Kurtosis and Skewness Test

Instructions:

Brainmass instructions:

1.) These Means tested stats need a simple Kurtosis and skewness Test.
Please add in the graphs as gif or Jpeg.

2.) Please add some explanatory commentary, if possible! Thanks!

3.) I need to know which software package (version and year) the graphs
came from!

THANKS!

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Ho is that there would be no significant difference between imagery
facilitated instructed group and the control group.

H1 is that there would be a significant difference between imagery
facilitated instructed group and the control group.

Table 1

Average Marking μ Score Results for Control Group and Imagery Group on
various making criteria relating to “Marking Results� in Appendix

Question / Marking Criterion Control Group Imagery Group


Day 1 Day 5 Day 1 Day 5

Overall Quality of 2.6 3 2.4 3.7

Movement from start to end.

Correct Postural Alignment 2.4 2.5 2.6 3.5

Correct Slow Speed 2.6 2.75 4.2 4.6

Correct Sequence 3.4 3.5 2.6 3.7

Fluidity of Movement 3.2 3.75 2.5 3.6

Observed ease of 3.2 3.5 2.4 3.7

implementing instruction

Observed Attention 4 4 2.6 3.5

Each figure represents the marking means taken from all participants of
each groups on that criteria on either day one or day five. :Mean μ=
sum ( of marks / N of participants in the group

Minimum score that could be reached was 1, maximal .score was 5, see
“objective observational marking of videoed evidence in Methodology

Solution.

Taking average marking score for Control Group and Imagery Group by μ1
and μ2, respectively.

The null hypothesis H0: μ 1= μ2; with H1: μ1 is not equal to μ2.
Data is non parametric which suggests Mann Whitney Means analysis.

Testing if there has been any difference between groups on day one.

Data Source Rank

2.4 2 1

2.4 1 2

2.4 2 3

2.5 2 4

2.6 2 5

2.6 2 6

2.6 2 7

2.6 1 8

2.6 1 9

3.2 1 10

3.2 1 11

3.4 1 12

4.0 1 13

4.2 2 14

The sizes of two samples are n1=n2=7

The sum of the rank for the sample�Control Group� , denoted by W1,
is

W1=2+8+9+10+11+12+13=65, and the sum of the rank for the
sample�Imagery Group� , denoted by W2 is W2=1+3+4+5+6+7+14=40. From
W1 and W2, take

U1=n1*n2+1/2*n1*(n1+1)-W1=7*7+1/2*7*8-65=49+28-65=12

U2= n1*n2+1/2*n2*(n2+1)-W1=7*7+1/2*7*8-40=49+28-40=37

U=Min(U1,U2)=Min(12,37)=12.

Given the significance level alpha=0.05, U_alpha is (7,7)=9. Since
U=12>9,

null hypothesis can not be rejected .

there is no difference between the average marking score for Control
Group and Imagery Group on day 1.

(2) Testing if there has been any difference between groups on day 5.

Data Source Rank

2.5 1 1

2.75 1 2

3 1 3

3.5 1 4

3.5 1 5

3.5 2 6

3.5 2 7

3.6 2 8

3.7 2 9

3.7 2 10

3.7 2 11

3.75 1 12

4.0 1 13

4.6 2 14

The sizes of two samples are n1=n2=7

Sum of the rank for the sample�Control Group� , denoted by W1 is

W1=1+2+3+4+5+12+13=40, and the sum of the rank for the sample�Imagery
Group� , denoted by W2 is W2=1/2*14*(14+1)-W1=65. From W1 and w2 it
follows that

U1=n1*n2+1/2*n1*(n1+1)-W1=7*7+1/2*7*8-40=49+28-40=37

U2= n1*n2+1/2*n2*(n2+1)-W1=7*7+1/2*7*8-65=49+28-65=12

U=Min(U1,U2)=Min(37,12)=12.

Given the significance level alpha=0.05, U_alpha is (7,7)=9. Since
U=12>9,

null hypothesis can not be rejected.

There is no difference between the average marking score for Control
Group and Imagery Group on day 5.

Testing if there has been any difference between groups overall.

Average of day1 and day5, and get the following

Table 1

Average Marking Score Results for Control Group and Imagery Group on
various making criteria relating to “Marking Results� in Appendix

Question / Marking Criterion Control Group Imagery Group


Day 1 Day 5 Day 1 Day 5

Overall Quality of (2.6+3)/2=2.8
(2.4 +3.7)/2=3.05

Movement from start to end.

Correct Postural Alignment 2.45 3.05

Correct Slow Speed 2.675 4.4

Correct Sequence 3.45 3.15

Fluidity of Movement 3.475 3.05

Observed ease of 3.35 3.05

implementing instruction

Observed Attention 4 3.05

We start by arranging the data jointly in ascending order and keep track
of which sample each point originated from the above table.

Data Source Rank

2.45 1 1

2.675 1 2

2.8 1 3

3.05 2 4

3.05 2 5

3.05 2 6

3.05 2 7

3.05 2 8

3.15 2 9

3.35 1 10

3.45 1 11

3.475 1 12

4.0 1 13

4.4 2 14

The sizes of two samples are n1=n2=7

From the sum of the rank for the sample�Control Group� , denoted by
W1,

W1=1+2+3+10+11+12+13=52, and the sum of the rank for the
sample�Imagery Group� denoted by W2, W2=1/2*14*(14+1)-W1=53. one
gets

U1=n1*n2+1/2*n1*(n1+1)-W1=7*7+1/2*7*8-52=49+28-52=25

U2= n1*n2+1/2*n2*(n2+1)-W1=7*7+1/2*7*8-53=49+28-53=24

U=Min(U1,U2)=Min(25,24)=24.

Given the significance level alpha=0.05, we have U_alpha(7,7)=9. Since
U=24>9,

the null hypothesis can not be rejected.

There is no difference between the average marking score for Control
Group and Imagery Group on overall.

From the analysis above, we conclude that there are no difference
between the average marking score for Control Group and Imagery Group.

________________________________________________________________________
________________________________________________________________________
__________

                     
Day1          Day1               
 Day5       Day5                    Overall

U-value                     9
9                       
9             9 9

U-observation         12               12   Â
 Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â 12 12 24 

                          M         

SD                     M             S
D                 M            SD

 

Control Group    3.06      
0.562                3.286       
 0.548              3.171       0.547

 

Imagery Group   2.76       0.634               
3.757         0.382              3.257      
0.505

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significance level alpha=0.05

 

There was no significant difference between the means of the Control
Group and the Imagery Group on day1, day5 and overall.