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Inferensi Statistika

Inferensi Statistika. Confidence Interval. Estimation Process. Population. Random Sample. I am 95% confident that  is between 40 & 60, which is a point estimate + or - margin of error. Mean X = 50. Mean,  , is unknown. Sample. Confidence Interval Estimation.

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Inferensi Statistika

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  1. Inferensi Statistika

  2. Confidence Interval

  3. Estimation Process Population Random Sample I am 95% confident that  is between 40 & 60, which is a point estimate + or - margin of error. Mean X = 50 Mean, , is unknown Sample

  4. Confidence Interval Estimation • Provides Range of Values • Based on Observations from 1 Sample • Gives Information about Closeness to Unknown Population Parameter • Stated in terms of Probability Never 100% Sure

  5. Elements of Confidence Interval Estimation A Probability (confidence level, denoted by C) That the Population Parameter Falls Somewhere Within the Interval. Sample Statistic Confidence Interval Confidence Limit (Lower) Confidence Limit (Upper)

  6. _  estimate Confidence Intervals _ X 90% Samples 95% Samples 99% Samples

  7. Level of Confidence • Probability that the unknown • population parameter falls within the • interval • Denoted by C = level of confidence e.g. C=90%, 95%, 99%.

  8. Intervals & Level of Confidence Sampling Distribution of the Mean  _ x (1-C)/2 (1-C)/2 Area=C _ X Intervals Extend from C proportion of Intervals Contains . 1-C proportion Does Not. to Confidence Intervals

  9. Factors Affecting Interval Width • Data Variation • measured by  • Sample Size • Level of Confidence C Intervals Extend from X - z* to X + z* x x

  10. Confidence Intervals (Known) • Assumptions • Population Standard Deviation Is Known • Population Is Normally Distributed • If Not Normal, use large samples • Confidence Interval Estimate

  11. Confidence Intervals (Unknown) • Assumptions • Population Standard Deviation Is Unknown • Population Must Be Normally Distributed • Use Student’s t Distribution • Confidence Interval Estimate

  12. Student’s t Distribution Standard Normal t (df = 13) Bell-Shaped Symmetric ‘Fatter’ Tails t (df = 5) Z t 0

  13. Student’s t Table  / 2=(1-C)/2 Assume: n = 3 df = n - 1 = 2  = .10/2 =.05 Upper Tail Area df .25 .10 .05 1 1.000 3.078 6.314 0.817 1.886 2 2.920 .05 3 0.765 1.638 2.353 0 t 2.920 t Values

  14. Example: Interval Estimation Unknown • A random sample ofn = 25has= 50and • s = 8.Set up a95%confidence interval estimate for. .   .  46 69 53 30

  15. Sample Size • Too Big: • Requires too • much resources • Too Small: • Won’t do • the job

  16. Example: Sample Size for Mean • What sample size is needed to be90%confident of being correct within± 5? A pilot study suggested that the standard deviation is45. 2 2 2 2 Z  1 . 645 45  n   219 . 2  220 2 2 m 5 Round Up

  17. Test of Significance • The second method of making statistical inference. • It is to test a Hypothesis about the population parameters

  18. What is a Hypothesis? I assume the mean GPA of this class is 3.5! • A hypothesis is an assumption about the population parameter. • A parameteris a Population mean or proportion • The parameter must be identified before analysis.

  19. The Null Hypothesis,H0 • States the Assumption (numerical) to be tested • e.g. The average # TV sets in US homes is at 3 (H0: m = 3) • Begin with the assumption that the null hypothesis is TRUE. (Similar to the notion ofinnocent until proven guilty) • Refers to theStatus Quo • TheNull Hypothesismay or may not berejected.

  20. The Alternative Hypothesis,Ha • Is theopposite of the null hypothesise.g. The average # TV sets in US homes is less than 3 (Ha: m < 3) or is NOT equal to 3 (Ha: m ¹3). • Challenges the Status Quo • Nevercontains the‘=‘ sign • TheAlternative Hypothesismay or may not beaccepted

  21. Identify the Problem • State the Null Hypothesis (H0: m= 3) • State the Alternative Hypothesis (Ha: m < 3 or Ha: m¹3) as the conclusion if H0 is not true. • Logically, if the alternative Ha: m < 3 is not true, we can accept either Ho: m> 3 or Ho: m = 3.

  22. Hypothesis Testing Process Assume the population mean age is 50. (Null Hypothesis) Population The Sample Mean Is20 No, not likely! REJECT Null Hypothesis Sample

  23. Reason for RejectingH0 Sampling Distribution It is unlikely that we would get a sample mean of this value ... ... Therefore, we reject the null hypothesis that m = 50. ... if in fact this were the population mean. m = 50 Sample Mean 20 H0

  24. Level of Significance,a • DefinesUnlikely Values of SampleStatistic if Null Hypothesis Is True • Called Rejection Region of Sampling Distribution • Designateda(alpha) • Typical values are 0.01, 0.05, 0.10 • Selectedby the Researcherat the Start • Provides theCritical Value(s)of the Test

  25. Level of Significance,a and the Rejection Region Critical Value(s) a H0: m= 3 H1: m < 3 Rejection Regions 0 a H0: m= 3 H1: m > 3 0 a/2 H0: m= 3 H1: m¹ 3 0

  26. Z-Test Statistics (s Known) • Convert Sample Statistic (e.g., ) to Standardized Z Variable • Compare to Critical Z Value(s) • If Z test Statistic falls in Critical Region, Reject H0; Otherwise Do Not Reject H0 X Test Statistic

  27. Hypothesis Testing: Steps • 1. State H0H0 :m = 3 • 2. State H1Ha : m < 3 • 3. Choose aa = .05 • 4. Choose n n = 100 • 5. Choose Test: Z Test (or p Value) Test the Assumption that the true mean # of TV sets in US homes is at least 3.

  28. Hypothesis Testing: Steps (continued) Test the Assumption that the average # of TV sets in US homes is at least 3. • 6. Set Up Critical Value(s)Z = -1.645 • 7. Collect Data 100 households surveyed • 8. Compute Test StatisticComputed Test Stat.= -2 • 9. Make Statistical DecisionReject Null Hypothesis • 10. Express DecisionThe true mean # of TV set is less than 3 in the US households.

  29. One-Tail Z Test for Mean (s Known) • Assumptions • Population IsNormally Distributed • If Not Normal,use large samples • Null Hypothesis Has£or³Sign Only • ZTest Statistic:

  30. Rejection Region H0: m ³ 0 H1: m < 0 H0: m £ 0 H1: m > 0 Reject H Reject H 0 0 a a Z Z 0 0 Must BeSignificantly Belowm = 0 Small values don’t contradict H0 Don’t Reject H0!

  31. Example: One Tail Test • Does an average box ofcerealcontain more than368grams of cereal? A random sample of 25 boxes showedX = 372.5.The company has specifiedsto be15grams. Test at thea=0.05level. _ 368 gm. H0: m = 368 H1: m > 368

  32. Finding Critical Values: One Tail Standardized Normal Probability Table (Portion) What Is Z Givena = 0.05? .50 -.05.45 .05 s Z .04 .06 = 1 Z 1.6 .5495 .5505 .5515 a = .05 1.7 .5591 .5599 .5608 0 1.645 Z 1.8 .5671 .5678 .5686 Critical Value = 1.645 .5738 .5750 .5744 1.9

  33. Example Solution: One Tail H0: m £ 368H1: m >368 Test Statistic: Decision: Conclusion: • a = 0.025 • n = 25 • Critical Value: 1.645 Reject Do Not Reject at a = .05 .05 No Evidence True Mean Is More than 368 0 Z 1.645

  34. p Value Approach -- used in the pdf Chapter • Probability of Obtaining a Test StatisticMore Extreme(£ or ³) than Actual Sample ValueGivenH0 Is True • CalledObserved Level of Significance • Smallest Value of a H0 Can Be Rejected • Used toMake Rejection Decision • If p value ³ a, Do Not Reject H0 • If p value < a, Reject H0

  35. p Value Solution p Value is P(Z³ 1.50) = 0.0668 Use the alternative hypothesis to find the direction of the test. p Value 1.0000-.9332.0668 .0668 .9332 0 Z 1.50 Z Value of Sample Statistic From Z Table: Lookup1.50

  36. p Value Solution (p Value = 0.0668)³ (a = 0.05). Do Not Reject. p Value = 0.0668 Reject a = 0.05 0 Z 1.50 Test StatisticIs In the Do Not Reject Region

  37. Example: Two Tail Test • Does an average box of cerealcontains 368grams of cereal? A random sample of 25 boxes showedX = 372.5.The company has specifiedsto be15grams. Test at thea=0.05level. 368 gm. H0: m = 368 H1: m ¹368

  38. Example Solution: Two Tail H0: m = 386H1: m ¹386 Test Statistic: Decision: Conclusion: • a = 0.05 • n = 25 • Critical Value: ±1.96 Reject Do Not Reject at a = .05 .025 .025 No Evidence that True Mean Is Not 368 -1.96 0 1.96 Z

  39. Connection to Confidence Intervals _ • For X = 372.5oz,s = 15and n = 25, • The 95%Confidence Interval is: • 372.5 - (1.96)15/ 25 to 372.5 + (1.96)15/ 25 • or • 366.62 £m£378.38 • If this interval contains the Hypothesized mean(368), we do not reject the null hypothesis. • It does. Do not reject.

  40. t-Test: s Unknown • Assumptions • Population isnormally distributed • If not normal, onlyslightly skewed& a large sample taken • Parametric test procedure • t test statistic

  41. Example: One Tail t-Test Does an average box of cerealcontain more than368grams of cereal? A random sample of36boxes showedX = 372.5, ands= 15. Test at thea=0.01level. 368 gm. H0: m £368 H1: m >368 s is not given,

  42. Example Solution: One Tail H0: m £ 368 H1: m >368 Test Statistic: Decision: Conclusion: • a = 0.01 • n = 36, df = 35 • Critical Value: 2.4377 Reject Do Not Reject at a = .01 .01 No Evidence that True Mean Is More than 368 0 Z 2.4377

  43. Errors in Making Decisions • Type I Error • Reject True Null Hypothesis • Has Serious Consequences • Probability of Type I Error Isa • CalledLevel of Significance • Type II Error • Do Not Reject False Null Hypothesis • Probability of Type II Error Isb (Beta)

  44. Result Possibilities H0: Innocent Hypothesis Test Jury Trial Actual Situation Actual Situation Innocent Guilty H True H False Verdict Decision 0 0 Do Not Type II Correct Error Innocent 1 - a Reject b ) Error ( H 0 Type I Power Reject Error Correct Guilty Error H (1 - ) b 0 ( ) a

  45. a &b Have an Inverse Relationship Reduce probability ofone errorand theother onegoes up. b a

  46. Factors Affecting Type II Error,b • True Value of Population Parameter • Increases WhenDifferenceBetween Hypothesized Parameter & True ValueDecreases • Significance Level a • Increases Whena Decreases • Population Standard Deviation s • Increases WhensIncreases • Sample Size n • IncreasesWhen n Decreases b a s b b n

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