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Chapter 12

Analysis of Variance. Chapter 12. 1. 2. 3. 4. Chapter Goals. When you have completed this chapter, you will be able to:. Discuss the general idea of analysis of variance. List the characteristics of the F distribution.

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Chapter 12

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  1. Analysis of Variance Chapter 12

  2. 1. 2. 3. 4. Chapter Goals When you have completed this chapter, you will be able to: Discuss the general idea of analysis of variance. List the characteristics of the F distribution. Conducta test of hypothesis to determine whether the variances of two populations are equal. Organize data into a one-way and a two-wayANOVA table. and...

  3. 5. 6. 7. Chapter Goals Define the terms treatments and blocks. Conducta test of hypothesis to determine whether three or more treatment means are equal. Develop multiple tests for difference between each pair oftreatment means.

  4. 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 Fcannot 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. 2 2 and are the sample variances for the two samples s s 1 2 Test for Equal Variances For the two tailed test,thetest statistic is given by: where The null hypothesis is rejected if the computed value of the test statistic is greater than the critical value

  6. Q uestion Solve 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 utilitystockswas 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. Hypothesis Testing Recall 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 Compute the value of the test statistic and make a decision Do not reject H0 Reject H0 and acceptH1

  8. 2 2 s £ s H : I U 0 2 2 s > s H : I 1 U Step 3 Identify the test statistic Step 4 Step 4 State the decision rule State the decision rule 2 s 2 ( 3 . 9 ) F 1 = = 2 s 2 ( 3 . 5 ) 2 Hypothesis Test Step 1 State the null and alternate hypotheses Step 2 Select the level of significance  = 0.05 The test statistic is the F distribution Reject H0 if F > 3.68The df are 9 in the numerator and 7 in the denominator. Step 5 Compute the test statistic and make a decision = 1.2416 Conclusion: Do not reject the null hypothesis; there is insufficient evidence to show more variation in the internet stocks.

  9. Underlying Assumptions for ANOVA TheFdistribution is also used for testing whether two or more sample means came from the sameor equalpopulations This this technique is called analysis of variance or ANOVA

  10. ANOVA requires thefollowing conditions… that …the sampled populations follow the normal distribution …the populations have equal standard deviations …the samples are randomly selected and are independent

  11. ANOVA Procedure TheNull Hypothesis(H0) is that the population means are the same The Alternative Hypothesis (H1) is that at least oneof the means is different The Test Statistic is theF distribution The Decision rule is to reject H0ifF(computed) is greater than F(table) with numerator and denominator df

  12. Analysis of Variance

  13. Terminology Total Variation …is the sum of the squared differences between each observation and the overall mean Treatment Variation …is the sum of the squared differences between each treatment mean and the overall mean Random Variation …is the sum of the squared differences between each observation and its treatment mean

  14. Analysis of Variance SST ( ) ) - k 1 ( n k - SSE F = Procedure • If there k populations being sampled, the numerator degrees of freedom is k – 1 • If there are a total of n observations the denominator degrees of freedom is n - k • The test statistic is computed by:

  15. Procedure Analysis of Variance 2 S ( X ) 2 = S - SS Total X n • SS Total is the total sum of squares

  16. Procedure Analysis of Variance ( ) æ ö 2 S 2 T X ç ÷ = S - c SST ç ÷ n n è ø c • SST is the treatment sum of squares TC is the column total, ncis the number of observations in each column, X the sum of all the observations, and n the total number of observations

  17. Procedure Analysis of Variance = SSE SS total - SST • SSE is thesum of squares error

  18. Procedure Analysis of Variance Example 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 meannumberof dinners sold per day at the Aynor, Loris, and Lander restaurants. Use the .05 significance level.

  19. Procedure Analysis of Variance Example …continued AynorLoris Lander 1310 18 1212 16 1413 17 1211 17 17 Tc5146 85 nc 4 4 5

  20. Procedure Analysis of Variance Example 2 S ( X ) …continued 2 = S - SS Total X n (182)2 • SS Total (is the total sum of squares) = 2634 - = 86 13

  21. Procedure Analysis of Variance ( ) æ ö 2 S 2 T X ç ÷ = S - Example c SST ç ÷ n n …continued è ø c ( ) ( ) ( ) æ ö 2 2 2 2 ( 182 ) 51 46 85 ç ÷ = + + - ç ÷ 4 4 5 13 è ø = 76.25 • SST is the treatment sum of squares

  22. Procedure Analysis of Variance Example …continued • SSE is thesum of squares error SSE = SS Total - SST 86 – 76.25 = 9.75

  23. m m m = = H : 1 2 3 0 H : 1 Step 3 Identify the test statistic Step 4 Step 4 State the decision rule State the decision rule SST ( ) ) - k 1 ( n k - SSE F = 2 76.25 = 10 9.75 Hypothesis Test Step 1 State the null and alternate hypotheses Treatment means are not all equal Step 2 Select the level of significance  = 0.05 The test statistic is the F distribution Reject H0 if F > 4.10The df are 2 in the numerator and 10 in the denominator. Step 5 Compute the test statistic and make a decision = 39.10

  24. Procedure Analysis of Variance Example …continued Conclusion: Decision • 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

  25. ANOVA Table …from the Minitab system Analysis of Variance Source DFSS 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

  26. Analysis of Variance in Excel

  27. Using Click on Tools See Click on DATA ANALYSIS See…

  28. Using See Highlight ANOVA: SINGLE FACTOR …Click OK See…

  29. Using Input the sample data in Columns A, B, C. See INPUT NEEDS Click on OK A1:C6 See…

  30. Using F test SST SSE SS Total

  31. Inferences About Treatment Means

  32. Confidence Interval Inferences About Treatment Means 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

  33. æ ö 1 1 ( ) ç ÷ - ± + X X t MSE 1 2 è ø n n 1 2 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)]

  34. Solve Confidence Interval for the Difference Between Two Means Example 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?

  35. 1 1 ( ) - ± + t MSE X X æ ö 1 2 n n 1 2 ç ÷ è ø ± ( 17-12.75) 2.228 æ ö 1 1 + ç ÷ . 975 è ø 4 5 ( 2.77, 5.73) ± Þ 4 . 25 1 . 48 Confidence Interval for the Difference Between Two Means MSE

  36. Example …continued Confidence Interval for the Difference Between Two Means • 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

  37. Two-Factor ANOVA 2 é ù B 2 ( ) S X r - = S SSB ê ú n k ë û For the two-factor ANOVA we test whether there is a significant difference between thetreatment effect and whether there is a difference in the blocking effect! …Let Brbe the block totals (r for rows) …Let SSBrepresent the sum of squares for the blocks

  38. Two-Factor ANOVA 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 unitsproduced when the employeeswork on various shifts. A sample of five workers is selected and their output recorded on each shift. At the .05 significancelevel, can we conclude there is a difference in themean production by shift and in the mean production by employee? Example

  39. Two-Factor ANOVA Employee Day Evening Night Output Output Output Example McCartney 31 25 35 …continued Neary 33 26 33 Schoen 28 24 30 Thompson 30 29 28 Wagner 28 26 27

  40. m m m = = H : 1 3 2 0 H : 1 Step 3 Identify the test statistic Step 4 Step 4 State the decision rule State the decision rule ( ) - SST k 1 F = ( ) - - ( )( ) SSE k 1 b 1 Solve Hypothesis Test Difference between various shifts? Step 1 State the null and alternate hypotheses Not all means are equal  = 0.05 Step 2 Select the level of significance The test statistic is the F distribution Reject H0 if F > 4.46. The df are 2 and 8 Step 5 Compute the test statistic and make a decision

  41. Two-Factor ANOVA Example Using to get these results …continued • 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

  42. Two-Factor ANOVA Example ( ) …continued - 62 . 53 3 1 = ( ( ) ( ) ) - - 43.47 3 1 5 1 ( ) - SST k 1 F = ( ) - - ( )( ) SSE k 1 b 1 Step 5 = 5.754 Since 5.754 > 4.46, H0 is rejected. Conclusion: There is a difference in the mean number of units produced on the different shifts.

  43. m m m = = H : 1 2 3 0 H : 1 Step 3 Identify the test statistic Step 4 Step 4 State the decision rule State the decision rule ( ) - SST k 1 F = ( ) - - ( )( ) SSE k 1 b 1 Solve Hypothesis Test Difference between various shifts? Step 1 State the null and alternate hypotheses Not all means are equal  = 0.05 Step 2 Select the level of significance The test statistic is the F distribution Reject H0 if F > 3.84The df are 4 and 8 Step 5 Compute the test statistic and make a decision

  44. Two-Factor ANOVA Example …continued 4 33.73 = ( ) ( ) 43.47 2 4 ( ) - SST k 1 F = ( ) - - ( )( ) SSE k 1 b 1 Step 5 = 1.55 Since 1.55 < 3.84, H0 is not rejected. Conclusion: There is no significant difference in the mean number of units produced by the various employees.

  45. Two-Factor ANOVA …from the Minitab system 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 Error8 43.47 5.43 Total 14 139.73

  46. Using Select INPUT DATA Highlight ANOVA: TWO FACTOR WITHOUT REPLICATION …Click OK See…

  47. Using There is no significant difference in the average number of units produced by the different employees. Conclusion: Since F(test) < F(critical), there is not sufficient evidence to reject H0 Ftest Fcritical SSB SST SSE SS Total

  48. www.mcgrawhill.ca/college/lind for quizzes extra content data sets searchable glossary access to Statistics Canada’s E-Stat data …and much more! Test your learning… Click on… Online Learning Centre

  49. This completes Chapter 12

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