1 / 70

Outline Chapter 6

Schaum’s Outline PROBABILTY and STATISTICS Chapter 6 ESTIMATION THEORY Presented by Professor Carol Dahl Examples from D. Salvitti J. Mazumdar C. Valencia. Outline Chapter 6. Trader in energy stocks random variable Y = value of share want estimates µ y , σ Y

cachet
Download Presentation

Outline Chapter 6

An Image/Link below is provided (as is) to download presentation Download Policy: Content on the Website is provided to you AS IS for your information and personal use and may not be sold / licensed / shared on other websites without getting consent from its author. Content is provided to you AS IS for your information and personal use only. Download presentation by click this link. While downloading, if for some reason you are not able to download a presentation, the publisher may have deleted the file from their server. During download, if you can't get a presentation, the file might be deleted by the publisher.

E N D

Presentation Transcript

  1. Schaum’s OutlinePROBABILTY and STATISTICS Chapter 6 ESTIMATION THEORYPresented by Professor Carol DahlExamples from D. Salvitti J. Mazumdar C. Valencia

  2. Outline Chapter 6 • Trader in energy stocks • random variable Y = value of share • want estimates µy, σY • Y = ß0 + ß1X+  • want estimates Ŷ, b0, b1 • Properties of estimators • unbiased estimates • efficient estimates

  3. Outline Chapter 6 • Types of estimators • Point estimates • µ = 7 • Interval estimates • µ = 7+/-2 • confidence interval

  4. Outline Chapter 6 • Population parameters and confidence intervals Means Large sample sizes Small sample sizes • Proportions • Differences and Sums • Variances • Variances ratios

  5. Properties of Estimators - Unbiased Unbiased Estimator of Population Parameter estimator expected value = to population parameter

  6. Unbiased Estimates Population Parameters: Sample Parameters: are unbiased estimates Expected value of standard deviation not unbiased

  7. Properties of Estimators - Efficient • Efficient Estimator – if distributions of two statistics same more efficient estimator = smaller variance efficient = smallest variance of all unbiased estimators

  8. Unbiased and Efficient Estimates Target Estimates which are efficient and unbiased Not always possible often us biased and inefficient easy to obtain

  9. Types of Estimates for Population Parameter Point Estimate single number Interval Estimate between two numbers.

  10. Estimates of Mean – Known VarianceLarge Sample or Finite with Replacement • X = value of share • sample mean is $32 • volatility is known σ2 = $4.00 • confidence interval for share value • Need • estimator for mean • need statistic with • mean of population • estimator

  11. Estimates of Mean- Sampling Statistic • P(-1.96 < <1.96) = 95% 2.5%

  12. Confidence Interval for Mean • P(-1.96 < <1.96) = 95% • P(-1.96 < <1.96 ) = 95% • P(-1.96 -X < -µ <1.96 - X ) = 95% • Change direction of inequality • P(+1.96 +X > µ > -1.96 + X ) = 95%

  13. Confidence Interval for Mean • P(+1.96 +X > µ > -1.96 + X ) = 95% • Rearrange • P(X - 1.96 < µ <X + 1.96 ) = 95% • Plug in sample values and drop probabilities • X = value of share, sample = 64 • sample mean is $32 • volatility is σ2 = $4 • {32 – 1.96*2/64, 32 + 1.96*2/64} = {31.51,32.49}

  14. Estimates for Mean for Normal • Take a sample • point estimate • compute sample mean • interval estimate – 0.95 (95%+) = (1 - 0.05) • X +/-1.96 • X +/-Zc • (Z<Zc) = 0.975 = (1 – 0.05/2) • 95% of intervals contain • 5% of intervals do not contain

  15. Estimates for Mean for Normal • interval estimate – 0.95 (95%+) = (1 - 0.05) • X +/-Zc • (Z<Zc) = 0.975 = (1 – 0.05/2) • interval estimate – (1-) % •  X +/-Zc • (Z<Zc) = (1 – /2) • % of intervals don’t contain • (1- )% of intervals do contain (Z<Zc) = 0.975 = (1 – 0.05/2)

  16. Confidence Interval Estimatesof Population Parameters Common values for corresponding to various confidence levels used in practice are:

  17. Confidence Interval Estimatesof Population Parameters Functions in EXCEL Menu Click on Insert Function or =confidence(,stdev,n) =confidence(0.05,2,64)= 0.49 X+/-confidence(0.05,2,64) =normsinv(1-/2) gives Zc value X+/-normsinv(1-/2) 32 +/- 1.96*2/64

  18. Confidence Intervals for MeansFinite Population (N) no Replacement Confidence interval Confidence level

  19. Example: Finite Population without replacement Evaluate density of oil in new reservoir 81 samples of oil (n) from population of 500 different wells samples density average is 29°API standard deviation is known to be 9 °API  = 0.05

  20. Confidence Intervals for MeansFinite Population (N) no ReplacementKnown Variance X = 29 , N= 500, n = 81 , σ = 9 ,  = 0.05 Zc = 1.96

  21. But don’t know Variance t-Distribution =N(0.1) = = tdf 2/df df

  22. Confidence Intervals of Meanst- distribution = = = =

  23. Confidence Intervals of MeansNormal compared to t- distribution t distribution Normal X +/-Zc X +/-tc

  24. Confidence Interval Unknown Variance Example: Eight independent measurements diameter of drill bit 3.236, 3.223, 3.242, 3.244, 3.228, 3.253, 3.253, 3.230 99% confidence interval for diameter of drill bit X +/-tc

  25. Confidence Intervals for MeansUnknown Variance X +/-tc X = ΣXi/n 3.236+3.223+3.242+3.244+3.228+3.253+3.253+3.230 8 X = 3.239 ŝ2 = Σ(Xi - X) = (3.236- X)2 + . . .(3.230 - X)2 (n-1) (8-1) ŝ = 0.0113

  26. Confidence Intervals for MeansUnknown Variance X +/-tc • X = 3.239, n = 8, ŝ= 0.0113, =0.01, • 1- /2=0.995 • From the t-table with 7 degrees of freedom, we find tc= t7,0.995=3.50 1-/2=.975 .005% -tc tc Find tc from Table of Excel

  27. Confidence Intervals for MeansUnknown Variance 1-/2=.975 /2= 0.005% Depends on Table -tc tc  GHJ /2 = 0.005  tc = 2. 499 Schaums 1- /2 = 0.995  tc = 2.35 Excel =tinv(0.01,7) = 3.499483

  28. Confidence Intervals for MeansUnknown Variance X +/-tc X = 3.239, n = 8, ŝ= 0.0113, =0.01,

  29. Confidence Intervals for Proportions Example 600 engineers surveyed 250 in favor of drilling a second exploratory well 95% confidence interval for proportion in favor of drilling the second well Approximate by Normal in large samples Solution: n=600, X=250 (successes), = 0.05 zc = 1.96 and

  30. Confidence Intervals for Proportions Example 600 engineers surveyed 250 in favor of drilling a second exploratory well. 95% confidence interval for proportion in favor of drilling the second well Approximate by Normal in large samples Solution: n=600, X=250 (successes), = 0.05 zc = 1.96 and

  31. Confidence Intervals for Proportions sampling from large population or finite onewith replacement

  32. Confidence Intervals Differences and SumsKnown Variances Samples are independent

  33. Confidence Interval for Differences and Sums – Known Variance Example sample of 200 steel milling balls average life of 350 days - standard deviation 25 days new model strengthened with molybdenum sample of 150 steel balls average life of 250 days - standard deviation 50 days samples independent Find 95% confidence interval for difference μ1-μ2

  34. Confidence Intervals for Differences and Sums Example Solution: X1=350, σ1=25, n1=200, X2=250, σ2=50, n2=150

  35. Confidence Intervals for Differences and Sums – Large Samples Where: P1, P2 two sample proportions, n1, n2 sizes of two samples

  36. Confidence Intervals for Differences and Sums Example random samples 200 drilled holes in mine 1, 150 found minerals 300 drilled holes in mine 2, 100 found minerals c Construct 95% confidence interval difference in proportions Solution: P1=150/200=0.75, n1=200, P2=100/300=0.33,n2=300 With 95% of confidence the difference of proportions {0.42, 0.08}

  37. Confidence Intervals Differences and Sums Example Solution: P1=150/200=0.75, n1=200, P2=100/300=0.33, n2=300 95% of confidence the difference of proportions [0.08,0.42]

  38. Confidence Intervals for Variances Need statistic with population parameter 2 estimate for population parameter ŝ2 its distribution - 2

  39. Confidence Intervals for Variances has a chi-squared distribution n-1 degrees of freedom. Find interval such that σ lies in the interval for 95% of samples 95% confidence interval

  40. Confidence Intervals for Variances Rearrange Take square root if want confidence interval for standard deviation

  41. Confidence Intervals for Variances and Standard Deviations Drop probabilities when substitute in sample values 1 -  confidence interval for variance 1 -  confidence interval for standard deviation

  42. Confidence Intervals for Variance Example Variance of amount of copper reserves 16 estimates chosen at random ŝ2 = 2.4 thousand million tons Find 99% confidence interval variance Solution: ŝ2=2.4, n=16, degrees of freedom = 16-1= 15

  43. How to get 2Critical Values Not symmetric /2 /2 2 lower 2 upper

  44. How to get 2Critical Values 1-/2 Not symmetric 1-/2 /2 /2 GHJ area above 20.995, 20.005 4.60092, 32.8013 Schaums area below 20.005, 20.995 4.60, 32.8 Excel = chiinv(0.995,15) = 4.60091559877155 Excel = chiinv(0.005,15) = 32.8013206461633

  45. Confidence Intervals for Variances and Standard Deviations Example 99% confidence interval variance of reserves Solution: ŝ=2.4 (n-1)=15 2lower = 4.60, 2upper = 32.8

  46. Confidence Intervals for Ratio of Variances Two independent random samples size m and n population variances estimated variances ŝ21, ŝ22 • interested in whether variances are the same • 21/ 22

  47. Confidence Intervals for Ratio of Variances Need statistic with population parameter 21/ 22 estimate for population parameter ŝ21/ ŝ22 its distribution - F

  48. F-Distribution df1 df2

  49. F-Distribution

  50. Confidence Intervals for Ratio of Variances Need statistic with population parameter 21/ 22 estimate for population parameter ŝ21/ ŝ22 its distribution - F

More Related