by shaimaa elkadi supervised by dr.amal fatani. from the previous one way anova but
TRANSCRIPT
By
Shaimaa Elkadi
Supervised by
Dr.Amal Fatani
From the previous One Way ANOVA
But
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 xi and xj at P = 0.05.
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% allowance= t √(S2(1/ni+1/nj))
Tabulated t value P=0.05DF= N-t (of S2)One orTwo tails
The pooled variance from ANOVA calculation
ni, nj
The number of observations from which means are determined
Apply on the previous example
Ranked meansB A,C I J F,G D,H E1 2.3 5 5.3 5.7 6.3 9.3
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
5% allowance= (Q/√2) √(S2(1/ni+1/nj))
Studentized range value atK= number of treatmentsDF= N-t (of S2)
Apply on the previous example
Ranked meansB A,C I J F,G D,H E1 2.3 5 5.3 5.7 6.3 9.3
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
5% allowance= td √(S2(1/ni+1/nj))
Dunnett’s td value atK= number of treatments – 1 =t-1DF= N-t (of S2)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
For the exclusion of the control group
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 meansB A,C I J F,G D,H E1 2.3 5 5.3 5.7 6.3 9.3
5% allowance = 3.55
B showed a statistically significant smaller weight gain than J
E showed a statistically significant larger weight gain than J
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 (td at P= 0.05) or one sided (td at P= 0.1) comparisons, according to the experiment design
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
Apply on the previous example
S2 = 2.17 DF = 20 ni, nj = 3 t (P=0.05,DF=20,two tailed)= 2.086
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 √(S2(1/ni+1/nj))
Apply on the previous example
K=10 DF = 20 Q = 5.01
So 5% allowance = 4.26
Thus any 2 means differ by more than 4.26 are significantly different from each other (differ in group)
5% allowance= (Q/√2) √(S2(1/ni+1/nj))
Apply on the previous example
K=10 – 1 = 9 DF = 20 td (P=0.05, two tailed)= 2.95
So 5% allowance = 3.55
5% allowance= td √(S2(1/ni+1/nj))
Source of variation
DF Sum of squares (SS)
Mean Square(SS/ DF)
F-ratio(BSS/WSS)
Between regimens(BSS)
t-1= 9 Calculated BSS
BSS/t -1 F calc
Within regimens(WSS)
N-t=20
Total SS - BS
WSS/N – t
=S2
=2.17total N-1=29 Total SS