統計學 Spring 2011 授課教師 : 統計系余清祥日期 :2011 年 3 月 22 日第十三章 : 變異數分析與實驗設計

Transcription

1 統計學 Spring 2011 授課教師 : 統計系余清祥日期 :2011 年 3 月 22 日第十三章 : 變異數分析與實驗設計

2 Chapter 13, Part A Analysis of Variance and Experimental Design Introduction to Analysis of Variance Analysis of Variance and the Completely Randomized Design Multiple Comparison Procedures

3 Father of Statistics arrange the plots deliberately at random Deliberately used chance in producing data Ronald A. Fisher

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

5 Observational vs. Experimental Observational Study( 觀察研究 ): Observes individuals and measures variables of interest but does not attempt to influence the responses. Experiment( 實驗 ): Deliberately imposes some treatment on individuals in order to observe their responses

6 An Introduction to Experimental Design and Analysis of Variance In this chapter three types of experimental designs are introduced. a completely randomized design ( 完全隨機設計 ) a randomized block design ( 隨機集區設計 ) a factorial experiment ( 多因子設計 )

7 An Introduction to Experimental Design A factor( 因子 ) is a variable that the experimenter has selected for investigation. A treatment( 處理 ) is a level of a factor. Experimental units( 實驗單位 ) are the objects of interest in the experiment. A completely randomized design is an experimental design in which the treatments are randomly assigned to the experimental units. If the experimental units are heterogeneous, blocking can be used to form homogeneous groups, resulting in a randomized block design.

8 Steps in an Experiment (Effects of brand and shelf location on coffee sales) Select factors to be included brand and shelf location Choose the treatments brand (2) and shelf location (3) combinations Determine the number of observations to be made for each treatment sales are recorded once a week for 18 weeks Plan how the treatments will be assigned to the experimental units

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

10 Introduction to Analysis of Variance H 0 : μ 1 = μ 2 = μ 3 =... = μ k H a : Not all population means are equal If H 0 is rejected, we cannot conclude that all population means are different. Rejecting H 0 means that at least two population means have different values.

11 Analysis of Variance: A Conceptual Overview Assumptions for Analysis of Variance For each population, the response variable is normally distributed. The variance of the response variable, denoted σ 2, is the same for all of the populations. The observations must be independent.

12 Introduction to Analysis of Variance Sampling Distribution of x Given H 0 is True Sample means are close together because there is only one sampling distribution when H 0 is true. 2 σ σ x = n 2 x 2 μ x 1 x x x 3

13 Introduction to Analysis of Variance Sampling Distribution of x Given H 0 is False Sample means come from different sampling distributions and are not as close together when H 0 is false. x 3 μ 3 x 1 x μ 1 μ 1 μ 2 x x 2 x

14 Analysis of Variance: Testing for the Equality of k Population Means Between-Treatments Estimate of Population Variance Within-Treatments Estimate of Population Variance Comparing the Variance Estimates: The F Test ANOVA Table

15 Analysis of Variance: Testing for the Equality of k Population Means Analysis of variance can be used to test for the equality of k population means. The hypotheses tested is H 0 : μ 1 = μ 2 = L = μ k H a : Not all population means are equal where mean of the jth population. μ j

16 Analysis of Variance: Testing for the Equality of k Population Means Sample data x ij n j x j 2 s j s j = value of observation i for treatment j = number of observations for treatment j = sample mean for treatment j = sample variance for treatment j = sample standard deviation for treatment j

17 Analysis of Variance: Testing for the Equality of k Population Means Statisitcs The sample mean for treatment j x j n j i = 1 = n x j ij The sample variance for treatment j s 2 j = n j i= 1 ( x n ij j x 1 j ) 2

18 Analysis of Variance: Testing for the Equality of k Population Means The overall sample mean x k n j= i= = 1 1 n j T x ij where n T = n 1 + n n k If the size of each sample is n, n T = kn then x = k n j j= 1 i= 1 kn x ij = k n j j= 1 i= 1 k x ij / n = k j= 1 x kn ij

19 Analysis of Variance: Testing for the Equality of k Population Means Between-Treatments Estimate of Population Variance The sum of squares due to treatments (SSTR) k j= 1 n j ( x j x 2 ) The mean square due to treatments (MSTR) MSTR = k j= 1 n j ( x j k 1 x) 2

20 Analysis of Variance: Testing for the Equality of k Population Means Within-Treatments Estimate of Population Variance The sum of squares due to error (SSE) j= 1 ( n j 1) s j The mean square due to error (MSE) k 2 MSE = k j=1 ( n n T j 1) s k 2 j

21 Between-Treatments Estimate of Population Variance A between-treatment estimate of σ 2 is called the mean square treatment and is denoted MSTR. MSTR = k j = 1 n ( x x ) j j k 1 2 Denominator represents the degrees of freedom associated with SSTR Numerator is the sum of squares due to treatments and is denoted SSTR

22 Within-Samples Estimate of Population Variance The estimate of σ 2 based on the variation of the sample observations within each sample is called the mean square error and is denoted by MSE. MSE = k j =1 ( n n T j 1) s k 2 j Denominator represents the degrees of freedom associated with SSE Numerator is the sum of squares due to error and is denoted SSE

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

24 Comparing the Variance Estimates: The F Test Sampling Distribution of MSTR/MSE Sampling Distribution of MSTR/MSE Reject H 0 Do Not Reject H 0 α F α Critical Value MSTR/MSE

25 ANOVA Table Source of Variation Sum of Squares Degrees of Freedom Mean Squares F Treatment Error Total SSTR SSE SST k 1 n T k n T - 1 MSTR MSE MSTR/MSE SST is partitioned into SSTR and SSE. SST s s degrees of freedom (d.f.).) are partitioned into SSTR s d.f.. and SSE s d.f.

26 ANOVA Table for a Completely Randomized Design SST divided by its degrees of freedom n T 1 1 is the overall sample variance that would be obtained if we treated the entire set of observations as one data set. With the entire data set as one sample, the formula for computing the total sum of squares, SST, is: n j k 2 SST = ( x ij x ) = SSTR + SSE j = 1 i = 1

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

28 Test for the Equality of k Population Means Hypotheses Test Statistic H 0 : μ 1 = μ 2 = μ 3 =... = μ k H a : Not all population means are equal F = MSTR/MSE

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

30 Completely Randomized Design Example: AutoShine, Inc. AutoShine, Inc. is considering marketing a longlasting car wax. Three different waxes (Type 1, Type 2, and Type 3) have been developed. 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.

31 Completely Randomized Design Example: AutoShine, Inc. The number of times each car went through the carwash is shown on the next slide. AutoShine, Inc. must decide which wax to market. Are the three waxes equally effective?

32 Completely Randomized Design Observation Wax Type 1 Wax Type 2 Wax Type Sample Mean Sample Variance

33 Completely Randomized Design Hypotheses where: H 0 : μ 1 = μ 2 = μ 3 H a : Not all the means are equal μ 1 = mean number of washes for Type 1 wax μ 2 = mean number of washes for Type 2 wax μ 3 = mean number of washes for Type 3 wax

34 Completely Randomized Design Mean Square Between Treatments Because the sample sizes are all equal: x = ( x + x + x )/3 = ( )/3 = SSTR = 5( )2 + 5( )2 + 5( )2 = 5.2 MSTR = 5.2/(3 1) = 2.6 Mean Square Error SSE = 4(2.5) + 4(3.3) + 4(2.5) = 33.2 MSE = 33.2/(15 3) = 2.77

35 Completely Randomized Design Rejection Rule p-value Approach: Reject H 0 if p-value <.05 Critical Value Approach: Reject H 0 if F > 3.89 where F.05 = 3.89 is based on an F distribution with 2 numerator degrees of freedom and 12 denominator degrees of freedom

36 Completely Randomized Design Test Statistic Conclusion F = MSTR/MSE = 2.6/2.77 =.939 The p-value is greater than.10, where F = (Excel provides a p-value of.42.) Therefore, we cannot reject H 0. There is insufficient evidence to conclude that the mean number of washes for the three wax types are not all the same.

37 Testing for the Equality of k Population Means: A Completely Randomized Design ANOVA Table Source of Variation Sum of Squares Degrees of Freedom Mean Squares F p-value Treatments Error Total

38 Test for the Equality of k Population Means Example: Reed Manufacturing 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).

39 Test for the Equality of k Population Means Example: Reed Manufacturing A simple random sample of five managers from each of the three plants was taken and the number of hours worked by each manager for the previous week is shown on the next slide. Conduct an F test using α =.05. Factor... Manufacturing plant Treatments... Buffalo, Pittsburgh, Detroit Experimental units... Managers Response variable... Number of hours worked

40 Test for the Equality of k Population Means Observation Sample Mean Sample Variance Plant 1 Buffalo Plant 2 Pittsburgh Plant 3 Detroit

41 Test for the Equality of k Population Means p -Value and Critical Value Approaches 1. Develop the hypotheses. H 0 : μ 1 = μ 2 = μ 3 H a : Not all the means are equal where: μ 1 = mean number of hours worked per week by the managers at Plant 1 μ 2 = mean number of hours worked per week by the managers at Plant 2 μ 3 = mean number of hours worked per week by the managers at Plant 3

42 Test for the Equality of k Population Means p -Value and Critical Value Approaches 2. Specify the level of significance. α = Compute the value of the test statistic. Mean Square Due to Treatments (Sample sizes are all equal.) x = ( )/3 = 60 SSTR = 5(55-60)2 + 5(68-60)2 + 5(57-60)2 = 490 MSTR = 490/(3-1) = 245

43 Test for the Equality of k Population Means p -Value and Critical Value Approaches 3. Compute the value of the test statistic. (continued) Mean Square Due to Error SSE = 4(26.0) + 4(26.5) + 4(24.5) = 308 MSE = 308/(15-3) = F = MSTR/MSE = 245/ = 9.55

44 Testing for the Equality of k Population Means: An Observational Study ANOVA Table Source of Variation Sum of Squares Degrees of Freedom Mean Square F p-value Treatment Error Total

45 Testing for the Equality of k Population Means: An Observational Study p Value Approach 4. Compute the p value. With 2 numerator d.f. and 12 denominator d.f., the p-value is.01 for F = Therefore, the p-value is less than.01 for F = Determine whether to reject H 0. The p-value <.05, so we 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.

46 Testing for the Equality of k Population Means: An Observational Study Critical Value Approach 4. Determine the critical value and rejection rule. Based on an F distribution with 2 numerator d.f. and 12 denominator d.f., F.05 = Reject H 0 if F > Determine whether to reject H 0. Because F = 9.55 > 3.89, we 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.

47 Multiple Comparison Procedures Suppose that 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.

48 Fisher s s LSD Procedure Hypotheses Test Statistic t = H H 0 a :μ i = μj :μ μ x i i x j MSE( 1 1 n + n ) i j j

49 Rejection Rule Fisher s s LSD Procedure p-value Approach: Reject H 0 if p-value < α Critical Value Approach: Reject H 0 if t < -t a/2 or t > t a/2 where the value of t a/2 is based on a t distribution with n T - k degrees of freedom.

50 Fisher s s LSD Procedure Based on the Test Statistic x i - x j Hypotheses H0 :μ i = μj H :μ μ a i j Test Statistic x i x j Rejection Rule where Reject H 0 if x x > LSD i LSD = t /2 MSE( 1 1 α n + n ) j i j

51 Fisher s LSD Procedure Based on the Test Statistic x i - x j Example: Reed Manufacturing 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 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.

52 Fisher s LSD Procedure Based on the Test Statistic x i - x j For α =.05 and n T - k = 15 3 = 12 degrees of freedom, t = LSD = t 1 1 α /2 MSE( n + n ) LSD = ( ) = 698. i j MSE value was computed earlier

53 Fisher s LSD Procedure Based on the Test Statistic x i - x j LSD for Plants 1 and 2 Hypotheses (A) Rejection Rule H : μ = μ H a 2 : μ1 μ Reject H 0 if x x > Test Statistic x1 x 2 = = 13 Conclusion The mean number of hours worked at Plant 1 is not equal to the mean number worked at Plant 2.

54 Fisher s LSD Procedure Based on the Test Statistic x i - x j LSD for Plants 1 and 3 Hypotheses (B) Rejection Rule H : μ = μ H a 3 : μ1 μ Reject H 0 if x x > Test Statistic x1 x 3 = = 2 Conclusion There is no significant difference between the mean number of hours worked at Plant 1 and the mean number of hours worked at Plant 3.

55 Fisher s LSD Procedure Based on the Test Statistic x i - x j LSD for Plants 2 and 3 Hypotheses (C) H : μ = μ H a 3 : μ2 μ Rejection Rule Reject H 0 if x x > Test Statistic x2 x 3 = = 11 Conclusion The mean number of hours worked at Plant 2 is not equal to the mean number worked at Plant 3.

56 Type I Error Rates The comparison-wise Type I error rate a indicates the level of significance associated with a single pairwise comparison. The experiment-wise Type I error rate a EW is the probability of making a Type I error on at least one of the (k 1)! pairwise comparisons. a EW = 1 (1 a) (k 1)! The experiment-wise Type I error rate gets larger for problems with more populations (larger k).

57 End of Chapter 13, Part A

58 Chapter 13, Part B Experimental Design and Analysis of Variance Randomized Block Design Factorial Experiment

59 Randomized Block Design Experimental units are the objects of interest in the experiment. A completely randomized design is an experimental design in which the treatments are randomly assigned to the experimental units. If the experimental units are heterogeneous, blocking can be used to form homogeneous groups, resulting in a randomized block design.

60 Randomized Block Design 應用範例 : 小明想要栽種產量較高的蕃茄, 他的朋友提供了幾個不同品種, 如何設計實驗以找出產量最高的品種? 註 : 植物的成長與陽光 水分 土壤肥沃等因素有關, 因此通常南向及最外圍的植物生長較快, 但這些因素應與品種是否為高產量區隔 應用範例 : 小華想要比較 A B 兩品牌球鞋的耐穿程度 如果讓每位球員只穿一種品牌, 個體差異可能會影響實驗結果

61 Randomized Block Design ANOVA Procedure For a randomized block design the sum of squares total (SST) is partitioned into three groups: sum of squares due to treatments, sum of squares due to blocks, and sum of squares due to error. SST = SSTR + SSBL + SSE The total degrees of freedom, n T - 1, are partitioned such that k - 1 degrees of freedom go to treatments, b - 1 go to blocks, and (k -1)(b - 1) go to the error term.

62 ANOVA Table Randomized Block Design Source of Variation Sum of Squares Degrees of Freedom Mean Square F p- Value Treatments Blocks Error SSTR SSE k - 1 SSBL b - 1 (k 1)(b 1) MSTR SSTR MSTR = k - 1 MSE MSBL = SSBL b-1 SSE MSE = ( k 1 )( b 1 ) Total SST n T - 1

63 RBD - ANOVA F-Test (Normal Data) Data Structure: (t Treatments, b Subjects/Blocks) Mean for Treatment i: Mean for Subject (Block) j: Overall Mean: y.. y i. Overall sample size: N = bt y. j ANOVA:Treatment, Block, and Error Sums of Squares TSS SST SSBL SSE = = = = b t i = 1 j = 1 t t i = 1 b ( y y ) ( y y ) j = 1 b i. ( y y ). j.. ( y y y + y ) ij ij i j df.. Total 2 df = T df = = B TSS bt t = b SST SSBL df E = ( b 1)( t 1)

64 Randomized Block Design Example: Crescent Oil Co. Crescent Oil has developed three new blends of gasoline and must decide which blend or blends to produce and distribute. A study of the miles per gallon ratings of the three blends is being conducted to determine if the mean ratings are the same for the three blends.

65 Randomized Block Design Example: Crescent Oil Co. Five automobiles have been tested using each of the three gasoline blends and the miles per gallon ratings are shown on the next slide. Factor... Gasoline blend Treatments... Blend X, Blend Y, Blend Z Blocks... Automobiles Response variable... Miles per gallon

66 Randomized Block Design Automobile (Block) Type of Gasoline (Treatment) Blend X Blend Y Blend Z Block Means Treatment Means

67 Randomized Block Design Mean Square Due to Treatments The overall sample mean is 29. Thus, SSTR = 5[( )2 + ( )2 + ( )2] = 5.2 MSTR = 5.2/(3 1) = 2.6 Mean Square Due to Blocks SSBL = 3[( ) ( )2] = MSBL = 51.33/(5 1) = 12.8 Mean Square Due to Error SSE = = 5.47 MSE = 5.47/[(3 1)(5 1)] =.68

68 Randomized Block Design ANOVA Table Source of Variation Sum of Squares Degrees of Freedom Mean Square F p-value Treatments Blocks Error Total

69 Randomized Block Design Rejection Rule p-value Approach: Reject H 0 if p-value <.05 Critical Value Approach: Reject H 0 if F > 4.46 For α =.05, F.05 = 4.46 (2 d.f. numerator and 8 d.f. denominator)

70 Randomized Block Design Test Statistic F = MSTR/MSE = 2.6/.68 = 3.82 Conclusion The p-value is greater than.05 (where F = 4.46) and less than.10 (where F = 3.11). (Excel provides a p-value of.07). Therefore, we cannot reject H 0. There is insufficient evidence to conclude that the miles per gallon ratings differ for the three gasoline blends.

71 Factorial Experiments In some experiments we want to draw conclusions about more than one variable or factor. Factorial experiments and their corresponding ANOVA computations are valuable designs when simultaneous conclusions about two or more factors are required. The term factorial is used because the experimental conditions include all possible combinations of the factors. For example, for a levels of factor A and b levels of factor B, the experiment will involve collecting data on ab treatment combinations.

72 Two-Factor Factorial Experiment ANOVA Procedure The ANOVA procedure for the two-factor factorial experiment is similar to the completely randomized experiment and the randomized block experiment. We again partition the sum of squares total (SST) into its sources. SST = SSA + SSB + SSAB + SSE The total degrees of freedom, n T 1, are partitioned such that (a 1) d.f go to Factor A, (b 1) d.f go to Factor B, (a 1)(b 1) d.f. go to Interaction, and ab(r 1) go to Error.

73 Two-Factor Factorial Experiment Source of Variation Sum of Squares Degrees of Freedom Mean Square F p- Value Factor A Factor B Interaction Error SSA SSE a - 1 SSB b - 1 MSA = SSA a-1 MSB = SSB b-1 SSAB (a 1)(b 1) MSAB SSAB = ( a 1)( b 1) ab (r 1) MSE = SSE ab ( r 1) MSA MSE MSB MSE MSAB MSE Total SST n T - 1

74 Two-Factor Factorial Experiment Step 1 Compute the total sum of squares a b r SST = ( x x ) i = 1 j = 1 k = 1 Step 2 Compute the sum of squares for factor A Step 3 Compute the sum of squares for factor B a i = 1 ijk SSA = br ( x. x ) b SSB = ar ( x. x ) j = 1 i j 2 2 2

75 Two-Factor Factorial Experiment Step 4 Compute the sum of squares for interaction a b SSAB = r ( x x. x. + x ) i = 1 j = 1 ij i j 2 Step 5 Compute the sum of squares due to error SSE = SST SSA SSB - SSAB

76 Two-Factor Factorial Experiment Example: State of Ohio Wage Survey A survey was conducted of hourly wages for a sample of workers in two industries at three locations in Ohio. Part of the purpose of the survey was to determine if differences exist in both industry type and location. The sample data are shown on the next slide.

77 Two-Factor Factorial Experiment Example: State of Ohio Wage Survey Industry Cincinnati Cleveland Columbus I I I II II II

78 Factors Two-Factor Factorial Experiment Factor A: Industry Type (2 levels) Factor B: Location (3 levels) Replications Each experimental condition is repeated 3 times

79 Two-Factor Factorial Experiment ANOVA Table Source of Variation Sum of Squares Degrees of Freedom Mean Square F p-value Factor A Error Factor B Interaction Total

80 Two-Factor Factorial Experiment Conclusions Using the p-value Approach Industries: p-value =.06 > α =.05 Mean wages do not differ by industry type. Locations: p-value =.03 < α =.05 Mean wages differ by location. Interaction: p-value =.25 > α =.05 Interaction is not significant. (p-values were found using Excel)

81 Two-Factor Factorial Experiment Conclusions Using the Critical Value Approach Industries: F = 4.19 < F α = 4.75 Mean wages do not differ by industry type. Locations: F = 4.69 > F α = 3.89 Mean wages differ by location. Interaction: F = 1.55 < F α = 3.89 Interaction is not significant.

82 End of Chapter 13