L5 - 5
Grafik Uji Seragam Tinggi Bahu Duduk
57.53 59.12
58.51 58.61
59.31 58.38
59.78 58.95
59.01 59.41
BKB = 56.92 BKA = 60.83
54 55
56 57
58 59
60 61
62
1 2
3 4
5 6
7 8
9 10
Subgroup ke - R
a ta
- r
a ta
Berdasarkan grafik di atas, dapat disimpulkan bahwa data seragam karena berada dalam batas BKA dan BKB.
Uji Cukup ∑ Xi
2
= 58
2
+ 59.8
2
+ ...... + 60.2
2
= 347842.46 ∑ Xi
2
= 58+ 59.8 + ...... + 60.2
2
= 5828.60
2
= 33972577.96
2 2
2
i i
i
x x
x N
c N
2
5828.6 6
33972577.9 46
. 347842
100 1
. 2
= 9.56 N = 100
N’ N → data yang diperoleh sudah cukup
3. Tinggi Siku Duduk
Uji Normal
k = 3.3 log n + 1
= 3.3 log 100 + 1 = 7.6 ≈ 8 kelas c = Xmax – Xmin k
= 32.5 – 19 7.6 = 1.78 ≈ 1.8
L5 - 6
Batas Kelas oi
Z1 Z2
PZ1 P Z2
PZ2 - PZ1
ei ei
gab oi
gab
2
18.95 -
∞ -1.904
0.028 0.028
2.847 10.819
10 0.062
18.95 – 20.75
10 -1.904
-1.236 0.028
0.108 0.080
7.972 20.75
– 22.55 23
-1.236 -0.569
0.108 0.285
0.177 17.663
17.663 23
1.613 22.55
– 24.35 24
-0.569 0.099
0.285 0.539
0.255 25.462
25.462 24
0.084 24.35
– 26.15 21
0.099 0.767
0.539 0.778
0.239 23.891
23.891 21
0.350 26.15
– 27.95 15
0.767 1.434
0.778 0.924
0.146 14.590
14.590 15
0.012 27.95
– 29.75 3
1.434 2.102
0.924 0.982
0.058 5.797
7.575 7
0.044 29.75
– 31.55 3
2.102 2.769
0.982 0.997
0.015 1.497
31.55 – 33.35
1 2.769
3.437 0.997
1.000 0.003
0.251 33.35
3.437 ∞
1.000 1
0.000 0.029
100 1
100 100
100 2.164
Tabel uji normal v = k
– r – 1 v = 6
– 2 – 1 = 3 α = 0.05
2
v, α
= 7.815
2
hasil perhitungan = 2.164
2
hasil perhitungan
2
v, α
sehingga mengikuti distribusi normal Uji Seragam
Subgroup 1
2 3
4 5
6 7
8 9
10 rata - rata
1 25.2
22.5 24.5
25.4 26.3
20.4 27.0
21.9 28.0
26.0 24.72
2 32.5
24.0 23.0
23.1 20.4
22.8 24.0
25.0 27.5
22.0 24.43
3 26.0
24.5 22.5
25.3 21.6
19.4 27.0
28.8 24.0
27.0 24.61
4 22.0
21.5 24.5
19.1 24.0
22.0 26.0
24.0 26.0
24.0 23.31
5 26.2
25.0 19.0
23.0 22.1
23.0 24.2
23.0 24.2
22.5 23.22
6 26.5
21.1 23.0
22.2 28.4
30.0 22.0
19.8 26.0
21.6 24.06
7 22.8
20.0 30.8
25.6 19.0
21.5 26.4
20.0 27.5
22.5 23.61
8 22.0
24.4 22.5
24.0 24.0
25.3 25.0
26.5 22.5
26.5 24.27
9 27.5
25.0 20.5
21.5 22.2
27.7 23.6
24.5 23.0
23.0 23.85
10 23.0
27.5 21.8
23.3 26.7
20.8 25.2
24.0 24.7
30.5 24.75
24.08 Rata-rata
L5 - 7
x
= 24.08 = 2.69
85 .
10 2.69
n
x
x
c x
BKB
= 24.08 – 20.85 = 22.38
x
c x
BKA
= 24.08 + 20.85 = 25.79
Grafik Uji Seragam Tinggi Siku Duduk
24.61 23.22
24.06 23.61
24.27 24.75
23.85 23.31
24.43 24.72
BKB = 22.38 BKA = 25.79
21 22
23 24
25 26
27
1 2
3 4
5 6
7 8
9 10
Subgroup ke - R
a ta
- r
a ta
Berdasarkan grafik di atas, dapat disimpulkan bahwa data seragam karena berada dalam batas BKA dan BKB.
Uji Cukup ∑ Xi
2
= 25.2
2
+ 32.5
2
+ ...... + 30.5
2
= 58718.75 ∑ Xi
2
= 25.2+ 32.5 + ...... + 30.5
2
= 2408.30
2
= 5799908.89
2 2
2
i i
i
x x
x N
c N
2
2408.30 5799908.89
58718.75 100
1 .
2
= 4.96
L5 - 8
N = 100 N’ N → data yang diperoleh sudah cukup
4. Tinggi Popliteal
