1. 1 Lecture 15 One Way Analysis of Variance  Designed experiments usually involve comparisons among more than two means.  The use of Z or t tests with more than two means is not efficient.  A more efficient approach is Analysis of Variance (Anova)

2. 2 Lecture 15 One Way Analysis of Variance Three approaches in this course: One Independent Variable  Completely randomized design (today)  analogous to independent groups Z and t test  Randomized block design  analogous to dependent pairs Z and t test Two Independent Variables  Two factor completely randomized Anova

3. 3 Lecture 15 Completely randomized design X2X2 Draw a sample Make an inference 11 X1X1 22 33 X3X3 44 X4X4

4. 4 Lecture 15 Completely randomized design The approach is to compare the differences among the treatment means (,,, … ) to the amount of error variability.  Differences among treatment means vs. error variability Our question is, are the differences among the treatment means so large that they could not be due to sampling error? X1X1 X2X2 X3X3 XPXP

5. 5 Lecture 15 Completely randomized design In order to answer our question, we need numerical measures of two things: 1. differences between the treatment means  how different from each other are the means? 2. sampling variability within each treatment  how different could the treatment means be just on the basis of chance?

6. 6 Lecture 15 1. Differences among treatment means Recall the formula for a variance: S 2 = Σ(X i - ) 2 (Eqn. 1) (n-1) The numerator of Eqn. 1 measures variability of individual scores around the sample mean. Compare with Eqn. 2 (next slide): X

7. 7 Lecture 15 1. Differences among treatment means S 2 = Σ(X i - ) 2 (Eqn. 1) (n-1) SST = Σn i ( - ) 2 (Eqn. 2) X X Individual treatment means Grand mean XiXi

8. 8 Lecture 15 1. Differences among treatment means In Eqn. 2, the sum (Σ) measures the variability among the treatment (sample) means.  SST is the Sum of Squared deviations of the Treatment means from the grand mean. In other words, SST is the Sum of Squares for Treatments. The more different the treatment means are from each other, the bigger SST will be.

9. 9 Lecture 15 2. Error variability SSE = Σ(X 1j – X 1 ) 2 + Σ(X 2j – X 2 ) 2 + … + Σ(X Pj – X P ) 2 One individual score In Treatment 1 sample Mean for Treatment 1

10. 10 Lecture 15 2. Error variability SSE = Σ(X 1j – X 1 ) 2 + Σ(X 2j – X 2 ) 2 + … + Σ(X Pj – X P ) 2 SSE = the Sum of Squares for Error. This measures the total variability of individual scores around their respective treatment means.  Key point : people who get the same treatment should all have the same score. Any deviation from that state (from X) reflects sampling error.

11. 11 Lecture 15 2. Sampling variability Important note: (S 1 ) 2 = Σ(X 1j – X 1 ) 2 (n 1 – 1) Therefore, (n 1 – 1) (S 1 ) 2 = Σ(X 1j – X 1 ) 2 Same is true for S 2, S 3, … S P. 22 2

12. 12 Lecture 15 2. Sampling variability Thus: SSE = (n 1 – 1) S 1 + (n 2 – 1) S 2 + … + (n P – 1) S P Now, we’re almost ready. One last issue:  SST is the sum of P terms (deviations)  SSE is the sum of N = Σn i terms 222

13. 13 Lecture 15 2. Sampling variability How can we compare SST to SSE? To make SST and SSE commensurable, we divide each by their degrees of freedom. SST = MST(Mean Square Treatment) P-1 SSE = MSE(Mean Square Error) N-P

14. 14 Lecture 15 The Analysis of Variance – F-test When there is a treatment effect, MST will be much larger than MSE. Therefore, the ratio of MST to MSE will be much larger than 1.0 F =MST MSE How much larger than 1 must F be for us to reject H 0 ? Check F table for α and d.f.

15. 15 Lecture 15 The Analysis of Variance – F-test d.f. numerator = p – 1 d.f. denominator = n – p Important note: for Anova, F test is always one- tailed.  You’re asking “is the treatment variance larger than the error variance?”

16. 16 Lecture 15 Computational Formulae CM = (ΣX i ) 2 N SSTotal = ΣX i 2 – CM SST = T 1 + T 2 + … + T P – CM n 1 n 2 n P SSE = SSTotal - SST 222

17. 17 Lecture 15 Example 1 You want to know whether different situations produce different amounts of stress. The amount of the hormone corticosterone circulating in the blood is a good measure of how stressed a person is. You randomly assign 15 students to three groups of five each. Subjects in Group 1 have their corticosterone levels measured immediately after returning from vacation (low stress). Group 2 subjects are measured after one week of classes (moderate stress). Group 3 is measured immediately before final exam week (high stress).

18. 18 Lecture 15 Example 1 All measurements are made at the same time of day. Scores in milligrams of corticosterone per 100 millilitres of blood are (α =.05): VacationClassFinal Exam XX 2 XX 2 XX 2 2486410100 391010013169 74974914196 2452513169 6 36 1010015225 ∑ =201024033865859

19. 19 Lecture 15 Example 1 n 1 = 5 n 2 = 5 n 3 = 5 X 1 = 4.0 X 2 = 8.0 X 3 = 13.0 ∑X = 125 ∑X 2 = 1299 X = X G = 125/15 = 8.33 Alternative names for the same quantity

20. 20 Lecture 15 Example 1 SS Total = ∑X 2 – CM = 1299 – 125 2 = 257.333 15 SS Treat = T 1 + T 2 + … + T P – CM n 1 n 2 n P = 20 2 +40 2 + 65 2 – 125 2 = 203.333 5 5 5 15 222

21. 21 Lecture 15 Example 1 SS Error = SS Total – SS Treat SS Error = 257.333 – 203.333 = 54 Now we are ready for our hypothesis test…

22. 22 Lecture 15 Example 1 – Hypothesis Test H O : μ 1 = μ 2 = μ 3 H A : At least one pair of means differ Test statistic:F = MS Treat MS Error Rejection region: F obt > F (2, 12, α =.05) = 3.88

23. 23 Lecture 15 Example 1 – Summary Table SourcedfSSMSF Treat2203.333101.6722.59 Error12544.5 Total14257.33 Decision: Reject H O. At least two of the situations differ in how much stress they produce.

24. 24 Lecture 15 Example 2 The first question we have to answer is, what is n 1 (n for the Rain group)? We are told that d.f. = 27. That means that n-1 = 27, so, therefore, n = 28. n 2 = 10 and n 3 = 8, so n 1 = 28 – 18 = 10.

25. 25 Lecture 15 Conceptual formulas - review We’ll need these conceptual formulas: SST = Σn i ( - ) 2 SSE = Σ(X 1j – X 1 ) 2 + Σ(X 2j – X 2 ) 2 + … + Σ(X Pj – X P ) 2 (See Slides 7 and 9 above) X XiXi

26. 26 Lecture 15 Example 2 Since we don’t have raw scores, we cannot use computational formulae. Therefore, we use the conceptual formulae. X = 10 (12.1) + 10 (22.7) + 8 (19.5) = 18.0 28 SS Treat = 10 (12.1 – 18.0) 2 + 10 (22.7 – 18.0) 2 + 8 (19.5 – 18.0) 2 = 587.0

27. 27 Lecture 15 Example 2 SS Error = 9 (4.0) 2 + 9 (5.1) 2 + 7 (6.9) 2 = 711.36 SS Total = 587 + 711.36 = 1298.36 Now we are ready for our hypothesis test.

28. 28 Lecture 15 Example 2 – Hypothesis Test H O : μ 1 = μ 2 = μ 3 H A : At least one pair of means differ Test statistic:F = MS Treat MS Error Rejection region: F obt > F (2, 25, α =.05) = 3.39

29. 29 Lecture 15 Example 2 – Summary Table SourcedfSSMSF Treat2587293.510.32 Error25711.3628.45 Total271298.36 Decision: Reject H O. At least two of the CDs differ significantly in length.

30. 30 Lecture 15 Example 3 Simple reaction times to green, red, and yellow instrument panel lights were compared. The three colors were randomly assigned to 31 different subjects who were instructed to press a button in response to the light. Shown below are average RTs in milliseconds for these subjects. GreenRedYellow X201215218 S2.93.53.4 n i 101110

31. 31 Lecture 15 Example 3 Is there an overall significant difference? (α =.05) SS Treat = ∑n i (X i – X) 2 X = [10 (201) + 11 (215) + 10 (218)] 3 = 211.45

32. 32 Lecture 15 Example 3 SS Treat = 10 (201 – 211.45) 2 + 11 (215 – 211.45) 2 = 1659.67 SS Error = 9 (2.9) 2 + 10 (3.5) 2 + 9 (3.4) 2 Now we’re ready for our hypothesis test.

33. 33 Lecture 15 Example 3 – Hypothesis Test H O : μ 1 = μ 2 = μ 3 H A : At least one pair of means differ Test statistic:F = MS Treat MS Error Rejection region: F obt > F (2, 28, α =.05) = 3.34

34. 34 Lecture 15 Example 3 – Summary Table SourcedfSSMSF Treat21659.67829.8476.91 Error28302.2310.79 Total301961.9 Decision: Reject H O. At least two of the treatments differ in average response time.