Tugas Individu Normality n Homogenity

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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 distributionScores 28-35 36-43 44-51 52-59 60-67 68-75 76-83 84-91 92-99 Total Students 2 5 9 14 12 13 15 19 11 100

Using Chi-SquareNo Kelas- Interval 1 2 3 4 5 6 7 28-35 36-43 44-51 52-59 60-67 68-75 76-83 xi 31,5 xi2 992.25 fi 2 5 9 14 12 13 15 fixi 63 197.5 427.5 777 762 929.5 fixi2 1984.5 7801.25 20306.25 43123.5 48387 66459.25

39,5 1560.25 47,5 2256.25 55,5 3080.25 63,5 4032.25 71,5 5112.25 79,5 6320.25

1192.5 94803.75

8 9

84-91 92-99 Total

87,5 7656.25 95,5 9120.25

19 11 100

1662.5 145468.8 1050.5 100322.8 7062 528657

Calculate the mean Calculate the standard deviation

Make a frequency distribution tableLimit of Class (x) 27.5 35.5 43.5 51.5 59.5 67.5 75.5 83.5 91.5 99.5 Z for limit of class Luas OZ 0.4936 0.4778 0.4418 0.3665 0.2389 0.0714 0.1103 0.2704 0.3869 0.4525 0.0158 0.036 0.0453 0.1276 0.1675 0.1817 0.1601 0.1165 0.0656 1.58 3.6 4.53 12.76 16.75 18.17 16.01 11.65 6.56 2 5 9 14 12 13 15 19 11 100 0.112 0.544 4.411 0.121 1.347 1.471 0.064 4.637 3.005 15.711 Large of Interval Class Frekuensi yang diharapkan (Ei) Observation Frequency (Oi) X2

-2.49 -2.03 -1.57 -1.11 -0.64 -0.18 0.28 0.74 1.21 1.67

TOTAL

Where:z! lim ofclass x s

Ei = Large Each of Class Interval v n.k

G2 ! i !1 2

(Oi Ei ) 2 Ei

G ! 15.711 With significant value so G 2 table = G 2 (1)(k-3)

= G 2 (0,95)(97) = 124,3. From this calculation,

we know that G 2 calculation< G 2 table so H0 accepted. The conclusion is sample is from normal distribution. Analysis using SPSSCase Processing Summary Cases Valid Name Value Students N 100 Percent 100.0% N 0 Missing Percent .0% N 100 Total Percent 100.0%

Descriptives Name Value Students Mean

a

Statistic 70.6100 67.1974 74.0226 70.9222 73.0000 295.796 1.71987E1 34.00 99.00 65.00 29.75 -.204 -.875

Std. Error 1.71987

95% Confidence Interval for Lower Bound Mean 5% Trimmed Mean Median Variance Std. Deviation Minimum Maximum Range Interquartile Range Skewness Kurtosis Upper Bound

.241 .478

Tests of Normalityb Kolmogorov-Smirnov Name Value Students Statistic .089 df 100a

Shapiro-Wilk Statistic .968 df 100 Sig. .016

Sig. .051

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 30 - 39 40 - 49 50 - 59 60 - 69 70 - 79 80 - 89 90 - 99 Total Students 2 5 8 8 11 11 5 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 30 - 39 40 - 49 50 - 59 60 - 69 70 - 79 80 - 89 90 - 99 Total Student 2 7 6 8 11 9 7 50

Calculation Data for Class A ( using Chi-Square) No 1 2 3 4 5 6 7 Kelas- Interval 30 - 39 40 - 49 50 - 59 60 - 69 70 - 79 80 - 89 90 - 99 xi 34.5 44.5 54.5 64.5 74.5 84.5 94.5 xi2 1190.25 1980.25 2970.25 4160.25 5550.25 7140.25 8930.25 50 fi 2 5 8 8 11 11 5 fixi 69 222.5 436 516 819.5 929.5 472.5 3465 fixi2 2380.5 9901.25 23762 33282 61052.75 78542.75 44651.25 253572.5

Total Calculation for mean

Calculation for standard deviation

Make a frequency distribution table Z for Large of Limit of limit of Luas OInterval Class (x) class Z Class 29.5 39.5 49.5 59.5 69.5 79.5 89.5 99.5 -2.40 -1.80 -1.20 -0.59 0.01 0.62 1.22 1.82 0.4918 0.4641 0.3849 0.2224 0.0040 0.2324 0.3888 0.4656 TOTAL Where:z! lim of class x s

Frekuensi yang diharapkan (Ei)

Obsevation Frequency (Oi) X2

0.0277 0.0792 0.1625 0.2264 0.2284 0.1564 0.0768

1.39 3.96 8.13 11.32 11.42 7.82 3.84 50

2 5 8 8 11 11 5

0.268 0.273 0.002 0.974 0.015 1.293 0.350 3.176

Ei = Large Each of Class Interval v n.k

G2 ! i !1 2

(Oi Ei ) 2 Ei

G ! 3.176 With significant value so G 2 table = G 2 (1)(k-3)

= G 2 (0,95)(47) = 67.5. From this calculation, we know

that G 2 calculation< G 2 table 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 1 2 3 4 5 6 7 Kelas- Interval 30 - 39 40 - 49 50 - 59 60 - 69 70 - 79 80 - 89 90 - 99 xi 34.5 44.5 54.5 64.5 74.5 84.5 94.5 xi2 1190.25 1980.25 2970.25 4160.25 5550.25 7140.25 8930.25 50 fi 2 7 6 8 11 9 7 fixi 69 311.5 327 516 819.5 760.5 661.5 3465 fixi2 2380.5 13861.75 17821.5 33282 61052.75 64262.25 62511.75 255172.5

Total Calculation for mean

Calculation for standard deviation

Make a frequency distribution table Z for limit Limit of Luas Oof class Class (x) Z 29.5 39.5 49.5 59.5 -2.27 -1.70 -1.13 -0.56 0.4884 0.4554 0.3708 0.2123 0.033 0.0846 0.1585 1.65 4.23 7.925 2 7 6 0.074 1.814 0.468 Large of Interval Class Frekuensi yang diharapkan (Ei) Obsevation Frequency (Oi) X2

69.5 79.5 89.5 99.5

0.01 0.58 1.15 1.72

0.0040 0.2190 0.3749 0.4573 TOTAL

0.2163 0.215 0.1559 0.0824

10.815 10.75 7.795 4.12

8 11 9 7 50

0.733 0.006 0.186 2.013 5.294

Where:z! lim of class x s

Ei = Large Each of Class Interval v n.

i !1

G ! 5.294 With significant value so G 2 table = G 2 (1)(k-3)

2

that G 2 calculation< G 2 table 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 havent the same variance NO Nilai (xi) 34 34 34 34 34 34 34 Class A Class B ( xi x) 2 900 900 900 900 900 900 900 Nilai (xi) 34 34 34 34 34 34 34

1 2 3 4 5 6 7

G2 !

k

(

i

i

i

)2

= G 2 (0,95)(47) = 67.5. From this calculation, we know

xi x-30 -30 -30 -30 -30 -30 -30

xi x-29.8 -29.8 -29.8 -29.8 -29.8 -29.8 -29.8

( xi x) 2 888.04 888.04 888.04 888.04 888.04 888.04 888.04

8 9 10 11 12 13 14 15 16 17 18 19 20 21 22 23 24 25 26 27 28 29 30 31 32 33 34 35 36 37 38 39 40 41 42 43 44 45 46

34 34 34 55 55 55 55 55 55 55 55 55 55 67 67 67 67 67 67 67 67 67 67 76 76 76 76 76 76 76 76 76 76 88 88 88 88 88 88

-30 -30 -30 -9 -9 -9 -9 -9 -9 -9 -9 -9 -9 3 3 3 3 3 3 3 3 3 3 12 12 12 12 12 12 12 12 12 12 24 24 24 24 24 24

900 900 900 81 81 81 81 81 81 81 81 81 81 9 9 9 9 9 9 9 9 9 9 144 144 144 144 144 144 144 144 144 144 576 576 576 576 576 576

34 34 34 54 54 54 54 54 54 54 54 54 54 68 68 68 68 68 68 68 68 68 68 76 76 76 76 76 76 76 76 76 76 87 87 87 87 87 87

-29.8 -29.8 -29.8 -9.8 -9.8 -9.8 -9.8 -9.8 -9.8 -9.8 -9.8 -9.8 -9.8 4.2 4.2 4.2 4.2 4.2 4.2 4.2 4.2 4.2 4.2 12.2 12.2 12.2 12.2 12.2 12.2 12.2 12.2 12.2 12.2 23.2 23.2 23.2 23.2 23.2 23.2

888.04 888.04 888.04 96.04 96.04 96.04 96.04 96.04 96.04 96.04 96.04 96.04 96.04 17.64 17.64 17.64 17.64 17.64 17.64 17.64 17.64 17.64 17.64 148.84 148.84 148.84 148.84 148.84 148.84 148.84 148.84 148.84 148.84 538.24 538.24 538.24 538.24 538.24 538.24

47 88 48 88 49 88 50 88 Total 3200 Average 64 Variance for Class A2

24 24 24 24 0

576 576 576 576 17100

87 87 87 87 3190 63.8

23.2 23.2 23.2 23.2 -4.3x10-13

538.24 538.24 538.24 538.24 16888

( xi x ) 2 17100 s ! ! ! 384.9 n 1 49

Variance for Class Bs2 ! F! ( x i x ) 2 16888 ! ! 344.6 n 1 49 Smallest Variance 384 .9 ! ! 1.116 igget Variance 344 .6

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 u 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 SPSSCase Processing Summary Cases Valid Class Value Class A Class B N 50 50 Percent 100.0% 100.0% N 0 0 Missing Percent .0% .0% N 50 50 Total Percent 100.0% 100.0%

Value

Descriptivesa Class Class A Mean 95% Confidence Interval for Mean 5% Trimmed Mean Lower Bound Upper Bound Statistic 70.7800 65.9337 75.6263 71.1111 Std. Error 2.41159

Median Variance Std. Deviation Minimum Maximum Range Interquartile Range Skewness Kurtosis Class B Mean 95% Confidence Interval for Mean 5% Trimmed Mean Median Variance Std. Deviation Minimum Maximum Range Interquartile Range Skewness Kurtosis Lower Bound Upper Bound

73.5000 290.787 1.70525E1 34.00 99.00 65.