L5 - 12
Grafik Uji Seragam Tebal Paha
14.51 14.18
14.75 14.19
15.22 14.86
14.32 15.62
15.18 14.64
BKB = 13.02 BKA = 16.49
12 13
14 15
16 17
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
= 13
2
+ 10
2
+ ...... + 18
2
= 22662.55 ∑ Xi
2
= 13+ 10 + ...... + 18
2
= 1446
2
= 2090916
2 2
2
i i
i
x x
x N
c N
2
1446 6
17569510.5 2090916
100 1
. 2
= 33.54 N = 100
N’ N → data yang diperoleh sudah cukup
6. Pantat Popliteal
Uji Normal
k = 3.3 log n + 1
= 3.3 log 100 + 1 = 7.6 ≈ 8 kelas c = Xmax – Xmin k
= 52.3 – 41.2 7.6 = 1.46 ≈ 1.5
L5 - 13
Batas Kelas oi
Z1 Z2
PZ1 P Z2
PZ2 - PZ1
ei ei
gab oi
gab
2
41.15 -
∞ -2.185
0.014 0.014
1.445 5.381
7 0.487
41.15 – 42.65
7 -2.185
-1.609 0.014
0.054 0.039
3.936 42.65
– 44.15 12
-1.609 -1.033
0.054 0.151
0.097 9.699
9.699 12
0.546 44.15
– 45.65 12
-1.033 -0.457
0.151 0.324
0.173 17.301
17.301 12
1.624 45.65
– 47.15 24
-0.457 0.119
0.324 0.547
0.224 22.350
22.350 24
0.122 47.15
– 48.65 19
0.119 0.695
0.547 0.756
0.209 20.910
20.910 19
0.175 48.65
– 50.15 14
0.695 1.271
0.756 0.898
0.142 14.168
14.168 14
0.002 50.15
– 51.65 11
1.271 1.847
0.898 0.968
0.070 6.951
10.191 12
0.321 51.65
– 53.15 1
1.847 2.423
0.968 0.992
0.025 2.469
53.15 2.423
∞ 0.992
1 0.008
0.770 100
1 100
100 100
3.277
Tabel uji normal v = k
– r – 1 v = 7
– 2 – 1 = 4 α = 0.05
2
v, α
= 9.488
2
hasil perhitungan = 3.277
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 45.0
51.5 46.0
45.5 44.1
48.5 49.0
48.0 49.0
44.9 47.15
2 42.5
47.0 43.0
46.8 45.8
42.3 51.0
46.3 48.4
46.3 45.94
3 48.0
47.8 49.0
43.5 47.4
47.1 50.5
49.5 46.5
44.8 47.18
4 45.2
51.0 43.5
45.5 43.9
42.6 46.4
46.0 47.0
45.0 45.61
5 46.8
49.0 50.0
49.2 48.5
50.5 46.3
47.0 49.0
43.0 47.93
6 50.0
51.0 47.0
43.2 51.1
46.6 43.1
49.0 48.3
48.8 47.81
7 43.0
46.3 47.5
48.0 48.4
44.9 48.4
44.5 41.6
47.8 46.04
8 48.6
50.3 46.6
46.3 51.2
47.8 44.9
46.0 46.3
41.2 46.92
9 42.5
45.0 45.5
49.9 47.5
52.3 47.0
48.0 41.3
48.4 46.74
10 46.0
49.7 43.8
47.3 44.1
43.8 46.5
50.3 51.0
48.7 47.12
46.84 Rata-rata
x
= 46.84 = 2.6
L5 - 14
82 .
10 2.6
n
x
x
c x
BKB
= 46.84 – 20.82 = 45.19
x
c x
BKA
= 46.84 + 20.82 = 48.49
Grafik Uji Seragam Pantat Popliteal
47.18 47.81
46.04 47.93
46.92 47.12
46.74 45.61
45.94 47.15
BKB = 45.19 BKA = 48.49
44 45
46 47
48 49
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
= 45
2
+ 42.5
2
+ ...... + 48.7
2
= 220323.31 ∑ Xi
2
= 45+ 42.5 + ..... + 48.7
2
= 4637.2
2
= 21503623.84
2 2
2
i i
i
x x
x N
c N
2
4637.2 4
21503623.8 220323.31
100 1
. 2
= 9.83 N = 100
N’ N → data yang diperoleh sudah cukup
L5 - 15
7. Panjang Sandaran
