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Sampling Methods Beberapa istilah

Learn about different sampling methods including simple random, systematic, stratified, and cluster sampling. Explore sample distributions, estimation methods, and confidence intervals. Discover how to select samples effectively.

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Sampling Methods Beberapa istilah

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  1. Sampling MethodsBeberapa istilah • Populasi • Sampel • Parameter • Statistik

  2. Some of reason for the sampling are : To contact the whole population would be time consuming The cost of studying all the items in a population may be prohibitive The physical impossibility of checking all items in the population. The destructive nature of some tests The sample results are adequate

  3. Methods to Select a Sample • Simple random sampling→ a sample selected so that each item or person in the population has the same chance of being included. • Systematic random sampling → a random starting point is selected, and then every kth member of the population is selected. • Stratified random sampling → a population is divided into subgroups, called strata, and a sample is randomly selected from each stratum. • Cluster sampling → a population is divided into clusters using naturally occurring geographic or other boundaries. Then, clusters are randomly selected and a sample is collected by randomly selecting from each cluster.

  4. Sampel random sederhanamemillih sampel dengan metode yang memberikan kesempatan yang sama kepada setiap calon anggota sampel dari anggota populasinya untuk menjadi anggota sampel. Misalnya dalam suatu proses recruiting karyawan baru terdapat 4 orang pelamar yaitu A,B,C, dan D. Dari pelamar ini akan dipilih 2 pelamar untuk mengikuti tes wawancara. Kombinasi yang mungkin sebanyak 2 yang dipilih dari 4 pelamar dengan kesempatan yang sama kepada setiap pelamar adalah pasangan AB,AC,AD,BC,CD. Karena terdapat 6 pasangan yang mungkin terjadi, maka masing-masing mempunyai kesempatan probabiltas yang sama yaitu 1/6

  5. Sampel sistematismemilih anggota sampel dari suatu populasi dengan interval sama, biasanya diukur dengan ukuran waktu, urutan, ranking atau tempat. Misal kita menginginkan informasi mengenai penghasilan rata-rata pedagang kaki lima dengan menggunakan interval urutan, terlebih dahulu ditentukan urutan yang ke berapa dari anggota populasi yang dipilih menjadi anggota sampel.Misalnya pemilihan menggunakan daftar nama pedagang kaki lima, kemudian akan dipilih secara random dimulai dari urutan kelima sebagai data pertama. Pengambilan anggota sampel berikutnya diambil pedagang kaki lima pada urutan kelima berikutnya pada daftar tersebut.

  6. Sampel bertingkatmemilih anggota sampel dengan cara membagi populasi menjadi beberapa lapisan, disebut strata, secara acak. Misalnya pada suatu penelitian untuk mengetahui minat masyarakat terhadap penggunaan ATM. masyarakat yang akan diteliti dibagi menjadi beberapa lapisan, misalnya pedagang,pegawai negeri, pegawai swasta. Anggota sampel yang digunakan dalam penelitian merupakan penjumlahan dari anggota masyarakat yang bekerja sebagai pedagang, pegawai negeri,dan pegawai swasta dipilih secara acak.

  7. Sampel berkelompokmemilih sampel dengan membuat populasi menjadi beberapa kelompok misal dalam satu penelitian bertujuan mengetahui pola perubahan pengeluaran masyarakat di kota Bengkulu. Maka kita bagi kota Bengkulu menjadi beberapa lokasi pemilihan sampel, yaitu Kec.Gading Cempaka, Kec. Selebar, Kec. Muara Bangkahulu, Kec Teluk Segara dst. Pada masing- masing kecamatan dipilih beberapa keluarga secara acak untuk membentuk sampel yang diperlukan.

  8. Sampling Distribution of The Sample Mean Sample distribution of the sample mean→ a probability distribution of all possible sample means of a given sample size Example : Tartus industries has seven production employees.The hourly of each employee are given in table :

  9. The Questions a.What is the populations mean? b.What is the sampling distribution of the sample mean for the samples of size 2? c.What the mean of the sampling distributions ? d.What the observasions can be made about the population and the sampling distribution?

  10. The Answer a.What is the populations mean? μ = $7 + $7+ $8 +$8 + $7 + $8 + $9 = $7.71 7 b. What is the sampling distribution of the sample mean for the samples of size 2 ? NCn = N! = 7! = 21 n!(N-n)! 2!(7-2)! c. What the mean of the sampling distributions ? μx = Sum of all sample means Total number of samples = $7.00+$7.50+…+$8.50 = $7.71 21

  11. Population Values 40 30 20 10 0 7 8 9 Distribution of sample mean 40 30 20 10 0 7 7.5 8 8.5 9 d. What the observasions can be made about the population and the sampling distribution?

  12. The Central Limit Theorem Central limit theorem→ if all samples of a particular size are selected from any population, the sampling distribution of the sample mean is approximately a normal distribution. This approximation improves with large samples.

  13. ESTIMATION AND CONFIDENCE INTERVALS Point Estimate→ the statistic, computed from sample information, which is used to estimate the population parameter. Confidence Interval → a range of value constructed from sample data so that the population parameter is likely to occur within that range at a specified probability. The specified probability is called the level of confidence. Confidence interval for the population mean (n > 30) = X + z s √ n

  14. Example : The American Management Assosiation wishes to have information on the income of middle managers in the retail idustry. A random sample of 256 managers reveals a sample mean of $45,420. The standard deviation of this sample is $2,050. The assosiation would like answer to the following

  15. Questions : a.What is the population mean? b.What is a reasonable range of values for the population mean? c.What do these results mean?

  16. The Answers : • What is the population mean ? in this case, we do not know. We do know the sample mean is $45,420. Hence, our best estimate of the unknown population value is the corresponding sample statistic. Thus the sample mean of $45, 420 is a point estimate of the unknown population mean.

  17. b.What is a reasonable range of values for the population mean? The association decides to use the 95 percent level of confidence. To determine the corresponding confidence interval we use : X + z s = $45,420 + 1.96 $ 2,050 √ n √ 256 = $45,420 + $251 the usual practice is to round these endpoints to $45,169 and $45,671. these endpoints called the confidence limits. The degree of confidence or the level of confidence is 95 percent and the confidence interval is from $45,169 to $45,671.

  18. c. What do these results mean? Suppose we select many samples of 256 managers, perhaps several hundred. For each sample,we compute the mean and the standard deviation and then construct a 95 percent confidence interval, such as we did in the previous section. We could expect about 95% of these confidence intervals to contain the population mean. About 5% of the intervals would not contain the population mean annual income, which is μ.

  19. A Confidence Interval for Proportion Proportion→ the fraction, ratio, or percent indicating the part of the sample or the population having a particular trait of interest. Sample Proportion : p = x n Confidence Interval for a Population Proportion p + zσp StandarError of The Sample Proportion σp = √ p(1-p) n

  20. Confidence Interval for A Population Proportion p + z = √ p(1-p) n Example : The union representing the Bottle Blowers of America (BBA) is considering a proposal to merge with the teamsters union. According to BBA union Bylaws, at least threefourths of the union Membership must approve any merger. A random Sample of 2,000 current BBA members reveals 1,600 plan to vote for the merger proposal.

  21. Questions: What is the estimate of the population proportion ? Develop a 95 % confidence interval for the Population proportion. Basing your decision on this Sample information, can you conclude that the Necessary of BBA member favor the merger, why?

  22. Answer : First, calculate the sample proportion from formula p = x = 1,600 = 0,80 n 2,000 Thus, we estimate that 80% of the population favor the merger proposal. The z value corresponding to the 95% level of confidence is 1.96 p + z √ p(1-p) = 0.80 + 1.96√ 0.80(1-0.80) n 2,000 = 0.80 + 0.018 the endpoints of the confidence interval are .782 and .818. the lower endpoint is greater than .75. Hence, we conclude that the merger proposal will likely pass because the interval estimate includes values greater than 75% of the unions memberships.

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