1/54 Statistics Analysis of Variance

2/54 Statistics in practice Introduction to Analysis of Variance Analysis of Variance: Testing for the Equality of k Population Means Multiple Comparison Procedures Contents

3/54 Burke Marketing Services, Inc., is one of the most experienced market research firms in the industry. In one study, a firm retained Burke to evaluate potential new versions of a children’s dry cereal. Analysis of variance was the statistical method used to study the data obtained from the taste tests. STATISTICS in PRACTICE

4/54 The experimental design employed by Burke and the subsequent analysis of variance were helpful in making a product design recommendation. STATISTICS in PRACTICE (Continue)

5/54 Introduction to Analysis of Variance (ANOVA) Analysis of Variance (ANOVA) can be used to test for the equality of three or more populations.. Data obtained from observational or experimental studies can be used for the analysis.

6/54 Introduction to Analysis of Variance (ANOVA) H 0 :  1  =  2  =  3  = ... =  k H a : Not all population means are equal We want to use the sample results to test the following hypotheses:

7/54 Introduction to Analysis of Variance (ANOVA) 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.

8/54 Sampling 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. Introduction to Analysis of Variance (ANOVA)

9/54 Sampling 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. Introduction to Analysis of Variance (ANOVA)

10/54 Introduction to Analysis of Variance (ANOVA) For each population, the response variable is normally distributed. For each population, the response variable is normally distributed. The variance of the response variable, denoted  2, is the same for all of the populations. The variance of the response variable, denoted  2, is the same for all of the populations. The observations must be independent. The observations must be independent.

11/54 Introduction to Analysis of Variance (ANOVA) Terminology: Treatment Treatment Between Treatment Between Treatment Within Treatment Within Treatment

12/54 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

13/54 Analysis of variance can be used to test for the equality of k population means. The hypotheses tested is H 0 : H a : Not all population means are equal where mean of the jth population. Analysis of Variance: Testing for the Equality of k Population Means

14/54 Analysis of Variance: Testing for the Equality of k Population Means Sample data = 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

15/54 statisitcs The sample mean for treatment j The sample variance for treatment j Analysis of Variance: Testing for the Equality of k Population Means

16/54 The overall sample mean where n T = n 1 + n n k If the size of each sample is n, n T = kn then Analysis of Variance: Testing for the Equality of k Population Means

17/54 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)

18/54 Analysis of Variance: Testing for the Equality of k Population Means Between-Treatments Estimate of Population Variance The mean square due to treatments (MSTR)

19/54 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 The mean square due to error (MSE)

20/54 Between-Treatments Estimate of Population Variance A between-treatment estimate of  2 is called the mean square treatment and is denoted MSTR. Denominator represents the degrees of freedom the degrees of freedom associated with SSTR associated with SSTR Numerator is the sum of squares sum of squares due to treatments due to treatments and is denoted SSTR

21/54 The estimate of  2 based on the variation of the sample observations within each sample is called the mean square error and is denoted by MSE. Within-Samples Estimate of Population Variance Denominator represents the degrees of freedom the degrees of freedom associated with SSE associated with SSE Numerator is the sum of squares sum of squares due to error and is denoted SSE

22/54 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.

23/54 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 Hypotheses Test Statistic

24/54 Rejection Rule: p-value Approach: Critical Value Approach: where the value of F α  is based on an F distribution with k - 1 numerator d.f. and nT - k denominator d.f. Reject H 0 if p-value <  Reject H 0 if F > F  Test for the Equality of k Population Means

25/54 Sampling Distribution of MSTR/MSE Rejection Region Do Not Reject H 0 Reject H 0 MSTR/MSE Critical Value FF FF Sampling Distribution of MSTR/MSE 

26/54 ANOVA Table Treatment Error Total SSTR SSE SST k– 1 n T n T – k nT nT nT nT - 1 MSTR MSE Source of Variation Sum of Squares Degrees of Freedom MeanSquares MSTR/MSE F 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.

27/54 SST divided by its degrees of freedom nT – 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: ANOVA Table

28/54 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. 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. ANOVA Table

29/54 Test for the Equality of k Population Means-- Example National Computer Products, Inc. (NCP), manufactures printers and fax machines at plants located in Atlanta, Dallas, and Seattle. Objective: To measure how much employees at these plants know about total quality management. A random sample of six employees was selected from each plant and given a quality awareness examination.

30/54 Test for the Equality of k Population Means--Example Data Let = mean examination score for population 1 = mean examination score for population 2 = mean examination score for population 3

31/54 Test for the Equality of k Population Means-- Example Hypotheses H 0 : = = H a : Not all population means are equal In this example 1. dependent or response variable : examination score 2. independent variable or factor : plant location 3. levels of the factor or treatments : the values of a factor selected for investigation, in the NCP example the three treatments or three population are Atlanta, Dallas, and Seattle.

32/54 Test for the Equality of k Population Means-- Example Three assumptions: 1. For each population, the response variable is normally distributed. The examination scores (response variable) must be normally distributed at each plant. 2. The variance of the response variable,, is the same for all of the populations. The variance of examination scores must be the same for all three plants. 3. The observations must be independent. The examination score for each employee must be independent of the examination score for any other employee.

33/54 Test for the Equality of k Population Means-- Example ANOVA Table p-value = < α =.05. We reject H 0.

34/54 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). Test for the Equality of k Population Means-- Example

35/54 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. Test for the Equality of k Population Means-- Example

36/ Plant 1 Buffalo Plant 2 Pittsburgh Plant 3 DetroitObservation Sample Mean Sample Variance Test for the Equality of k Population Means-- Example

37/54 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  2 = mean number of hours worked per week by the managers at Plant 2  3 = mean number of hours worked per week by the managers at Plant 3 1. Develop the hypotheses. p -Value and Critical Value Approaches Test for the Equality of k Population Means-- Example

38/54 2. Specify the level of significance.  = Compute the value of the test statistic. MSTR = 490/(3 - 1) = 245 SSTR = 5( )2 + 5( )2 + 5( )2 = 490 (Sample sizes are all equal.) Mean Square Due to Treatments = ( )/3 = 60 MSE = 308/(15 - 3) = SSE = 4(26.0) + 4(26.5) + 4(24.5) = 308 Mean Square Due to Error F = MSTR/MSE = 245/ = 9.55 Test for the Equality of k Population Means-- Example

39/54 Treatment Error Total Source of Variation Sum of Squares Degrees of Freedom MeanSquares 9.55 F ANOVA Table Test for the Equality of k Population Means-- Example

40/54 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 = Therefore, the p-value is less than.01 for F = Compute the p –value. p -Value Approaches Test for the Equality of k Population Means-- Example

41/54 5. Determine whether to reject H 0. Because F = 9.55 > 3.89, we reject H 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 = Critical Value Approaches Test for the Equality of k Population Means-- Example

42/54 Test for the Equality of k Population Means Summary

43/54 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.

44/54 Fisher’s LSD Procedure Test Statistic Hypotheses

45/54 Fisher’s LSD Procedure where the value of t a /2 is based on a t distribution with n T - k degrees of freedom. n Rejection Rule Reject H 0 if p-value < a p-value Approach: Critical Value Approach: Reject H 0 if t t a /2

46/54 Test Statistic Fisher’s LSD Procedure Based on the Test Statistic x i - x j Reject H 0 if > LSD Hypotheses Rejection Rule

47/54 Fisher’s LSD Procedure Based on the Test Statistic x i – x j -- Example 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.

48/54 Fisher’s LSD Procedure Based on the Test Statistic x i – x j -- Example 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.

49/54 For  =.05 and n T - k = 15 – 3 = 12 degrees of freedom, t.025 = MSE value was computed earlier Fisher’s LSD Procedure Based on the Test Statistic x i – x j -- Example

50/54 LSD for Plants 1 and 2 Conclusion: Test Statistic = | | = 13 Reject H 0 if > 6.98 Rejection Rule Hypotheses (A) The mean number of hours worked at Plant 1 is not equal to the mean number worked at Plant 2. Fisher’s LSD Procedure Based on the Test Statistic x i – x j -- Example

51/54 LSD for Plants 1 and 3 Conclusion: Test Statistic = |55  57| = 2 Reject H 0 if > 6.98 Rejection Rule Hypotheses (B) 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. Fisher’s LSD Procedure Based on the Test Statistic x i – x j -- Example

52/54 LSD for Plants 2 and 3 Conclusion: Test Statistic = | | = 11 Reject H 0 if > 6.98 Rejection Rule Hypotheses (C) The mean number of hours worked at Plant 2 is not equal to the mean number worked at Plant 3. Fisher’s LSD Procedure Based on the Test Statistic x i – x j -- Example

53/54 Type I Error Rates  EW = 1 – (1 –  ) ( k – 1)! The comparisonwise Type I error rate  indicates the level of significance associated with a single pairwise comparison. The experimentwise Type I error rate  EW is the probability of making a Type I error on at least one of the (k – 1)! pairwise comparisons.

54/54 The experimentwise Type I error rate gets larger for problems with more populations (larger k). Type I Error Rates