P: Probability Kemunmgkinan Terjadi Error————— 01/01/2002 2:02:35 ————————————————————

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Descriptive Statistics: ExpHealth

Variable N Mean Median TrMean StDev SE MeanExpHealt 151 473384 50000 108007 3911550 318317

Variable Minimum Maximum Q1 Q3ExpHealt 0 48000000 0 200000

One-Sample Z: ExpHealth

Test of mu = 50000 vs mu > 50000The assumed sigma = 3911550

Variable N Mean StDev SE MeanExpHealth 151 473384 3911550 318317

Variable 97,5% Lower Bound Z PExpHealth -150507 1,33 0,092

Descriptive Statistics: ExpFood

Variable N N* Mean Median TrMean StDevExpFood 148 3 3555669 3000000 3216448 2948179

Variable SE Mean Minimum Maximum Q1 Q3ExpFood 242339 120000 18000000 1800000 4500000

One-Sample Z: ExpFood

Test of mu = 3000000 vs mu > 3000000The assumed sigma = 2948179

Variable N Mean StDev SE MeanExpFood 148 3555669 2948179 242339

Variable 97,5% Lower Bound Z PExpFood 3080694 2,29 0,011

2.

Test and CI for One Proportion: Origin

Test of p = 0,66 vs p < 0,66

Success = 1

ExactVariable X N Sample p 95,0% Upper Bound P-ValueOrigin 86 151 0,569536 0,637640 0,013

Rata-rata yang diobservasi……

P value < α

3. di bagi berdasarkan kecamatan …menggunakan sample t untuk membedakan 2 object

H0 = M1 = M2 -> M1 – M2 = 0Ha = M1 ≠ M2 -> M1 – M2 ≠00.326 > 0.001

Karena nillai P value > alfa maka hipotesa awal bias diterima didapat nilai rata2 pengeluaran untuk makanan dimasing-masing 2 kecamatan adalah sama ditingkat confidence = 99%

Two-Sample T-Test and CI: ExpFood; Kecam

Two-sample T for ExpFood

Kecam N Mean StDev SE Mean0 50 3899580 3102049 4386961 98 3380204 2866883 289599

Difference = mu (0) - mu (1)Estimate for difference: 51937699% CI for difference: (-863291; 1902043)T-Test of difference = 0 (vs not =): T-Value = 0,99 P-Value = 0,326 DF = 92

4. Pengujian Beda Proporsi Dua Populasi:Ujilah apakah proporsi dari variabel kategorik I adalah sama di dua kecamatan (kecam=0, kecamatan Barat; kecam=1, kecamatan Timur)? Gunakan =2%.

Test and CI for Two Proportions: Origin; Kecam

Success = 1

Kecam X N Sample p0 16 51 0,3137251 70 100 0,700000

Estimate for p(0) - p(1): -0,38627598% CI for p(0) - p(1): (-0,571239; -0,201310)Test for p(0) - p(1) = 0 (vs not = 0): Z = -4,86 P-Value = 0,000