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1 1 Slide © 2005 Thomson/South-Western Chapter 13, Part A Analysis of Variance and Experimental Design n Introduction to Analysis of Variance n Analysis.

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Presentation on theme: "1 1 Slide © 2005 Thomson/South-Western Chapter 13, Part A Analysis of Variance and Experimental Design n Introduction to Analysis of Variance n Analysis."— Presentation transcript:

1 1 1 Slide © 2005 Thomson/South-Western Chapter 13, Part A Analysis of Variance and Experimental Design n Introduction to Analysis of Variance n Analysis of Variance: Testing for the Equality of k Population Means k Population Means n Multiple Comparison Procedures

2 2 2 Slide © 2005 Thomson/South-Western Introduction to Analysis of Variance 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

3 3 3 Slide © 2005 Thomson/South-Western Introduction to Analysis of Variance 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.

4 4 4 Slide © 2005 Thomson/South-Western n Sampling Distribution of Given H 0 is True Introduction to Analysis of Variance   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.

5 5 5 Slide © 2005 Thomson/South-Western Introduction to Analysis of Variance n 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.

6 6 6 Slide © 2005 Thomson/South-Western For each population, the response variable is For each population, the response variable is normally distributed. normally distributed. For each population, the response variable is For each population, the response variable is normally distributed. normally distributed. Assumptions for Analysis of Variance 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.

7 7 7 Slide © 2005 Thomson/South-Western Analysis of Variance: Testing for the Equality of k Population Means n Between-Treatments Estimate of Population Variance n Within-Treatments Estimate of Population Variance n Comparing the Variance Estimates: The F Test n ANOVA Table

8 8 8 Slide © 2005 Thomson/South-Western Between-Treatments Estimate of Population Variance A between-treatment estimate of  2 is called the mean square treatment and is denoted MSTR. 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

9 9 9 Slide © 2005 Thomson/South-Western 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. 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

10 10 Slide © 2005 Thomson/South-Western 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.

11 11 Slide © 2005 Thomson/South-Western 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 n Hypotheses n Test Statistic

12 12 Slide © 2005 Thomson/South-Western Test for the Equality of k Population Means n Rejection 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 

13 13 Slide © 2005 Thomson/South-Western Sampling Distribution of MSTR/MSE n Rejection Region Do Not Reject H 0 Reject H 0 MSTR/MSE Critical Value FF FF Sampling Distribution of MSTR/MSE 

14 14 Slide © 2005 Thomson/South-Western ANOVA Table 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. 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

15 15 Slide © 2005 Thomson/South-Western 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:

16 16 Slide © 2005 Thomson/South-Western 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.

17 17 Slide © 2005 Thomson/South-Western n Example: Reed Manufacturing Test for the Equality of k Population Means 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).

18 18 Slide © 2005 Thomson/South-Western n Example: Reed Manufacturing Test for the Equality of k Population Means 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 for the previous week is shown on the next slide. Conduct an F test using  =.05. Conduct an F test using  =.05.

19 19 Slide © 2005 Thomson/South-Western 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 Test for the Equality of k Population Means

20 20 Slide © 2005 Thomson/South-Western Test for the Equality of k Population Means 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

21 21 Slide © 2005 Thomson/South-Western 2. Specify the level of significance.  =.05 Test for the Equality of k Population Means 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

22 22 Slide © 2005 Thomson/South-Western 3. Compute the value of the test statistic. Test for the Equality of k Population Means MSE = 308/(15 - 3) = 25.667 SSE = 4(26.0) + 4(26.5) + 4(24.5) = 308 Mean Square Due to Error (continued) F = MSTR/MSE = 245/25.667 = 9.55 p -Value and Critical Value Approaches p -Value and Critical Value Approaches

23 23 Slide © 2005 Thomson/South-Western Treatment Error Total 490 308 798 2 12 14 245 25.667 Source of Variation Sum of Squares Degrees of Freedom MeanSquares 9.55 F Test for the Equality of k Population Means ANOVA Table ANOVA Table

24 24 Slide © 2005 Thomson/South-Western Test for the Equality of k Population Means 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.

25 25 Slide © 2005 Thomson/South-Western 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 Test for the Equality of k Population Means 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.

26 26 Slide © 2005 Thomson/South-Western Multiple Comparison Procedures n Suppose that analysis of variance has provided statistical evidence to reject the null hypothesis of equal population means. n Fisher’s least significant difference (LSD) procedure can be used to determine where the differences occur.

27 27 Slide © 2005 Thomson/South-Western Fisher’s LSD Procedure n Test Statistic n Hypotheses

28 28 Slide © 2005 Thomson/South-Western 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. n Rejection Rule Reject H 0 if p -value <  p -value Approach: Critical Value Approach: Reject H 0 if t t a /2

29 29 Slide © 2005 Thomson/South-Western n Test Statistic Fisher’s LSD Procedure Based on the Test Statistic x i - x j __ where Reject H 0 if > LSD n Hypotheses n Rejection Rule

30 30 Slide © 2005 Thomson/South-Western Fisher’s LSD Procedure Based on the Test Statistic x i - x j n Example: 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.

31 31 Slide © 2005 Thomson/South-Western 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

32 32 Slide © 2005 Thomson/South-Western n LSD 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.

33 33 Slide © 2005 Thomson/South-Western n LSD 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.

34 34 Slide © 2005 Thomson/South-Western n LSD 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.

35 35 Slide © 2005 Thomson/South-Western n The experimentwise Type I error rate gets larger for problems with more populations (larger k ). 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 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. 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.

36 36 Slide © 2005 Thomson/South-Western End of Chapter 13, Part A


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