L5 - 10
Uji Cukup ∑ Xi
2
= 42
2
+ 48.8
2
+ ...... + 43.2
2
= 183272.32 ∑ Xi
2
= 42+ 48.8 + ...... + 43.2
2
= 4191.60
2
= 17569510.56
2 2
2
i i
i
x x
x N
c N
2
4191.60 6
17569510.5 183272.32
100 1
. 2
= 17.25 N = 100
N’ N → data yang diperoleh sudah cukup
5. Tebal Paha
Uji Normal
k = 3.3 log n + 1
= 3.3 log 100 + 1 = 7.6 ≈ 8 kelas c = Xmax – Xmin k
= 21.8 – 8.6 7.6 = 1.74 ≈ 1.7
Batas Kelas oi
Z1 Z2
PZ1 P Z2
PZ2 - PZ1
ei ei
gab oi
gab
2
8.55 -
∞ -2.267
0.012 0.012
1.170 15.275
15 0.005
8.55 – 10.25
7 -2.267
-1.646 0.012
0.050 0.038
3.821 10.25
– 11.95 8
-1.646 -1.025
0.050 0.153
0.103 10.284
11.95 – 13.65
18 -1.025
-0.404 0.153
0.343 0.190
19.047 19.047
18 0.058
13.65 – 15.35
24 -0.404
0.217 0.343
0.586 0.243
24.281 24.281
24 0.003
15.35 – 17.05
24 0.217
0.838 0.586
0.799 0.213
21.306 21.306
24 0.341
17.05 – 18.75
12 0.838
1.459 0.799
0.928 0.129
12.869 12.869
12 0.059
18.75 – 20.45
5 1.459
2.080 0.928
0.981 0.053
5.349 7.223
7 0.007
20.45 – 22.15
2 2.080
2.701 0.981
0.997 0.015
1.529 22.15
2.701 ∞
0.997 1
0.003 0.345
100 1
100 100
100 0.472
L5 - 11
Tabel uji normal v = k
– r – 1 v = 6
– 2 – 1 = 3 α = 0.05
2
v, α
= 7.815
2
hasil perhitungan = 0.472
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 13.0
16.5 17.0
14.5 13.2
11.0 15.0
11.7 16.5
18.0 14.64
2 10.0
10.0 15.0
15.4 12.2
16.1 18.5
20.0 19.5
15.1 15.18
3 8.6
12.1 14.0
15.5 18.0
14.5 18.3
18.5 17.0
15.7 15.22
4 15.0
14.0 17.0
17.2 10.5
17.6 18.2
16.0 16.2
14.5 15.62
5 17.5
18.0 14.0
16.2 15.9
13.5 13.2
13.4 11.8
10.5 14.51
6 17.0
18.5 12.5
19.0 16.3
12.0 9.2
20.5 11.2
12.0 14.19
7 14.2
13.5 15.5
16.6 14.4
10.3 12.7
14.7 13.5
16.4 14.18
8 14.5
12.4 14.2
14.7 18.8
14.8 15.8
15.8 17.0
9.5 14.75
9 9.3
12.5 13.1
14.3 15.0
21.8 14.1
14.7 13.3
15.1 14.32
10 11.0
19.0 16.6
15.6 13.6
9.6 16.4
14.5 14.3
18.0 14.86
14.75 Rata-rata
x
= 14.75 = 2.74
87 .
10 2.74
n
x
x
c x
BKB
= 14.75 – 20.87 = 13.02
x
c x
BKA
= 14.75 + 20.87 = 16.49
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
