n 1 observations from the first population and n 2 from the second.  Test statistic for comparing two variances is F o s 1 squared and s 2 squared are samples variances. o Use n 1 -1 and n 2 -1 degrees of freedom in Table in App B.4. o The larger sample variance is placed in numerator.

9. Hypothesis Testing (Two-Tailed) H 0 : σ 1 2 = σ 2 2 H 1 : σ 1 2 ≠ σ 2 2 α/2 Critical Value Accept H 0 If F is less than the Critical Value, then we fail to reject H 0  By putting the larger variance value in the numerator, we are forcing the F ratio to be at least 1. This allows us to always use the right tail of distribution.

10. Hypothesis Testing (One-Tailed) H 0 : σ 1 2 ≤ σ 2 2 H 1 : σ 1 2 > σ 2 2 α Critical Value Accept H 0 If F is less than the Critical Value, then we fail to reject H 0

11. Example 4, page 412. First Population: A random sample of five observations resulted in a standard deviation of 12. Second Population: A random sample of seven observations resulted in a standard deviation of 7. At the 0.01 significance level, is there more variation in the first population? H 0 : σ 1 2 ≤ σ 2 2 H 1 : σ 1 2 > σ 2 2

12. Example 4, page 412. 0.01 Critical Value=9.15 Accept H 0 F is less than the Critical Value, then we fail to reject H 0 H 0 : σ 1 2 ≤ σ 2 2 H 1 : σ 1 2 > σ 2 2 Calculating the Critical Value: o α=0.01 o one-tailed test o df numerator = 5-1 = 4 o df denominator = 7-1 = 6 o Using App B.4, CV=9.15

13. We also use F statistic in ANOVA to compare three or more population means to determine whether they could be equal.  Assumptions for using ANOVA:  Populations follow normal distribution.  Populations have equal SD.  Populations are independent.  Is there a difference in the means between populations (treatments)? ANOVA (Analysis of Variance)

14. Consider, ANOVA (Analysis of Variance) H 0 : μ 1 = μ 2 = μ 3 = … = μ n H 1 : At least two of the means are not equal

15. Example A manager of a regional financial center wishes to compare the productivity, as measured by the number of customers served, among three employees. Four days are randomly selected and the number of customers served by each employee is recorded. Is there a difference in the mean number of customers served? WolfWhiteKorosa 556647 547651 596746 567148

16. Comparing productivity of 3 employees (different population means)

17. If the population means are the same

18. The ANOVA Test  Objective: to determine whether the various sample means came from a single population or populations with different means.  How does ANOVA work?  we compare sample means through their variances.  Use the ANOVA assumption that the populations SD’s are equal.  Estimate the population variance two ways:  If ratio is equal to 1, we conclude that population means are the same. If quite different from 1, conclude that population means are different.  ANOVA tests were first developed for applications in agriculture.  This is reflected in the use of the term treatment to identify the different populations being studied.

19.  State the null and alternate hypothesis.  Given the significance level α and using the F distribution, find the Critical Value.  Identify the Accept and Reject Regions.  Compute the statistic F using an ANOVA table.  Check if the statistic F falls in Accept or Reject Region. Hypothesis testing using ANOVA H 0 : μ 1 = μ 2 = μ 3 = … = μ n H 1 : At least two of the means are not equal

20. F statistic: The first estimate of the population variance is based on the treatments, that is, the difference between the means. The second estimate of the population variance is the estimate within the treatments. The F statistic

21.  Let be the overall grand mean.  Let be the mean for treatment c.  Compute three measures of variance (SS total, SSE and SST)  SS total: The sum of the squared error between each sample and the overall mean.  SSE: The sum of the squared error between each sample and its treatment (group) mean. ANOVA (Computing F).

22.  SST: The sum of the square error of each treatment mean and the overall mean weighted by the number of observations in each treatment.  SS total = SST + SSE ANOVA (Computing F).

23. SST WolfWhiteKoros a 556647 547651 596746 567148 567048

24. ANOVA (Computing F). SSE WolfWhiteKorosa (55-56) 2 (66-70) 2 (47-48) 2 (54-56) 2 (76-70) 2 (51-48) 2 (59-56) 2 (67-70) 2 (46-48) 2 (56-56) 2 (71-70) 2 (48-48) 2 Sum = 14Sum=62Sum=14

25. ANOVA (Computing F). SS total WolfWhiteKorosa (55-58) 2 (66-58) 2 (47-58) 2 (54-58) 2 (76-58) 2 (51-58) 2 (59-58) 2 (67-58) 2 (46-58) 2 (56-58) 2 (71-58) 2 (48-58) 2

26. ANOVA (Computing F). ANOVA Table Sum of Squares Degrees of freedom Mean SquareF SSTk-1SST/(k-1)=MSTMST/MSE SSEn-kSSE/(n-k)=MSE SS totaln-1 k = the number of treatments n = the overall sample size k-1 = degrees of freedom in the numerator n-k = degrees of freedom in the denominator

27. ANOVA (Computing F). ANOVA Table Sum of Squares Degrees of freedom Mean SquareF 992k-1=2992/2=496496/10=49.6 90n-k=990/9=10 108211 k =3 n = 12

28. Example 8, page 421. T1: 9 7 11 9 12 10 T2: 13 20 14 13 T3: 10 9 15 14 15 Use a 0.05 significance level. H 0 : μ 1 = μ 2 = μ 3 H 1 : At least two of the means are not equal

29. Example 8, page 421. Using a 0.05 significance level. => Critical Value = 3.89 H 0 : μ 1 = μ 2 = μ 3 H 1 : At least two of the means are not equal k = the number of treatments = 3 n = the overall sample size = 15 k-1 = degrees of freedom in the numerator = 2 n-k = degrees of freedom in the denominator = 12 0.05 Critical Value = 3.89 Accept H 0 F(2,12)

30. Example 8, page 421. SSE = (9-9.67) 2 + (7-9.67) 2 + (11-9.67) 2 + (9-9.67) 2 + (12-9.67) 2 + (10-9.67) 2 + (13-15) 2 + (20-15) 2 + (14-15) 2 + (13-15) 2 + (13-12.6) 2 + (13-12.6) 2 + (13-12.6) 2 + (13-12.6) 2 + (13-12.6) 2 => SSE = 82.53 SST = 6(12.07-9.67) 2 + 4(12.07-15) 2 + 5(12.07-12.6) 2 = 70.40 SS total = 82.53+70.40 = 152.93

31. ANOVA Table Sum of Squares Degrees of freedom Mean SquareF SST=70.4k-1 =2SST/(k-1)=MST = 35.2 MST/MSE = 5.12 SSE=82.53n-k = 15-3 =12SSE/(n-k)=MSE = 6.88 SS total=152.93 n-1=14 Example 8, page 421. 5.12 > CV= 3.89, so we reject the null hypothesis. At least two of the treatments population means are not the same.

32. Inferences about Pairs of Treatment Means  When we reject the null hypothesis and conclude that all the treatment (or population) means are not equal, we may want to know which treatment means differ.  Students opinions ex.: if the students opinion do differ, the question is: Between which groups do the treatment means differ?  The simplest way to answer this question is through confidence intervals.

33.  To check if there is a difference between μ 1 and μ 2, we construct a Confidence interval using, is the mean of 1 st sample; is the mean of 2 nd sample; t is obtained from Appendix B.2 with n-k degrees of freedom; MSE is obtained from ANOVA table [SSE/(n-k)]; n 1 and n 2 are number of observations in 1 st and 2 nd sample. Inferences about Pairs of Treatment Means (Confidence interval for the difference in treatment means)

34.  We compute confidence interval limits.  If confidence interval includes zero, we conclude that there is no difference between the treatment means.  If CI does not include zero, we conclude that there is a difference between the treatment means. Inferences about Pairs of Treatment Means

35. Example 11, page 425. T1: 8 11 10 T2: 3 2 1 3 2 T3: 3 4 5 4 Use a 0.05 significance level. H 0 : μ 1 = μ 2 = μ 3 H 1 : At least two of the means are not equal

36. Example 11, page 425. Using a 0.05 significance level. => Critical Value = 4.26 H 0 : μ 1 = μ 2 = μ 3 H 1 : At least two of the means are not equal k = the number of treatments = 3 n = the overall sample size = 12 k-1 = degrees of freedom in the numerator = 2 n-k = degrees of freedom in the denominator = 9 0.05 Critical Value = 4.26 Accept H 0 F(2,9)

37. Example 8, page 421. SS total = 9.47 + 107.2 = 116.67

38. ANOVA Table Sum of Squares Degrees of freedom Mean SquareF SST=107.2k-1 =2SST/(k-1)=MST = 53.6 MST/MSE = 51.05 SSE=9.47n-k = 15-3 =9SSE/(n-k)=MSE = 1.05 SS total=116.67 n-1=14 Example 8, page 421. 51.05 > CV =4.26, so we reject the null hypothesis. At least two of the treatments population means are not the same.

39. Example 8, page 421. Can we conclude that treatment 1 and treatment 2 differ using a 95% level of confidence? Confidence Interval: (t has n-k = 9 df => t=2.262) CI does not include 0, so we can conclude that T1 and T2 have different means.