Raxit - Anova.xlsx
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- Pages: 13
Age Satisfaction Level
1<
>1
<5
>5
23
0
0
1
0
24
1
0
0
0
25
0
0
0
0
26
0
0
0
0
27
0
1
0
0
28
0
0
0
0
29
0
0
0
0
30
2
0
1
0
31
0
2
0
0
32
0
3
1
0
33
3
1
1
1
34
1
2
0
0
35
3
3
1
2
Average
0.23
0.23
0.08
0.15
Degrees of freedom Between Sample = Number of Columns (-) 1 and Within Samples = Number of items (-) Number of Columns
The above table shows that the calculated value of F is 1.61 which is lesser than the table value of 2.80 at 5% level with v1 = 3 and v2 = 48 and hence could have arisen due to chance. The analysis support the null hypothesis. A higher P-Value P = 0.20,(P ≥ 0.05) indicates strong evidence in favour of null hypothesis. So the null hypothesis is accepted
H0: There is no significant between Work Experience and Satisfaction level H1: There is significant between Work Experience and Satisfaction level
Anova: Single Factor Average 0.25
SUMMARY
0.25
Groups
Count
Sum Average
Variance
0.00
Column 1
13
10
0.77
1.36
0.00
Column 2
13
12
0.92
1.41
0.25
Column 3
13
5
0.38
0.26
0.00
Column 4
13
3
0.23
0.36
SS
df
MS
F 1.61
0.00 0.75 0.50 1.00
ANOVA Source of Variation
1.50
Between Groups
4.0769
3
1.36
0.75
Within Groups
40.6154
48
0.85
Total
44.6923
51
2.25 0.17
P-value F crit 0.20
2.80
Age Satisfaction Level
20-25
26-30
31-36
Average
23
0
0
1
0.33
24
1
0
0
0.33
25
0
0
0
0.00
26
0
0
0
0.00
27
1
0
0
0.33
28
0
0
0
0.00
29
0
0
0
0.00
30
2
1
0
1.00
31
0
2
0
0.67
32
3
1
0
1.33
33
3
1
2
2.00
34
3
0
0
1.00
35
3
5
1
3.00
Average
0.23
0.38
0.08
0.23
Degrees of freedom Between Sample = Number of Columns (-) 1 and Within Samples = Number of items (-) Number of Columns
The above table shows that the calculated value of F is 1.94 which is lesser than the table value of 3.26 at 5% level with v1 = 2 and v2 = 36 and hence could have arisen due to chance. The analysis support the null hypothesis. A higher P-Value P = 0.16,(P ≥ 0.05) indicates strong evidence in favour of null hypothesis. So the null hypothesis is accepted
H0: There is no significant between Age and Satisfaction level H1: There is significant between Age and Satisfaction level
Anova: Single Factor SUMMARY Groups
Count
Sum
Average
Variance
Column 1
13
16
1.23
1.86
Column 2
13
10
0.77
2.03
Column 3
13
4
0.31
0.40
ANOVA Source of Variation Between Groups
SS
df
MS
5.54
2
2.77
Within Groups
51.38
36
1.43
Total
56.92
38
F
P-value 1.94
0.16
F crit 3.26
Age Engagement Level
25
26
27
28
% Employees falling under each Engagement level 39
0
1
0
0
40
0
1
1
0
41
2
1
0
1
42
1
3
0
4
43
1
2
4
4
44
0
1
1
3
45
1
0
1
0
46
0
0
0
0
47
0
0
0
1
Sum
5.00
9.00
7.00
13.00
Average
0.56
1.00
0.78
1.44
: Calculation : Calculation of Sum of Squares of variance between samples i.e. Particulars
Mean Mean of Sample Mean Mean (-) Mean of Sample Means (Mean (-) Mean of Sample Means)2 n(Mean (-) Mean of Sample Means)2
2004
2005
0.56 0.96 -0.41 0.17 1.49
2006
1.00 0.96 0.04 0.00 0.01
0.78 0.96 -0.19 0.03 0.31
SS Between is equal to S(n(Mean (-) Mean of Sample Means)2) = Calculation of Variance or Mean Square between Sample =
Calculation of Variance or Mean Square Within Sample =
Source of Variation Between Sample
1.44 0.96 0.48 0.23 2.09 4.148
MS Between = k = Number of Columns = 6 (Number of Years) MS Within n= number of items in all sample observationss
SS 4.15
2007
ANOVA TABLE MS
d.f
(6-1) =
5
0.83
Within Sample
65.78
Total
69.93
(54-6) =
48
1.37
53
Degrees of freedom Between Sample = Number of Columns (-) 1 and Within Samples = Number of items (-) Number of
The above table shows that the calculated value of F is 0.61 which is lesser than the table value of 2.41 at 5% level with v 1 = 5 hence could have arisen due to chance. The analysis support the null hypothesis of no difference of sample means. A higher P-Value (P ≥ 0.05) indicates strong evidence in favour of null hypothesis. So the null hypothesis is accepted
Age 29
30
Average
2004
der each Engagement level
2005
2006
2007
Square of deviations of items taken from respective Means
0
0
0.17
0.31
0.00
0.60
2.09
0
0
0.33
0.31
0.00
0.05
2.09
0
1
0.83
2.09
0.00
0.60
0.20
2
0
1.67
0.20
4.00
0.60
6.53
3
0
2.33
0.20
1.00
10.38
6.53
1
2
1.33
0.31
0.00
0.05
2.42
1
2
0.83
0.20
1.00
0.05
2.09
1
1
0.33
0.31
1.00
0.60
2.09
2
2
0.83
0.31
1.00
0.60
0.20
4.22
8.00
13.56
24.22
10.00
8.00
8.67
1.11
0.89
0.96
Sum->
Sum total of Sum of Square of Deviations of items taken from respective Means
SS Within
65.78
Calculation :
of variance between samples i.e. SS Between 2008
1.11 0.96 0.15 0.02 0.20
2009
0.89 Mean of Sample Means 0.96 -0.07 0.01 0.05 SS Between =
=
0.96
4.15
n=9; Number of Schemes
MS Between = SS between/k-1 = ns = 6 (Number of Years)
=(4.15)/(6-1)
=
0.83
= SS Within/(n - k) = ms in all sample observationss
= (65.78)/(54-6)
=
1.37
ANOVA TABLE MS 0.83
F-Ratio (F Value)
0.83/1.37
5% F-Limit (From the F-Table)
0.61 F(5,48) =
2.41
1.37
= Number of items (-) Number of Columns
alue of 2.41 at 5% level with v 1 = 5 and v2 = 48 and
rence of sample means.
ull hypothesis is accepted
2008
2009
taken from respective Means 1.23
0.79
1.23
0.79
1.23
0.01
0.79
0.79
3.57
0.79
0.01
1.23
0.01
1.23
0.01
0.01
0.79
1.23
8.89
6.89
s taken from respective Means
5.78
You can use the F statistic when deciding to support or reject the null hypothesis. In your F test results, you’ll have both an F value and an F critical value. The F critical value is what is referred to as the F statistic. In general, if your F statistic in a test is smaller than your F value, you can reject the null hypothesis. However, the statistic is only one measure of significance in an F Test. You should also consider the p value. The p value is determined by the F statistic and is the probability your results could have happened by chance.
The p-value is a number between 0 and 1 and interpreted in the following way: A small p-
value (typically ≤ 0.05) indicates strong evidence against the null hypothesis, so you reject the null hypothesis.
SUMMARY Standard Deviation
Co efficient of Variation
Consistancy Ranking
0.53
0.73
130.77
V
1.00
1.00
1.00
100.00
II
7
0.78
1.69
1.30
167.36
VI
9
13
1.44
3.03
1.74
120.47
IV
29
9
10
1.11
1.11
1.05
94.87
I
30
9
8
0.89
0.86
0.93
104.40
III
Age
Count
Sum
Average Variance
25
9
5
0.56
26
9
9
27
9
28
Interpretation
Since the CV is the most, the employees of 27 years of age shows consistancy with their engagement level at the least.
Since the CV is least, the employees of 29 years of age shows consistancy with their engagement level at the most.
1<
>1
<5
>5
23
0
0
1
0
24
1
0
0
0
25
0
0
0
0
26
0
0
0
0
27
0
1
0
0
28
0
0
0
0
29
0
0
0
0
30
2
0
1
0
31
0
2
0
0
32
0
3
1
0
33
3
1
1
1
34
1
2
0
0
35
3
3
1
2
Average
0.23
0.23
0.08
0.15
Degrees of freedom Between Sample = Number of Columns (-) 1 and Within Samples = Number of items (-) Number of Columns
The above table shows that the calculated value of F is 1.61 which is lesser than the table value of 2.80 at 5% level with v1 = 3 and v2 = 48 and hence could have arisen due to chance. The analysis support the null hypothesis. A higher P-Value P = 0.20,(P ≥ 0.05) indicates strong evidence in favour of null hypothesis. So the null hypothesis is accepted
H0: There is no significant between Work Experience and Satisfaction level H1: There is significant between Work Experience and Satisfaction level
Anova: Single Factor Average 0.25
SUMMARY
0.25
Groups
Count
Sum Average
Variance
0.00
Column 1
13
10
0.77
1.36
0.00
Column 2
13
12
0.92
1.41
0.25
Column 3
13
5
0.38
0.26
0.00
Column 4
13
3
0.23
0.36
SS
df
MS
F 1.61
0.00 0.75 0.50 1.00
ANOVA Source of Variation
1.50
Between Groups
4.0769
3
1.36
0.75
Within Groups
40.6154
48
0.85
Total
44.6923
51
2.25 0.17
P-value F crit 0.20
2.80
Age Satisfaction Level
20-25
26-30
31-36
Average
23
0
0
1
0.33
24
1
0
0
0.33
25
0
0
0
0.00
26
0
0
0
0.00
27
1
0
0
0.33
28
0
0
0
0.00
29
0
0
0
0.00
30
2
1
0
1.00
31
0
2
0
0.67
32
3
1
0
1.33
33
3
1
2
2.00
34
3
0
0
1.00
35
3
5
1
3.00
Average
0.23
0.38
0.08
0.23
Degrees of freedom Between Sample = Number of Columns (-) 1 and Within Samples = Number of items (-) Number of Columns
The above table shows that the calculated value of F is 1.94 which is lesser than the table value of 3.26 at 5% level with v1 = 2 and v2 = 36 and hence could have arisen due to chance. The analysis support the null hypothesis. A higher P-Value P = 0.16,(P ≥ 0.05) indicates strong evidence in favour of null hypothesis. So the null hypothesis is accepted
H0: There is no significant between Age and Satisfaction level H1: There is significant between Age and Satisfaction level
Anova: Single Factor SUMMARY Groups
Count
Sum
Average
Variance
Column 1
13
16
1.23
1.86
Column 2
13
10
0.77
2.03
Column 3
13
4
0.31
0.40
ANOVA Source of Variation Between Groups
SS
df
MS
5.54
2
2.77
Within Groups
51.38
36
1.43
Total
56.92
38
F
P-value 1.94
0.16
F crit 3.26
Age Engagement Level
25
26
27
28
% Employees falling under each Engagement level 39
0
1
0
0
40
0
1
1
0
41
2
1
0
1
42
1
3
0
4
43
1
2
4
4
44
0
1
1
3
45
1
0
1
0
46
0
0
0
0
47
0
0
0
1
Sum
5.00
9.00
7.00
13.00
Average
0.56
1.00
0.78
1.44
: Calculation : Calculation of Sum of Squares of variance between samples i.e. Particulars
Mean Mean of Sample Mean Mean (-) Mean of Sample Means (Mean (-) Mean of Sample Means)2 n(Mean (-) Mean of Sample Means)2
2004
2005
0.56 0.96 -0.41 0.17 1.49
2006
1.00 0.96 0.04 0.00 0.01
0.78 0.96 -0.19 0.03 0.31
SS Between is equal to S(n(Mean (-) Mean of Sample Means)2) = Calculation of Variance or Mean Square between Sample =
Calculation of Variance or Mean Square Within Sample =
Source of Variation Between Sample
1.44 0.96 0.48 0.23 2.09 4.148
MS Between = k = Number of Columns = 6 (Number of Years) MS Within n= number of items in all sample observationss
SS 4.15
2007
ANOVA TABLE MS
d.f
(6-1) =
5
0.83
Within Sample
65.78
Total
69.93
(54-6) =
48
1.37
53
Degrees of freedom Between Sample = Number of Columns (-) 1 and Within Samples = Number of items (-) Number of
The above table shows that the calculated value of F is 0.61 which is lesser than the table value of 2.41 at 5% level with v 1 = 5 hence could have arisen due to chance. The analysis support the null hypothesis of no difference of sample means. A higher P-Value (P ≥ 0.05) indicates strong evidence in favour of null hypothesis. So the null hypothesis is accepted
Age 29
30
Average
2004
der each Engagement level
2005
2006
2007
Square of deviations of items taken from respective Means
0
0
0.17
0.31
0.00
0.60
2.09
0
0
0.33
0.31
0.00
0.05
2.09
0
1
0.83
2.09
0.00
0.60
0.20
2
0
1.67
0.20
4.00
0.60
6.53
3
0
2.33
0.20
1.00
10.38
6.53
1
2
1.33
0.31
0.00
0.05
2.42
1
2
0.83
0.20
1.00
0.05
2.09
1
1
0.33
0.31
1.00
0.60
2.09
2
2
0.83
0.31
1.00
0.60
0.20
4.22
8.00
13.56
24.22
10.00
8.00
8.67
1.11
0.89
0.96
Sum->
Sum total of Sum of Square of Deviations of items taken from respective Means
SS Within
65.78
Calculation :
of variance between samples i.e. SS Between 2008
1.11 0.96 0.15 0.02 0.20
2009
0.89 Mean of Sample Means 0.96 -0.07 0.01 0.05 SS Between =
=
0.96
4.15
n=9; Number of Schemes
MS Between = SS between/k-1 = ns = 6 (Number of Years)
=(4.15)/(6-1)
=
0.83
= SS Within/(n - k) = ms in all sample observationss
= (65.78)/(54-6)
=
1.37
ANOVA TABLE MS 0.83
F-Ratio (F Value)
0.83/1.37
5% F-Limit (From the F-Table)
0.61 F(5,48) =
2.41
1.37
= Number of items (-) Number of Columns
alue of 2.41 at 5% level with v 1 = 5 and v2 = 48 and
rence of sample means.
ull hypothesis is accepted
2008
2009
taken from respective Means 1.23
0.79
1.23
0.79
1.23
0.01
0.79
0.79
3.57
0.79
0.01
1.23
0.01
1.23
0.01
0.01
0.79
1.23
8.89
6.89
s taken from respective Means
5.78
You can use the F statistic when deciding to support or reject the null hypothesis. In your F test results, you’ll have both an F value and an F critical value. The F critical value is what is referred to as the F statistic. In general, if your F statistic in a test is smaller than your F value, you can reject the null hypothesis. However, the statistic is only one measure of significance in an F Test. You should also consider the p value. The p value is determined by the F statistic and is the probability your results could have happened by chance.
The p-value is a number between 0 and 1 and interpreted in the following way: A small p-
value (typically ≤ 0.05) indicates strong evidence against the null hypothesis, so you reject the null hypothesis.
SUMMARY Standard Deviation
Co efficient of Variation
Consistancy Ranking
0.53
0.73
130.77
V
1.00
1.00
1.00
100.00
II
7
0.78
1.69
1.30
167.36
VI
9
13
1.44
3.03
1.74
120.47
IV
29
9
10
1.11
1.11
1.05
94.87
I
30
9
8
0.89
0.86
0.93
104.40
III
Age
Count
Sum
Average Variance
25
9
5
0.56
26
9
9
27
9
28
Interpretation
Since the CV is the most, the employees of 27 years of age shows consistancy with their engagement level at the least.
Since the CV is least, the employees of 29 years of age shows consistancy with their engagement level at the most.
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