Download docx - Tugas Individu Normality n Homogenity

I. Testing Normality for scores at class A

Example:

A lecturer was recapitalize test scores of learners. The total number of learners is 100

people. He make a frequency table to see how much value obtained by learners. But before, he

wanted to test normality of the data obtained

Solution:

Testing Hypotheses

Ho : Data is from normal distribution

H1 : Data is not from normal distribution

Scores Students

28-35 2

36-43 5

44-51 9

52-59 14

60-67 12

68-75 13

76-83 15

84-91 19

92-99 11

Total 100

Using Chi-Square

N

oKelas- Interval xi xi

2 fi fixi fixi2

1 28-35 31,5 992.25 2 63 1984.5

2 36-43 39,5 1560.25

5 197.5 7801.25

3 44-51 47,5 2256.25

9 427.5 20306.25

4 52-59 55,5 3080.25

14 777 43123.5

5 60-67 63,5 4032.25

12 762 48387

6 68-75 71,5 5112.25

13 929.5 66459.25

7 76-83 79,5 6320.25

15 1192.5 94803.75

8 84-91 87,5 7656.25

19 1662.5 145468.8

9 92-99 95,5 9120.25

11 1050.5 100322.8

Total 100 7062 528657

Calculate the mean

x=∑ f i. x i

∑ f i

x=7062100

=70.62

Calculate the standard deviation

S=√ N∑ f X i2−(∑ f X i )

2

N (N−1)

S=√ 100(528657)− (7062 )2

100 x 99

S=√ 52865700−498718449900

S=√ 29938569900

=17.3

Make a frequency distribution table

Limit of

Class (x)

Z for

limit of

class

Luas O-

Z

Large of

Interval Class

Frekuensi yang

diharapkan (Ei)

Observation

Frequency (Oi)X2

27.5 -2.49 0.4936

35.5 -2.03 0.4778 0.0158 1.58 20.112

43.5 -1.57 0.4418 0.036 3.6 50.544

51.5 -1.11 0.3665 0.0453 4.53 94.411

59.5 -0.64 0.2389 0.1276 12.76 140.121

67.5 -0.18 0.0714 0.1675 16.75 121.347

75.5 0.28 0.1103 0.1817 18.17 131.471

83.5 0.74 0.2704 0.1601 16.01 150.064

91.5 1.21 0.3869 0.1165 11.65 194.637

99.5 1.67 0.4525 0.0656 6.56 113.005

TOTAL 100 15.711

Where:

z=lim ofclass −xs

Ei = Large Each of Class Interval ¿ n.

χ2=∑i=1

k (Oi−Ei )2

Ei

χ2=15 .711

With significant value so χ2table = χ

2(1-α)(k-3) = χ2

(0,95)(97) = 124,3. From this calculation, we

know that χ2calculation< χ

2table so H0 accepted. The conclusion is sample is from normal

distribution.

Analysis using SPSS

Case Processing Summary

Name

Cases

Valid Missing Total

N Percent N Percent N Percent

Value Students 100 100.0% 0 .0% 100 100.0%

Descriptivesa

Name Statistic Std. Error

Value Students Mean 70.6100 1.71987

95% Confidence Interval for

Mean

Lower Bound 67.1974

Upper Bound 74.0226

5% Trimmed Mean 70.9222

Median 73.0000

Variance 295.796

Std. Deviation 1.71987E1

Minimum 34.00

Maximum 99.00

Range 65.00

Interquartile Range 29.75

Skewness -.204 .241

Kurtosis -.875 .478

Tests of Normalityb

Name

Kolmogorov-Smirnova Shapiro-Wilk

Statistic df Sig. Statistic df Sig.

Value Students .089 100 .051 .968 100 .016

a. Lilliefors Significance Correction

b. There are no valid cases for Value when Name = .000. Statistics cannot be computed for this

level.

We can see the results of SPSS analysis by Kolmogorov-Smirnov significance level> 0.05

ie 0.051 but for the Shapiro-Wilk stated significance level< 0.05 is 0.016. We ca use analysis by

Kolmogorov-Smirnov. The criteria receipt of H0 if the level of significance> 0.05. So in this

case H0 received stated that the sample data derived from a normal distribution population.

II. Testing Normality and Homogenity of two samples(Class A and Class B)

Example:

A lecturer was to compare the value of class A and class B. Lecturers gave equal treatment

to each class, and then recapitalize the value. Now the lecturer will conduct normality test of the

value of both classes.

Solution:

Testing Hypotheses of class A



Data of statistics score in Class A

Score Students

30 - 39 2

40 - 49 5

50 - 59 8

60 - 69 8

70 - 79 11

80 - 89 11

90 - 99 5

Total 50

Testing Hypotheses of class B


H1 :Data is not from normal distribution

Data of statistics score in Class B

Score Student

30 - 39 2

40 - 49 7

50 - 59 6

60 - 69 8

70 - 79 11

80 - 89 9

90 - 99 7

Total 50

Calculation Data for Class A ( using Chi-Square)

No Kelas- Interval xi xi2 fi fixi fixi

2

1 30 - 39 34.5 1190.25 2 69 2380.5

2 40 - 49 44.5 1980.25 5 222.5 9901.25

3 50 - 59 54.5 2970.25 8 436 23762

4 60 - 69 64.5 4160.25 8 516 33282

5 70 - 79 74.5 5550.25 11 819.5 61052.75

6 80 - 89 84.5 7140.25 11 929.5 78542.75

7 90 - 99 94.5 8930.25 5 472.5 44651.25

Total 50 3465 253572.5

Calculation for mean

x A=∑ f i . x i

∑ f i

x A=3465

50=69.3

Calculation for standard deviation

sA=√ N ∑ f X i2−(∑ f X i )

2

N (N−1)

sA=√ 50 (253572.5)−(3465 )2

50 x 49

sA=√ 12678600−120062252450

sA=√ 6723752450

=16.56


Limit of

Class (x)

Z for

limit of

class

Luas O-

Z

Large of

Interval

Class

Frekuensi

yang

diharapkan

(Ei)

Obsevation

Frequency

(Oi)

X2

29.5 -2.40 0.4918

39.5 -1.80 0.4641 0.0277 1.39 2 0.26849.5 -1.20 0.3849 0.0792 3.96 5 0.27359.5 -0.59 0.2224 0.1625 8.13 8 0.00269.5 0.01 0.0040 0.2264 11.32 8 0.97479.5 0.62 0.2324 0.2284 11.42 11 0.01589.5 1.22 0.3888 0.1564 7.82 11 1.29399.5 1.82 0.4656 0.0768 3.84 5 0.350

TOTAL 50 3.176

Where:

z=lim of class −xs


χ2=∑i=1

k (Oi−Ei )2

Ei

χ2=3 .176


2(1-α)(k-3) = χ2

(0,95)(47) = 67.5. From this calculation, we know

that χ2calculation< χ

2table so H0 accepted. The conclusion is sample of score statistics at class A is

from normal distribution.

Calculation Data for Class B ( using Chi-Square)


2

1 30 - 39 34.5 1190.25 2 69 2380.52 40 - 49 44.5 1980.25 7 311.5 13861.753 50 - 59 54.5 2970.25 6 327 17821.54 60 - 69 64.5 4160.25 8 516 332825 70 - 79 74.5 5550.25 11 819.5 61052.756 80 - 89 84.5 7140.25 9 760.5 64262.257 90 - 99 94.5 8930.25 7 661.5 62511.75

Total 50 3465 255172.5

Calculation for mean

xB=∑ f i . x i

∑ f i

xB=3465

50=69.3


sB=√ N ∑ f X i2−(∑ f X i )

2

N (N−1)

sB=√ 50 (255172.5)−(3465 )2

50 x49

sB=√ 12758625−120062252450

sB=√ 7524002450

=17.52


Limit of

Class (x)

Z for limit

of classLuas O-

Z

Large of

Interval

Class

Frekuensi yang

diharapkan (Ei)

Obsevation

Frequency

(Oi)

X2

29.5 -2.27 0.4884

39.5 -1.70 0.4554 0.033 1.65 2 0.07449.5 -1.13 0.3708 0.0846 4.23 7 1.81459.5 -0.56 0.2123 0.1585 7.925 6 0.46869.5 0.01 0.0040 0.2163 10.815 8 0.73379.5 0.58 0.2190 0.215 10.75 11 0.00689.5 1.15 0.3749 0.1559 7.795 9 0.18699.5 1.72 0.4573 0.0824 4.12 7 2.013

TOTAL 50 5.294

Where:



χ2=∑i=1

k (Oi−Ei )2

Ei

χ2=5 .294


2(1-α)(k-3) = χ2


that χ2calculation< χ

2table so H0 accepted. The conclusion is sample of score statistics at class B is


III. Testing Homogeneity Class A and Class B

Example:

Now a professor of tests done before, lecturers will compare the value of kedual class and

will conduct a second test of homogeneity of data

Solution:

H0 : Population have the same variance

H1 : Population haven’t the same variance

NO Class A Class BNilai (xi)

x i−x ( x i−x )2 Nilai (xi) x i−x ( x i−x )2

1 34 -30 900 34 -29.8 888.042 34 -30 900 34 -29.8 888.043 34 -30 900 34 -29.8 888.044 34 -30 900 34 -29.8 888.045 34 -30 900 34 -29.8 888.046 34 -30 900 34 -29.8 888.047 34 -30 900 34 -29.8 888.048 34 -30 900 34 -29.8 888.049 34 -30 900 34 -29.8 888.0410 34 -30 900 34 -29.8 888.0411 55 -9 81 54 -9.8 96.0412 55 -9 81 54 -9.8 96.0413 55 -9 81 54 -9.8 96.0414 55 -9 81 54 -9.8 96.0415 55 -9 81 54 -9.8 96.0416 55 -9 81 54 -9.8 96.0417 55 -9 81 54 -9.8 96.0418 55 -9 81 54 -9.8 96.0419 55 -9 81 54 -9.8 96.0420 55 -9 81 54 -9.8 96.0421 67 3 9 68 4.2 17.6422 67 3 9 68 4.2 17.6423 67 3 9 68 4.2 17.6424 67 3 9 68 4.2 17.6425 67 3 9 68 4.2 17.6426 67 3 9 68 4.2 17.6427 67 3 9 68 4.2 17.6428 67 3 9 68 4.2 17.6429 67 3 9 68 4.2 17.6430 67 3 9 68 4.2 17.6431 76 12 144 76 12.2 148.8432 76 12 144 76 12.2 148.8433 76 12 144 76 12.2 148.8434 76 12 144 76 12.2 148.8435 76 12 144 76 12.2 148.8436 76 12 144 76 12.2 148.8437 76 12 144 76 12.2 148.8438 76 12 144 76 12.2 148.8439 76 12 144 76 12.2 148.8440 76 12 144 76 12.2 148.8441 88 24 576 87 23.2 538.24

42 88 24 576 87 23.2 538.2443 88 24 576 87 23.2 538.2444 88 24 576 87 23.2 538.2445 88 24 576 87 23.2 538.2446 88 24 576 87 23.2 538.2447 88 24 576 87 23.2 538.2448 88 24 576 87 23.2 538.2449 88 24 576 87 23.2 538.2450 88 24 576 87 23.2 538.24

Total 3200 0 17100 3190 -4.3x10-13 16888Average 64 63.8

Variance for Class A

s2=∑ ( xi−x )2

n−1=17100

49=384 .9

Variance for Class B

s2=∑ ( x i−x )2

n−1=16888

49=344 . 6

F=Smallest VarianceBigget Variance

=384 . 9344 .6

=1 .116

With significant value 0,05 dan v1 = v2 = 49 so Ftable is F(0,05)(49,49) = 1,61. The testing criteria

is H0 rejected jika F¿ F(0,05)(49,49) , for others H0 daccepted. From calculation F = 1.116, while at

table = 1,60. It means that F < F(0,05)(49,49). So H0 is accepted. The conclusion is Population have

the same variance,

Analysis using SPSS


Class

Cases

Valid Missing Total


Value Class A 50 100.0% 0 .0% 50 100.0%

Class B 50 100.0% 0 .0% 50 100.0%

Descriptivesa

Class Statistic Std. Error

Value Class A Mean 70.7800 2.41159


Mean

Lower Bound 65.9337

Upper Bound 75.6263


Median 73.5000

Variance 290.787


Minimum 34.00

Maximum 99.00

Range 65.00


Skewness -.269 .337

Kurtosis -.828 .662

Class B Mean 70.4400 2.47702


Mean

Lower Bound 65.4622

Upper Bound 75.4178


Median 71.5000

Variance 306.782


Minimum 34.00

Maximum 99.00

Range 65.00


Skewness -.149 .337

Kurtosis -.882 .662

a. There are no valid cases for Value when Class = .000. Statistics cannot be computed for this level.

Tests of Normalityb

Class



Value Class A .109 50 .188 .964 50 .127

Class B .081 50 .200* .969 50 .206


We can see the results for Class A of SPSS analysis by Kolmogorov-Smirnov significance

level> 0.05 ie 0.188 as well as the Shapiro-Wilk stated significance level> 0.05 is 0.127.

The criteria receipt of H0 if the level of significance> 0.05. So in this case H0 received

stated that the sample data of Class A derived from a normal distribution population.

Than see the results for Class B of SPSS analysis by Kolmogorov-Smirnov significance

level> 0.05 ie 0.200 as well as the Shapiro-Wilk stated significance level> 0.05 is 0.206.

The criteria receipt of H0 if the level of significance> 0.05. So in this case H0 received

stated that the sample data of Class B derived from a normal distribution population.

Test of Homogeneity of Variancea

Levene Statistic df1 df2 Sig.

Value Based on Mean .013 1 98 .910

Based on Median .024 1 98 .876

Based on Median and with

adjusted df.024 1 97.998 .876

Based on trimmed mean .014 1 98 .905

a. There are no valid cases for Value when Class = .000. Statistics cannot be computed for

this level.

Test Criteria:

Value Sig. Or significance or probability value <0.05, data derived from populations that have a

variance that is not the same.

Value Sig. Or significance or probability value> 0.05, data derived from populations that have

the same variance.

In the output above shows that the level of significance or probability value of the mean

(average) which is above 0.05 (0.910 bigger than 0.05). So soundly if the basic measurement is

the median data, numbers Sig. Is 0.876, which is still above 0.05. So in this case H0 is accepted

and stated that these populations have the same variance.

IV. Testing Normality and Homogeneity of three samples

(Subject A, Subject B and Subject C)

Example:

A student recapitalize the value he can from the three eyes of different subjects. He made a

table for subject A, subject B and subject C Therefore, students will conduct the third test against

normality test data from three different subjects.

Solution:

Data for subject A

Testing Hypotheses of

class A

Ho : Data is from normal

distribution

H1 : Data is not from

normal distribution

Score Student

34-44 4

45-55 3

56-66 5

67-77 6

78-88 8

89-99 4

Total 30

Data for subject B


class B


distribution


normal distribution

Score Student

39-48 3

49-58 7

59-68 4

69-78 6

79-88 5

89-98 5

Total 30

Data for subject C


class C


distribution


normal distribution

Score Student

34-44 2

45-55 6

56-66 7

67-77 13

78-88 6

89-99 6

Total 40Testing Normality for Subject A


2

1 34-44 39 1521 4 156 6084

2 45-55 50 2500 3 150 7500

3 56-66 61 3721 5 305 18605

4 67-77 72 5184 6 432 31104

5 78-88 83 6889 8 664 55112

6 89-99 94 8836 4 376 35344

Total 30 2083 153749Calculation for mean

x A=∑ f i . x i

∑ f i

x A=2083

30=69.4


sA=√ N ∑ f X i2−(∑ f X i )

2

N (N−1)

sA=√ 30 (153749)−(2083 )2

30 x29

sA=√ 4612470−4338889870

sA=√ 273581870

=17.73

Make a frequency distribution table Limit of

Class (x)

Z for limit

of class

Luas O-Z Large of

Interval

Class

Frekuensi

yang

diharapkan

Obsevation

Frequency

(Oi)

X2

(Ei)

33.5 -2.02 0.4783

44.5 -1.40 0.4192 0.0591 1.773 4 2.797

55.5 -0.78 0.2823 0.1369 4.107 3 0.298

66.5 -0.16 0.0636 0.2187 6.561 5 0.371

77.5 0.46 0.1772 0.2408 7.224 6 0.207

88.5 1.08 0.3599 0.1827 5.481 8 1.158

99.5 1.70 0.4554 0.0955 2.865 4 0.450

TOTAL 30 5.28176

Where:



χ2=∑i=1

k (Oi−Ei )2

Ei

χ2=5 .28176


2(1-α)(k-3) = χ2


that χ2

calculation< χ2

table so H0 accepted. The conclusion is sample of score statistics at subject A is


Testing Normality for Subject B


2

1 39-48 43.5 1892.25 3 130.5 5676.75

2 49-58 53.5 2862.25 7 374.5 20035.75

3 59-68 63.5 4032.25 4 254 16129

4 69-78 73.5 5402.25 6 441 32413.5

5 79-88 83.5 6972.25 5 417.5 34861.25

6 89-98 93.5 8742.25 5 467.5 43711.25

Total 30 2085 152827.5Calculation for mean

xB=∑ f i . x i

∑ f i

xB=2085

30=69.5


sB=√ N ∑ f X i2−(∑ f X i )

2

N (N−1)

sB=√ 30 (152827.5)−(2085 )2

30 x29

sB=√ 4584825−4347225870

sB=√ 237600870

=16.52


Limit of

Class (x)

Z for limit

of classLuas O-

Z

Large of

Interval

Class

Frekuensi

yang

diharapkan

(Ei)

Obsevation

Frequency

(Oi)

X2

38.5 -1.88 0.4699

48.5 -1.27 0.3980 0.0719 0.993 3 0.32958.5 -0.67 0.2486 0.1494 1.656 7 1.41568.5 -0.06 0.0239 0.2247 2.589 4 1.11578.5 0.54 0.2054 0.2293 3.369 6 0.11288.5 1.15 0.3749 0.1695 4.098 5 0.00198.5 1.76 0.4608 0.0859 4.278 5 2.278

TOTAL 30 5.251

Where:



χ2=∑i=1

k (Oi−Ei )2

Ei

χ2=5 .251


2(1-α)(k-3) = χ2


that χ2calculation< χ2

table so H0 rejected. The conclusion is sample of score statistics at subject B is


Testing Normality for Subject C


2

1 34-44 39 1521 2 78 3042

2 45-55 50 2500 6 300 15000

3 56-66 61 3721 7 427 26047

4 67-77 72 5184 13 936 67392

5 78-88 83 6889 6 498 41334

6 89-99 94 8836 6 564 53016

Total 40 2803 205831Calculation for mean

xC=∑ f i. x i

∑ f i

xC=280340

=70.07


sC=√ N∑ f X i2−(∑ f X i )

2

N (N−1)

sC=√ 40 (205831)−(2803 )2

40 x39

sC=√ 8233240−78568091560

sC=√ 3764311560

=15.53


Limit of

Class (x)

Z for limit

of class Luas O-Z

Large of

Interval

Class

Frekuensi

yang

diharapkan

(Ei)

Obsevation

Frequency

(Oi)

X2

33.5 -2.35 0.4906

44.5 -1.65 0.4505 0.0401 1.203 2 0.528

55.5 -0.94 0.3264 0.1241 3.723 6 1.393

66.5 -0.23 0.0910 0.2354 7.062 7 0.001

77.5 0.48 0.1844 0.2754 8.262 13 2.717

88.5 1.19 0.3830 0.1986 5.958 6 0.000

99.5 1.90 0.4713 0.0883 2.649 6 4.239

TOTAL 40 8.878

Where:



χ2=∑i=1

k (Oi−Ei )2

Ei

χ2=8 .878


2(1-α)(k-3) = χ2


that χ2

calculation< χ2

table so H0 accepted. The conclusion is sample of score statistics at subject C is


V. Test Homogeneity of subject A, subject B, and Subject C

Example:

After testing normality tests, these students are now testing the homogeneity test of the

three data values obtained from three Subject

Solution:

To test the homogeneity by manual we use Barthlett testing

The step as follow

1) Formulating hypothesis

H0 : σ 12 = σ 2

2 = σ 32

H1 : σ 12 ≠ σ 2

2 = σ 32

σ 12 = σ 2

2 ≠ σ 32

2) Determine α and x2 of table, with df = n of sample population -1.

α = 5 %, df = 3-1= 2

X (1−α ) (2 )2 =X (0.95 ) (2)

2 =5.99

3) Criteria of testing

X2 is accepted if X2 < 5.99

X2 is rejected if X2 ≥ 5.99

4) Statistics testing

Find the mean (x ) of each population, with the formula x=∑ x

n

We now that nA = 30, nB = 30, nC = 40

From the calculation, we get:

x A=69.4

xB=69.5

xC=70.07

Find the variants of each sample population, with with the formula:

s2=∑ ¿ x−x ¿2

n−1

From the calculation :

sA2 = 314.35

sB2 = 272.91

sC2 = 241.18

Make the table :

Sampl

e(n-1) 1/(n-1) si 2 log si2 (n-1)*log si2 (n-1)*si2

1 29 0.034 314.35 2.497 72.425 9116.15

2 29 0.034 272.91 2.436 70.645 7914.39

3 39 0.025 241.18 2.382 92.911 9406.02

∑ 97 235.981 26436.560

So, the total variants for 3 sample population :

S2=∑ (ni−1)S i2

∑ (n−1)=26436.560

97=272.54

B = (log s2 ) ∑(n – 1)

= log (272.54). 97

= 236.23

X2 = ( ln 10 ) { B- ∑ (n - 1) log S2}

= 2.3026. (236.32 – 235.981)

= 0.78

5) Make conclusion

X2< X (0.95 )( 2)2

0.78 < 5.99, so H0 is accepted

So the data is homogeny

VI. Liliefors test

Example:

Another way to test the normality test is by using the method liliefors. In this case students

want to do a test for normality with liliefors method. He took one of its data, ie data values

subject B

Solution:



Data (Xi)

f fkbz=

x i−x

s

F(x) S(x) T=F(x)- S(x) S(x)S(x)

T=|F(x)-S(x)|

39 1 1 -1.87 0.0306

0.0333 -0.0027 0.002746 2 3 -1.46 0.072

70.1000 -0.0273 0.0273

49 2 5 -1.28 0.1007

0.1667 -0.0660 0.0660

54 2 7 -0.98 0.1634

0.2333 -0.0699 0.0699

55 1 8 -0.92 0.1785

0.2667 -0.0881 0.0881

57 1 9 -0.80 0.2112

0.3000 -0.0888 0.0888

58 1 10 -0.74 0.2288

0.3333 -0.1045 0.1045

64 2 12 -0.39 0.3497

0.4000 -0.0503 0.0503

67 2 14 -0.21 0.4176

0.4667 -0.0490 0.0490

70 1 15 -0.03 0.4881

0.5000 -0.0119 0.0119

73 1 16 0.15 0.5590

0.5333 0.0257 0.0257

74 1 17 0.21 0.5824

0.5667 0.0157 0.0157

76 3 20 0.33 0.6281

0.6667 -0.0386 0.0386

80 1 21 0.56 0.7138

0.7000 0.0138 0.0138

85 1 22 0.86 0.8055

0.7333 0.0722 0.0722

87 2 24 0.98 0.8366

0.8000 0.0366 0.0366

88 1 25 1.04 0.8508

0.8333 0.0175 0.0175

89 2 27 1.10 0.8642

0.9000 -0.0358 0.0358

97 2 29 1.57 0.9423

0.9667 -0.0243 0.0243

98 1 30 1.63 0.9489

1.0000 -0.0511 0.0511

Total 30Average 70.5

Std. Dev 16.83With compare the Lacoount with Ltable if Lacoount < Ltable, so Ho accepted. In our data the value of

Lacoount is 0.1045 and the value of Ltable (with df = n = 30 and significant level is 0.05 is 0.161).

because 0.1045 < 0.161, so Ho rejected and data is from normal distribution population.

Analysis SPSS


Class

Cases

Valid Missing Total


Value Subject A 30 100.0% 0 .0% 30 100.0%

Subject B 30 100.0% 0 .0% 30 100.0%

Subject C 40 100.0% 0 .0% 40 100.0%

Descriptivesa

Class Statistic Std. Error

Value Subject A Mean 70.9333 3.33078


Class

Cases

Valid Missing Total


Value Subject A 30 100.0% 0 .0% 30 100.0%

Subject B 30 100.0% 0 .0% 30 100.0%


Mean

Lower Bound 64.1211

Upper Bound 77.7455


Median 76.0000

Variance 332.823


Minimum 34.00

Maximum 99.00

Range 65.00


Skewness -.454 .427

Kurtosis -.835 .833

Subject B Mean 70.3667 3.11743


Mean

Lower Bound 63.9908

Upper Bound 76.7425


Median 71.5000

Variance 291.551


Minimum 39.00

Maximum 98.00

Range 59.00


Class

Cases

Valid Missing Total


Value Subject A 30 100.0% 0 .0% 30 100.0%

Subject B 30 100.0% 0 .0% 30 100.0%


Skewness -.040 .427

Kurtosis -1.099 .833

Subject C Mean 70.5500 2.67609


Mean

Lower Bound 65.1371

Upper Bound 75.9629


Median 71.5000

Variance 286.459


Minimum 34.00

Maximum 99.00

Range 65.00


Skewness -.122 .374

Kurtosis -.643 .733

a. There are no valid cases for Value when Class = .000. Statistics cannot be computed for this level.

Tests of Normalityb

Class



Value Subject A .143 30 .122 .945 30 .126

Subject B .104 30 .200* .958 30 .270

Subject C .062 40 .200* .977 40 .578


*. This is a lower bound of the true significance.

b. There are no valid cases for Value when Class = .000. Statistics cannot be computed for this

level.

We can see the results for Subject A of SPSS analysis by Kolmogorov-Smirnov

significance level> 0.05 ie 0.122 as well as the Shapiro-Wilk stated significance level> 0.05 is

0.126. The criteria receipt of H0 if the level of significance> 0.05. So in this case H0 received

stated that the sample data of Subject A derived from a normal distribution population.

Than see the results for Subject B of SPSS analysis by Kolmogorov-Smirnov significance

level> 0.05 ie 0.200 as well as the Shapiro-Wilk stated significance level> 0.05 is 0.270. The

criteria receipt of H0 if the level of significance> 0.05. So in this case H0 received stated that the

sample data of Class B derived from a normal distribution population.

We can see the results for Subject C of SPSS analysis by Kolmogorov-Smirnov

significance level> 0.05 ie 0.200 as well as the Shapiro-Wilk stated significance level> 0.05 is

0.578. The criteria receipt of H0 if the level of significance> 0.05. So in this case H0 received

stated that the sample data of Subject C derived from a normal distribution population.

Test of Homogeneity of Variancea

Levene Statistic df1 df2 Sig.

Value Based on Mean .264 2 97 .769

Based on Median .149 2 97 .861

Based on Median and with

adjusted df.149 2 93.438 .861

Based on trimmed mean .252 2 97 .777

Test Criteria:

Value Sig. Or significance or probability value <0.05, data derived from populations that have

a variance that is not the same.

Value Sig. Or significance or probability value> 0.05, data derived from populations that have

the same variance.

In the output above shows that the level of significance or probability value of the mean

(average) which is above 0.05 (0.769 bigger than 0.05). So soundly if the basic measurement is

the median data, numbers Sig. Is 0.861, which is still above 0.05. So in this case H0 is accepted

and stated that these populations have the same variance.