1. CHAPTER 3 Analysis of Variance (ANOVA) PART 2 =TWO- WAY ANOVA WITHOUT REPLICATION

2. Two-Way ANOVA without Replication (Randomized Block Design) Known as Randomized Block Design (RBD) For RBD there is one factor or variable that is of primary interest. However, there are also several other nuisance factor. Nuisance factors are those that may affect the measured result, but not of primary interest. For example, in applying a treatment, nuisance factors might be the specific operator who prepared the treatment, the time of day the experiment was run and the room temperature. So, to control this, the important technique known as blocking.

3. Randomized Block Design, cont. Effects model for RBD: Two Sets of Hypothesis: Treatment Effect: H 0 :  1 =  2 =... =  t =0 H 1 :  j  0 at least one j Block Effect: H 0 :  i = 0 for each value of i through n H 1 :  i ≠ 0 at least one i

4. Randomized Block Design, cont. Format for data: Data appear in a table, where location in a specific row and a specific column is important. Calculations: – Sum of squares total (SST) = sum of squared differences between each individual data value (regardless of group membership) minus the grand mean,, across all data... total variation in the data (not variance).

5. Randomized Block Design, cont. Calculations, cont.: – Sum of squares treatment (SSTR) = sum of squared differences between each treatment group mean and the grand mean, balanced by sample size... between-treatment-groups variation (not variance). – Sum of squares block (SSBL) = sum of squared differences between each block group mean and the grand mean, balanced by sample size... between-block-groups variation (not variance). – Sum of squares error (SSE): SSE = SST – SSTR – SSBL

6. Randomized Block Design, cont. Calculations, cont.: – Mean square treatment (MSTR) = SSTR/( t – 1), where t is the number of treatment groups. – Mean square block (MSBL) = SSBL/( n – 1), where n is the number of block groups. Controls the size of SSE by removing variation that is explained by the blocking categories. – Mean square error (MSE):

7. Randomized Block Design, cont. Calculations, cont.: Test Statistics, F -Ratios: ( F test) – F -Ratio, Treatment = MSTR/MSE, This F -ratio is the test statistic for the hypothesis that the treatment group means are equal. - F -Ratio, Block = MSBL/MSE, This F -ratio is the test statistic for the hypothesis that the block group means are equal.

8. Randomized Block Design, cont. ANOVA Table Source of Variation Sum of Squares Degrees of Freedom Mean Square Fp -Value Treatments SSTRt-1 Blocks SSBLn-1 Error SSE(t-1)(n-1) Total SSTtn-1

9. l F  value - Treatment effect - F , t -1,(t-1)(n-1) v 1 = t -1; v 2 = (t-1)(n-1) = table (ms 30) - Block effect - F , n -1,(t-1)(n-1) v 1 = n -1; v 2 = (t-1)(n-1)= table (ms 30) l If F -Ratio (F test) > F  or p-value < , reject H 0 at the  level l Conclusion * To reject the null hypothesis means that at least one treatment group had a different effect than the rest. *To reject the null hypothesis means that at least one block group had a different effect on the dependent variable than the rest. Randomized Block Design, cont.

10. EXAMPLE : Crescent Oil Co. Crescent Oil has developed three new blends of gasoline and must decide which blend or blends to produce and distribute. A study of the miles per gallon ratings of the three blends is being conducted to determine if the mean ratings are the same for the three blends.

11. Randomized Block Design, cont. Example: Crescent Oil Co. Five automobiles have been tested using each of the three gasoline blends and the miles per gallon ratings are shown on the next slide. Factor... Gasoline blend Treatments... Blend X, Blend Y, Blend Z Blocks... Automobiles Response variable... Miles per gallon

12. Randomized Block Design, cont. 29.8 28.8 28.4 Treatment Means 1234512345 31 30 29 33 26 30 29 31 25 30 29 28 29 26 30.333 29.333 28.667 31.000 25.667 Type of Gasoline (Treatment) Block Means Blend XBlend YBlend Z Automobile (Block)

13. SOLUTION l 1. Hypothesis: H 0 :  1 =  2 =  3 =0 H 1 :  j  0 at least one j l  1 = Blend X l  2 = Blend Y l  3 = Blend Z

14.  Mean Square Due to Error: Randomized Block Design, cont. MSE = 5.47/[(3 - 1)(5 - 1)] = 0.68 iv. SSE = SST – SSTR – SSBL = 62 - 5.2 - 51.33 = 5.47 MSBL = 51.33/(5 - 1) = 12.8 ii. SSBL = 3[(30.333 - 29) 2 +... + (25.667 - 29) 2 ] = 51.33 MSTR = 5.2/(3 - 1) = 2.6 i. SSTR = 5[(29.8 - 29) 2 + (28.8 - 29) 2 + (28.4 - 29) 2 ] = 5.2 The overall sample mean is 29. Thus, n Mean Square Due to Treatments: n Mean Square Due to Blocks: 2. Test Statistics ( F test) iii. SST = (31-29) 2 + ………………+ ( 26 -29) 2 = 62.00

15. Source of Variation Sum of Squares Degrees of Freedom Mean Square F Treatments Error Total 2 14 5.20 5.47 62.00 8 2.60 0.68 3.82 n v. ANOVA Table Randomized Block Design, cont. Blocks51.3312.804 p -Value 0.07 ***

16. 3. Rejection Region ( Draw picture) Randomized Block Design, cont. Treatment : For  = 0.05, F 0.05,2,8 = 4.46 (2 d.f. numerator and 8 d.f. denominator) p -Value Approach: Do not Reject H 0 Since p -value (0.07) > 0.05 Critical Value Approach: Do not Reject H 0 Since F Test > F alfa ; 3.82 < 4.46 2. F (alfa) value – critical value

17. 5. Conclusion : Randomized Block Design, cont. 2. There is sufficient evidence to conclude that the mean ratings are the same for the three blends. 1. The p -value is greater than.05 (Excel provides a p - value of 0.07). or since F Test > F alfa ; 3.82 < 4.46. Therefore, we cannot reject H 0.

18. EXERCISE TEXT BOOK PAGE 148 (Example 9.5) 149 ( 1 & 2) 5 IMPORTANT STEP: 1.HYPOTHESIS TESTING 2.TEST STATISTIC – F TEST 3.F (alfa) – VALUE (CRITICAL VALUE) 4.REJECTION REGION 5.CONCLUSION