2. Objektif Pembelajaran
• Untuk digunakan dalam pengujian hipotesis
apabila tidak boleh membuat sebarang
anggapan terhadap taburan yang kita ambil
• Untuk mengetahui ujian untuk taburan bebas
yang digunakan dalam keadaan tertentu
• Untuk menggunakan dan menjelaskan enam
jenis pengujian hipotesis tak berparameter
• Ujian mengetahui kelemahan dan kelebihan
ujian tak berparameter
3. Statistik Berparameter vs Tidak
Berparameter
• Statistik Berparameter adalah teknik statistik
berdasarkan kepada andaian berkaitan populasi
dimana sampel data adalah dipungut.
– Andaian dimana data yang dianalisis adalah
dipilih secara rawak dari populasi yang
bertaburan normal.
– Memerlukan ukuran kuantitatif yang
menghasilkan data bertaraf interval atau
perkadaran.
4. Statistik Berparameter vs Tidak
Berparameter
• Statistik Tidak Berparameter adalah
berdasarkan andaian yang kurang populasi
dan parameter.
– Kadangkala dipanggil sebagai statistik
“tidak mempunyai taburan”.
– Berbagai-bagai jenis statistik tidak
berparameter yang ada untuk digunakan
dengan data bertaraf nominal atau ordinal.
5. Kebaikan Teknik Tidak Berparameter
• Kadangkala tidak terdapat teknik berparameter
alternatif untuk digunakan berbanding teknik tidak
berparameter.
• Beberapa ujian tidak berparameter boleh digunakan
untuk menganalisis data nominal.
• Beberapa ujian tidak berparameter boleh digunakan
untuk menganalisis data ordinal.
• Pengiraan statistik tidak berparameter kurang rumit
berbanding kaedah berparameter, terutama untuk
sampel yang kecil.
• Pernyataan kebarangkalian yang diperolehi dari
kebanyakan ujian tidak berparameter adalah
kebarangkalian yang tepat.
6. Kelemahan Statistik Tidak
Berparameter
• Ujian tidak berparameter boleh membazirkan data
jika ujian berparaeter boleh digunakan untuk data
tersebut.
• Ujian tidak berparameter biasanya tidak
digunakan dengan meluas dan kurang dikenali
berbanding ujian berparameter.
• Untuk sampel yang besar, pengiraan bagi
kebanyakan ujian tidak berparameter boleh
mengelirukan.
8. Runs Test
• Test for randomness - is the order or sequence of
observations in a sample random or not
• Each sample item possesses one of two possible
characteristics
• Run - a succession of observations which possess
the same characteristic
• Example with two runs: F, F, F, F, F, F, F, F, M,
M, M, M, M, M, M
• Example with fifteen runs: F, M, F, M, F, M, F,
M, F, M, F, M, F, M, F
9. Runs Test: Sample Size
Consideration
• Sample size: n
• Number of sample member possessing
the first characteristic: n1
• Number of sample members possessing
the second characteristic: n2
• n = n1 + n2
• If both n1 and n2 are ≤ 20, the small
sample runs test is appropriate.
10. Runs Test: Small Sample
H : The observations in Example
0 the sample are randomly generated.
H : The observations in the sample are not randomly generated.
a
α = .05
n1 = 18
n2 = 8
If 7 ≤ R ≤ 17, do not reject H0
Otherwise, reject H0.
1 2 3 4 5 6 7 8 9 10 11 12
D CCCCC D CC D CCCC D C D CCC DDD CCC
R = 12
Since 7 ≤ R = 12 ≤ 17, do not reject H0
11. Runs Test: Large Sample
2n n
If either n or n is > 20, µ = +1 1 2
1
the sampling
2
R + n n
1 2
distribution of R is
approximately normal.
2 n1 n2 (2 n1 n2 − n1 − n2 )
σ =
(n1+ n2)
R 2
+ (n1 + n2 − 1)
R − µR
Z=
σ R
12. Runs Test: Large Sample
0
Example
H : : The observations in the sample are randomly generated.
H The observations in the sample are randomly generated.
0
H : : The observations in the sample are not randomly generated.
H The observations in the sample are not randomly generated.
a
a
α = .05
α = .05
n1 = 40
n = 40
n21= 10
n2 = 10
If -1.96 ≤ Z ≤ 1.96, do not reject H0
If -1.96 ≤ Z ≤ 1.96, do not reject H0
Otherwise, reject H0. .
Otherwise, reject H0
1
1
1
1 2
2 3 4 5 6
3 4 5 6 7
7 8
8 9
9 0
0 11
11
NNN
NNN F NNNNNNN F NN FF NNNNNN
F NNNNNNN F NN FF NNNNNN F NNNN
F NNNN F NNNNN
F NNNNN
12
12 13
13
FFFF NNNNNNNNNNNN
FFFF NNNNNNNNNNNN R = 13
R = 13
13. Runs Test: Large Sample
2n n Example n n − n − n )
2 n n (2
µ = 1 2
+1 σR= 1 2 1 2 1 2
R
n1 + n2 (n1+ n2)
2
+ (n1 + n2 − 1)
2(40)(10)
= +1 2(40)(10)[ 2(40)(10) − (40) − (10)]
40 + 10 =
(40+10)
2
= 17 + (40 + 10 − 1)
= 2.213
R − µR 13 − 17
-1.96 ≤ Z = -1.81 ≤ 1.96,
Z= = = −181
. do not reject H0
σ R
2.213
15. Mann-Whitney U Test
• Nonparametric counterpart of the t test for
independent samples
• Does not require normally distributed populations
• May be applied to ordinal data
• Assumptions
– Independent Samples
– At Least Ordinal Data
16. Mann-Whitney U Test:
Sample Size Consideration
• Size of sample one: n1
• Size of sample two: n2
• If both n1 and n2 are ≤ 10, the small sample
procedure is appropriate.
• If either n1 or n2 is greater than 10, the large
sample procedure is appropriate.
17. Mann-Whitney U Test:
Small Sample Example
H0: The health service Health Educational
population is identical to the Service Service
educational service 20.10 26.19
population on employee 19.80 23.88
compensation 22.36 25.50
Ha: The health service 18.75 21.64
population is not identical to 21.90 24.85
the educational service 22.96 25.30
population on employee 20.75 24.12
compensation 23.45
18. Mann-Whitney U Test:
α = .05
Small Sample Example
Compensation Rank Group
18.75 1 H
If the final p-value < .05, reject H0. 19.80 2 H
20.10 3 H
20.75 4 H
21.64 5 E
21.90 6 H
W1 = 1 + 2 + 3 + 4 + 6 + 7 + 8 22.36 7 H
22.96 8 H
= 31 23.45 9 E
23.88 10 E
W2 = 5 + 9 + 10 + 11 + 12 + 13 + 14 + 15 24.12 11 E
24.85 12 E
= 89 25.30 13 E
25.50 14 E
26.19 15 E
19. Mann-Whitney U Test:
Small Sample Example
n (n + 1) − Since U2 < U1, U = 3.
U =n n
1 1 2
+ 1
2
1
W1
(7)(8) p-value = .0011 < .05, reject H0.
= (7)(8) + − 31
2
= 53
n (n + 1)
U =n n
2 1 2
+ 2 2
2
−W 2
(8)(9)
= (7)(8) n1 n2 + − 89
2
=3
20. Mann-Whitney U Test:
Formulas for Large Sample Case
U = n1 n2 n ( n + 1) − n ⋅n
µ = 2 1 2
+
2
1 1
W 1 U
where : n1 = number in group 1 n ⋅n ( n + n )
+1
σU
=
1 2 1
12
2
n 2
= number in group 2
U − µU
Z=
W 1
= sum or the ranks of σ U
values in group 1
21. Incomes of PBS PBS Non-PBS
and Non-PBS Viewers 24,500
39,400
41,000
32,500
Ho: The incomes for PBS viewers 36,800 33,000
44,300 21,000
and non-PBS viewers are
57,960 40,500
identical 32,000 32,400
Ha: The incomes for PBS viewers 61,000 16,000
and non-PBS viewers are not 34,000 21,500
identical 43,500 39,500
55,000 27,600
n1 = 14
α =.05 39,000 43,500
62,500 51,900
If Z < −1.96 or Z > 1.96, reject Ho
n2 = 13 61,400 27,800
53,000
22. Ranks of Income from Combined
Groups of PBS and Non-PBS
Income Rank Viewers Rank Group
Group Income
16,000 1 Non-PBS 39,500 15 Non-PBS
21,000 2 Non-PBS 40,500 16 Non-PBS
21,500 3 Non-PBS 41,000 17 Non-PBS
24,500 4 PBS 43,000 18 PBS
27,600 5 Non-PBS 43,500 19.5 PBS
27,800 6 Non-PBS 43,500 19.5 Non-PBS
32,000 7 PBS 51,900 21 Non-PBS
32,400 8 Non-PBS 53,000 22 PBS
32,500 9 Non-PBS 55,000 23 PBS
33,000 10 Non-PBS 57,960 24 PBS
34,000 11 PBS 61,000 25 PBS
36,800 12 PBS 61,400 26 PBS
39,000 13 PBS 62,500 27 PBS
39,400 14 PBS
26. Wilcoxon Matched-Pairs
Signed Rank Test
• A nonparametric alternative to the t test for related
samples
• Before and After studies
• Studies in which measures are taken on the same
person or object under different conditions
• Studies or twins or other relatives
27. Wilcoxon Matched-Pairs
Signed Rank Test
• Differences of the scores of the two matched
samples
• Differences are ranked, ignoring the sign
• Ranks are given the sign of the difference
• Positive ranks are summed
• Negative ranks are summed
• T is the smaller sum of ranks
28. Wilcoxon Matched-Pairs Signed
Rank Test: Sample Size
Consideration
• n is the number of matched pairs
• If n > 15, T is approximately normally
distributed, and a Z test is used.
• If n ≤ 15, a special “small sample” procedure is
followed.
– The paired data are randomly selected.
– The underlying distributions are symmetrical.
29. Wilcoxon Matched-Pairs Signed
Rank Test: Small Sample
H: M =0
0 d
Example
Family
Ha: Md ≠ 0 Pair Pittsburgh Oakland
1 1,950 1,760
n=6 2 1,840 1,870
3 2,015 1,810
α =0.05 4 1,580 1,660
5 1,790 1,340
6 1,925 1,765
If Tobserved ≤ 1, reject H0.
30. Wilcoxon Matched-Pairs Signed
Rank Test: Small Sample
Family
Example d Rank
Pair Pittsburgh Oakland
1 1,950 1,760 190 +4
2 1,840 1,870 -30 -1
3 2,015 1,810 205 +5
4 1,580 1,660 -80 -2
5 1,790 1,340 450 +6
6 1,925 1,765 160 +3
T = minimum(T+, T-) T = 3 > Tcrit = 1, do not reject H0.
T+ = 4 + 5 + 6 + 3= 18
T- = 1 + 2 = 3
T=3
31. Wilcoxon Matched-Pairs Signed
Rank Test: Large Sample
Formulas
( n )( n + 1)
µ T 4
=
n( n + 1)( 2n + 1)
σT= 24
T−µ
Z= T
σ T
where : n = number of pairs
T = total ranks for either + or - differences, whichever is less
36. Kruskal-Wallis Test
• A nonparametric alternative to one-way analysis
of variance
• May used to analyze ordinal data
• No assumed population shape
• Assumes that the C groups are independent
• Assumes random selection of individual items
37. Kruskal-Wallis K Statistic
12 T j
C 2
− 3( n + 1)
K= ∑
n( n + 1) j =1 n j
where : C = number of groups
n = total number of items
T j
= total of ranks in a group
n j = number of items in a group
K ≈ χ 2 , with df = C - 1
38. Number of Patients per Day
per Physician in Three Organizational
Categories
Ho: The three populations are identical
Ha: At least one of the three populations is different
Three or
Two More
α = 0.05 Partners Partners HMO
df = C − 1 = 3 − 1 = 2 13 24 26
15 16 22
χ
2
.05, 2
= 5.991 20 19 31
18 22 27
If K > 5.991, reject Ho. 23 25 28
14 33
17
42. Friedman Test
• A nonparametric alternative to the randomized
block design
• Assumptions
– The blocks are independent.
– There is no interaction between blocks and
treatments.
– Observations within each block can be ranked.
• Hypotheses
– Ho: The treatment populations are equal
– Ha: At least one treatment population
yields larger values than at least one
other treatment population
43. Friedman Test
C
12
χ ∑ R j − 3b(C + 1)
2 2
=
r bC (C + 1) j =1
where : C = number of treatment levels (columns)
b = number of blocks (rows)
R j = total ranks for a particular treatment level
j = particular treatment level
χ ≈χ
2 2
, with df = C - 1
r
44. Friedman Test: Tensile Strength
of Plastic Housings
Ho: The supplier populations are equal
Ha: At least one supplier population yields larger
values than at least one other supplier population
Supplier 1 Supplier 2 Supplier 3 Supplier 4
Monday 62 63 57 61
Tuesday 63 61 59 65
Wednesday 61 62 56 63
Thursday 62 60 57 64
Friday 64 63 58 66
45. Friedman Test: Tensile Strength
of Plastic Housings
α = 0.05
df = C − 1 = 4 − 1 = 3
χ
2
.05, 3
= 7.81473
χ
2
If r
> 7.81473, reject Ho.
49. Spearman’s Rank Correlation
• Analyze the degree of association of two
variables
• Applicable to ordinal level data (ranks)
6∑ d
2
r = 1−
s
( n − 1)
n
2
where: n = number of pairs being correlated
d = the difference in the ranks of each pair
