KULIAH KE 9 Elementary Statistics Eleventh Edition and the Triola Statistics Series by Mario F. Triola Copyright © 2010, 2007, 2004 Pearson Education, Inc. All Rights Reserved.

Chapter 9 Inferences from Two Samples 9-1 Review and Preview 9-2 Inferences About Two Proportions 9-3 Inferences About Two Means: Independent Samples 9-4 Inferences from Dependent Samples 9-5 Comparing Variation in Two Samples Copyright © 2010, 2007, 2004 Pearson Education, Inc. All Rights Reserved.

Pengujian Tentang Dua Proporsi Section 9-2 Pengujian Tentang Dua Proporsi Copyright © 2010 Pearson Education Copyright © 2010, 2007, 2004 Pearson Education, Inc. All Rights Reserved.

Key Concept Pada bagian ini disajikan metode untuk (1) menguji pernyataan yang dibuat tentang dua proporsi populasi dan (2) membangun kepercayaan estimasi interval perbedaan antara dua proporsi populasi. Copyright © 2010 Pearson Education Copyright © 2010, 2007, 2004 Pearson Education, Inc. All Rights Reserved.

Notation for Two Proportions For population 1, we let: p1 = population proportion n1 = size of the sample x1 = number of successes in the sample n1 x1 ^ p1 = (the sample proportion) q1 = 1 – p1 ^ The corresponding notations apply to p2, n2 , x2 , p2. and q2 , which come from population 2. ^ Copyright © 2010, 2007, 2004 Pearson Education, Inc. All Rights Reserved. Copyright © 2010 Pearson Education

Test Statistic for Two Proportions + z = ( p1 – p2 ) ^ n1 pq n2 Example given at the bottom of page 458-460 of Elementary Statistics, 10th Edition. Copyright © 2010, 2007, 2004 Pearson Education, Inc. All Rights Reserved. Copyright © 2010 Pearson Education

Test Statistic for Two Proportions = p1 ^ x1 n1 p2 x2 n2 and and q = 1 – p n1 + n2 p = x1 + x2 Copyright © 2010, 2007, 2004 Pearson Education, Inc. All Rights Reserved. Copyright © 2010 Pearson Education

Pengujian Hipotesis beda dua proporsi Prosedur Pengujian. 1 Pengujian Hipotesis beda dua proporsi Prosedur Pengujian . 1.Formulasi Hipotesis : a. Ho : P₁ = P₂ b. Ho : P₁ = P₂ c. Ho : P₁ = P₂ Hi : P₁ > P₂ Hi : P₁ < P₂ Hi : P₁ ‡ P₂ 2.Nilai Taraf nyata berdasarkan tabel Z 3.Kriteria pengujian a. a.Untuk Ho : P₁ = P₂ dan Hi : P₁ > P₂ Ho diterima , jika Z₀ ≤ Zα Ho ditolak , jika Z₀ > Zα b. Untuk Ho : P₁ = P₂ dan Hi : P₁ < P₂ Ho diterima , jika Z₀ ≥ - Zα Ho ditolak , jika Z₀ < -Zα c. Untuk Ho : P₁ = P₂ dan Hi : P₁ ‡ P₂ Ho diterima , jika -Zα/₂ ≤ Zo ≤ Zα/₂ Ho ditolak , jika Z₀ > Zα/₂ atau Zo < - Zα/₂

4.Uji Statistik. Z₀ = (P₁ - P₂) : VP(1 – P) ( 1/n₁ + 1/n₂ ) Dimana : P₁ =X₁/n₁ P₂ = X₂/n₂ P = (X₁ +X₂) : (n₁ + n₂) 5.Kesimpulan. Kesimpulan merupakan penerimaan atau penolakan dari Ho. Contoh Soal : * Suatu pemungutan suara akan dilakukan di antara penduduk kota M dan sekitarnya mengenai pendapat mereka tentang pembangunan gedung serba guna. Untuk mengetahui apakah ada perbedaan antara proporsi penduduk kota dengan penduduk disekitarnya yang menyetujui rencana itu, diambil sampel acak yakni 200 penduduk kota dan 500 penduduk di sekitarnya. Ternyata ada 120 penduduk kota dan 240 penduduk di sekitarnya yang setuju. Apakah anda setuju jika dikatakan bahwa proporsi penduduk kota yang setuju lebih tinggi dari penduduk sekitarnya ? Gunakan taraf nyata 1 %

Jawab: n₁ = 200 ; n₂ = 500 ; P₁ = 120/200 =0,6 X₁ = 120 ; X ₂= 240 ; P₂ = 240/500 = 0,48 1.Formulasi Hipotesis : Ho : P₁ = P₂ Hi : P₁ > P₂ 2.Taraf Nyata : α = 1 % = 0,01 t₀‚₀₁ = 2,33 3.Kriteria pengujian : Ho diterima , jika Z₀ ≤ 2,33 Ho ditolak ,jika Zo > 2,33 4.Uji Statistik . P = (120 + 240 ) : (200 + 500) = 0,51 Z₀ = (0,6 – 0,48) : V(0,51)(0,49) .( 1/200 + 1/500) = 2,9 5.Kesimpulan. Karena Z₀ > 2,33 , maka Ho ditolak , Jadi setuju dengan pendapat bahwa proporsi penduduk kota yang setuju rencana itu lebih kecil daripada proporsi penduduk disekitarnya.

Contoh Soal: 1. Sebuah pabrik rokok memproduksi dua merek rokok yang berbeda.Ternyata 56 orang dari 200 perokok suka merek B , dan 29 orang dari 150 perokok suka merek D. Dapatkah kita menyimpulkan pada taraf nyata 10 %, bahwa merek D terjual lebih banyak dari pada merek B ? 2. Dari sampel random yang diambil dari turis yg mengunjungi kota X , diketahui 105 dari 325 turis laki laki dan 245 dari 400 turis wanita membeli cendera mata. Ujilah dengan taraf nyata 10 % , pernyataan diatas dengan alternatif proporsi wanita lebih besar dari laki-laki.

Inferences About Two Means: Independent Samples Section 9-3 Inferences About Two Means: Independent Samples Copyright © 2010 Pearson Education Copyright © 2010, 2007, 2004 Pearson Education, Inc. All Rights Reserved.

Key Concept Bagian ini menyajikan metode untuk menggunakan data sampel dari dua sampel independen untuk menguji hipotesis dua populasi yang berarti untuk membangun perkiraan interval kepercayaan dari perbedaan antara dua populasi berarti. Copyright © 2010 Pearson Education Copyright © 2010, 2007, 2004 Pearson Education, Inc. All Rights Reserved.

Notation 1 = population mean σ1 = population standard deviation n1 = size of the first sample = sample mean s1 = sample standard deviation Corresponding notations for 2, σ2, s2, and n2 apply to population 2. page 470 of Elementary Statistics, 10th Edition Copyright © 2010, 2007, 2004 Pearson Education, Inc. All Rights Reserved. Copyright © 2010 Pearson Education

Hypothesis Test for Two Means: Independent Samples (where 1 – 2 is often assumed to be 0) Copyright © 2010 Pearson Education Copyright © 2010, 2007, 2004 Pearson Education, Inc. All Rights Reserved.

PENGUJIAN HIPOTESIS BEDA DUA RATA RATA & BEDA DUA PROPORSI

1. Pengujian Beda Dua Rata rata A 1.Pengujian Beda Dua Rata rata A.Sampel Besar ( n > 30) Prosedur Pengujian : 1.Formulasi Hipotesis : a. Ho : μ₁ = μ₂ b. Ho : μ₁ = μ₂ c. Ho : μ₁ = μ₂ Hi : μ₁ > μ₂ Hi : μ₁ < μ₂ Hi : μ₁ ‡ μ₂ 2.Penentuan Nilai α dan nilai Z 3.Kriteria Pengujian a.Untuk Ho : μ₁ = μ₂ dan Hi : μ₁ > μ₂ Ho diterima , jika Z₀ ≤ Zα Ho ditolak , jika Z₀ > Zα b. Untuk Ho : μ₁ = μ₂ dan Hi : μ₁ < μ₂ Ho diterima , jika Z₀ ≥ - Zα Ho ditolak , jika Z₀ < -Zα c. Untuk Ho : μ₁ = μ₂ dan Hi : μ₁ ‡ μ₂ Ho diterima , jika -Zα/₂ ≤ Zo ≤ Zα/₂ Ho ditolak , jika Z₀ > Zα/₂ atau Zo < - Zα/₂

4.Uji Statistik: a.Jika simpangan baku diketahui : Zo = (Ẍ₁ - Ẍ₂) : V σ₁²/n₁ + σ₂²/n₂ b.Jika simpangan baku tidak diketahui, maka : Zo = (Ẍ₁ - Ẍ₂) : V s₁²/n₁ + s₂²/n₂ 5.Kesimpulan. Kesimpulan pengujian merupakan penerimaan atau penolakan Ho.

B. Sampel Kecil ( n ≤ 30 ) Prosedur pengujian : 1 B.Sampel Kecil ( n ≤ 30 ) Prosedur pengujian : 1.Formulasi Hipotesis : a. Ho : μ₁ = μ₂ b. Ho : μ₁ = μ₂ c. Ho : μ₁ = μ₂ Hi : μ₁ > μ₂ Hi : μ₁ < μ₂ Hi : μ₁ ‡ μ₂ 2.Penentuan Nilai α dan gunakan tabel t 3.Kriteria Pengujian a.Untuk Ho : μ₁ = μ₂ dan Hi : μ₁ > μ₂ Ho diterima , jika t₀ ≤ tα Ho ditolak , jika t₀ > tα b. Untuk Ho : μ₁ = μ₂ dan Hi : μ₁ < μ₂ Ho diterima , jika t₀ ≥ - tα Ho ditolak , jika t₀ < - tα c. Untuk Ho : μ₁ = μ₂ dan Hi : μ₁ ‡ μ₂ Ho diterima , jika -tα/₂ ≤ to ≤ tα/₂ Ho ditolak , jika t₀ > tα/₂ atau to < - tα/₂

Contoh Soal . Dinas tenaga kerja berpendapat bahwa rata rata jam kerja buruh di daerah A dan daerah B sama, dengan alternatif A lebih besar dari B. Untuk itu diambil sampel dikedua daerah masing masing 100 dan 70 dengan rata rata dan simpangan baku 38 dan 9 jam perminggu serta 35 dan 7 jam perminggu. Ujilah pendapat tersebut dengan taraf nyata 5 %. Jawab : n₁ = 100 ; Ẍ₁ = 38 ; σ₁ = 9 n₂ = 70 ; Ẍ₂ = 35 ; σ₂ = 7 1,Formulasi Hipotesis : Ho : μ₁ = μ₂ Hi : μ₁ > μ₂ 2.Taraf Nyata : α = 5 % = 0,05 tα = 1,64 3.Kriteria Pengujian : Ho diterima , jika Z₀ ≤ 1,64 Ho ditolak , jika z₀ > 1,64 4.Uji Statistik. Z₀ = (38 – 35 ) : V 9²/100 + 7²/70 = 2,44 5.Kesimpulan: Karena Z₀ > Zα , maka Ho ditolak . Jadi rata rata jam kerja buruh di daerah A dan daerah B adalah tidak sama

4.Uji Statistik. A.Untuk pengamatan tidak berpasangan t₀ = (Ẍ₁ - Ẍ₂) : V (n₁ - 1)s₁² + (n₂ - 1)s₂² : (n₁ +n₂ - 2) . (1/n₁ + 1/n₂) B.Untuk pengamatan berpasangan : t₀ = ḋ : (sd/Vn) Dimana : ḋ =∑ d/n sd² = ∑d²/n-1 - (∑d)²/ n(n-1). n = banyaknya pasangan t₀ memiliki distribusi dengan db = n - 1

Contoh Soal : 1.Sebuah perusahan mengadakan pelatihan teknik pemasaran.Sampel sebanyak 12 orang dengan metode biasa dan 10 orang dengan metode terprogram.Pada akhir pelatihan diberi evaluasi dengan materi yang sama. Kelas pertama mencapai nilai rata rata 80 dengan simpangan baku 4 dan kelas ledua dengan nilai rata rata 75 dengan simpangan baku 4,5. Ujilah hipotesis kedua metode pelatihan dengan alternatif keduanya tidak sama. Gunakan taraf nyata 10 %. Jawab : n₁ = 12 ; Ẍ₁ = 80 ; σ₁ = 4 n₂ = 10 ; Ẍ₂ = 75 ; σ₂ = 4,5 1.Formulasi Hipotesis : Ho : μ₁ = μ₂ Hi : μ₁ ‡ μ₂ 2.Taraf Nyata : α = 10 % = 0,1 α/₂ = 0,05 ; db = 12 + 10 – 2 = 20 t₀‚₀₅ -₂₀ = 1,725

3.Kriteria pengujian Ho diterima , jika - 1,725≤ t₀≤ 1,725 Ho ditolak ,jika t₀ > 1,725 atau t₀ < - 1,725 4. Uji statistik : t₀ = (80-75) : V (12-1).4² + (10-1).4,5² : (12+10-2) . (1/12 + 1/10) = 2,76 5,Kesimpulan . Karena t₀ > tα , maka Ho ditolak. Jadi kedua metode yang digunakan dalam pelatihan tidak sama hasilnya.

Contoh Soal : * Seorang pemilik perusahaan produksi Bohlam berpendapat bahwa bohlam A dan Bohlam B tidak memiliki perbedaan rata rata lamanya menyala dengan alternatif tidak sama. Untuk menguji pendapatnya, dilakukan percobaan dengan menyalakan 75 bohlam A dan 40 bohlam B sebagai sampel random. Ternyata diperoleh hasil bahwa rata rata menyala bohlam A adalah 945 jam dan bohlam B 993 jam, dengan simpangan baku 88 jam dan 97 jam. Ujilah pendapat itu dengan taraf nyata 5 %.

Hypothesis Test Statistic for Matched Pairs to = d sd n where degrees of freedom = n – 1 đ = ∑ d : n S²d = ∑d² : n-1 - (∑ d)² : n(n-1) page 485 of Elementary Statistics, 10th Edition Copyright © 2010 Pearson Education Copyright © 2010, 2007, 2004 Pearson Education, Inc. All Rights Reserved.

Contoh soal pengamatanberpasangan Contoh soal pengamatanberpasangan. * Untuk mengetahui apakah keanggotaan dalam organisasi mahasiswa memiliki akibat baik atau buruk terhadap prestasi akademik seseorang, diadakan penelitian mutu rata rata prestasi akademik dala 5 tahun sbb : Ujilah pada taraf nyata 1 % ,apakah keanggotaan dalam organisasi mahasiswa berakibat baik pada prestasi akademik ? TAHUN 1 2 3 4 5 Anggota Bukan Anggota 7,0 7,2 6,9 7,3 7,5 7,1 7,4

Contoh Soal : * Sebuah sampel random diambil dari 6 salesmen untuk diselidiki hasil penjualannya pada semester I dan II suatu produk tertentu. Hasil sbb : Ujilah pada taraf nyata 5 %, apakah hasil penjualan semester I lebih baik daripada semester II. Salesmen Penjualan Semester I Semester II A B C D E F 146 166 189 162 159 165 145 154 180 170 161

Jawab : 1. Formulasi Hipotesis : Ho : μ₁ = μ₂ Hi : μ₁ < μ₂ 2 Jawab : 1.Formulasi Hipotesis : Ho : μ₁ = μ₂ Hi : μ₁ < μ₂ 2.Taraf Nyata : α = 1 % = 0,01 ; db = 5 – 1 = 4 t₀‚₀₁ _₄ = 3,747 3.kriteria Pengujian : Ho diterima , jika t₀ ≥ -3,747 ho ditolak ,jika t₀ < - 3,747 Uji statistik : ḋ = - 0,5/ 5 = - 0,1 s²d = 0,13/ 4 - (-0,5)²/20 = 0,02 sd = 0,14 t₀ = - 0,1 : 0,14/V5 = - 1,6 5.Kesimpulan Karena t₀ > -tα , maka Ho diterima .Jadi keanggotaan organisasi kemahasiswaan tidak memberikan pengaruh buruk terhadap prestasi akademik. ANGGOTA BUKAN ANGGOTA d d² 7,0 7,3 7,1 7,4 7,2 6,9 7,5 -0,2 0,1 0,0 0,04 0,01 0,00 Jumlah - 0,5 0,13

2.Suatu sampel random yang terdiri dari 6 orang dewasa diklasifikasikan berdasarkan absensi ketidakhadiran. Dengan menggunakan taraf nyata 10 %, ujilah hipotesis bahwa ketidakhadiran thn.2015 lebih baik. Nama pekerja Thn.2014 Thn.2015 Andy Yanto Bambang Charles Dony Reyhan 5 3 6 8 7 1 4 Copyright © 2010, 2007, 2004 Pearson Education, Inc. All Rights Reserved.

Independent Samples with σ1 and σ2 Known. Copyright © 2010 Pearson Education Copyright © 2010, 2007, 2004 Pearson Education, Inc. All Rights Reserved.

Hypothesis Test for Two Means: Independent Samples with σ1 and σ2 Both Known (x1 – x2) – (µ1 – µ2) z = σ1 σ2 2 2 + n1 n2 P-values and critical values: Refer to Table A-2. Copyright © 2010, 2007, 2004 Pearson Education, Inc. All Rights Reserved. Copyright © 2010 Pearson Education

Hypothesis Test Statistic for Two Means: Independent Samples and σ1 = σ2 = (n1 – 1) + (n2 -1) (x1 – x2) – (µ1 – µ2) t = n1 n2 + sp 2 sp. Where s1 s2 (n1 – 1) + (n2 – 1) and the number of degrees of freedom is df = n1 + n2 - 2 Copyright © 2010 Pearson Education Copyright © 2010, 2007, 2004 Pearson Education, Inc. All Rights Reserved.

Hypothesis Test Statistic for Matched Pairs sd n where degrees of freedom = n – 1 page 485 of Elementary Statistics, 10th Edition Copyright © 2010 Pearson Education Copyright © 2010, 2007, 2004 Pearson Education, Inc. All Rights Reserved.

P-values and Critical Values Use Table A-3 (t-distribution). page 485 of Elementary Statistics, 10th Edition Hypothesis example given on this page Copyright © 2010, 2007, 2004 Pearson Education, Inc. All Rights Reserved. Copyright © 2010 Pearson Education

Confidence Intervals for Matched Pairs d – E < µd < d + E where E = t/2 sd n Critical values of tα/2 : Use Table A-3 with n – 1 degrees of freedom. Copyright © 2010, 2007, 2004 Pearson Education, Inc. All Rights Reserved. Copyright © 2010 Pearson Education

Example: Step 6: find values of d and sd differences are: –1, –1, 4, –2, 1 d = 0.2 and sd = 2.4 now find the test statistic Table A-3: df = n – 1, area in two tails is 0.05, yields a critical value t = ±2.776 Copyright © 2010, 2007, 2004 Pearson Education, Inc. All Rights Reserved. Copyright © 2010 Pearson Education

Example: Confidence Interval method: Construct a 95% confidence interval estimate of d , which is the mean of the “April–September” weight differences of college students in their freshman year. = 0.2, sd = 2.4, n = 5, ta/2 = 2.776 Find the margin of error, E Copyright © 2010 Pearson Education Copyright © 2010, 2007, 2004 Pearson Education, Inc. All Rights Reserved.

Example: Construct the confidence interval: We have 95% confidence that the limits of 2.8 kg and 3.2 kg contain the true value of the mean weight change from September to April. In the long run, 95% of such samples will lead to confidence interval limits that actually do contain the true population mean of the differences. Copyright © 2010, 2007, 2004 Pearson Education, Inc. All Rights Reserved. Copyright © 2010 Pearson Education

Recap In this section we have discussed: Requirements for inferences from matched pairs. Notation. Hypothesis test. Confidence intervals. Copyright © 2010 Pearson Education Copyright © 2010, 2007, 2004 Pearson Education, Inc. All Rights Reserved.

Finding Critical F Values To find a critical F value corresponding to a 0.05 significance level, refer to Table A-5 and use the right-tail are of 0.025 or 0.05, depending on the type of test: Two-tailed test: use 0.025 in right tail page 496 of Elementary Statistics, 10th Edition One-tailed test: use 0.05 in right tail Copyright © 2010, 2007, 2004 Pearson Education, Inc. All Rights Reserved. Copyright © 2010 Pearson Education

Finding Critical F Values page 496 of Elementary Statistics, 10th Edition Copyright © 2010 Pearson Education Copyright © 2010, 2007, 2004 Pearson Education, Inc. All Rights Reserved.

Properties of the F Distribution - continued If the two populations do have equal variances, then F = will be close to 1 because and are close in value. s1 2 s2 s1 2 s 2 page 497 of Elementary Statistics, 10th Edition Give some numerical examples of the computation of F close to 1. Copyright © 2010, 2007, 2004 Pearson Education, Inc. All Rights Reserved. Copyright © 2010 Pearson Education

Properties of the F Distribution - continued If the two populations have radically different variances, then F will be a large number. Remember, the larger sample variance will be s1 . 2 Give some numerical examples of an F value that is not close to 1. Copyright © 2010 Pearson Education Copyright © 2010, 2007, 2004 Pearson Education, Inc. All Rights Reserved.

Conclusions from the F Distribution Consequently, a value of F near 1 will be evidence in favor of the conclusion that 1 = 2 . 2 But a large value of F will be evidence against the conclusion of equality of the population variances. Copyright © 2010 Pearson Education Copyright © 2010, 2007, 2004 Pearson Education, Inc. All Rights Reserved.

Example: Data Set 20 in Appendix B includes weights (in g) of quarters made before 1964 and weights of quarters made after 1964. Sample statistics are listed below. When designing coin vending machines, we must consider the standard deviations of pre-1964 quarters and post-1964 quarters. Use a 0.05 significance level to test the claim that the weights of pre-1964 quarters and the weights of post-1964 quarters are from populations with the same standard deviation. Copyright © 2010 Pearson Education Copyright © 2010, 2007, 2004 Pearson Education, Inc. All Rights Reserved.

Example: Requirements are satisfied: populations are independent; simple random samples; from populations with normal distributions Copyright © 2010 Pearson Education Copyright © 2010, 2007, 2004 Pearson Education, Inc. All Rights Reserved.