1. 1 Pertemuan Kesembilan Sampling and Data Collection Matakuliah: MN J0412/ Riset Pemasaran Tahun: 2007 Versi:

2. 2 Learning Outcomes Pada akhir pertemuan ini, diharapkan mahasiswa akan mampu : Mahasiswa dapat menyisihkan sampel dan populasi tanpa menghilangkan unsur

3. 3 Outline Materi Materi 1: Required Steps in Sampling Materi 2: Types of Sampling Plans Materi 3: Non Probability Samples Materi 4:Probability Samples Materi 5: Sample Size

4. 4 6-Step Procedure for Drawing a Sample Define the Target Population Identify the Sampling Frame Select a Sampling Procedure Determine the Sample Size Select the Sample Elements Collect the Data from the Designated Elements Step 1 Step 2 Step 4 Step 3 Step 5 Step 6

5. 5 Classification of Sampling Techniques Sampling Designs Nonprobability Samples Convenience Judgment Quota Probability Samples Simple Random Statified Proportionate Disproportionate Cluster Systematic Area

6. 6 Distribution of Sample Means for Different Samples Sizes, Population Shapes n = 2 Populations n = 5 Sampling Distributions of mean n = 30

7. 7 Relationship Between Parameters of Parent Population and Derived Population ABCDEFABCDEF ElementTotal Parent Population 3 6 9 12 15 18 Mean:  3 + 6 + 9 + 12 + 15 + 18 6  X i N Variance:  2   (X i -  ) 2 (3-10.5) 2 +... +(18-10.5) 2 N 6 = 157.5 = 26.25 6 = = = = = 10.5

8. 8 Derived Population of All Possible Distinguishable Samples 1 2 3 4 5 6 7 8 9 10 11 12 13 14 15 16 17 18 19 20 ABC ABD ABE ABF ACD ACE ACF ADE ADF AEF BCD BCE BCF BDE BDF BEF CDE CDF CEF DEF 18 21 24 27 24 27 30 33 36 27 30 33 36 39 36 39 42 45 6 7 8 9 8 9 10 11 12 9 10 11 12 13 12 13 14 15 ElementTotalMean X j

9. 9 Derived Population Mean: =(6 -10.5) 2 + (7-10.5) 2 +... + (15-10.5) 2 20 Variance:  2  = 5.25 E( x )  x L  x j j =1 L 6 + 7 +... + 15 105 L  ( x j j =1 L - E(x)) 2 20 == = = 10.5 = _x_x

10. 10 Distribution of Variables in Parent and Derived Population 1 0 1 0 2 3 Values of x Values of X Absolute Frequency Absolute Frequency

11. 11 Example Calculations of Generating a Confidence Interval with a Stratified Sample III IIIIV n 1 = 100 x 1 =  X i1 = 3.2 n1n1 x 1 ) 2  (X i1 =.14 n1n1 s 1 2 = ^ n 3 = 100 x 3 =  X i3 = 5.8 n3n3 x 3 ) 2  (X i3 =.20 n3n3 s 3 2 = ^ n 4 = 100 x 4 =  X i4 = 7.2 n4n4 x 4 ) 2  (X i4 =.18 n4n4 s 4 2 = ^ n 2 = 100 x 2 =  X i2 = 4.6 n2n2 x 2 ) 2  (X i2 =.12 n2n2 s 2 2 = ^

12. 12 Size and Strata in Parent Population N 1 = N 2 = N 3 = N 4 = N = 50000 x st = = 1/10(3.2) + 5/10(4.6) + 3/10(5.8) + 1/10(7.2) = 5.08 = (1/10) 2 (.14) + (5/10) 2 (.12) + (3/10) 2 (.20) 100 + (1/10) 2 (.18) =.000512 100 =.0226 NhxhNhxh N  L h=1 WhxhWhxh  L s x st =  L h=1 Whs2xhWhs2xh = NhNh N 2 (s h ) 2 nhnh ^  L h=1 2 s x st = I. II. III. IV. 5000 25000 15000 5000 ()