1. 1 1 Slide Analysis of Variance Chapter 13 BA 303

2. 2 2 Slide An Introduction to Analysis of Variance n nA factor is a variable that the experimenter has selected for investigation. n nA treatment is a level of a factor. n nExample:   A company president is interested in the number of hours worked by managers at three different plants: Gulfport, Long Beach, and Biloxi.   Factor: Plant   Treatment: Gulfport, Long Beach, and Biloxi

3. 3 3 Slide Analysis of Variance: A Conceptual Overview 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 :  1  =  2  =  3  = ... =  k H a : Not all population means are equal

4. 4 4 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 population means are different. Rejecting H 0 means that at least two population means have different values. Analysis of Variance: A Conceptual Overview

5. 5 5 Slide For each population, the response (dependent) variable is normally distributed. The variance of the response variable, denoted  2, is the same for all of the populations. The observations must be independent. nAssumptions for Analysis of Variance Analysis of Variance: A Conceptual Overview

6. 6 6 Slide nSampling Distribution of Given H 0 is True   Sample means are close together because there is only one sampling distribution when H 0 is true. Analysis of Variance: A Conceptual Overview

7. 7 7 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. Analysis of Variance: A Conceptual Overview

8. 8 8 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

9. 9 9 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) The estimate of  2 based on the variation of the sample means is called the mean square due to treatments and is denoted by MSTR.

10. 10 Slide 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-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)

11. 11 Slide F Test Statistic

12. 12 Slide 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  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.

13. 13 Slide F Distribution

14. 14 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

15. 15 Slide Source of Variation Sum of Squares Degrees of Freedom Mean Square 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 for a Completely Randomized Design p - Value

16. 16 Slide SST divided by its degrees of freedom n T – 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 for a Completely Randomized Design

17. 17 Slide 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 for a Completely Randomized Design

18. 18 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

19. 19 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 

20. 20 Slide 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). An F test will be conducted using  =.05. Testing for the Equality of k Population Means

21. 21 Slide 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 in the previous week is shown on the next slide. Testing for the Equality of k Population Means Factor... Manufacturing plant Treatments... Buffalo, Pittsburgh, Detroit Experimental units... Managers Response variable... Number of hours worked

22. 22 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

23. 23 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  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 p -Value and Critical Value Approaches Testing for the Equality of k Population Means

24. 24 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 (Sample sizes are all equal.) Mean Square Due to Treatments Testing for the Equality of k Population Means = (55 + 68 + 57)/3 = 60

25. 25 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

26. 26 Slide Treatment Error Total 490 308 798 2 12 14 245 25.667 Source of Variation Sum of Squares Degrees of Freedom Mean Square 9.55 F ANOVA Table ANOVA Table Testing for the Equality of k Population Means p -Value.0033

27. 27 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

28. 28 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

29. 29 Slide ANOVA PRACTICE

30. 30 Slide Perform an ANOVA Using the data in the table on the next slide, compute the mean and standard deviation for the Direct treatment. Compute the sum of the squares due to treatments (SSTR) and the mean square due to treatments (MSTR). Compute the sum of the squares due to the error (SSE) and the mean square due to the error (MSE). Compute the F test statistic. Using  =0.05, what do you conclude about the null hypothesis that the means of the treatments are equal? Determine the approximate p-value. Present the results in an ANOVA table.

31. 31 Slide DirectIndirectCombination 117.016.614.4425.20.04 218.522.23.2424.01.00 315.820.50.0121.512.25 418.218.34.4126.83.24 520.224.214.4427.56.25 616.019.80.3625.80.64 713.321.20.6424.20.64 20.425.0 6.264.01