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By Shaimaa Elkadi Supervised by Dr.Amal Fatani. From the previous One Way ANOVA But.

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Presentation on theme: "By Shaimaa Elkadi Supervised by Dr.Amal Fatani. From the previous One Way ANOVA But."— Presentation transcript:

1 By Shaimaa Elkadi Supervised by Dr.Amal Fatani

2 From the previous One Way ANOVA But

3 Least significant test procedure Studentized Range procedure Dunnett’s Procedure

4  List the means of treatments in a single list in which they are ranked from lowest to highest  Underline (or put in brackets) the means that are not statistically significantly different from each other (referring to the limit stated) i.e. for the generally used P= 0.05 we calculate the 5% allowance. 5% allowance The critical difference between means which allows one to reject the null hypothesis for any two sample means x i and x j at P =  Means are divided into groups  Within a group means differ by less than the 5% allowance  In between groups the difference is more than 5% allowance  Means are divided into groups  Within a group means differ by less than the 5% allowance  In between groups the difference is more than 5% allowance

5 5% allowance= t √(S 2 (1/n i +1/n j )) Tabulated t value P=0.05 DF= N-t (of S 2 ) One orTwo tails The pooled variance from ANOVA calculation n i, n j The number of observations from which means are determined

6 Apply on the previous example  Ranked means BA,CIJF,GD,HE  5% allowance = 2.51  Grouped means (B A,C) (I JF,G D,H) E  So for high weight gain choose regimen E  And for low weight gain choose regimen B, A or C

7 5% allowance= (Q/√2) √(S 2 (1/n i +1/n j )) Studentized range value at K= number of treatments DF= N-t (of S 2 )

8 Apply on the previous example  Ranked means BA,CIJF,GD,HE  5% allowance = 4.26  Grouped means (B A,C I) (A,C I J F,G D,H) (JF,G D,H E)  This method is more conservative why? since it need more difference to exist between means to declare significance

9 5% allowance= t d √(S 2 (1/n i +1/n j )) Dunnett’s td value at K= number of treatments – 1 =t-1 DF= N-t (of S 2 ) One or two tail This is applied when One of the groups represents the control group while other groups represent the tested treatments For the exclusion of the control group

10 Apply on the previous example  Consider J is the standard regimen (control group) {We want to know which regimen will cause weight gain less or more than J (i.e. two tail)}  Ranked means BA,CIJF,GD,HE  5% allowance = 3.55  B showed a statistically significant smaller weight gain than J  E showed a statistically significant larger weight gain than J

11  This method is the most conservative why? It ensures that probability of one or more comparisons between treatments and control judged significant by chance alone is 5 %  Test is used for both two sided (t d at P= 0.05) or one sided (t d at P= 0.1) comparisons, according to the experiment design

12  Make a list the means of the treatments from lowest to highest value  Choose the procedure  Calculate 5% allowance  Rank the means in groups  Make your conclusion

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14 AApply on the previous example SS 2 = 2.17 DDF = 20 nn i, n j = 3 tt (P=0.05,DF=20,two tailed)= SSo 5% allowance = 2.51 Thus any 2 means differ by more than 2.51 are significantly different from each other (differ in group) 5% allowance= t √(S 2 (1/n i +1/n j ))

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16 AApply on the previous example KK=10 DDF = 20 QQ = 5.01 SSo 5% allowance = 4.26 TThus any 2 means differ by more than 4.26 are significantly different from each other (differ in group) 5% allowance= (Q/√2) √(S 2 (1/n i +1/n j ))

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18 AApply on the previous example KK=10 – 1 = 9 DDF = 20 tt d (P=0.05, two tailed)= 2.95 SSo 5% allowance = % allowance= t d √(S 2 (1/n i +1/n j ))

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20 Source of variation DFSum of squares (SS) Mean Square (SS/ DF) F-ratio (BSS/WSS) Between regimens (BSS) t-1= 9Calculated BSS BSS/t -1 F calc Within regimens (WSS) N-t=20 Total SS - BS WSS/N – t =S 2 = 2.17 totalN-1=29Total SS

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