I use the following small data set with three groups:
DATA
Group one Group two Group three
2 12 17
4 13 18
6 14 19
8 15 21
16 22
23
Overall descriptive statistics
Group N Sum Mean Std. Deviation Variance SS*
ALL 15 210 14. 6.536 42.714 598
*SS=Sum of Squares of deviations from the mean=(N-1)*Variance
GROUP STATISTICS
Group N Sum Mean Std Deviation Variance SS
One 4 20 5. 2.582 6.667 20
Two 5 70 14. 1.581 2.5 10
Three 6 120 20. 2.366 5.6 28
ANOVA
Response
Sum of Squares df Mean Square F Sig.
Between Groups 540. 2 270 55.862 .000
Within Groups 58 12 4.833
Total 598. 14
The total sum of squares (of deviations from the overall mean) = 598
The within group sum of squares = 20 + 10 + 28 = 58
This can be thought of as the noise or random part, because there is no other explanation why the 4 observations in group 1 are (2 4 6 8) rather than (5 5 5 5) etc. for the other groups.
The between groups sum of squares = 598 – 58 = 540
It can also be written as 4(5-14)**2 +5 (14-14)**2 + 6(20-14)**2=540
The reason why the 3 group means are (5 14 20) rather than (14 14 14) is the fact that they belong to different groups.
We say the total sum of squares is split into Between Group and Within Group sums of squares.
The degrees of freedom between groups is 3-1=number of groups-1
The degrees of freedom within groups = (4-1)+(5-1)+(6-1)=12
The total degrees of freedom is 15-1=14
Mean Square = Sum of squares/df
MS Between groups: 540/2=270
MS Within groups: 58/12 = 4.833
F=270/4.833 = 55.862
This F can be thought of as a Signal to Noise ratio
I use the following small data set with three groups:
DATA
Group one Group two Group three
2 12 17
4 13 18
6 14 19
8 15 21
16 22
23
Overall descriptive statistics
Group N Sum Mean Std. Deviation Variance SS*
ALL 15 210 14. 6.536 42.714 598
*SS=Sum of Squares of deviations from the mean=(N-1)*Variance
GROUP STATISTICS
Group N Sum Mean Std Deviation Variance SS
One 4 20 5. 2.582 6.667 20
Two 5 70 14. 1.581 2.5 10
Three 6 120 20. 2.366 5.6 28
ANOVA
Response
Sum of Squares df Mean Square F Sig.
Between Groups 540. 2 270 55.862 .000
Within Groups 58 12 4.833
Total 598. 14
The total sum of squares (of deviations from the overall mean) = 598
The within group sum of squares = 20 + 10 + 28 = 58
This can be thought of as the noise or random part, because there is no other explanation why the 4 observations in group 1 are (2 4 6 8) rather than (5 5 5 5) etc. for the other groups.
The between groups sum of squares = 598 – 58 = 540
It can also be written as 4(5-14)**2 +5 (14-14)**2 + 6(20-14)**2=540
The reason why the 3 group means are (5 14 20) rather than (14 14 14) is the fact that they belong to different groups.
We say the total sum of squares is split into Between Group and Within Group sums of squares.
The degrees of freedom between groups is 3-1=number of groups-1
The degrees of freedom within groups = (4-1)+(5-1)+(6-1)=12
The total degrees of freedom is 15-1=14
Mean Square = Sum of squares/df
MS Between groups: 540/2=270
MS Within groups: 58/12 = 4.833
F=270/4.833 = 55.862
This F can be thought of as a Signal to Noise ratio
When u have two vaiables,
one catagorial independent variable with three or or more distinct catagories.this can be a continuous variable that has been recoded to a given three equal groups
----and one continuous dependent variable
then u can use one way between group annova with post-hoc testa.
----it will tell u whether there are significant difference in the mean scores on the dependent variable across the three groups
- Badges
- Science topic