This file was generated by AnalyzeEverything (www.norrsken-data-teknik.com)

The Original Data Tables:

NoteLeistung [%]AnzahlDatumMartin
624.2122003-06-16
528.722003-06-265
276.712003-06-16
359.312003-06-16
441.922003-06-16
192.602003-06-16
6 62003-06-24
5 22003-06-245
4 22003-06-24
3 42003-06-24
2 22003-06-24
1 22003-06-24
6 22004-05-05
5 12004-05-05
4 32004-05-05
3 42004-05-05
2 42004-05-05
1 02004-05-05


The Analyzing Results:


The Chi-Square Test

The data distribution Σ Anzahl = f ( Note ) has been tested against the binomial distribution with the parameters
nmax. class index=18
p0=0.1844
sumΣ Anzahl=50.0

(This has been calculated by the parameters
mean-argNote=4.32
standard-deviationNote=1.6424)

The chi-square test with the
significance level 0.95
(error probability 5.0%)
and the degree of freedom m = n -1
(manually changed the number of classes, class borders)
mNote = 5
(reduced by 2 because the mean-value and the standard-deviation are unknown and hence calculated)
produces the chi-square values
chi2Σ Anzahl = 34.291
chi20.05, 3 = 0.37.

If the condition
chi2Σ Anzahl < chi20.05, 3
is true, the assumption that the distribution is binomial can be accepted.

So the theory, Note is randomly distributed, must be rejected.

The data distribution is

NoteΣ Anzahl [%]Σ Anzahlestimated:
Σ Anzahl
1.04.02.01.2739
2.014.07.05.1858
3.018.09.09.9689
4.014.07.012.024
5.010.05.010.198
6.040.020.06.4576



 Image: Hanisch_0p.jpg
 

 Image: Hanisch_0f.jpg
 

 Image: Hanisch_0a.jpg


The Chi-Square Test

The data distribution Σ Anzahl = f ( Note ) has been tested against the binomial distribution with the parameters
nmax. class index=18
p0=0.1844
sumΣ Anzahl=50.0

(This has been calculated by the parameters
mean-argNote=4.32
standard-deviationNote=1.6424)

The assumption is now: The data is not randomly distributed.

The chi-square test with the
significance level 0.95
(error probability 5.0%)
and the degree of freedom m = n -1
(manually changed the number of classes, class borders)
mNote = 5
(reduced by 2 because the mean-value and the standard-deviation are unknown and hence calculated)
produces the chi-square values
chi2Σ Anzahl = 34.291
chi20.95, 3 = 7.82.

If the condition
chi20.95, 3 < chi2Σ Anzahl
is true, the assumption that the distribution is not binomial can be accepted.

So the theory, Note is not randomly distributed, can be accepted.

The data distribution is

NoteΣ Anzahl [%]Σ Anzahlestimated:
Σ Anzahl
1.04.02.01.2739
2.014.07.05.1858
3.018.09.09.9689
4.014.07.012.024
5.010.05.010.198
6.040.020.06.4576



 Image: Hanisch_1p.jpg
 

 Image: Hanisch_1f.jpg
 

 Image: Hanisch_1a.jpg


The STUDENT t-test

The data distribution Σ Anzahl = f ( Note ) has been tested against the binomial distribution with the parameters
nmax. class index=18
p0=0.1844
sumΣ Anzahl=50.0

(This has been calculated by the parameters
mean-argNote=4.32
standard-deviationNote=1.6424)

The STUDENT t-test with the
significance level 0.95
(error probability 5.0%)
and the degree of freedom m = n -1
(manually changed the number of classes, class borders)
mNote = 5
produces the t-test values
TΣ Anzahl = 76.677
t0.95, 5 = 2.59.

If the condition
TΣ Anzahl < t 0.95, 5
is true, the assumption that the distribution is binomial can be accepted.

So the theory, Note is randomly distributed, must be rejected.

The data distribution is

NoteΣ Anzahl [%]Σ Anzahlestimated:
Σ Anzahl
1.04.02.01.2739
2.014.07.05.1858
3.018.09.09.9689
4.014.07.012.024
5.010.05.010.198
6.040.020.06.4576



 Image: Hanisch_2p.jpg
 

 Image: Hanisch_2f.jpg
 

 Image: Hanisch_2a.jpg