kurva normal s2 biomed
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KURVA NORMAL Sumedi Sudarsono 23 Feb 2015
Rujuk pada Tabel Normal (A1).
1. Kurva normal = Kurva simetris = Mesokurtic.
2. Besar nilai Mean = nilai Median = nilai Mode.
3. Wilayah di kiri (50%) = wilayah di kanan (50%).
s
4. Besar c.o.v. = ------------ x 100% = < 30%
mean
-.-.-.-.-.-.-.-.-.-.-.-.-.-.-.-.-.-.-.-.-.-.-.-.-.-.-.-.-.-.-.-.-.-
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Luas wilayah pd ekor Kurva Normal
Luas wilayah antara Mean dg 1 (satu) s dpt
dilihat pada Tabel Kurva Normal ( Table A1 )
0.1587 Luas wilayah pd ekor tsb = 15.87 %
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Luas wilayah antara Mean dg 1.20 s dpt
dilihat pada Tabel Kurva Normal ( Table A1 )
1.20 s 0.1151 = 11.51 %
1.25 s 0.1056 = 10.56 %
1.96 s 0.0250 = 2½ %
1.960 dinamakan Zα, utk α = 5% ( Table A3 )
2.576 dinamakan Zα, utk α = 1% dst.
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Alpha (α) Zα
---------------------------
0.05 1.96
0.02 2.326
0.01 2.576
0.001 3.291
--------------------------
( Rujuk pada Table A3 )
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NILAI UJIAN MAHASISWA
Rata-rata mean nilai ujian mhs = 75
dengan simpang baku (SD) = s = 5
Jumlah peserta ujian ada 150 mhs = n
( Apakah data tsb distribusi Normal ? )
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Rata-rata mean ujian mhs = 75
dengan simpang baku SD = s = 5
( Apakah data distribusi Normal ? )
Hitunglah c.o.v.
s 5
cov = ------------ (100%) = -------- (100%) = 6.67%
mean 75
cov < 30% simpulan data distribusi Normal.
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(1). Berapa persen mhs yg mempunyai nilai
82 atau lebih ?
Selisih nilai 82 dg nilai Mean (75)
dinyatakan dalam berapa s
x – Mean
µ = -----------------
s
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Mean = 75 s = 5
Nilai = x ( 82 atau lebih )
x – Mean 82 - 75
µ = ----------------- = ----------- = 1.40
s 5
Lihat Tabel 0.0808 = 8.08 %
Ada 8.08 % mhs. memp. nilai 82 atau lebih.
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(2). Ada berapa jumlah mhs yg mempunyai
nilai 82 atau lebih?
Jadi ada 8.08 % x (150 mhs) = 12 mhs.
Ingat, jumlah mhs data diskret.
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(3). Ada berapa orang mhs. yg mempunyai nilai
antara 66 dan 72 ?
x - mean
µ = ---------------
s
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(3). Antara 66 dan 72
72 - 75
Utk 72 µ = --------------- = - 0.60 0.2743
5
66 - 75
Utk 66 µ = --------------- = - 1.80 0.0359
5 --------------- -
0.2384
23.84 % x (150) = 36 orang mahasiswa.
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(4). Bila 5% mhs dg nilai tertinggi akan diberi
hadiah dan piagam perhargaan,
berapa batas nilai tsb ?
x - mean
µ = ---------------
s
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Lima % mhs dg nilai tertinggi 0.05
Lihat Tabel 0.0505 0.0495
0.05 0.05
----------------------------
0.0005 0.0005
(1). Digunakan 0.0505
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0.0505 maka µ = 1.64
x - 75
µ = 1.64 = ----------
5
1.64 (5) = x – 75
x = 8.20 + 75 83.20
Sebagai batas nilai.
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(2). Digunakan 0.0495
0.0495 maka µ = 1.65
x - 75
µ = 1.65 = ----------
5
1.65 (5) = x – 75
x = 8.25 + 75 83.25
Sebagai batas nilai.
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0.0505 83.20
0.0495 83.25
----------------------- (+)
166.45
½ x 166.45 83.225
Sebagai batas nilai tinggi.
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NILAI BATAS LULUS UJIAN
(5). Bila sebesar 90% mhs akan diluluskan,
berapa nilai batas lulus ujian ?
x - mean
µ = ---------------
s
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Yang diluluskan 90% maka
artinya yg tidak lulus = 10% = 0.100
Nilai ini terletak di ekor kiri .
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Diluluskan = 90 % Tidak lulus = 10 % = 0.100
Lihat Tabel 0.1003 dan 0.0985
0.1000 0.1000
-------------------------------------- beda
0.0003 0.0015
Digunakan 0.1003 maka µ = 1.28
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x - 75
µ = - 1.28 = ------------
5
(-1.28) (5) = x – 75
x = 75 – 6.4 = 68.6
Sbg nilai batas lulus
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Menghitung besar mean Populasi = µ
berdasarkan besar mean sampel.
µ = Mean ± Zα ( SE )
Untuk α = 0.05 Zα = 1.96 (Dari Tabel)
α = 0.05 artinya Confidence interval = 95%
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Diketahui besar sampel = n = 16 mhs
Dihitung Rata-rata nilai ujian (Mean = 75)
Dihitung besar simpang baku (SD = s = 5)
Dengan kepercayaan 95% ingin diketahui
berapa besar mean populsi ( µ ) ?
Digunakan rumus sbb :
µ = Mean ± Zα ( SE )
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Alpha (α) Zα
---------------------------
0.05 1.960
0.02 2.326
0.01 2.576
0.001 3.291
--------------------------
( Lihat pada Table A3 )
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µ = Mean ± Zα ( SE )
SE = s /√ n
µ = 75 ± 1.960 (5/√ 16)
µ = 75 ± 1.960 (1.25) = 75 ± 2.45
72.55 < µ < 77.45
Dg kepercayaan 95%, kita yakin bahwa
Mean populasi ( µ ) antara 72.55 dan 77.45
=====o0o=====
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Data tidak Normal
Dilakukan transformasi data ialah suatu
upaya untuk menormalkan data.
1. Menggunakan fungsi log.
2. Akar.
3. Kuadrat.
4. 1/x. atau
5. Fungsi lain-lain.
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Uji Normalitas (SPSS)
Test for Normality.
Analyze ….. Descriptive statistics … Explore
[ Masukkan variabel Age ke dl Dependent List ]
Pada Display pilih Both.
Click pd Plots.
Click pd Factor level together.
Click pd Histogram.
Click pd Normality plots with test.
Click Continue…..click OK.
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Hasilnya dilihat pada :
(1). Uji Kolmogorov-Smirnov bila n > 50
Uji Shapiro-Wilk bila n ≤ 50.
(2). Bila nilai p > 0.05 Distribusi Normal.
bila p < 0.05 Distribusi Tidak Normal.
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Distribusi Normal Uji Statistik Parametrik
Distribusi Tidak Normal Uji Non-parametrik.
==========o0o========== selesai.