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Sampling Theory

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

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Sampling Theory

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

  2. Outline Chapter 5 • Population X • mean and variance - µ, 2 • Sample • mean and variance X, ^s2 • Sample Statistics • X mean and variance • ^s2 mean and variance

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

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

  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)

  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

  7. Mean of Population average experience mean:

  8. Variance of Population variance:

  9. Standard Deviation of Population standard deviation:

  10. Sample Statistics • What if don’t 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

  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

  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)

  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)

  14. Population of Averages Average experience or sample means are: Xi (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)

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

  16. Variance of Sample Means variance of sampling distribution of means (Xi -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

  17. Variance of Sample Means Calculating values: 16 12.25 9 1 0.25 12.25 9 2.25 0.25 1 9 2.25 0 1 6.25 1 0.25 1 4 12.25 0.25 1 6.25 12.25 25

  18. Variance of Sample Means variance is: Therefore standard deviation is

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

  20. Math Proof X mean • X = X1 + X2 + X3 + . . . Xn • n • E(X) = E(X1) + E(X2)+ E(X3) + . . . E(Xn) • n • E(X) =  +  +  + . . .  • n • E(X) =

  21. Math Proof X variance • X = X1 + X2 + X3 + . . . Xn • n • Var(X) = 2x = 2x + 2x + 2x + . . . 2x • n2 • =

  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.

  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:

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

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

  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] = x2 = 2/n • 2: variance of population

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

  28. Theorems on Sampling Distributions • Theorem IV: population normally distributed • mean , variance 2 • then sample mean normally distributed • mean , variance 2/n

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

  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

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

  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

  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)=?

  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

  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

  36. Sampling Distribution of Sums & Differences • Suppose we have two populations. • Population XA XB • Sample of size nA nB • Compute statistic SA SB • Samples are independent • Sampling distribution for SA and SB gives • mean: SA SB • variance:SA2 SB2

  37. Sampling Distribution of Sums and Differences • combination of 2 samples from 2 populations • sampling distribution of differences • S = SA +/- SB • For new sampling distribution we have: • mean: S = SA +/- SB • variance: S2 = SA2 + SB2

  38. Sampling Distribution of Sums and Differences • two populations XA and XB • SA= XAand SB = XB sample means • mean: XA+XB = XA + XB = A + B • variance: • Sampling from infinite population • Sampling with replacement

  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 XA: 300, 700 million barrels Company B two oil fields production XB: 500, 1100 million barrels

  40. Population Means • Average oil field size of company A: • Average oil field size of company B:

  41. Population Variances Company A - two oil fields production XA: 300, 700 million barrels Company B two oil fields production XB: 500, 1100 million barrels XA2 = (300 – 500)2 + (700 – 500)2/2 = 40,000 XB2 = (500 – 800)2 + (1100 – 800)2/2 = 90,000

  42. Example Sampling Distribution of Sums Interested in total production: XA + XB Compute all possible leases assignments Two choices XA, Two choices XB XAi XBi {300, 500} {300, 1100} {700, 500} {700, 1100}

  43. Example Sampling Distribution of Sums XAi XBi {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

  44. Example Sampling Distribution of Sums

  45. Example Sampling Distribution of Sums

  46. Compute Sum and Means of each sample

  47. Compute Sum and Means of each Sample

  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} _______ XAi+XBi = (800 + 1100 + 1000 + 1300 + 1100 + 1400 + 1300 + 1600 + 1000+ 1300 + 1200 + 1500 + 1300 + 1600 + 1500 + 1800) 16 = 1300

  49. Mean of Sum of Sample Means This illustrates theorem on means _____  (XA+XB)= 1300= XA+XB = 500 + 800 = 1300 _____ What about variances of XA+XB

  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

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