1. 12 - 1 Copyright © 2004 by The McGraw-Hill Companies, Inc. All rights reserved.

2. 12 - 2 Copyright © 2004 by The McGraw-Hill Companies, Inc. All rights reserved. 1. Discuss the general idea of analysis of variance. 2. List the characteristics of the F distribution. When you have completed this chapter, you will be able to: Organize data into a one-way and a two-way ANOVA table. 3. Conduct a test of hypothesis to determine whether the variances of two populations are equal. 4.

3. 12 - 3 Copyright © 2004 by The McGraw-Hill Companies, Inc. All rights reserved. 5. Define the terms treatments and blocks. 6. Conduct a test of hypothesis to determine whether three or more treatment means are equal. 7. Develop multiple tests for difference between each pair of treatment means.

4. 12 - 4 Copyright © 2004 by The McGraw-Hill Companies, Inc. All rights reserved. Characteristics of the F-Distribution There is a “family of F-Distributions: Each member of the family is determined by two parameters: …the numerator degrees of freedom, and the … denominator degrees of freedom F can not be negative, and it is a continuous distribution The F distribution is positively skewed Its values range from 0 to  as F  , the curve approaches the X-axis

5. 12 - 5 Copyright © 2004 by The McGraw-Hill Companies, Inc. All rights reserved. Test for Equal Variances For the two tailed test, the test statistic is given by: The null hypothesis is rejected if the computed value of the test statistic is greater than the critical value and are the sample variances for the two samples 2 1 s 2 2 s

6. 12 - 6 Copyright © 2004 by The McGraw-Hill Companies, Inc. All rights reserved. Colin, a stockbroker at Critical Securities, reported that the mean rate of return on a sample of 10 internet stocks was 12.6 percent with a standard deviation of 3.9 percent. The mean rate of return on a sample of 8 utility stocks was 10.9 percent with a standard deviation of 3.5 percent. At the.05 significance level, can Colin conclude that there is more variation in the internet stocks?

7. 12 - 7 Copyright © 2004 by The McGraw-Hill Companies, Inc. All rights reserved. Do not reject H 0 Reject H 0 and accept H 1 State the null and alternate hypotheses Step 1 Select the level of significance Step 2 Identify the test statistic Step 3 State the decision rule Step 4 Step 5 Hypothesis Testing Compute the value of the test statistic and make a decision

8. 12 - 8 Copyright © 2004 by The McGraw-Hill Companies, Inc. All rights reserved. Hypothesis Test State the null and alternate hypotheses Step 1 Select the level of significance Step 2 Identify the test statistic Step 3 State the decision rule Step 4  = 0.05 The test statistic is the F distribution State the decision rule Step 4 Compute the test statistic and make a decision Step 5 Reject H 0 if F > 3.68 The df are 9 in the numerator and 7 in the denominator. Do not reject the null hypothesis; there is insufficient evidence to show more variation in the internet stocks. = 1.2416 F 2 2 2 1 s s   2 2 )5.3( )9.3( 22 0 : U I H  U 22 1 : I H 

9. 12 - 9 Copyright © 2004 by The McGraw-Hill Companies, Inc. All rights reserved. This this technique is called analysis of variance or ANOVA The F distribution is also used for testing whether two or more sample means came from the same or equal populations ANOVA

10. 12 - 10 Copyright © 2004 by The McGraw-Hill Companies, Inc. All rights reserved. …the populations have equal standard deviations …the samples are randomly selected and are independent …the sampled populations follow the normal distribution ANOVA requires the following conditions…

11. 12 - 11 Copyright © 2004 by The McGraw-Hill Companies, Inc. All rights reserved. The Null Hypothesis (H 0 ) is that the population means are the same The Alternative Hypothesis (H 1 ) is that at least one of the means is different ANOVA Procedure The Test Statistic is the F distribution The Decision rule is to reject H 0 if F (computed) is greater than F (table) with numerator and denominator df

12. 12 - 12 Copyright © 2004 by The McGraw-Hill Companies, Inc. All rights reserved.

13. 12 - 13 Copyright © 2004 by The McGraw-Hill Companies, Inc. All rights reserved. T erminology Total Variation …is the sum of the squared differences between each observation and the overall mean Random Variation …is the sum of the squared differences between each observation and its treatment mean Treatment Variation …is the sum of the squared differences between each treatment mean and the overall mean

14. 12 - 14 Copyright © 2004 by The McGraw-Hill Companies, Inc. All rights reserved. SSE SST F   kn    k  1  If there are a total of n observations the denominator degrees of freedom is n - k  The test statistic is computed by:  If there k populations being sampled, the numerator degrees of freedom is k – 1

15. 12 - 15 Copyright © 2004 by The McGraw-Hill Companies, Inc. All rights reserved. SS Total is the total sum of squares n X X 2 2 )( TotalSS  

16. 12 - 16 Copyright © 2004 by The McGraw-Hill Companies, Inc. All rights reserved. SST is the treatment sum of squares  n X n T SST c c 2 2            T C is the column total, n c is the number of observations in each column,  X the sum of all the observations, and n the total number of observations

17. 12 - 17 Copyright © 2004 by The McGraw-Hill Companies, Inc. All rights reserved. SSE is the sum of squares error SST - totalSS SSE 

18. 12 - 18 Copyright © 2004 by The McGraw-Hill Companies, Inc. All rights reserved. Easy Meals Restaurants specialize in meals for senior citizens. Katy Smith, President, recently developed a new meat loaf dinner. Before making it a part of the regular menu she decides to test it in several of her restaurants. She would like to know if there is a difference in the mean number of dinners sold per day at the Aynor, Loris, and Lander restaurants. Use the.05 significance level.

19. 12 - 19 Copyright © 2004 by The McGraw-Hill Companies, Inc. All rights reserved. Aynor Loris Lander 13 10 18 12 12 16 14 13 17 12 11 17 17 T c 51 46 85 n c 4 4 5 T c 51 46 85 n c 4 4 5 …continued

20. 12 - 20 Copyright © 2004 by The McGraw-Hill Companies, Inc. All rights reserved. …continued SS Total (is the total sum of squares) = 86 13 = 2634 - )( TotalSS 2 2     n X X (182) 2

21. 12 - 21 Copyright © 2004 by The McGraw-Hill Companies, Inc. All rights reserved. …continued  n X n T SST c c 2 2            SST is the treatment sum of squares  = 76.25 13 )182( 5 85 4 46 4 51 2 22 2          

22. 12 - 22 Copyright © 2004 by The McGraw-Hill Companies, Inc. All rights reserved. SSE is the sum of squares error …continued SSE = SS Total - SST 86 – 76.25 = 9.75

23. 12 - 23 Copyright © 2004 by The McGraw-Hill Companies, Inc. All rights reserved. Hypothesis Test State the null and alternate hypotheses Step 1 Select the level of significance Step 2 Identify the test statistic Step 3 State the decision rule Step 4  = 0.05 The test statistic is the F distribution State the decision rule Step 4 Compute the test statistic and make a decision Step 5 Reject H 0 if F > 4.10 The df are 2 in the numerator and 10 in the denominator. = 39.10 1 :H 0 :H 1  2  == 3  Treatment means are not all equal SSE SST F   kn    k  1 10 9.75 2  76.25

24. 12 - 24 Copyright © 2004 by The McGraw-Hill Companies, Inc. All rights reserved.  The decision is to reject the null hypothesis  The treatment means are not the same  The mean number of meals sold at the three locations is not the same …continued

25. 12 - 25 Copyright © 2004 by The McGraw-Hill Companies, Inc. All rights reserved. Analysis of Variance Source DF SS MS F P Factor 2 76.250 38.125 39.10 0.000 Error 10 9.750 0.975 Total 12 86.000 Individual 95% CIs For Mean Based on Pooled St.Dev Level N Mean St.Dev ---------+--------- +---------+------- Aynor 4 12.750 0.957 (---*---) Loris 4 11.500 1.291 (---*---) Lander 5 17.000 0.707 (---*---) ---------+---------+---------+------- Pooled St.Dev = 0.987 12.5 15.0 17.5 ANOVA Table …from the Minitab system

26. 12 - 26 Copyright © 2004 by The McGraw-Hill Companies, Inc. All rights reserved. Analysis of Variance in Excel

27. 12 - 27 Copyright © 2004 by The McGraw-Hill Companies, Inc. All rights reserved. See Using Click on DATA ANALYSIS See… Click on Tools

28. 12 - 28 Copyright © 2004 by The McGraw-Hill Companies, Inc. All rights reserved. Highlight ANOVA: SINGLE FACTOR …Click OK Using See… See

29. 12 - 29 Copyright © 2004 by The McGraw-Hill Companies, Inc. All rights reserved. Using INPUT NEEDS A1:C6 See Click on OK See… Input the sample data in Columns A, B, C.

30. 12 - 30 Copyright © 2004 by The McGraw-Hill Companies, Inc. All rights reserved. Using SS Total SST SSE F test

31. 12 - 31 Copyright © 2004 by The McGraw-Hill Companies, Inc. All rights reserved. Inferences About Treatment Means

32. 12 - 32 Copyright © 2004 by The McGraw-Hill Companies, Inc. All rights reserved. When we reject the null hypothesis that the means are equal, we may want to know which treatment means differ One of the simplest procedures is through the use of confidence intervals Inferences About Treatment Means Confidence Interval

33. 12 - 33 Copyright © 2004 by The McGraw-Hill Companies, Inc. All rights reserved. Confidence Interval for the Difference Between Two Means where t is obtained from the t table with degrees of freedom (n - k). MSE = [ SSE/(n - k) ] where t is obtained from the t table with degrees of freedom (n - k). MSE = [ SSE/(n - k) ]  XX 12  t  MSE nn 12 11       

34. 12 - 34 Copyright © 2004 by The McGraw-Hill Companies, Inc. All rights reserved. Develop a 95% confidence interval for the difference in the mean number of meat loaf dinners sold in Lander and Aynor. Can Katy conclude that there is a difference between the two restaurants? Confidence Interval for the Difference Between Two Means

35. 12 - 35 Copyright © 2004 by The McGraw-Hill Companies, Inc. All rights reserved. MSE  XX 12  t  nn 12 11        (17-12.75) 2.228 .975 1 4 1 5       ..425148  ( 2.77,5.73) Confidence Interval for the Difference Between Two Means

36. 12 - 36 Copyright © 2004 by The McGraw-Hill Companies, Inc. All rights reserved.  Because zero is not in the interval, we conclude that this pair of means differs  The mean number of meals sold in Aynor is different from Lander Confidence Interval for the Difference Between Two Means …continued

37. 12 - 37 Copyright © 2004 by The McGraw-Hill Companies, Inc. All rights reserved. For the two-factor ANOVA we test whether there is a significant difference between the treatment effect and whether there is a difference in the blocking effect! …Let B r be the block totals ( r for rows) …Let SSB represent the sum of squares for the blocks ANOVA SSB B k X n r           2 2 ( )

38. 12 - 38 Copyright © 2004 by The McGraw-Hill Companies, Inc. All rights reserved. The Bieber Manufacturing Co. operates 24 hours a day, five days a week. The workers rotate shifts each week. Todd Bieber, the owner, is interested in whether there is a difference in the number of units produced when the employees work on various shifts. A sample of five workers is selected and their output recorded on each shift. At the.05 significance level, can we conclude there is a difference in the mean production by shift and in the mean production by employee? ANOVA

39. 12 - 39 Copyright © 2004 by The McGraw-Hill Companies, Inc. All rights reserved. ANOVA EmployeeDay Output Evening Output Night Output McCartney312535 Neary332633 Schoen282430 Thompson302928 Wagner282627 …continued

40. 12 - 40 Copyright © 2004 by The McGraw-Hill Companies, Inc. All rights reserved. Hypothesis Test State the null and alternate hypotheses Step 1 Select the level of significance Step 2 Identify the test statistic Step 3 State the decision rule Step 4  = 0.05 The test statistic is the F distribution State the decision rule Step 4 Compute the test statistic and make a decision Step 5 Reject H 0 if F > 4.46. The df are 2 and 8 1 :H 0 :H 1  2  == 3  Not all means are equal    ) 1 )( 1 ( 1   bkSSE kSST F Difference between various shifts?

41. 12 - 41 Copyright © 2004 by The McGraw-Hill Companies, Inc. All rights reserved. Compute the various sum of squares: SS(total) = 139.73 SST = 62.53 SSB = 33.73 SSE = 43.47 df(block) = 4, df(treatment) = 2 df(error)=8 ANOVA …continued Using to get these results Using to get these results

42. 12 - 42 Copyright © 2004 by The McGraw-Hill Companies, Inc. All rights reserved. Since 5.754 > 4.46, H 0 is rejected.   151343.47 1353.62    ANOVA …continued Step 5 There is a difference in the mean number of units produced on the different shifts.    ) 1 )( 1 ( 1   bkSSE kSST F = 5.754

43. 12 - 43 Copyright © 2004 by The McGraw-Hill Companies, Inc. All rights reserved. Hypothesis Test State the null and alternate hypotheses Step 1 Select the level of significance Step 2 Identify the test statistic Step 3 State the decision rule Step 4  = 0.05 The test statistic is the F distribution State the decision rule Step 4 Compute the test statistic and make a decision Step 5 1 :H 0 :H 1  2  == 3  Not all means are equal    ) 1 )( 1 ( 1   bkSSE kSST F Difference between various shifts? Reject H 0 if F > 3.84 The df are 4 and 8

44. 12 - 44 Copyright © 2004 by The McGraw-Hill Companies, Inc. All rights reserved. ANOVA …continued Step 5    ) 1 )( 1 ( 1   bkSSE kSST F Since 1.55 < 3.84, H 0 is not rejected. = 1.55  4243.47 4 33.73  There is no significant difference in the mean number of units produced by the various employees.

45. 12 - 45 Copyright © 2004 by The McGraw-Hill Companies, Inc. All rights reserved. Units versus Worker, Shift Analysis of Variance for Units Source DF SS MS F P Worker 4 33.73 8.43 1.55 0.276 Shift 2 62.53 31.27 5.75 0.028 Error 8 43.47 5.43 Total 14 139.73 Units versus Worker, Shift Analysis of Variance for Units Source DF SS MS F P Worker 4 33.73 8.43 1.55 0.276 Shift 2 62.53 31.27 5.75 0.028 Error 8 43.47 5.43 Total 14 139.73 …from the Minitab system ANOVA

46. 12 - 46 Copyright © 2004 by The McGraw-Hill Companies, Inc. All rights reserved. Using See… Highlight ANOVA: TWO FACTOR WITHOUT REPLICATION …Click OK Select INPUT DATA

47. 12 - 47 Copyright © 2004 by The McGraw-Hill Companies, Inc. All rights reserved. SS Total SST SSE SSB Ftest Fcritical Using Since F(test) < F(critical), there is not sufficient evidence to reject H 0 There is no significant difference in the average number of units produced by the different employees.

48. 12 - 48 Copyright © 2004 by The McGraw-Hill Companies, Inc. All rights reserved. Test your learning … www.mcgrawhill.ca/college/lind Click on… Online Learning Centre for quizzes extra content data sets searchable glossary access to Statistics Canada’s E-Stat data …and much more!

49. 12 - 49 Copyright © 2004 by The McGraw-Hill Companies, Inc. All rights reserved. This completes Chapter 12