 5-1 Chapter 5 Theory & Problems of Probability & Statistics Murray R. Spiegel Sampling Theory.

Presentation on theme: "5-1 Chapter 5 Theory & Problems of Probability & Statistics Murray R. Spiegel Sampling Theory."— Presentation transcript:

5-1 Chapter 5 Theory & Problems of Probability & Statistics Murray R. Spiegel Sampling Theory

5-2 Outline Chapter 5 Population X mean and variance - µ, 2 Sample mean and variance X, ^s 2 Sample Statistics X mean and variance ^s 2 mean and variance

5-3 Outline Chapter 5 Distributions Population Samples Statistics Mean Proportions Differences and Sums Variances Ratios of Variances

5-4 Outline Chapter 5 Other ways to organize samples Frequency Distributions Relative Frequency Distributions Computation Statistics for Grouped Data mean variance standard deviation

5-5 Population Parameters A population - random variable X probability distribution (function) f(x) probability function - discrete variable f(x) density function - continuous variable f(x) function of several parameters, i.e.: mean:, variance: 2 want to know parameters for each f(x)

5-6 Example of a Population 5 project engineers in department total experience of (X) 2, 3, 6, 8, 11 years company performing statistical report employees expertise based on experience survey must include: average experience variance standard deviation

5-7 Mean of Population average experience mean:

5-8 Variance of Population variance:

5-9 Standard Deviation of Population standard deviation:

5-10 Sample Statistics What if dont have whole population Take random samples from population estimate population parameters make inferences lets see how How much experience in company hire for feasibility study performance study

5-11 Sampling Example manager assigns engineers at random each time chooses first engineer she sees same engineer could do both lets say she picks (2,2) mean of sample X = (2+2)/2 = 2 you want to make inferences about true µ

5-12 Samples of 2 replacement she will go to project department twice pick engineer randomly potentially 25 possible teams 25 samples of size two 5 * 5 = 25 order matters (6, 11) is different from (11, 6)

5-13 Population of Samples All possible combinations are: (2,2)(2,3)(2,6)(2,8)(2,11) (3,2)(3,3)(3,6)(3,8)(3,11) (6,2)(6,3)(6,6)(6,8)(6,11) (8,2)(8,3)(8,6)(8,8)(8,11) (11,2)(11,3)(11,6)(11,8) (11,11)

5-14 Population of Averages Average experience or sample means are: X i (2)(2.5)(3)(5)(6.5) (2.5)(3)(4.5)(5.5)(7) (3)(4.5)(6)(7)(8.5) (5)(5.5)(7)(8)(9.5) (6.5)(7)(8.5)(9.5) (11)

5-15 Mean of Population Means And mean of sampling distribution of means is : This confirms theorem that states:

5-16 Variance of Sample Means variance of sampling distribution of means ( X i - X) 2 (2-6) 2 (2.5-6) 2 (3-6) 2 (5-6) 2 (6.5-6) 2 (2.5-6) 2 (3-6) 2 (4.5-6) 2 (5.5-6) 2 (7-6) 2 (3-6 ) (4.5-6) 2 (6-6) 2 (7-6) 2 (8.5-6) 2 (5-6 ) 2 (5.5-6) 2 (7-6) 2 (8-6) 2 (9.5-6) 2 (6.5-6 ) 2 (7-6) 2 (8.5-6) 2 (9.5-6 ) 2 (11-6) 2

5-17 Variance of Sample Means Calculating values: 1612.2591 0.25 12.25 92.250.251 9 2.25 0 1 6.25 1 0.25 1 4 12.25 0.25 1 6.2512.25 25

5-18 Variance of Sample Means variance is: Therefore standard deviation is

5-19 Variance of Sample Means These results hold for theorem: Where n is size of samples. Then we see that:

5-20 Math Proof X mean X = X 1 + X 2 + X 3 +... X n n E( X) = E(X 1 ) + E(X 2 )+ E(X 3 ) +... E(X n ) n E( X) = + + +... n E( X) =

5-21 Math Proof X variance X = X 1 + X 2 + X 3 +... X n n Var( X) = 2 x = 2 x + 2 x + 2 x +... 2 x n 2 =

5-22 Sampling Means No Replacement manager picks two engineers at same time order doesn't matter order (6, 11) is same as order (11, 6) 10 choose 2 5!/(2!)(5-2)! = 10 10 possible teams, or 10 samples of size two.

5-23 Sampling Means No Replacement All possible combinations are: (2,3) (2,6) (2,8) (2,11) (3,6) (3,8) (3,11) (6,8) (6,11) (8,11) corresponding sample means are: (2.5) (3)(5)(6.5) (4.5) (5.5)(7)(7)(8.5)(9.5) mean of corresponding sample of means is:

5-24 Sampling Variance No Replacement variance of sampling distribution of means is: standard deviation is:

5-25 Theorems on Sampling Distributions with No Replacements 1. 2.

5-26 Sum Up Theorems on Sampling Distributions Theorem I: Expected values sample mean = population mean E( X ) = x = : mean of population Theorem II: infinite population or sampling with replacement variance of sample is E[( X- ) 2 ] = x 2 = 2 /n 2 : variance of population

5-27 Theorems on Sampling Distributions Theorem III: population size is N sampling with no replacement sample size is n then sample variance is:

5-28 Theorems on Sampling Distributions Theorem IV: population normally distributed mean, variance 2 then sample mean normally distributed mean, variance 2 /n

5-29 Theorems on Sampling Distributions Theorem V: samples are taken from distribution mean, variance 2 (not necessarily normal distributed) standardized variables asymptotically normal

5-30 Sampling Distribution of Proportions Population properties: * Infinite * Binomially Distributed ( p success; q=1-p fail) Consider all possible samples of size n statistic for each sample = proportion P of success

5-31 Sampling Distribution of Proportions Sampling distribution of proportions of: mean: std. deviation:

5-32 Sampling Distribution of Proportions large values of n (n>30) sample distribution for P approximates normal distribution finite population sample without replacing standardized P is

5-33 Example Proportions Oil service company explores for oil according to geological department 37% chances of finding oil drill 150 wells P(0.4<P<0.6)=?

5-34 Example Proportions P(0.4<P<0.6)=? P(0.4-0.37 < P-.37 < 0.6-0.37) =? (.37*.63/150).5 (pq/n ).5 (.37*.63/150).5

5-35 Example Proportions P(0.4<P<0.6)=P(0.24<Z<1.84) =normsdist(1.84)-normsdist(0.24)= 0.372 Think about mean, variance and distribution of np the number of successes

5-36 Sampling Distribution of Sums & Differences Suppose we have two populations. PopulationX A X B Sample of size n A n B Compute statistic S A S B Samples are independent Sampling distribution for S A and S B gives mean: SA SB variance: SA 2 SB 2

5-37 Sampling Distribution of Sums and Differences combination of 2 samples from 2 populations sampling distribution of differences S = S A +/- S B For new sampling distribution we have: mean: S = SA +/- SB variance: S 2 = SA 2 + SB 2

5-38 Sampling Distribution of Sums and Differences two populations X A and X B S A = X A and S B = X B sample means mean: XA+XB = XA + XB = A + B variance: Sampling from infinite population Sampling with replacement

5-39 Example Sampling Distribution of Sums You are leasing oil fields from two companies for two years lease expires at end of each year randomly assigned a new lease for next year Company A - two oil fields production X A : 300, 700 million barrels Company B two oil fields production X B : 500, 1100 million barrels

5-40 Population Means Average oil field size of company A: Average oil field size of company B:

5-41 Population Variances Company A - two oil fields production X A : 300, 700 million barrels Company B two oil fields production X B : 500, 1100 million barrels XA 2 = (300 – 500) 2 + (700 – 500) 2 /2 = 40,000 XB 2 = (500 – 800)2 + (1100 – 800) 2 /2 = 90,000

5-42 Example Sampling Distribution of Sums Interested in total production: X A + X B Compute all possible leases assignments Two choices X A, Two choices X B X Ai X Bi {300, 500} {300, 1100} {700, 500} {700, 1100}

5-43 Example Sampling Distribution of Sums X Ai X Bi {300, 500} {300, 1100} {700, 500} {700, 1100} Then for each of the 4 possibilities – 4 choices year 1, four choices year 2 = 4*4 samples

5-44 Example Sampling Distribution of Sums SamplesX Ai X Bi X Ai X Bi Year 13005003001100 Year 2300500300500 Year 13005003001100 Year 230011003001100 Year 13005003001100 Year 2700500700500 Year 13005003001100 Year 270011007001100

5-45 Example Sampling Distribution of Sums SamplesX Ai X Bi X Ai X Bi Year 17005007001100 Year 2300500300500 Year 17005007001100 Year 230011003001100 Year 17005007001100 Year 2700500700500 Year 17005007001100 Year 270011007001100

5-46 Compute Sum and Means of each sample MeansX Ai +X Bi MeanX Ai +X Bi Mean Year 1800 14001100 Year 2800 Year 180011001400 Year 21400 Year 1800100014001300 Year 21200 Year 1800130014001600 Year 21800

5-47 Compute Sum and Means of each Sample MeansX Ai +X Bi MeanX Ai +X Bi Mean Year 11200100018001300 Year 2800 Year 11200130018001600 Year 21400 Year 11200 18001500 Year 21200 Year 1120015001800 Year 21800

5-48 Mean of Sum of Sample Means Population of Samples {800, 1100, 1000, 1300, 1100, 1400, 1300, 1600, 1000, 1300, 1200, 1500, 1300, 1600, 1500, 1800} _______ X Ai +X Bi = (800 + 1100 + 1000 + 1300 + 1100 + 1400 + 1300 + 1600 + 1000 + 1300 + 1200 + 1500 + 1300 + 1600 + 1500 + 1800) 16 = 1300

5-49 Mean of Sum of Sample Means This illustrates theorem on means _____ (XA+XB) = 1300= XA + XB = 500 + 800 = 1300 _____ What about variances of X A +X B

5-50 Variance of Sum of Means Population of samples {800, 1100, 1000, 1300, 1100, 1400, 1300, 1600, 1000, 1300, 1200, 1500, 1300, 1600, 1500, 1800} 2 = {(800 - 1300) 2 + (1100 - 1300) 2 + (1000 - 1300) 2 + (1300 - 1300) 2 + (1100 - 1300) 2 + (1400 - 1300) 2 + (1300- 1300) 2 + (1600 - 1300) 2 + (1000 - 1300) 2 + (1300 - 1300) 2 + (1200 - 1300) 2 + (1500 - 1300) 2 + (1300 - 1300) 2 + (1600 - 1300) 2 + (1500 - 1300) 2 + (1800 - 1300) 2 }/16 = 65,000

5-51 Variance of Sum of Means This illustrates theorem on variances

5-52 Normalize to Make Inferences on Means

5-53 Estimators for Variance use for populations unbiased better for smaller samples Two choices

5-54 Sampling Distribution of Variances All possible random samples of size n each sample has a variance all possible variances give sampling distribution of variances sampling distribution of related random variable

5-55 Example Population of Samples All possible teams are: (2,2)(2,3)(2,6)(2,8)(2,11) (3,2)(3,3)(3,6)(3,8)(3,11) (6,2)(6,3)(6,6)(6,8)(6,11) (8,2)(8,3)(8,6)(8,8)(8,11) (11,2)(11,3)(11,6)(11,8) (11,11)

5-56 Compute Variance for Each Sample sample variance corresponding to each of 25 possible choice that manager makes are: ^s 2 0 0.2549 20.25.25 02.256.25 16 4 2.2501 6.25 9 6.2510 2.25 20.25 166.252.25 0

5-57 Sampling Distribution of Variance Population of Variances mean variance distribution (n-1)s 2 / 2 2 n-1

5-58 What if Unknown Population Variance? X is Normal (, 2 ) to make inference on means we normalize

5-59 Unknown Population Variance

5-60 Unknown Population Variance Use in the same way as for normal except use different Tables α = 0.05 n = 25, =tinv(0.05,24)= 2.0639 2.06-2.06

5-61 Uses t -statistics Will use for testing means, sums, and differences of means small samples when variable is normal substitute sample variance in for true

5-62 Uses t -statistics sums and differences of means unknown variance

5-63 Uses 2 statistic Inference on Variance Large sample test

5-64 Inferences F Statistic 2 df1 /df1 = 2 df2 /df2

5-65 F Statistic Other tests groups of coefficients

5-66 Other Statistics. Medians. n > 30, sample distribution of medians nearly normal if X is normal

5-67 Frequency Distributions If sample or population is large difficult to compute statistics (i.e. mean, variance, etc) Organizing RAW DATA is useful arrange into CLASSES or categories determine number in each class Class Frequency or Frequency Distribution

5-68 Frequency Distributions - Example Example of Frequency Distribution: middle size oil company portfolio of 100 small oil reservoirs reserves vary from 89 to 300 million barrels

5-69 Frequency Distributions - Example arrange data into categories create table showing ranges of reservoirs sizes number of reservoirs in each range

5-70 Frequency Distributions - Example Class intervals are in ranges of 50 million barrels Each class interval represented by median value e.g. 200 up to 250 will be represented by 225 Can plot data histogram polygon This plot is represents frequency distribution

5-71 Frequency Distributions Plotted - Example

5-72 Relative Frequency Distributions and Ogives number of individuals - frequency distribution - empirical probability distribution percentage of individual - relative frequency distribution empirical cumulative probability distribution - ogive

5-73 Percent Ogives OGIVE for oil company portfolio of reservoirs Shows percent reservoirs < than x reserves

5-74 Computation of Statistics for Grouped Data can calculate mean and variance from grouped data

5-75 Computation of Statistics for Grouped Data take 420 samples of an ore body measure % concentration of Zinc (Zn) frequency distribution of lab results

5-76 Computation of Statistics for Grouped Data

5-77 Computation of Statistics for Grouped Data mean will then be: And in our example:

5-78 Computation of Statistics for Grouped Data variance will then be:

5-79 Computation of Statistics for Grouped Data And in our example:

5-80 Computation of Statistics for Grouped Data Similar formula are available for higher moments:

5-81 Sum up Chapter 5 Population X mean and variance - µ, σ 2 distribution A Sample statistic from sample usually mean and variance X, ^s 2

5-82 Sum up Chapter 5 Sample Statistics X mean and variance x, x 2 ^s 2 mean and variance ^s 2, ^s 2 Distribution

5-83 Sum Up Chapter 5 Samples Statistics Mean X ~ µ, σ 2 /n Distribution

5-84 Sum Up Chapter 5 Samples Statistics Proportions P ~ p, p(1-p)/n n>30 Distribution

5-85 Sum Up Chapter 5 Samples Statistics Differences and Sums X 1 +/- X 2 ~ 1 + 2, 1 2 /n 1 + 2 2 /n 2 Distribution

5-86 Sum Up Chapter 5 Samples Statistics Variances Distribution Mean = n-1 Variance = 2(n-1)

5-87 Sum Up Chapter 5 Samples Statistics Ratios of Variances

5-88 Sum up Chapter 5 Other ways to organize samples Frequency Distributions Relative Frequency Distributions Computation Statistics for Grouped Data mean variance standard deviation

5-89 THATS ALL FOR CHAPTER 5 THANK YOU!!

Download ppt "5-1 Chapter 5 Theory & Problems of Probability & Statistics Murray R. Spiegel Sampling Theory."

Similar presentations