1 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) Lecture 15

2 One Way Analysis of Variance Three approaches in this course: 1.One variable completely randomized analogous to independent groups Z and t test 2.One variable randomized block analogous to dependent pairs Z and t test 3.Two variable completely randomized Lecture 15

3 µ2 µ1 µ4 µ3

4 Lecture 15 X2 X1 X4 X3

5 Completely randomized design Compares the variability among the treatment means (X 1, X 2, X 3, … X P ) to error variability. Is the variability among the treatment means so large that it could not be due to sampling error? Lecture 15

6 Completely randomized design To answer that question, we need to measure two things: 1. Variability among the treatment means how different from each other are the means? 2. Error variability how different could the treatment means be just on the basis of chance? Lecture 15

7 Variability among treatment means Recall the formula for a variance: S 2 = Σ(X i – X) 2 (1) (n-1) The numerator measures the variability of individual scores around the sample mean. Lecture 15

8 Variability among treatment means S 2 = Σ(X i – X) 2 (1) (n-1) SST = Σn i (X i – X) 2 (2) Lecture 15 Individual treatment means Grand mean X = X G

9 Variability among treatment means SST = Σn i (X i – X) 2 (2) Here, we’re finding the difference between each element in a set of scores and the mean of that set – but this time, the elements are themselves sample means Lecture 15

10 Variability among treatment means In Eqn. 2, the sum (Σ) measures the variability among the 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 is. Lecture 15

11 Lecture 15 X1X1 X3X3 X2X2 X4X4 X

12 Error variability SSE = Σ(X 1j – X 1 ) 2 + Σ(X 2j – X 2 ) 2 + … + Σ(X Pj – X P ) 2 Lecture 15 Mean for Sample 1 One individual score in Sample 1 Once again, we’re finding the difference between each element in a set of scores and the mean of that set – but this time we’re working at the level of individual samples

13 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. Lecture 15

14 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 2, S 3 2, … S P 2. Lecture 15

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

16 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 Lecture 15

17 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. Lecture 15

18 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?” Lecture 15

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

20 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). Lecture 15

21 Example 1 All measurements are made at the same time of day. Scores in milligrams of corticosterone per 100 milliliters of blood are (α =.05): VacationClassFinal Exam XX 2 XX 2 XX Lecture 15

22 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 Lecture 15

23 Example 1 SS Total = ∑X 2 – CM = 1299 – = SS Treat = T T … + T P 2 – CM n 1 n 2 n P = – = Lecture 15

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

25 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 Lecture 15

26 Example 1 – Summary Table SourcedfSSMSF Treat Error Total Decision: Reject H O. At least two of the situations differ in how much stress they produce. Lecture 15

27 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 n = 28. n 2 = 10 and n 3 = 8 and = 18 so n 1 = 28 – 18 = 10. Lecture 15

28 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) = SS Treat = 10 (12.1 – 18.0) (22.7 – 18.0) (19.5 – 18.0) 2 = Lecture 15

29 Example 2 SS Error = 9 (4.0) (5.1) (6.9) 2 = SS Total = = Now we are ready for our hypothesis test. Lecture 15

30 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 Lecture 15

31 Example 2 – Summary Table SourcedfSSMSF Treat Error Total Decision: Reject H O. At least two of the CDs differ significantly in length. Lecture 15

32 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 X S n i Lecture 15

33 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 = Lecture 15

34 Example 3 SS Treat = 10 (201 – ) (215 – ) (218 – ) 2 = SS Error = 9 (2.9) (3.5) (3.4) 2 Now we’re ready for our hypothesis test. Lecture 15 Why?

35 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 Lecture 15

36 Example 3 – Summary Table SourcedfSSMSF Treat Error Total Decision: Reject H O. At least two of the treatments differ in average response time. Lecture 15