Kuliah ke.7 Elementary Statistics Eleventh Edition

Kuliah ke.7 Elementary Statistics Eleventh Edition
and the Triola Statistics Series by Mario F. Triola

Chapter 7 Estimates and Sample Sizes
7-1 Review and Preview 7-2 Estimating a Population Proportion 7-3 Estimating a Population Mean: σ Known 7-4 Estimating a Population Mean: σ Not Known 7-5 Estimating a Population Variance

Section 7-1 Review and Preview

Review Chapters 2 & 3 we used “descriptive statistics” when we summarized data using tools such as graphs, and statistics such as the mean and standard deviation. Chapter 6 we introduced critical values: z denotes the z score with an area of  to its right. If  = 0.025, the critical value is z0.025 = That is, the critical value z0.025 = 1.96 has an area of to its right.

Estimating a Population Proportion
Section 7-2 Estimating a Population Proportion

Key Concept Bagian ini menyajikan metode menggunakan proporsi sampel untuk memperkirakan nilai dari proporsi populasi.. Proporsi sampel adalah estimasi titik dari proporsi populasi. Kita dapat menggunakan proporsi sampel untuk membangun sebuah interval kepercayaan untuk memperkirakan nilai sebenarnya dari proporsi populasi, dan kita harus tahu bagaimana menafsirkan interval kepercayaan tersebut. Kita harus tahu bagaimana menemukan ukuran sampel yang diperlukan untuk memperkirakan proporsi populasi.

Definition ˆ Proporsi sampel p adalah estimasi dari proporsi populasi p.

Definition Sebuah estimasi titik adalah nilai tunggal (atau titik) digunakan untuk memperkirakan suatu parameter populasi..

Pendugaan/estimasi merupakan suatu pernyataan mengenai parameter populasi yang diketahui berdasarkan informasi dari sampel. Jadi dengan pendugaan ini, keadaan parameter populasi dapat diketahui. Jenis jenis Pendugaan. 1.Pendugaan Tunggal, yakni pendugaan yang hanya mempunyai atau menyebutkan satu nilai Contoh : Pendugaan untuk μ adalah rata rata dari sampel Ẍ, yang dirumusksn dengan Ẍ = (x1 + x2 + x3 …+ xn ) : n Penduga untuk p adalah P, yakni proporsi dalam sampel yang dirumuskan dengan P = X / n

2.Pendugaan Interval adalah pendugaan yang mempunyai dua nilai sebagaipembatasan atau daerah pembatasan. Pada pendugaan interval digunakan tingkat keyakinan terhadap daerah yang nilai sebenarnya/parameternya akan berada. Bentuk umunya : p – Zα/2 . σ < P < p + Zα/2. σ Dimana : p = proporsi sampel ( p = X/n) Z α/2 = koefisien yg sesuai dengan interval keyakinan dan nilainya diambil dari tabel kurva normal σ = simpangan baku P = pendugaan Tingkat keyakinan /koefisien keyakinan dilambangkan dengan : C = 1 - α

Pendugaan Interval untuk Proporsi 1
Pendugaan Interval untuk Proporsi 1.Untuk sampel besar ( n> 30) Dugunakan Rumus : p – tα/₂ . V p(1-p)/n < P < p + tα/₂ . V p(1-p)/n Contoh. Sebuah peti kemas diperiksa untuk ditaksir persentase barang yang rusak. Untuk itu diambil 60 buah barang yang ada dan diperoleh 9 buah rusak. Dugalah persentase barang yang rusak dengan interval keyakinan 99 %. Jawab. n= ; X = 9 ; p = 9/60=0,15 ; α = 1% =0, t α/₂ = t₀‚₀₀₅ = 2,58 0,15 - 2,58 .V0,15(1-0,15)/60 < P < 0,15 +2,58 .V 0,15(1-0,15)/ , < P < 0, ,11 % < P < 26,89 % Jadi persentase kerusakan barang pada interval keyakinan 99 % adalah 3,11 % sampai 26,89 %

2. Untuk sampel kecil (n < 30 ) Rumus yang digunakan : p – tα/₂
2. Untuk sampel kecil (n < 30 ) Rumus yang digunakan : p – tα/₂ . V p(1-p)/n < P < p + tα/₂ . V p(1-p)/n Contoh . *Dari 20 karyawan , 6 diantaranya memiliki mobil. Dengan interval keyakinan 95 %, tentukan proporsi karyawan yang memiliki mobil. Jawab. n= ; X =6 ; p = 6/20 = 0,3 ; α = 5 % = 0, α/₂ = 0,025 ; n-1=20-1=19 ; t₀‚₀₂₅ : 19 = 2, ,3 - 2,093 .V(0,3)(0,7)/20 < P < 0,3 + 2,093 . V(0,3)(0,7)/ ,0855 < P < 0, ,55 % < P < 51,45 % Jadi proporsi karyawan yang punya mobil antara 8,55 % sampai 51,45 %.

Contoh Soal : 1. Sebuah peti kemas milik PT
Contoh Soal : 1.Sebuah peti kemas milik PT.Gombal diperiksa untuk menaksir persentase barang yang rusak. Untuk keperluan itu, diambil 50 buah barang yang ada dalam peti kemas itu dan diperoleh 8 buah yang rusak. Dugalah persentase barang yang rusak dalam peti tersebut, dengan interval keyakinan 95 %. 2.Penelitian terhadap sampel sebanyak 12 mahasiswa, ternyata 6 mahasiswa diantaranya kekampus dengan bersepeda .Dengan interval keyakinan 95 %, tentukan proporsi mahasiswa yang berkendaraan sepeda.

Definition Kebanyakan pilihan tingkat kepercayaan
Tingkat kepercayaan probabilitas adalah C=1 - α (sering dinyatakan sebagai nilai persentase setara) bahwa selang kepercayaan sebenarnya mengandung parameter populasi, dengan asumsi bahwa proses estimasi diulang berkali kali. (Tingkat kepercayaan disebut juga koefisien kepercayaan.) Kebanyakan pilihan tingkat kepercayaan adalah 90%, 95%, atau 99%.. ( = 10%), ( = 5%), ( = 1%)

Critical Values Skor z standar dapat digunakan untuk membedakan antara statistik sampel yang mungkin Nilai z disebut nilai kritis. nilai kritis didasarkan pada observasi berikut: 1. Dalam kondisi tertentu, distribusi sampling dari proporsi sampel dapat didekati dengan distribusi normal.

Critical Values 2. Sebuah skor z terkait dengan proporsi sampel memiliki probabilitas α / 2 jatuh di ekor kanan-kiri

z2 The Critical Value

Finding z2 for a 95% Confidence Level
 = 5%  2 = 2.5% = .025 -z2 z2 Critical Values

Finding z2 for a 95% Confidence Level - cont
 = 0.05 Use Table A-2 to find a z score of 1.96 z2 =  

Margin of Error for Proportions

Confidence Interval for Estimating a Population Proportion p
p = population proportion = sample proportion n = number of sample values E = margin of error z/2 = z score separating an area of /2 in the right tail of the standard normal distribution

p – E < < E ˆ p where

p – E < < E p + E p ˆ ˆ (p – E, p + E) ˆ

Example: In the Chapter Problem we noted that a Pew Research Center poll of 1501 randomly selected U.S. adults showed that 70% of the respondents believe in global warming. The sample results are n = 1501, and a. Find the margin of error E that corresponds to a 95% confidence level. b. Find the 95% confidence interval estimate of the population proportion p. c. Based on the results, can we safely conclude that the majority of adults believe in global warming? d. Assuming that you are a newspaper reporter, write a brief statement that accurately describes the results and includes all of the relevant information.

Example: Requirement check: simple random sample; fixed number of trials, 1501; trials are independent; two categories of outcomes (believes or does not); probability remains constant. a) Use the formula to find the margin of error.

Example: b) The 95% confidence interval:

Sample Size Suppose we want to collect sample data in order to estimate some population proportion. The question is how many sample items must be obtained?

Determining Sample Size
p q ˆ n (solve for n by algebra) ( )2 ˆ p q a / 2 Z n = E 2

Sample Size for Estimating Proportion p
When an estimate of p is known: ˆ ˆ ( )2 p q n = E 2 a / 2 z When no estimate of p is known: ( )2 0.25 n = E 2 a / 2 z ˆ

Example: The Internet is affecting us all in many different ways, so there are many reasons for estimating the proportion of adults who use it. Assume that a manager for E-Bay wants to determine the current percentage of U.S. adults who now use the Internet. How many adults must be surveyed in order to be 95% confident that the sample percentage is in error by no more than three percentage points? a. In 2006, 73% of adults used the Internet. b. No known possible value of the proportion.

Example: a) Use To be 95% confident that our sample percentage is within three percentage points of the true percentage for all adults, we should obtain a simple random sample of 842 adults.

Example: b) Use To be 95% confident that our sample percentage is within three percentage points of the true percentage for all adults, we should obtain a simple random sample of 1068 adults.

Recap In this section we have discussed: Point estimates.
Confidence intervals. Confidence levels. Critical values. Margin of error. Determining sample sizes.

Estimating a Population Mean:  Known
Section 7-3 Estimating a Population Mean:  Known

Key Concept This section presents methods for estimating a population mean. In addition to knowing the values of the sample data or statistics, we must also know the value of the population standard deviation, . Here are three key concepts that should be learned in this section:

Key Concept 1. We should know that the sample mean is the best point estimate of the population mean . 2. We should learn how to use sample data to construct a confidence interval for estimating the value of a population mean, and we should know how to interpret such confidence intervals. 3. We should develop the ability to determine the sample size necessary to estimate a population mean.

Point Estimate of the Population Mean
The sample mean x is the best point estimate of the population mean µ.

Confidence Interval for Estimating a Population Mean (with  Known)
 = population standard deviation = sample mean n = number of sample values page 338 of Elementary Statistics, 10th Edition E = margin of error z/2 = z score separating an area of a/2 in the right tail of the standard normal distribution

1. The sample is a simple random sample. (All samples of the same size have an equal chance of being selected.) 2. The value of the population standard deviation  is known. 3. Either or both of these conditions is satisfied: The population is normally distributed or n > 30. page 338 of Elementary Statistics, 10th Edition

page 338 of Elementary Statistics, 10th Edition

Sample Mean 1. For all populations, the sample mean x is an unbiased estimator of the population mean , meaning that the distribution of sample means tends to center about the value of the population mean . 2. For many populations, the distribution of sample means x tends to be more consistent (with less variation) than the distributions of other sample statistics. page 339 of Elementary Statistics, 10th Edition

Procedure for Constructing a Confidence Interval for µ (with Known )
1. Verify that the requirements are satisfied. 2. Refer to Table A-2 or use technology to find the critical value z2 that corresponds to the desired confidence level. 3. Evaluate the margin of error 4. Find the values of Substitute those values in the general format of the confidence interval: page 340 of Elementary Statistics, 10th Edition 5. Round using the confidence intervals round-off rules.

Round-Off Rule for Confidence Intervals Used to Estimate µ
When using the original set of data, round the confidence interval limits to one more decimal place than used in original set of data. When the original set of data is unknown and only the summary statistics (n, x, s) are used, round the confidence interval limits to the same number of decimal places used for the sample mean.

Example: People have died in boat and aircraft accidents because an obsolete estimate of the mean weight of men was used. In recent decades, the mean weight of men has increased considerably, so we need to update our estimate of that mean so that boats, aircraft, elevators, and other such devices do not become dangerously overloaded. Using the weights of men from Data Set 1 in Appendix B, we obtain these sample statistics for the simple random sample: n = 40 and = lb. Research from several other sources suggests that the population of weights of men has a standard deviation given by  = 26 lb.

Example: a. Find the best point estimate of the mean weight of the population of all men. b. Construct a 95% confidence interval estimate of the mean weight of all men. c. What do the results suggest about the mean weight of lb that was used to determine the safe passenger capacity of water vessels in 1960 (as given in the National Transportation and Safety Board safety recommendation M-04-04)?

Example: a. The sample mean of lb is the best point estimate of the mean weight of the population of all men. b. A 95% confidence interval or 0.95 implies  = 0.05, so z/2 = Calculate the margin of error. Construct the confidence interval.

Example: c. Based on the confidence interval, it is possible that the mean weight of lb used in 1960 could be the mean weight of men today. However, the best point estimate of lb suggests that the mean weight of men is now considerably greater than lb. Considering that an underestimate of the mean weight of men could result in lives lost through overloaded boats and aircraft, these results strongly suggest that additional data should be collected. (Additional data have been collected, and the assumed mean weight of men has been increased.)

Finding a Sample Size for Estimating a Population Mean
σ = population standard deviation = population standard deviation E = desired margin of error zα/2 = z score separating an area of /2 in the right tail of the standard normal distribution (z/2)   n = E 2

Round-Off Rule for Sample Size n
If the computed sample size n is not a whole number, round the value of n up to the next larger whole number. page 343 of Elementary Statistics, 10th Edition

Finding the Sample Size n When  is Unknown
1. Use the range rule of thumb (see Section 3-3) to estimate the standard deviation as follows:   range/4. 2. Start the sample collection process without knowing  and, using the first several values, calculate the sample standard deviation s and use it in place of  . The estimated value of  can then be improved as more sample data are obtained, and the sample size can be refined accordingly. page 344 of Elementary Statistics, 10th Edition 3. Estimate the value of  by using the results of some other study that was done earlier.

Example: Assume that we want to estimate the mean IQ score for the population of statistics students. How many statistics students must be randomly selected for IQ tests if we want 95% confidence that the sample mean is within 3 IQ points of the population mean?  = 0.05 /2 = 0.025 z / 2 = 1.96 E = 3  = 15 n = • = = 97 3 2 Example on page 344 of Elementary Statistics, 10th Edition With a simple random sample of only 97 statistics students, we will be 95% confident that the sample mean is within 3 IQ points of the true population mean .

Recap In this section we have discussed: Margin of error.
Confidence interval estimate of the population mean with σ known. Round off rules. Sample size for estimating the mean μ.

Estimating a Population Mean:  Not Known
Section 7-4 Estimating a Population Mean:  Not Known

Sample Mean Sampel rata adalah estimasi titik terbaik dari mean populasi. Smaller samples will have means that are likely to vary more. The greater variation is accounted for by the t distribution. Students usually like hearing about the history of how the Student t distribution got its name. page 350 of text See Note to Instructor in margin for other ideas.

Pendugaan Interval untuk Rata rata a
Pendugaan Interval untuk Rata rata a.Untuk Sampel Besar ( n > 30 ) Rumus perhitungannya : Ẍ - Z α/2 . σ/vn < μ < Ẍ + Zα/2 . σ/Vn Contoh : Sebuah restauran mengadakan penelitian dengan mengambil sampel 300 karyawan.Ternyata rata rata pengeluaran untuk membeli makanan adalah Rp setahun dengan simpangan baku Rp Dugalah rata rata pengeluaran karyawan untuk membeli makanan dalam setahun dengan interval keyakinan 95 % Jawab . n = ; Ẍ = ; σ = ; α = 5 % = 0, – Z 0, /V300 < μ < Z0, /V (1,96)(9.526, < μ < (1,96)(9.526,28) , < μ < ,51 Artinya: Dugaan bahwa rata rata pengeluaran karyawan yg berada antara Rp ,49 sampai Rp ,51 , akan benar 95 %

B. Untuk Sampel Kecil , n < 30 Rumus yang digunakan : Ẍ - tα/2
B.Untuk Sampel Kecil , n < 30 Rumus yang digunakan : Ẍ - tα/2 . s/Vn < μ < Ẍ + tα/2 . s/Vn s = V ∑ X²/n (∑X)²/n(n-1) Contoh .Dari 9 orang karyawan , memiliki waktu yang diperlukan untuk menyelesaikan sebuah pekerjaan yakni 14; 17; 15;18;18;14;15;19 dan 15 menit. Dugalah rata rata waktu yang digunakan bagi karyawan tersebut dengan interval keyakinan 99 %. Jawab. n=9 ; ∑ X = ; ∑ X² = ; Ẍ =145/9 = 16, α = 1-99% = 1%= 0,01 ; n-1 = 9 – 1 = 8 ; tα/2= 3, s = V 2.365/8 - (145)²/ = V 3,61 = 1, ,11 – (3,355) . (1,9/3) < μ < 16,11 + (3,355). (1,9/3) , < μ < 18,235 Jadi Rata rata waktu yang digunakan oleh karyawan dengan interval keyakinan 99 % berkisar antara 13,985 menit sampai 18,235 menit.

Contoh Soal : 1.PT.Mekar, mengadakan penelitian mengenai IQ para karyawannya. Untuk itu diambil sampel karyawan secara acak.Jika diketahui rata rata IQ sampel adalah 109 dengan simpangan baku , buatlah pendugaan interval rata rata dengan tingkat kepercayaan 95 % 2.Lima karyawan PT. Teliti dipilih secara acak, kemudian diukur beratnya. Data diperoleh sbb : ; 67 ; 70 ; 65 dan 60 Kg. Buatlah pendugaan interval rata rata nya dengan tingkat keyakinan %

Definition degrees of freedom = n – 1 in this section.
The number of degrees of freedom for a collection of sample data is the number of sample values that can vary after certain restrictions have been imposed on all data values. The degree of freedom is often abbreviated df. degrees of freedom = n – 1 in this section. Explain that other statistical procedures may have different formulas for degrees of freedom.

Margin of Error E for Estimate of  (With σ Not Known)
Formula 7-6 where t/2 has n – 1 degrees of freedom. n s E = t/ 2 Table A-3 lists values for tα/2

Notation  = population mean = sample mean
s = sample standard deviation n = number of sample values E = margin of error t/2 = critical t value separating an area of /2 in the right tail of the t distribution Smaller samples will have means that are likely to vary more. The greater variation is accounted for by the t distribution. Students usually like hearing about the history of how the Student t distribution got its name. page 350 of text See Note to Instructor in margin for other ideas.

Confidence Interval for the Estimate of μ (With σ Not Known)
x – E < µ < x + E where E = t/2 n s df = n – 1 t/2 found in Table A-3

Penentuan Ukuran Sampel Pendugaan Contoh : Tentukan besar sampel (n) yang harus diambil untuk menyelidiki waktu rata rata yg digunakan oleh mahasiswa untuk sebuah soal ujian statistik, jika digunakan interval keyakinan 95 % dengan kesalahan duga 0.08 menit dan simpangan baku 0,7 menit. Jawab : n s E = t/ 2

Example: A common claim is that garlic lowers cholesterol levels. In a test of the effectiveness of garlic, 49 subjects were treated with doses of raw garlic, and their cholesterol levels were measured before and after the treatment. The changes in their levels of LDL cholesterol (in mg/dL) have a mean of 0.4 and a standard deviation of Use the sample statistics of n = 49, = 0.4 and s = 21.0 to construct a 95% confidence interval estimate of the mean net change in LDL cholesterol after the garlic treatment. What does the confidence interval suggest about the effectiveness of garlic in reducing LDL cholesterol?

Example: Requirements are satisfied: simple random sample and n = 49 (i.e., n > 30). 95% implies a = With n = 49 margin of error is:

Construct the confidence interval:
Example: Construct the confidence interval: We are 95% confident that the limits of –5.6 and 6.4 actually do contain the value of , the mean of the changes in LDL cholesterol for the population. Because the confidence interval limits contain the value of 0, it is very possible that the mean of the changes in LDL cholesterol is equal to 0, suggesting that the garlic treatment did not affect the LDL cholesterol levels. It does not appear that the garlic treatment is effective in lowering LDL cholesterol.

Student t Distributions for n = 3 and n = 12
Student t distributions have the same general shape and symmetry as the standard normal distribution, but reflect a greater variability that is expected with small samples. Figure 7-5

Choosing the Appropriate Distribution

Estimating a Population Variance
Section 7-5 Estimating a Population Variance

Key Concept Bagian ini kami memperkenalkan distribusi probabilitas chi-square sehingga kita dapat membangun perkiraan interval kepercayaan dari populasi standar deviasi atau varians. Kami juga menyajikan metode untuk menentukan ukuran sampel yang diperlukan untuk memperkirakan populasi standar deviasi atau varians..

Chi-Square Distribution
(n – 1) s2  2 = 2 where n = sample size s 2 = sample variance 2 = population variance One place where students tend to make errors with this formula is switching the values for s and sigma. Careful reading of the problem should eliminate this difficulty. page 364 of Elementary Statistics, 10th Edition degrees of freedom = n – 1

Properties of the Distribution of the Chi-Square Statistic
1. The chi-square distribution is not symmetric, unlike the normal and Student t distributions. As the number of degrees of freedom increases, the distribution becomes more symmetric. Different degrees of freedom produce different shaped chi-square distributions. Chi-Square Distribution Chi-Square Distribution for df = 10 and df = 20

Critical Values of the Chi-Square Distribution
Example Critical Values of the Chi-Square Distribution

Estimators of 2 The sample variance s2 is the best point estimate of the population variance 2.

Confidence Interval for Estimating a Population Standard Deviation or Variance
s = sample standard deviation n = number of sample values = left-tailed critical value of 2  2 = population variance s 2 = sample variance E = margin of error Remind students that we have started with the variance which is the square of the standard deviation. Therefore, taking the square root of the variance formula will produce the standard deviation formula. = right-tailed critical value of 2

Requirements: 1. The sample is a simple random sample. 2. The population must have normally distributed values (even if the sample is large). Remind students that we have started with the variance which is the square of the standard deviation. Therefore, taking the square root of the variance formula will produce the standard deviation formula.

Confidence Interval for the Population Variance 2 Remind students that we have started with the variance which is the square of the standard deviation. Therefore, taking the square root of the variance formula will produce the standard deviation formula.

Confidence Interval for the Population Standard Deviation  Remind students that we have started with the variance which is the square of the standard deviation. Therefore, taking the square root of the variance formula will produce the standard deviation formula.

Example: The proper operation of typical home appliances requires voltage levels that do not vary much. Listed below are ten voltage levels (in volts) recorded in the author’s home on ten different days. These ten values have a standard deviation of s = 0.15 volt. Use the sample data to construct a 95% confidence interval estimate of the standard deviation of all voltage levels.

Example: n = 10 so df = 10 – 1 = 9 Use table A-4 to find:
Construct the confidence interval: n = 10, s = 0.15

Pendugaan interval Variance 2
Remind students that we have started with the variance which is the square of the standard deviation. Therefore, taking the square root of the variance formula will produce the standard deviation formula.

Pendugaan interval Standard Deviation 
Remind students that we have started with the variance which is the square of the standard deviation. Therefore, taking the square root of the variance formula will produce the standard deviation formula.

Contoh Soal : Seorang ahli pemasaran ingin mengetahui batas batas varians (keragaman ) harga barang “X” di suatu daerah. Untuk itu dia mengambil sampel dengan n = 15 , yang menghasilkan s² = 5,96. Dengan interval keyakinan 95 %, dugalah batas batas varians dan simpangan baku dari harga barang “X” tersebut. Penyelesaian : n=15 ; s²= 5,96 ; n-1 = 15-1=14 ; α = 0,05 ; α/2 =0,025 ; X²R = X² 0,025 = 26,119 ; X²L= X²0,975 = 5,63 (15-1).5,96 < σ² < (15-1).5,96 __________ _________ 26, , , < σ² < 14, , < σ < 3, s² = ∑X² - (∑X)² n-1 n(n-1)

Recap In this section we have discussed: The chi-square distribution.
Using Table A-4. Confidence intervals for the population variance and standard deviation. Determining sample sizes.

TUGAS III: 1.Probabilitas seorang mahasiswa lulus ujian sarjana adalah 0,6. Jika terdapat 8 orang mahasiswa yang akan mengikuti ujian, Hitunglah probabilitas paling banyak 2 orang yang lulus. 2. Sebuah uang logam dilempar 12 kali .Tentukan probabilitas untuk mendapatkan 8 kali muncul Gambar ( gunakan distribusi binomial dan distribusi Normal). 3.Penelitian terhadap sampel 25 pasien kelainan darah tertentu di sebuah rumah sakit, 3 diantaranya mengidap kanker darah. Dengan interval keyakinan 95 %. Tentukan proporsi orang yang mengidap kanker darah tersebut.

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