Analysis Of Varience In Two-way Classification Model.pptx

Analysis Of Variance in Two-way Classification Model

1. Analysis Of Variance In Two-way Classification Model VIVEK GIRI MSc. Microbiology Sem 2

2. Introduction The analysis of variance (ANOVA) technique is developed by R.A. Fisher in 1920s. ANOVA is a statistical technique specially designed to test whether the means of more than two quantitative populations are equal or they have a significant difference The most common types of ANOVA are: One way ANOVA Two way ANOVA

3. One way ANOVA Determine how one factor imacts a response variable Factor Response Variable Study technique Exam score

4. Two way ANOVA Determines how more than one factor impacts a response variable The data are classified according to two different criteria or factors Petrol mileage may be affected by the type of car driven, the way it is driven, road conditions and other factors Exersise Diet Body weight

5. One way vs two way anova

6. ANOVA TABLE SSC = Sum of squares between columns SSR = Sum of squares between rows SSE = Sum of squares due to errors SST = total sum of squares Source of variation Sum of squares Degrees of freedom Mean sum of squares Ratio of F Between samples SSC (c-1) MSC= SSC/(c-1) MSC/MSE Between rows SSR (r-1) MSR=SSR/(r-1) MSR/MSE Residual or error SSE (c-1) (r-1) MSE=SSE/(c-1) (r-1) Total SST n-1

7. The sum of squares for the source ‘Residual’ is obtained by subtracting from the total sum of squares between columns and rows SSE = SST – [SSC + SSR] The total number of degree of freedom = n-1 or cr-1 Where c refers to number of columns and r refers to number of rows. number of degrees of freedom between columns = (c-1) number of degree of freedom between rows = (r-1) number of degree of freedom for residual = (c-1)(r-1)

8. The calculated values of F are compared with the table values. If calculated of F is greater than the table value at pre assigned level of significance, the null hypothesis is rejected, otherwise accepted. ‘Residual’ is the measuring rod for testing significance. It represents the magnitude of variation due to forces called ‘chance’.

9. The following data represent the number of units of production per day turned out by 5 different workers using 4 different types of machines: Test whether the mean productivity is the same for the different machine types. Test whether the 5 men differ with respect to mean productivity. Machine type Workers A B C D 1 44 38 47 36 2 46 40 52 43 3 34 36 44 32 4 43 38 46 33 5 38 42 49 39

10. Let us take the hypothesis that The mean productivity is the same for four different machines The 5 men do not differ with respect to mean productivity to simplify the calculation 40 is subtracted correction factor = T 2 /N = 400/20 = 20 Sum of squares between machines this will be obtained by squaring up the machines totals, dividing each total by the number of items included in it, adding these figures and then subtracting the correction factor from them = (5) 2 + (-6) 2 + (38) 2 + (-17) 2 - correction factor 5 5 5 5 = (5+7.2+288.8+57.8)-20 = 338.8 v = (c-1) = (4-1) = 3

11. residual or remainder = total sum of squares – (sum of squares between machines – sum of squares between workers) = 574-(338.8+161.5) = 73.7 degree of freedom for remainder= 19-(3+4) = 12 (c-1)(r-1) = (3 x 4) = 12

12. ANALYSIS OF VARIANCE TABLE Source of variation S.S. d.f. M.S. Variance Ratio or F Between Machine types 338.8 3 112.933 112.933 = 18.387 6.142 Between workers 161.5 4 40.375 40.375 = 6.574 6.142 Remainder or Residual 73.7 12 6.142 Total 574 19

13. For (3, 12) d.f. , F 0.05 = 3.49 since the calculated value (18.4) is greater than the table value, we conclude that the productivity is not same for the four different types of machines. For (4, 12) d.f. , F 0.05 =3.26 the calculated value (6.58) is greater than the table value. Hence the workers differ with respect to mean productivity.

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