1. 1 1 Slide Slides by JOHN LOUCKS St. Edward’s University

2. 2 2 Slide Chapter 13, Part A Experimental Design and Analysis of Variance nIntroduction to Experimental Design and Analysis of Variance and Analysis of Variance nAnalysis of Variance and the Completely Randomized Design and the Completely Randomized Design nMultiple Comparison Procedures

3. 3 3 Slide nStatistical studies can be classified as being either experimental or observational. nIn an experimental study, one or more factors are controlled so that data can be obtained about how the factors influence the variables of interest. nIn an observational study, no attempt is made to control the factors. nCause-and-effect relationships are easier to establish in experimental studies than in observational studies. An Introduction to Experimental Design and Analysis of Variance nAnalysis of variance (ANOVA) can be used to analyze the data obtained from experimental or observational studies.

4. 4 4 Slide An Introduction to Experimental Design and Analysis of Variance nIn this chapter three types of experimental designs are introduced. a completely randomized design a completely randomized design a randomized block design a randomized block design a factorial experiment a factorial experiment

5. 5 5 Slide An Introduction to Experimental Design and Analysis of Variance nA factor （因子） is a variable that the experimenter has selected for investigation. nA treatment （处理或因子水平 ） is a level of a factor. nExperimental units are the objects of interest in the experiment. nA completely randomized design is an experimental design in which the treatments are randomly assigned to the experimental units.

6. 6 6 Slide Analysis of Variance: A Conceptual Overview Analysis of Variance (ANOVA) can be used to test Analysis of Variance (ANOVA) can be used to test for the equality of three or more population means. for the equality of three or more population means. Analysis of Variance (ANOVA) can be used to test Analysis of Variance (ANOVA) can be used to test for the equality of three or more population means. for the equality of three or more population means. Data obtained from observational or experimental Data obtained from observational or experimental studies can be used for the analysis. studies can be used for the analysis. Data obtained from observational or experimental Data obtained from observational or experimental studies can be used for the analysis. studies can be used for the analysis. We want to use the sample results to test the We want to use the sample results to test the following hypotheses: following hypotheses: We want to use the sample results to test the We want to use the sample results to test the following hypotheses: following hypotheses: H 0 :  1  =  2  =  3  = ... =  k H a : Not all population means are equal

7. 7 7 Slide H 0 :  1  =  2  =  3  = ... =  k H a : Not all population means are equal If H 0 is rejected, we cannot conclude that all If H 0 is rejected, we cannot conclude that all population means are different. population means are different. If H 0 is rejected, we cannot conclude that all If H 0 is rejected, we cannot conclude that all population means are different. population means are different. Rejecting H 0 means that at least two population Rejecting H 0 means that at least two population means have different values. means have different values. Rejecting H 0 means that at least two population Rejecting H 0 means that at least two population means have different values. means have different values. Analysis of Variance: A Conceptual Overview

8. 8 8 Slide For each population, the response (dependent) For each population, the response (dependent) （反应变量） variable is normally distributed. For each population, the response (dependent) For each population, the response (dependent) （反应变量） variable is normally distributed. The variance of the response variable, denoted  2, The variance of the response variable, denoted  2, is the same for all of the populations. is the same for all of the populations. The variance of the response variable, denoted  2, The variance of the response variable, denoted  2, is the same for all of the populations. is the same for all of the populations. The observations must be independent. The observations must be independent. nAssumptions for Analysis of Variance Analysis of Variance: A Conceptual Overview

9. 9 9 Slide nSampling Distribution of Given H 0 is True   Sample means are close together because there is only because there is only one sampling distribution one sampling distribution when H 0 is true. when H 0 is true. Analysis of Variance: A Conceptual Overview

10. 10 Slide nSampling Distribution of Given H 0 is False 33 33 11 11 22 22 Sample means come from different sampling distributions and are not as close together when H 0 is false. when H 0 is false. Analysis of Variance: A Conceptual Overview

11. 11 Slide Analysis of Variance nBetween-Treatments Estimate of Population Variance nWithin-Treatments Estimate of Population Variance nComparing the Variance Estimates: The F Test nANOVA Table

12. 12 Slide Between-Treatments Estimate of Population Variance  2 Denominator is the degrees of freedom associated with SSTR Numerator is called the sum of squares due to treatments (SSTR) （处理平方和） The estimate of  2 based on the variation of the The estimate of  2 based on the variation of the sample means is called the mean square due to sample means is called the mean square due to treatments （处理均方） and is denoted by MSTR. treatments （处理均方） and is denoted by MSTR.

13. 13 Slide The estimate of  2 based on the variation of the sample observations within each sample is called the mean square error （误差均值） and is denoted by MSE. The estimate of  2 based on the variation of the sample observations within each sample is called the mean square error （误差均值） and is denoted by MSE. Within-Treatments Estimate of Population Variance  2 Denominator is the degrees of freedom associated with SSE Numerator is called the sum of squares due to error (SSE) （误差平方和）

14. 14 Slide Comparing the Variance Estimates: The F Test If the null hypothesis is true and the ANOVA If the null hypothesis is true and the ANOVA assumptions are valid, the sampling distribution of assumptions are valid, the sampling distribution of MSTR/MSE is an F distribution with MSTR d.f. MSTR/MSE is an F distribution with MSTR d.f. equal to k - 1 and MSE d.f. equal to n T - k. equal to k - 1 and MSE d.f. equal to n T - k. If the means of the k populations are not equal, the If the means of the k populations are not equal, the value of MSTR/MSE will be inflated because MSTR value of MSTR/MSE will be inflated because MSTR overestimates  2. overestimates  2. Hence, we will reject H 0 if the resulting value of Hence, we will reject H 0 if the resulting value of MSTR/MSE appears to be too large to have been MSTR/MSE appears to be too large to have been selected at random from the appropriate F selected at random from the appropriate F distribution. distribution.

15. 15 Slide nSampling Distribution of MSTR/MSE Do Not Reject H 0 Reject H 0 MSTR/MSE Critical Value FF FF Sampling Distribution of MSTR/MSE  Comparing the Variance Estimates: The F Test

16. 16 Slide Source of Variation Sum of Squares Degrees of Freedom MeanSquare F Treatments Error Total k - 1 n T - 1 SSTR SSE SST n T - k SST is partitioned into SSTR and SSE. SST’s degrees of freedom (d.f.) are partitioned into SSTR’s d.f. and SSE’s d.f. ANOVA Table （方差分析表） p - Value

17. 17 Slide ANOVA Table SST divided by its degrees of freedom n T – 1 is the SST divided by its degrees of freedom n T – 1 is the overall sample variance that would be obtained if we overall sample variance that would be obtained if we treated the entire set of observations as one data set. treated the entire set of observations as one data set. SST divided by its degrees of freedom n T – 1 is the SST divided by its degrees of freedom n T – 1 is the overall sample variance that would be obtained if we overall sample variance that would be obtained if we treated the entire set of observations as one data set. treated the entire set of observations as one data set. With the entire data set as one sample, the formula With the entire data set as one sample, the formula for computing the total sum of squares, SST, is: for computing the total sum of squares, SST, is: With the entire data set as one sample, the formula With the entire data set as one sample, the formula for computing the total sum of squares, SST, is: for computing the total sum of squares, SST, is:

18. 18 Slide ANOVA Table ANOVA can be viewed as the process of partitioning ANOVA can be viewed as the process of partitioning the total sum of squares and the degrees of freedom the total sum of squares and the degrees of freedom into their corresponding sources: treatments and error. into their corresponding sources: treatments and error. ANOVA can be viewed as the process of partitioning ANOVA can be viewed as the process of partitioning the total sum of squares and the degrees of freedom the total sum of squares and the degrees of freedom into their corresponding sources: treatments and error. into their corresponding sources: treatments and error. Dividing the sum of squares by the appropriate Dividing the sum of squares by the appropriate degrees of freedom provides the variance estimates degrees of freedom provides the variance estimates and the F value used to test the hypothesis of equal and the F value used to test the hypothesis of equal population means. population means. Dividing the sum of squares by the appropriate Dividing the sum of squares by the appropriate degrees of freedom provides the variance estimates degrees of freedom provides the variance estimates and the F value used to test the hypothesis of equal and the F value used to test the hypothesis of equal population means. population means.

19. 19 Slide Test for the Equality of k Population Means F = MSTR/MSE H 0 :  1  =  2  =  3  = ... =  k  H a : Not all population means are equal nHypotheses nTest Statistic

20. 20 Slide Test for the Equality of k Population Means nRejection Rule where the value of F  is based on an F distribution with k - 1 numerator d.f. and n T - k denominator d.f. Reject H 0 if p -value <  p -value Approach: Critical Value Approach: Reject H 0 if F > F 

21. 21 Slide AutoShine, Inc. is considering marketing a long- AutoShine, Inc. is considering marketing a long- lasting car wax. Three different waxes (Type 1, Type 2, and Type 3) have been developed. n Example: AutoShine, Inc. In order to test the durability In order to test the durability of these waxes, 5 new cars were waxed with Type 1, 5 with Type 2, and 5 with Type 3. Each car was then repeatedly run through an automatic carwash until the wax coating showed signs of deterioration. Testing for the Equality of k Population Means: A Completely Randomized Experimental Design

22. 22 Slide The number of times each car went through the The number of times each car went through the carwash before its wax deteriorated is shown on the next slide. AutoShine, Inc. must decide which wax to market. Are the three waxes equally effective? n Example: AutoShine, Inc. Testing for the Equality of k Population Means: A Completely Randomized Experimental Design Factor... Car wax Treatments... Type I, Type 2, Type 3 Experimental units... Cars Response variable... Number of washes

23. 23 Slide 12345273029283133283130302928303231 Sample Mean Sample Variance Observation Wax Type 1 Wax Type 2 Wax Type 3 2.5 3.3 2.5 2.5 3.3 2.5 29.0 30.4 30.0 Testing for the Equality of k Population Means: A Completely Randomized Experimental Design

24. 24 Slide nHypotheses where:  1 = mean number of washes using Type 1 wax  2 = mean number of washes using Type 2 wax  3 = mean number of washes using Type 3 wax H 0 :  1  =  2  =  3  H a : Not all the means are equal Testing for the Equality of k Population Means: A Completely Randomized Experimental Design

25. 25 Slide Because the sample sizes are all equal: Because the sample sizes are all equal: MSE = 33.2/(15 - 3) = 2.77 MSTR = 5.2/(3 - 1) = 2.6 SSE = 4(2.5) + 4(3.3) + 4(2.5) = 33.2 SSTR = 5(29–29.8) 2 + 5(30.4–29.8) 2 + 5(30–29.8) 2 = 5.2 nMean Square Error nMean Square Between Treatments = (29 + 30.4 + 30)/3 = 29.8 Testing for the Equality of k Population Means: A Completely Randomized Experimental Design

26. 26 Slide nRejection Rule where F.05 = 3.89 is based on an F distribution with 2 numerator degrees of freedom and 12 denominator degrees of freedom p -Value Approach: Reject H 0 if p -value <.05 Critical Value Approach: Reject H 0 if F > 3.89 Testing for the Equality of k Population Means: A Completely Randomized Experimental Design

27. 27 Slide nTest Statistic There is insufficient evidence to conclude that the mean number of washes for the three wax types are not all the same. nConclusion F = MSTR/MSE = 2.60/2.77 =.939 The p -value is greater than.10, where F = 2.81. (Excel provides a p -value of.42.) (Excel provides a p -value of.42.) Therefore, we cannot reject H 0. Therefore, we cannot reject H 0. Testing for the Equality of k Population Means: A Completely Randomized Experimental Design

28. 28 Slide Source of Variation Sum of Squares Degrees of Freedom MeanSquares F Treatments Error Total 2 14 5.2 33.2 38.4 12 2.60 2.77.939 nANOVA Table Testing for the Equality of k Population Means: A Completely Randomized Experimental Design p -Value.42

29. 29 Slide nExample: Reed Manufacturing Janet Reed would like to know if Janet Reed would like to know if there is any significant difference in the mean number of hours worked per week for the department managers at her three manufacturing plants (in Buffalo, Pittsburgh, and Detroit). An F test will be conducted using An F test will be conducted using  =.05. Testing for the Equality of k Population Means: An Observational Study

30. 30 Slide nExample: Reed Manufacturing A simple random sample of five A simple random sample of five managers from each of the three plants was taken and the number of hours worked by each manager in the previous week is shown on the next slide. Testing for the Equality of k Population Means: An Observational Study Factor... Manufacturing plant Treatments... Buffalo, Pittsburgh, Detroit Experimental units... Managers Response variable... Number of hours worked

31. 31 Slide 1 2 3 4 5 48 54 57 54 62 73 63 66 64 74 51 63 61 54 56 Plant 1 Buffalo Plant 2 Pittsburgh Plant 3 Detroit Observation Sample Mean Sample Variance 55 68 57 26.0 26.5 24.5 Testing for the Equality of k Population Means: An Observational Study

32. 32 Slide H 0 :  1  =  2  =  3  H a : Not all the means are equal where:  1 = mean number of hours worked per week by the managers at Plant 1 week by the managers at Plant 1  2 = mean number of hours worked per  2 = mean number of hours worked per week by the managers at Plant 2 week by the managers at Plant 2  3 = mean number of hours worked per week by the managers at Plant 3 week by the managers at Plant 3 1. Develop the hypotheses. p -Value and Critical Value Approaches p -Value and Critical Value Approaches Testing for the Equality of k Population Means: An Observational Study

33. 33 Slide 2. Specify the level of significance.  =.05 p -Value and Critical Value Approaches p -Value and Critical Value Approaches 3. Compute the value of the test statistic. MSTR = 490/(3 - 1) = 245 SSTR = 5(55 - 60) 2 + 5(68 - 60) 2 + 5(57 - 60) 2 = 490 = (55 + 68 + 57)/3 = 60 (Sample sizes are all equal.) Mean Square Due to Treatments Testing for the Equality of k Population Means: An Observational Study

34. 34 Slide 3. Compute the value of the test statistic. MSE = 308/(15 - 3) = 25.667 SSE = 4(26.0) + 4(26.5) + 4(24.5) = 308 Mean Square Due to Error (con’t.) F = MSTR/MSE = 245/25.667 = 9.55 p -Value and Critical Value Approaches p -Value and Critical Value Approaches Testing for the Equality of k Population Means: An Observational Study

35. 35 Slide Treatment Error Total 490 308 798 2 12 14 245 25.667 Source of Variation Sum of Squares Degrees of Freedom MeanSquare 9.55 F ANOVA Table ANOVA Table Testing for the Equality of k Population Means: An Observational Study p -Value.0033

36. 36 Slide 5. Determine whether to reject H 0. We have sufficient evidence to conclude that the mean number of hours worked per week by department managers is not the same at all 3 plant. The p -value <.05, so we reject H 0. With 2 numerator d.f. and 12 denominator d.f., the p -value is.01 for F = 6.93. Therefore, the p -value is less than.01 for F = 9.55. p –Value Approach p –Value Approach 4. Compute the p –value. Testing for the Equality of k Population Means: An Observational Study

37. 37 Slide 5. Determine whether to reject H 0. Because F = 9.55 > 3.89, we reject H 0. Critical Value Approach Critical Value Approach 4. Determine the critical value and rejection rule. Reject H 0 if F > 3.89 We have sufficient evidence to conclude that the mean number of hours worked per week by department managers is not the same at all 3 plant. Based on an F distribution with 2 numerator d.f. and 12 denominator d.f., F.05 = 3.89. Testing for the Equality of k Population Means: An Observational Study

38. 38 Slide Multiple Comparison Procedures （多重比较方法） nSuppose that analysis of variance has provided statistical evidence to reject the null hypothesis of equal population means. nFisher’s least significant difference (LSD) （最小显著 差异） procedure can be used to determine where the differences occur.

39. 39 Slide Fisher’s LSD Procedure nTest Statistic nHypotheses

40. 40 Slide Fisher’s LSD Procedure where the value of t a /2 is based on a where the value of t a /2 is based on a t distribution with n T - k degrees of freedom. nRejection Rule Reject H 0 if p -value <  p -value Approach: Critical Value Approach: Reject H 0 if t t a /2

41. 41 Slide nTest Statistic Fisher’s LSD Procedure Based on the Test Statistic x i - x j __ where Reject H 0 if > LSD nHypotheses nRejection Rule

42. 42 Slide Fisher’s LSD Procedure Based on the Test Statistic x i - x j nExample: Reed Manufacturing Recall that Janet Reed wants to know Recall that Janet Reed wants to know if there is any significant difference in the mean number of hours worked per week for the department managers at her three manufacturing plants. Analysis of variance has provided Analysis of variance has provided statistical evidence to reject the null hypothesis of equal population means. Fisher’s least significant difference (LSD) procedure can be used to determine where the differences occur.

43. 43 Slide For  =.05 and n T - k = 15 – 3 = 12 degrees of freedom, t. 025 = 2.179 degrees of freedom, t. 025 = 2.179 MSE value was computed earlier Fisher’s LSD Procedure Based on the Test Statistic x i - x j

44. 44 Slide nLSD for Plants 1 and 2 Fisher’s LSD Procedure Based on the Test Statistic x i - x j Conclusion Conclusion Test Statistic Test Statistic = |55  68| = 13 Reject H 0 if > 6.98 Rejection Rule Rejection Rule Hypotheses (A) Hypotheses (A) The mean number of hours worked at Plant 1 is The mean number of hours worked at Plant 1 is not equal to the mean number worked at Plant 2.

45. 45 Slide nLSD for Plants 1 and 3 Fisher’s LSD Procedure Based on the Test Statistic x i - x j Conclusion Conclusion Test Statistic Test Statistic = |55  57| = 2 Reject H 0 if > 6.98 Rejection Rule Rejection Rule Hypotheses (B) Hypotheses (B) There is no significant difference between the mean There is no significant difference between the mean number of hours worked at Plant 1 and the mean number of hours worked at Plant 1 and the mean number of hours worked at Plant 3. number of hours worked at Plant 3.

46. 46 Slide nLSD for Plants 2 and 3 Fisher’s LSD Procedure Based on the Test Statistic x i - x j Conclusion Conclusion Test Statistic Test Statistic = |68  57| = 11 Reject H 0 if > 6.98 Rejection Rule Rejection Rule Hypotheses (C) Hypotheses (C) The mean number of hours worked at Plant 2 is The mean number of hours worked at Plant 2 is not equal to the mean number worked at Plant 3. not equal to the mean number worked at Plant 3.

47. 47 Slide nThe experiment-wise Type I error rate gets larger for problems with more populations (larger k ). Type I Error Rates  EW = 1 – (1 –  ) ( k – 1)! The comparison-wise Type I error rate （比较性第一 类错误概率）  indicates the level of significance associated with a single pairwise comparison. The comparison-wise Type I error rate （比较性第一 类错误概率）  indicates the level of significance associated with a single pairwise comparison. The experiment-wise Type I error rate( 试验性第一类 错误）  EW is the probability of making a Type I error on at least one of the ( k – 1)! pairwise comparisons. The experiment-wise Type I error rate( 试验性第一类 错误）  EW is the probability of making a Type I error on at least one of the ( k – 1)! pairwise comparisons.

48. 48 Slide End of Chapter 13, Part A