L5 - 24
Uji Cukup ∑ Xi
2
= 24
2
+ 28
2
+ ...... + 26
2
= 66509.92 ∑ Xi
2
= 24+ 28 + ..... + 26
2
= 2570.4
2
= 6606956.16
2 2
2
i i
i
x x
x N
c N
2
2570.4 6606956.16
92 .
6509 6
100 1
. 2
= 2.67 N = 100
N’ N → data yang diperoleh sudah cukup
11. Tinggi Badan Tegak
Uji Normal
k = 3.3 log n + 1
= 3.3 log 100 + 1 = 7.6 ≈ 8 kelas c = Xmax – Xmin k
= 177.3 – 155 7.6 = 2.934 ≈ 2.9
Batas Kelas oi
Z1 Z2
PZ1 P Z2
PZ2 - PZ1
ei ei
gab oi
gab
2
154.95 -
∞ -2.008
0.022 0.022
2.233 6.705
4 1.091
154.95 - 157.85 4
-2.008 -1.498
0.022 0.067
0.045 4.472
157.85 - 160.75 14
-1.498 -0.988
0.067 0.161
0.094 9.443
9.443 14
2.200 160.75 - 163.65
15 -0.988
-0.479 0.161
0.316 0.155
15.460 15.460
15 0.014
163.65 - 166.55 19
-0.479 0.031
0.316 0.512
0.196 19.630
19.630 19
0.020 166.55 - 169.45
17 0.031
0.541 0.512
0.706 0.193
19.329 19.329
17 0.281
169.45 - 172.35 12
0.541 1.050
0.706 0.853
0.148 14.759
14.759 12
0.516 172.35 - 175.25
12 1.050
1.560 0.853
0.941 0.087
8.739 8.739
12 1.217
175.25 - 178.15 7
1.560 2.070
0.941 0.981
0.040 4.013
5.936 7
0.191 178.5
2.070 ∞
0.981 1
0.019 1.923
100 1
100 100
100 5.528
L5 - 25
Tabel uji normal v = k
– r – 1 v = 8
– 2 – 1 = 5 α = 0.05
2
v, α
= 11.07
2
hasil perhitungan = 5.528
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
1 160.0 168.0 169.0 177.3 162.3 164.5 161.0 176.5 168.5 171.7
167.88
2 162.5 164.0 167.0 165.6 166.5 169.5 175.0 165.0 163.0 164.8
166.29
3 156.3 156.4 166.0 170.3 175.0 159.0 173.0 173.0 167.5 165.0
166.15
4 165.2 166.0 171.0 168.6 170.0 168.0 174.0 171.4 169.5 175.5
169.76
5 162.5 167.0 156.5 161.4 168.0 165.0 164.0 168.8 163.0 159.3
162.97
6 167.0 164.0 176.5 163.2 169.0 170.0 168.2 175.0 158.0 158.0
166.89
7 173.0 159.0 173.0 161.0 158.5 168.0 173.0 162.0 173.0 161.0
166.15
8 163.0 164.0 164.5 155.0 168.0 177.0 172.0 167.0 162.0 159.0
165.15
9 160.0 169.0 165.0 158.0 171.5 160.0 160.0 166.0 164.0 163.5
163.70
10 165.0 159.0 177.2 159.4 173.0 173.0 176.5 171.0 162.0 171.9
168.80 166.37
Rata-rata
x
= 166.37 = 5.69
78 .
1 10
5.69
n
x
x
c x
BKB
= 166.37 – 21.78 = 162.97
x
c x
BKA
= 10.26 + 21.78 = 169.97
L5 - 26
Grafik Uji Seragam Tinggi Badan Tegak
166.15 166.89
166.15 165.15
167.88 166.29
169.76
162.97 163.70
168.80
BKB = 162.97 BKA = 169.97
162 164
166 168
170
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
= 160
2
+ 162.5
2
+ ...... + 171.9
2
= 2773664 ∑ Xi
2
= 160+ 162.5 + ..... + 171.9
2
= 16304.60
2
= 265839981.16
2 2
2
i i
i
x x
x N
c N
2
16304.6 16
. 265839981
2773664 100
1 .
2
= 17.34
N = 100 N’ N → data yang diperoleh sudah cukup
12. Pangkal ke Tangan