Post on 26-Nov-2014
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
Ho : Data is from normal distribution
H1 : Data is not from normal distribution
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
Ho : Data is from normal distribution
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
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)
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
Ei = Large Each of Class Interval ¿ n.
χ2=∑i=1
k (Oi−Ei )2
Ei
χ2=3 .176
With significant value so χ2table = χ
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)
No Kelas- Interval xi xi2 fi fixi fixi
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
Calculation for standard deviation
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
Make a frequency distribution table
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:
z=lim of class −xs
Ei = Large Each of Class Interval ¿ n.
χ2=∑i=1
k (Oi−Ei )2
Ei
χ2=5 .294
With significant value so χ2table = χ
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 B is
from normal distribution.
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
Case Processing Summary
Class
Cases
Valid Missing Total
N Percent N Percent N Percent
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
95% Confidence Interval for
Mean
Lower Bound 65.9337
Upper Bound 75.6263
5% Trimmed Mean 71.1111
Median 73.5000
Variance 290.787
Std. Deviation 1.70525E1
Minimum 34.00
Maximum 99.00
Range 65.00
Interquartile Range 29.25
Skewness -.269 .337
Kurtosis -.828 .662
Class B Mean 70.4400 2.47702
95% Confidence Interval for
Mean
Lower Bound 65.4622
Upper Bound 75.4178
5% Trimmed Mean 70.7111
Median 71.5000
Variance 306.782
Std. Deviation 1.75152E1
Minimum 34.00
Maximum 99.00
Range 65.00
Interquartile Range 29.50
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
Kolmogorov-Smirnova Shapiro-Wilk
Statistic df Sig. Statistic df Sig.
Value Class A .109 50 .188 .964 50 .127
Class B .081 50 .200* .969 50 .206
a. Lilliefors Significance Correction
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
Testing Hypotheses of
class B
Ho : Data is from normal
distribution
H1 : Data is not from
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
Testing Hypotheses of
class C
Ho : Data is from normal
distribution
H1 : Data is not from
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
No Kelas- Interval xi xi2 fi fixi fixi
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
Calculation for standard deviation
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:
z=lim of class −xs
Ei = Large Each of Class Interval ¿ n.
χ2=∑i=1
k (Oi−Ei )2
Ei
χ2=5 .28176
With significant value so χ2table = χ
2(1-α)(k-3) = χ2
(0,95)(27) = 40.1. From this calculation, we know
that χ2
calculation< χ2
table so H0 accepted. The conclusion is sample of score statistics at subject A is
from normal distribution.
Testing Normality for Subject B
No Kelas- Interval xi xi2 fi fixi fixi
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
Calculation for standard deviation
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
Make a frequency distribution table
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:
z=lim of class −xs
Ei = Large Each of Class Interval ¿ n.
χ2=∑i=1
k (Oi−Ei )2
Ei
χ2=5 .251
With significant value so χ2table = χ
2(1-α)(k-3) = χ2
(0,95)(27) = 40.1. From this calculation, we know
that χ2calculation< χ2
table so H0 rejected. The conclusion is sample of score statistics at subject B is
from normal distribution.
Testing Normality for Subject C
No Kelas- Interval xi xi2 fi fixi fixi
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
Calculation for standard deviation
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
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)
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:
z=lim of class −xs
Ei = Large Each of Class Interval ¿ n.
χ2=∑i=1
k (Oi−Ei )2
Ei
χ2=8 .878
With significant value so χ2table = χ
2(1-α)(k-3) = χ2
(0,95)(37) = 55.8. From this calculation, we know
that χ2
calculation< χ2
table so H0 accepted. The conclusion is sample of score statistics at subject C is
from normal distribution.
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:
Ho : Data is from normal distribution
H1 : Data is not from normal distribution
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
Case Processing Summary
Class
Cases
Valid Missing Total
N Percent N Percent N Percent
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
Case Processing Summary
Class
Cases
Valid Missing Total
N Percent N Percent N Percent
Value Subject A 30 100.0% 0 .0% 30 100.0%
Subject B 30 100.0% 0 .0% 30 100.0%
95% Confidence Interval for
Mean
Lower Bound 64.1211
Upper Bound 77.7455
5% Trimmed Mean 71.4074
Median 76.0000
Variance 332.823
Std. Deviation 1.82434E1
Minimum 34.00
Maximum 99.00
Range 65.00
Interquartile Range 29.75
Skewness -.454 .427
Kurtosis -.835 .833
Subject B Mean 70.3667 3.11743
95% Confidence Interval for
Mean
Lower Bound 63.9908
Upper Bound 76.7425
5% Trimmed Mean 70.4630
Median 71.5000
Variance 291.551
Std. Deviation 1.70749E1
Minimum 39.00
Maximum 98.00
Range 59.00
Case Processing Summary
Class
Cases
Valid Missing Total
N Percent N Percent N Percent
Value Subject A 30 100.0% 0 .0% 30 100.0%
Subject B 30 100.0% 0 .0% 30 100.0%
Interquartile Range 32.25
Skewness -.040 .427
Kurtosis -1.099 .833
Subject C Mean 70.5500 2.67609
95% Confidence Interval for
Mean
Lower Bound 65.1371
Upper Bound 75.9629
5% Trimmed Mean 70.8333
Median 71.5000
Variance 286.459
Std. Deviation 1.69251E1
Minimum 34.00
Maximum 99.00
Range 65.00
Interquartile Range 28.00
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
Kolmogorov-Smirnova Shapiro-Wilk
Statistic df Sig. Statistic df Sig.
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
a. Lilliefors Significance Correction
*. 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.