Class 16. Daniel B. Rowe, Ph.D. Department of Mathematics, Statistics, and Computer Science. Marquette University MATH 1700

Transcription

1 Class 16 Daniel B. Rowe, Ph.D. Department of Mathematics, Statistics, and Computer Science Copyright 013 by D.B. Rowe 1

2 Agenda: Recap Chapter Lecture Chapter Review Chapter 6. Problem Solving Session.

3 Recap Chapter

4 7: Sample Variability 7. The Sampling Distribution of Sample Means Sample distribution of sample means (SDSM): If random samples of size n, are taken from ANY population with mean μ and standard deviation σ, then the SDSM: 1. A mean equal to μ. A standard deviation equal to n Central Limit Theorem (CLT): The sampling distribution of sample means will more closely resemble the normal distribution as the sample size n increases. 5

5 7: Sample Variability 7. The Sampling Distribution of Sample Means Eample: N=5 balls in bucket, select n=1 with replacement. 0,, 4, 6, P( ) 0 1/ 5 1/ 5 4 1/ 5 6 1/ 5 P() /

6 7: Sample Variability 7. The Sampling Distribution of Sample Means Eample: N=5 balls in bucket, select n= with replacement. P( ) 0,, 4, 6, / 5 1 / 5 3 / / / / / 5 7 / 5 8 1/ 5 P() 7

7 7: Sample Variability 7. The Sampling Distribution of Sample Means from n=1 distribution from n= distribution P() 0. 4, 4, / ? from n=3 distribution Looks like? Figure from Johnson & Kuby, 01. from n=4 distribution 4, / 3 4, / 4? Looks like? n large? 4 Looks like? n 1

8 7: Sample Variability 7. The Sampling Distribution of Sample Means Uniform(0,00) Normal(100,(57.7) ) f( ) 1 00 f( ) e μ= μ=

9 7: Sample Variability 7. The Sampling Distribution of Sample Means Generate n million random observations,..., n from the Uniform(a=0,b=00) and also from the Normal(μ=100,σ =(57.7) ) distributions, for each of n=1,, 3, 4, 5, and data sets of Uniform and Normal random observations 14

10 7: Sample Variability 7. The Sampling Distribution of Sample Means n= observations scale changes observations are uniform 6 5 observations are normal limited range

11 7: Sample Variability 7. The Sampling Distribution of Sample Means n= 10 6 observations scale changes observations are uniform 1. 1 observations are normal limited range

12 7: Sample Variability 7. The Sampling Distribution of Sample Means n= observations scale changes observations are uniform 1.5 observations are normal 1 1 limited range

13 7: Sample Variability 7. The Sampling Distribution of Sample Means n= observations scale changes observations are uniform.5 observations are normal limited range

14 7: Sample Variability 7. The Sampling Distribution of Sample Means n= observations scale changes observations are uniform 3.5 observations are normal limited range

15 7: Sample Variability 7. The Sampling Distribution of Sample Means n= observations scale changes observations are uniform observations are normal 5 5 limited range

16 7: Sample Variability 7. The Sampling Distribution of Sample Means Sample means and standard deviations from each of the n million observations from the Uniform(a=0,b=00) and Normal(μ=100,σ =(57.7) ) distributions. i.e. n=5, Groups of n=5 Mean of groups 1 1,, 3, 4, data sets of Uniform and Normal , 7 8, 9, ,.,.,., s Histogram of 's 1

17 7: Sample Variability 7. The Sampling Distribution of Sample Means n= means scale same Histogram of means from uniform Histogram of means from normal s 1 s limited range same classes same classes

18 7: Sample Variability 7. The Sampling Distribution of Sample Means n= means scale same Histogram of means from uniform Histogram of means from normal s 1 s limited range same classes same classes 3

19 7: Sample Variability 7. The Sampling Distribution of Sample Means n= means scale same Histogram of means from uniform Histogram of means from normal s 1 s limited range same classes same classes 4

20 7: Sample Variability 7. The Sampling Distribution of Sample Means n= means scale same Histogram of means from uniform Histogram of means from normal 1 1 s s same classes same classes 5

21 7: Sample Variability 7. The Sampling Distribution of Sample Means n= means scale same Histogram of means from uniform Histogram of means from normal 1 1 s s same classes same classes 6

22 7: Sample Variability 7. The Sampling Distribution of Sample Means n= means scale same Histogram of means from uniform Histogram of means from normal 1 1 s s same classes same classes 7

23 7: Sample Variability 7.3 Application of the Sampling Distribution of Sample Means Now that we believe that the mean from a sample of n=15 is normally distributed with mean and standard deviation n, we can find probabilities. a b b a P( a b) P( b ) P( a) 40

24 7: Sample Variability 7.3 Application of the Sampling Distribution of Sample Means To find these probabilities, we first convert to z scores a b b a P( a b) P( b ) P( c z d) same area P( d z) same area P( a) P( z c) same area z a b, c, d and use the table in book. 9

25 7: Sample Variability Questions? Homework: Chapter 7 # 6, 1, 3, 9, 33, 35 34

26 Lecture Chapter

27 Chapter 8: Introduction to Statistical Inference Daniel B. Rowe, Ph.D. Department of Mathematics, Statistics, and Computer Science 36

28 8: Introduction to Statistical Inference 8.1 The Nature of Estimation The purpose of Statistical Inference is to use the info in a sample of data to increase knowledge of a population. Figure from Johnson & Kuby,

29 8: Introduction to Statistical Inference 8.1 The Nature of Estimation We discussed how if we compute a quantity from a population of data then it is called a parameter and if we estimate it from a sample of data then it is called a statistic. Recall: Chapter 1 definitions. Parameter: A numerical value summarizing all the data of an entire population. Statistic: A numerical value summarizing the sample data. 38

30 8: Introduction to Statistical Inference 8.1 The Nature of Estimation More precisely, a single number to estimate a parameter is called a point estimate. Point estimate for a parameter: A single number designed to estimate a quantitative parameter of a population, usually the value of the corresponding sample statistic. i.e. is a point estimate for μ 39

31 8: Introduction to Statistical Inference 8.1 The Nature of Estimation Interval estimate: An interval bounded by two values and used to estimate the value of a population parameter. The values that bound this interval are statistics calculated from the sample that is being used as the basis for estimation. i.e. ( some amount) is an interval estimate for μ. The interval estimate will be of the form point estimate some amount 40

32 8: Introduction to Statistical Inference 8.1 The Nature of Estimation Interval Estimate for μ. Significance Level: Pre assigned probability of a parameter being outside our interval estimate, α. P( not in some amount) = i.e P( some amount < < some amount) = 41

33 8: Introduction to Statistical Inference 8.1 The Nature of Estimation Take many samples and for each calculate interval estimate then. Level of Confidence 1-α: The proportion of all interval estimates that include the parameter being estimated. i.e μ P( in some amount) =1- P( some amount < < some amount) = 1 4

34 8: Introduction to Statistical Inference 8.1 The Nature of Estimation Confidence Interval: An interval estimate with a specified level of confidence. A range of values for the parameter with a level of confidence attached. (i.e. 95% confident) point estimator some amount(1- ) some amount that depends on The general form for a confidence interval is point estimate margin of error confidence level 1-43

35 8: Introduction to Statistical Inference 8. Estimation of Mean μ (σ Known) The assumption for estimating mean μ using a known σ: The sampling distribution of has a normal distribution. Recall from Chapter 6 that for the standard normal distribution, P( 1.96 z 1.96)= 0.95 From the CLT in Chapter 7, we know that when n is large, the sample mean is approimately normally distributed with, n 44

36 8: Introduction to Statistical Inference 8. Estimation of Mean μ (σ Known) What this implies is that has an approimate standard normal distribution! z P( 1.96 z 1.96)= 0.95 Or more generally, 1.95 P( z( / ) z z( / ) )= 1. z( / ) called the confidence coefficient. -z( / ) z( / )

37 8: Introduction to Statistical Inference 8. Estimation of Mean μ (σ Known) P( z( / ) z z( / ) )= 1 With some algebra, we can see that. (fill in) 46

38 8: Introduction to Statistical Inference 8. Estimation of Mean μ (σ Known) Thus, a (1-α) 100% confidence interval for μ is σ σ - z( / ) + z( / ) n n which if α=0.05, a 95% confidence interval for μ is σ σ n n Confidence Interval for Mean: σ z( / ) σ to z( / ) (8.1) n n 49

39 8: Introduction to Statistical Inference 8. Estimation of Mean μ (σ Known) THE CONFIDENCE INTERVAL: A FIVE STEP PROCESS Step 1 The Set-UP: Step Confidence Interval Criteria: Step 3 The Sample Evidence: Step 4 The Confidence Interval: Step 5 The Results: Your Book describes as a 5 step process. Read this. Important. 50

40 8: Introduction to Statistical Inference 8. Estimation of Mean μ (σ Known) Philosophically, μ is fied and the interval varies. If we take a sample of data,,..., 1 n and determine a confidence interval from it, we get. ± z( / ) σ n μ each sample a different interval of same width If we had a different sample of data, y,..., 1 yn we would have determined a different confidence interval. y ± z( / ) σ n Figure from Johnson & Kuby,

41 8: Introduction to Statistical Inference 8. Estimation of Mean μ (σ Known) We never truly know if our CI from our sample of data will contain the true population mean μ. But we do know that there is a (1-α) 100% chance that a confidence interval from a sample of data will contain μ. 5

42 8: Introduction to Statistical Inference 8. Estimation of Mean μ (σ Known) Eample: Make normal μ=100,σ=57.7 random values True Random s Sample mean and variance from the 5 million values Here is a histogram of the 5 million values. 53

43 8: Introduction to Statistical Inference 8. Estimation of Mean μ (σ Known) Form s ' (means) from successive n=5 random numbers True Random s Sample mean and variance from the 1 million means., / Here is a histogram of the 1 million means. n 54

44 8: Introduction to Statistical Inference 8. Estimation of Mean μ (σ Known) s ' Form 1 million U and L values from s '. n=5 and 57.7 U = σ / n U = σ / n L = σ / n Random, s ' L U insert each L = σ / n insert true We will also get 1 million L s and U s that we can use to make histograms. 56

45 8: Introduction to Statistical Inference 8. Estimation of Mean μ (σ Known) Form 1 million U and L values from s '. L s U s U = σ / n L = σ / n True Random L U L U Histogram of the 1 million L s and U s. 57

46 8: Introduction to Statistical Inference 8. Estimation of Mean μ (σ Known) Form 1 million U and L values from s '. L s U s U = σ / n L = σ / n True Random L U L U Histogram of the 1 million L s and U s..5% of time upper less than 100.5% of time lower greater than 100 5% of time μ=100 not in interval P( not in 1.96 )= n 58

47 8: Introduction to Statistical Inference 8. Estimation of Mean μ (σ Known) Eample: A random sample of size n=15 student heights was taken from this class. Assume that we know that σ=4. Construct a 95% confidence Interval for μ. σ σ - z( / ) to + z( / ) n n to z(.05) ,69,67,66,71,74,75,70,67,73,64,65,68,67, to 70.7 z(.05) 63

48 Chapter 8: Introduction to Statistical Inference Questions? Homework: Chapter 8 # 5, 15,, 4, 35, 47, 57, 59, 81, 93, 97, 106, 109, 119, 145,

49 Review Chapter 6 65

50 6: Normal Probability Distributions 6.1 Normal Probability Distributions The mathematical formula for the normal distribution is (p 69): f( ) e 1 where e = π = μ = population mean σ = population std. deviation f(), 0 We will not use this formula. Figure from Johnson & Kuby,

51 6: Normal Probability Distributions 6. The Standard Normal Probability Distributions Eample: Here is a normal distribution with μ = 5 and σ = 4. f() σ= Let s say we want to know the red area under the normal distribution between 1 =.8 and = 9.8. μ= What is the area under the normal distribution between these two values? 67

52 6: Normal Probability Distributions 6. The Standard Normal Probability Distributions Eample: Here is a normal distribution with μ = 5 and σ = 4. z f() f(z) 1 =.8 = 9.8 σ= z 1 1 z σ=1 Area between 1 and is the same as area between z 1 and z. We find z 1 = ( 1 - μ)/σ = and z = ( - μ)/σ=.14? μ= μ= 68 z

53 6: Normal Probability Distributions 6. The Standard Normal Probability Distributions Standard normal curve μ = 0 and σ = 1. Now we can simply look up the z areas in a table. f(z) Appendi B Table 3 Page 716. σ=1 μ= z 69

54 First Decimal Place in z 6: Normal Probability Distributions Appendi B Table 3 Page 716 Second Decimal Place in z z 1 = z =.14 70

55 6: Normal Probability Distributions Appendi B, Table 3, Page 716 z 1 = z =.14 P(z<-1.36)=Area less than We get this from Table 3. Row labeled -1.3 over to column Labeled.06. z 71

56 6: Normal Probability Distributions Appendi B, Table 3, Page 717 z 1 = z =.14 (continued) P(z<.14)=Area less than.14. We get this from Table 3. Row labeled.1 over to column Labeled.04. z 7

57 6: Normal Probability Distributions Appendi B, Table 3, Page z 1 = z =.14 P(-1.36<z<.14) = P(z<.14) - P(z<-1.36) = = - z Red Area= z 73 z

58 6: Normal Probability Distributions 6. The Standard Normal Probability Distributions z 1 = z =.14 Eample: Here is a normal distribution with μ = 5 and σ = 4. f() f(z) Area between 1 and is same as the area between z 1 and z. z 74

59 6: Normal Probability Distributions 6.3 Applications of Normal Distributions Eample: Assume that IQ scores are normally distributed with a mean μ of 100 and a standard deviation σ of 16. If a person is picked at random, what is the probability that his or her IQ is between 100 and 115? i.e. P( )? μ μ Figures from Johnson & Kuby,

60 6: Normal Probability Distributions 6.3 Applications of Normal Distributions IQ scores normally distributed μ=100 and σ=16. P( ) z z z Figures from Johnson & Kuby,

61 6: Normal Probability Distributions 6.3 Applications of Normal Distributions Now we can use the table. = - P(0 z 0.94) P( z 0.94) P( z 0) Figures from Johnson & Kuby,

62 6: Normal Probability Distributions 6.4 Notation Eample: Let α=0.05. Let s find z(0.05). P(z>z(0.05))=0.05. P(z>z(0.05)) Same as finding P(z<z(0.05))= z =P(z>z(0.05)) Figures from Johnson & Kuby,

63 6: Normal Probability Distributions 6.4 Notation Eample: Same as finding P(z<z(0.05))= Figures from Johnson & Kuby,

64 6: Normal Probability Distributions 6.5 Normal Approimation of the Binomial Distribution Approimate binomial probabilities with normal areas. Use a normal with np, np(1 p) n=14 p=1/ (14)(.5) 7 (14)(.5)(1.5) 3.5 Figures from Johnson & Kuby,

65 6: Normal Probability Distributions 6.5 Normal Approimation of the Binomial Distribution n=14, p=1/ We then approimate binomial probabilities with normal areas. P ( 4) from the binomial formula is approimately P( ) from the normal with 7, 3.5 the ±.5 is called a continuity correction Figures from Johnson & Kuby,

66 6: Normal Probability Distributions 6.5 Normal Approimation of the Binomial Distribution From the binomial formula 14! P(4) (.5) (1.5) 4!(14 4)! P ( 4) P( 1.87 z 1.34) P( 1.87 z 1.34) Pz ( 1.34) From the Normal Distribution P( ) z z Pz n=14, p=1/ 7, 3.5 ( 1.87)

67 6: Normal Probability Distributions Questions? Homework: Chapter 6 # 7, 9, 13, 17, 19, 9, 31, 33, 41, 45, 47, 53, 61, 75, 95, 99 83

68 Problem Solving Session 84