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