STATISTICAL INFERENCE PART II CONFIDENCE INTERVALS AND HYPOTHESIS TESTING

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1 STATISTICAL INFERENCE PART II CONFIDENCE INTERVALS AND HYPOTHESIS TESTING 1

2 LOCATION PARAMETER Let f(x) be any pdf. The family of pdfs f(x) indexed by parameter, is called the location family with standard pdf f(x) and is the location parameter for the family. is a location parameter for f(x) iff the distribution of X does not depend on.

3 LOCATION PARAMETER Let X 1,X,,X n be a r.s. of a distribution with pdf (or pmf); f(x; );. An estimator t(x 1,,x n ) is defined to be a location equivariant iff t(x 1 +c,,x n +c)= t(x 1,,x n ) +c for all values of x 1,,x n and a constant c. t(x 1,,x n ) is location invariant iff t(x 1 +c,,x n +c)= t(x 1,,x n ) for all values of x 1,,x n and a constant c. 3

4 SCALE PARAMETER Let f(x) be any pdf. The family of pdfs f(x/)/ for >0, indexed by parameter, is called the scale family with standard pdf f(x) and is the scale parameter for the family. is a scale parameter for f(x) iff the distribution of X/ does not depend on. 4

5 SCALE PARAMETER Let X 1,X,,X n be a r.s. of a distribution with pdf (or pmf); f(x; );. An estimator t(x 1,,x n ) is defined to be a scale equivariant iff t(cx 1,,cx n )= ct(x 1,,x n ) for all values of x 1,,x n and a constant c>0. t(x 1,,x n ) is scale invariant iff t(cx 1,,cx n )= t(x 1,,x n ) for all values of x 1,,x n and a constant c>0. 5

6 LOATION-SCALE PARAMETER Let f(x) be any pdf. The family of pdfs f(x)/ for >0, indexed by parameter (,), is called the location-scale family with standard pdf f(x) and is a location parameter and is the scale parameter for the family. is a location parameter and is a scale parameter for f(x) iff the distribution of (X)/ does not depend on and. 6

7 LOCATION-SCALE PARAMETER Let X 1,X,,X n be a r.s. of a distribution with pdf (or pmf); f(x; );. An estimator t(x 1,,x n ) is defined to be a location-scale equivariant iff t(cx 1 +d,,cx n +d)= ct(x 1,,x n )+d for all values of x 1,,x n and a constant c>0. t(x 1,,x n ) is location-scale invariant iff t(cx 1 +d,,cx n +d)= t(x 1,,x n ) for all values of x 1,,x n and a constant c>0. 7

8 INTERVAL ESTIMATION Point estimation of : The inference is a guess of a single value as the value of. No accuracy associated with it. Interval estimation for : Specify an interval in which the unknown parameter, is likely to lie. It contains measure of accuracy through variance. 8

9 INTERVAL ESTIMATION An interval with random end points is called a random interval. 5X 5X, 8 3 5X 5X Pr is a random interval that contains the true value of with probability

10 INTERVAL ESTIMATION An interval (l(x 1,x,,x n ), u(x 1,x,,x n )) is called a 100 % confidence interval (CI) for if Pr l x, x,, x u x, x,, x 1 n 1 where 0<<1. The observed values l(x 1,x,,x n ) is a lower confidence limit and u(x 1,x,,x n ) is an upper confidence limit. The probability is called the confidence coefficient or the confidence level. n 10

11 INTERVAL ESTIMATION If Pr(l(x 1,x,,x n ))=, then l(x 1,x,,x n ) is called a one-sided lower 100 % confidence limit for. If Pr( u(x 1,x,,x n ))=, then u(x 1,x,,x n ) is called a one-sided upper 100 % confidence limit for. 11

12 METHODS OF FINDING PIVOTAL QUANTITIES PIVOTAL QUANTITY METHOD: If Q=q(x 1,x,,x n ) is a r.v. that is a function of only X 1,,X n and, then Q is called a pivotal quantity if its distribution does not depend on or any other unknown parameters (nuisance parameters). 1

13 PIVOTAL QUANTITY METHOD Theorem: Let X 1,X,,X n be a r.s. from a distribution with pdf f(x;) for and assume that an MLE (or ss) of, ˆ exists. If is a location parameter, then Q= ˆ is a pivotal quantity. If is a scale parameter, then Q= ˆ / is a pivotal quantity. If 1 and are location and scale parameters respectively, then ˆ 1 ˆ 1 and ˆ are PQs for 1 and. 13

14 CONSTRUCTION OF CI USING PIVOTAL QUANTITIES If Q is a PQ for a parameter and if percentiles of Q say q 1 and q are available such that Pr{q 1 Q q }=, Then for an observed sample x 1,x,,x n ; a 100% confidence region for is the set of that satisfy q 1 q(x 1,x,,x n ;)q. 14

15 EXAMPLE Let X 1,X,,X n be a r.s. of Exp(), >0. Find a 100 % CI for. Interpret the result. 15

16 EXAMPLE Let X 1,X,,X n be a r.s. of N(, ). Find a 100 % CI for and. Interpret the results. 16

17 EXAMPLE Let X 1,X,,X n be a r.s. of Uniform(,1), 0<<1. a) Show that Z=(X (1) -1)/(-1) is a PQ for where X (1) is the first order statistic. b) Find a 90% CI for with equal tail probabilities. 17

18 APPROXIMATE CI USING CLT Let X 1,X,,X n be a r.s. By CLT, X E X X V X / n d The approximate 100(1 )% random interval for : N 0,1 PX z/ X z/ 1 n n The approximate 100(1 )% CI for : x z x z n / / n 18

19 APPROXIMATE CI USING CLT Usually, is unknown. So, the approximate 100(1)% CI for : x t s x t n /, n1 /, n1 When the sample size n=30, t /,n-1 ~N(0,1). s n x z s x z n / / s n 19

20 The Confidence Interval for ( is known) This leads to the following equivalent statement (x z x z ) 1 n n P The confidence interval 0

21 Interpreting the Confidence Interval for x 1 of all the values of obtained in repeated sampling from a given distribution, construct an interval x z, x z n that includes (covers) the expected value of the population. n 1

22 Graphical Demonstration of the Confidence Interval for 1 - Confidence level x z n x x z n Lower confidence limit z n Upper confidence limit

23 The Confidence Interval for ( is known) Example: Estimate the mean value of the distribution resulting from the throw of a fair die. It is known that = Use a 90% confidence level, and 100 repeated throws of the die Solution: The confidence interval is 1.71 x z x x. 8 n 100 The mean values obtained in repeated draws of samples of size 100 result in interval estimators of the form [sample mean -.8, Sample mean +.8], 90% of which cover the real mean of the distribution. 3

24 The Confidence Interval for ( is known) Recalculate the confidence interval for 95% confidence level Solution: x z x 1.96 x. 34 n x.8 x.34 x.8 x.34 4

25 The Confidence Interval for ( is known) The width of the 90% confidence interval = (.8) =.56 The width of the 95% confidence interval = (.34) =.68 Because the 95% confidence interval is wider, it is more likely to include the value of. 5

26 Information and the Width of the Interval Wide interval estimator provides little information. Where is??????????????? 6

27 Information and the Width of the Interval Wide interval estimator provides little information. Where is Ahaaa! Here is a much narrower interval. If the confidence level remains unchanged, the narrower interval provides more meaningful information. 7

28 The Width of the Confidence Interval The width of the confidence interval is affected by the population standard deviation () the confidence level (1-) the sample size (n). 8

29 The Affects of on the interval width / =.05 90% Confidence level z. 05 (1.645) n n z 05 (1.645) n n. / =.05 Suppose the standard deviation has increased by 50%. To maintain a certain level of confidence, a larger standard deviation requires a larger confidence interval. 9

30 The Affects of Changing the Confidence Level / =.5% / = 5% Confidence level 90% 95% / = 5% / =.5% z. 05 (1.645) n n z. 05 (1.96) n n Let us increase the confidence level from 90% to 95%. Larger confidence level produces a wider confidence interval 30

31 The Affects of Changing the Sample Size 90% Confidence level z. 05 (1.645) n n Increasing the sample size decreases the width of the confidence interval while the confidence level can remain unchanged. 31

32 Selecting the Sample Size The required sample size to estimate the mean is n z w 3

33 Selecting the Sample Size Example To estimate the amount of lumber that can be harvested in a tract of land, the mean diameter of trees in the tract must be estimated to within one inch with 99% confidence. What sample size should be taken? Assume that diameters are normally distributed with = 6 inches. 33

34 Solution Selecting the Sample Size The estimate accuracy is +/-1 inch. That is w = 1. The confidence level 99% leads to =.01, thus z / = z.005 =.575. We compute n z w.575(6) 1 39 If the standard deviation is really 6 inches, the interval resulting from the random sampling will be of the form x 1. If the standard deviation is greater than 6 inches the actual interval will be wider than +/-1. 34

35 Inference About the Population Mean when is Unknown The Student t Distribution Standard Normal Student t 0 35

36 Effect of the Degrees of Freedom on the t Density Function Student t with DF Student t with 30 DF Student t with 10 DF 0 The degrees of freedom, (a function of the sample size) determine how spread the distribution is compared to the normal distribution. 36

37 Finding t-scores Under a t- Distribution (t-tables) Degrees of Freedom t.100 t.05 t.05 t.01 t t t 0.05, 10 =

38 EXAMPLE A new breakfast cereal is test-marked for 1 month at stores of a large supermarket chain. The result for a sample of 16 stores indicate average sales of $100 with a sample standard deviation of $180. Set up 99% confidence interval estimate of the true average sales of this new breakfast cereal. Assume normality. n 16,x $100,s $180, 0.01 t t.947 /,n ,15 38

39 ANSWER 99% CI for : s 180 x t /,n1 n 16 ( , ) With 99% confidence, the limits and cover the true average sales of the new breakfast cereal. 39

40 Example An investor is trying to estimate the return on investment in companies that won quality awards last year. A random sample of 83 such companies is selected, and the return on investment is calculated had he invested in them. Construct a 95% confidence interval for the mean return. 40

41 Solution (solving by hand) The problem objective is to describe the population of annual returns from buying shares of quality award-winners. The data are interval. Solving by hand From the data file we determine x t and s 8.31 s 8.31, n n 83 t.05,8 t.05,80 x ,

42 Checking the required conditions We need to check that the population is normally distributed, or at least not extremely nonnormal. There are statistical methods to test for normality From the sample histograms we see More 4

43 TESTS OF HYPOTHESIS A hypothesis is a statement about a population parameter. The goal of a hypothesis test is to decide which of two complementary hypothesis is true, based on a sample from a population. 43

44 TESTS OF HYPOTHESIS STATISTICAL TEST: The statistical procedure to draw an appropriate conclusion from sample data about a population parameter. HYPOTHESIS: Any statement concerning an unknown population parameter. Aim of a statistical test: test an hypothesis concerning the values of one or more population parameters. 44

45 NULL AND ALTERNATIVE HYPOTHESIS NULL HYPOTHESIS=H 0 states that a treatment has no effect or there is no change compared with the previous situation. The parameter is equal to a single value. ALTERNATIVE HYPOTHESIS=H A states that a treatment has a significant effect or there is development compared with the previous situation. The parameter can be greater than or less than or different than the value shown in H 0. 45

46 TESTS OF HYPOTHESIS Sample Space, A: Set of all possible values of sample values x 1,x,,x n. (x 1,x,,x n ) A Parameter Space, : Set of all possible values of the parameters. =Parameter Space of Null Hypothesis Parameter Space of Alternative Hypothesis =

47 TESTS OF HYPOTHESIS A=CC C C C A Reject H 0 Not reject H 0 H 0 : 0 H 1 : 1 47

48 TESTS OF HYPOTHESIS Critical Region, C is a subset of A which leads to rejection region of H 0. Reject H 0 if (x 1,x,,x n )C Not Reject H 0 if (x 1,x,,x n )C A test defines a critical region A test is a rule which leads to a decision to accept or reject H 0 on the basis of the sample information. 48

49 TEST STATISTIC AND REJECTION REGION TEST STATISTIC: The sample statistic on which we base our decision to reject or not reject the null hypothesis. REJECTION REGION: Range of values such that, if the test statistic falls in that range, we will decide to reject the null hypothesis, otherwise, we will not reject the null hypothesis. The probability that the (standardized) test statistic falls in the rejection region is the PROBABILITY OF TYPE I ERROR or SIGNIFICANCE LEVEL FOR THE TEST, which is known as. 49

50 TESTS OF HYPOTHESIS If the hypothesis completely specify the distribution, then it is called a simple hypothesis. Otherwise, it is composite hypothesis. =( 1, ) H 0 : 1 =3f(x;3, ) H 1 : 1 =5f(x;5, ) If is known, simple hypothesis. Composite Hypothesis 50

51 TESTS OF HYPOTHESIS Reject H 0 Do not reject H 0 H 0 is True Type I error P(Type I error) = Correct Decision 1- H 0 is False Correct Decision 1- Type II error P(Type II error) = Tests are based on the following principle: Fix, minimize. ()=Power function of the test for all. = P(Reject H 0 )=P((x 1,x,,x n )C) 51

52 TESTS OF HYPOTHESIS Reject H H is true 0 I error P Type P 0 0 Type I error=rejecting H 0 when H 0 is true max 0 max. prob. of Type I error PReject H H is true 1 1 P max Not Reject H H is true max. prob. of Type II error 5

53 PROCEDURE OF STATISTICAL TEST 1. Determining H 0 and H A.. Choosing the best test statistic. 3. Deciding the rejection region (Decision Rule). 4. Conclusion. 53

54 POWER OF THE TEST AND P-VALUE = Type I error = Significance level of the test. It measures the weight of the evidence favoring rejection of H = Power of the test = P(Reject H 0 H 0 is not true) p-value = Observed significance level = The smallest level of significance at which the null hypothesis can be rejected OR the maximum value of that you are willing to tolerate. 54

55 HYPOTHESIS TEST FOR POPULATION MEAN, KNOWN AND X~N(, ) OR LARGE SAMPLE CASE: Two-sided Test Test Statistic Rejecting Area H 0 : = 0 H A : 0 z x 0 1- / / n / -z / z / Reject H o if z < -z / or z > z /. Reject H 0 Reject H p Do not reject H 0 55

56 HYPOTHESIS TEST FOR POPULATION MEAN, One-sided Tests Test Statistic Rejecting Area 1. H 0 : = 0 x 0 H A : > 0 z 1- Reject H o if z > z.. H 0 : = 0 H A : < 0 Reject H o if z < - z. z / n x 0 / n Do not reject H 0z Reject H 0 - z 1- Reject H 0 Do not reject H 0 56

57 CALCULATION OF P-VALUE Determine the value of the test statistics, For One-Tailed Test: p-value= P(z > z 0 ) if H A : > 0 z 0 x 0 / n p-value p-value= P(z < z 0 ) if H A : < 0 p-value z 0 For Two-Tailed Test p=p-value =.P(z>z 0 ) for z 0 >0 p=p-value =.P(z<z 0 ) for z 0 <0 p/ z 0 p/ -z 0 z 0 57

58 DECISION RULE BY USING P-VALUES REJECT H 0 IF p-value < p-value DO NOT REJECT H 0 IF p-value 58

59 EXAMPLES The weights of pots of jam made by a standard process is normally distributed with mean =345gr and =.8gr. A pot produced just before the process closed for the day weight 338.5gr. Is the process working correctly? = 0.01 H 0 : = 345 H A : 345 x =.8 n=1 z z.575 / Decision Rule: Reject H o if z < -z / or z > z /. 59

60 The test statistic= x z.31 / n REJECT DO NOT REJECT REJECT 60

61 CONCLUSION: DO NOT REJECT H 0 AT 1% SIGNIFICANCE LEVEL. THE PROCESS IS WORKING CORRECTLY. p-value =.P(z<-.31) =.( ) = Since p-value = > 0.01, we cannot reject H 0 at 1% significance level 61

62 Example Do the contents of bottles of catsup have a net weight below an advertised threshold of 16 ounces? To test this 5 bottles of catsup were selected. They gave a net sample mean weight of. It is known that the standard deviation is. We want to test this at significance levels 1% and 5%. X

63 Computer Output Excel Output Test of Hypothesis About MU (SIGMA Known) Test of MU = 16 Vs MU less than 16 SIGMA = 0.4 Sample mean = 15.9 Test Statistic: z = -1.5 P-Value = Minitab Output: Z-Test Test of mu = vs mu < The assumed sigma = Variable N Mean StDev SE Mean Z P Catsup SO DON T REJECT THE NULL HYPOTHESIS IN THIS CASE 63

64 CALCULATIONS The z-score is: Z The p-value is the probability of getting a score worse than this (relative to the alternative hypothesis) i.e., P(Z 1.5).1056 Compare the p-value to the significance level. Since it is bigger than both 1% and 5%, we do not reject the null hypothesis. 64

65 P-value for this one-tailed Test The p-value for this test is Thus, do not reject H 0 at 1% and 5% significance level. The contents of bottles of catsup have a net weight of 16 ounces. 65

66 Test of Hypothesis for the Population Mean ( unknown) For samples of size n drawn from a Normal Population, the test statistic: t x- has a Student t-distribution with n-1 degrees of freedom s/ n 66

67 EXAMPLE 5 measurements of the tar content of a certain kind of cigarette yielded 14.5, 14., 14.4, 14.3 and 14.6 mg per cigarette. Show the difference between the mean of this sample x 14.4and the average tar content claimed by the manufacturer, =14.0 is significance at = x x ) i i 1( ( )... ( ) s n s

68 SOLUTION H 0 : = 14.0 H A : 14.0 x t 5.66 s / n / 5 t t.766 /,n1 0.05,4 Decision Rule: Reject H o if t<-t / or t> t /. 68

69 CONCLUSION Reject H 0 at = Difference is significant Reject H 0 Reject H 0 69

70 P-value of This Test p-value =.P(t > 5.66) = (0.004)= Since p-value = < = 0.05, reject H 0. Minitab Output T-Test of the Mean Test of mu = vs mu not = Variable N Mean StDev SE Mean T P-Value C

71 CONCLUSION USING THE CONFIDENCE INTERVALS MINITAB OUTPUT: Confidence Intervals Variable N Mean StDev SE Mean 95.0 % C.I. C ( , ) Since 14 is not in the interval, reject H 0. 71

72 EXAMPLE Current output of a (chemical) corporation is 800 liters/hour of sulfuric acid. An experiment yields a sample of 16 (hourly outputs of the acid) under alternate conditions. X 8,110 and H 0 : 800 H A : 800 s=70.5 7

73 ANSWER The value of the test statistic is: t Rejection region (=.05): t<-t,n-1 =-t.05,15 = Conclusion: Do NOT reject H 0 since is NOT in the rejection region 73

74 P-VALUE p-value = P(t < -1.33) = Since p-value = > 0.05, do not reject H 0. 74

75 EXAMPLE Problem: At a certain production facility that assembles computer keyboards, the assembly time is known (from experience) to follow a normal distribution with mean of 130 seconds and standard deviation of 15 seconds. The production supervisor suspects that the average time to assemble the keyboards does not quite follow the specified value. To examine this problem, he measures the times for 100 assemblies and found that the sample mean assembly time ( x) is 16.8 seconds. Can the supervisor conclude at the 5% level of significance that the mean assembly time of 130 seconds is incorrect? 75

76 We want to prove that the time required to do the assembly is different from what experience dictates: H A : 130 X 16.8 Since the standard deviation is 15, The standardized test statistic value is: Z

77 Two-Tail Hypothesis: H 0 : 130 H A : 130 Type I Error Probability z=test statistic values -z 1-0 z Reject H 0 (z<-z Do not Reject H 0 (-z z<z Reject H 0 (z>z 77

78 Test Statistic: z = X - n = = -.13 Rejection Region z z.0 1. Z 78

79 CONCLUSION Since.13<-1.96, it falls in the rejection region. Hence, we reject the null hypothesis that the time required to do the assembly is still 130 seconds. The evidence suggests that the task now takes either more or less than 130 seconds. 79

80 DECISION RULE Reject H o if z < or z > In terms of X, reject H 0 if 15 X = or X =

81 P-VALUE In our example, the p-value is p value.p(z.13) (0.0166) So, since < 0.05, we reject the null. 81

82 Calculating the Probability of Type II Error H o : = 130 H A : 130 Suppose we would like to compute the probability of not rejecting H 0 given that the null hypothesis is false (for instance =135 instead of 130), i.e. =P(not rejecting H o H o is false). Assuming =135 this statement becomes: P(17.06 x ) P( Z ) 15/ / 100 P(-5.9 Z -1.37)

83 Probability of Type II Error = = =135 83

84 EXAMPLE Consider the test H o : = 400 H A : > 400 n=50, s=00 and = 0.05 Test Statistic: x z x 400 z 00/ Rejection Region: z > z / = or 00 x

85 TYPE II ERROR If the actual is A =45, then X P(X ) P( ) 00/ 50 00/ 50 P(Z 0.76) X 85

86 TESTING HYPOTHESIS ABOUT POPULATION PROPORTION, p ASSUMPTIONS: 1. The experiment is binomial.. The sample size is large enough. x: The number of success The sample proportion is x pq ˆp ~ N(p, ) n n approximately for large n (np 5 and nq 5 ). 86

87 HYPOTHESIS TEST FOR p Two-sided Test Test Statistic Rejecting Area H 0 : p = p 0 H A : p p 0 ˆp p z pq / n / 1- / - z / z / Reject H o if z < -z / or z > z /. Reject H 0 Do not reject H 0 Reject H 0 87

88 HYPOTHESIS TEST FOR p One-sided Tests Test Statistic Rejecting Area 1. H 0 : p= p 0 H A : p > p 0 z ˆp p pq / n 1- z Reject H o if z > z.. H 0 : p = p 0 H A : p < p 0 z ˆp p pq / n Do not reject H 0 Reject H z Reject H o if z < - z. Reject H 0 Do not reject H 0 88

89 EXAMPLE Mom s Home Cokin claims that 70% of the customers are able to dine for less than $5. Mom wishes to test this claim at the 9% level of confidence. A random sample of 110 patrons revealed that 66 paid less than $5 for lunch. H o : p = 0.70 H A : p

90 ANSWER x = 66, n = 110 and p = 0.70 = 0.08, z / = z 0.04 = 1.75 Test Statistic: x 66 ˆp 0.6 n z.89 (0.7)(0.3)/110 90

91 CONCLUSION DECISION RULE: Reject H 0 if z < or z > CONCLUSION: Reject H 0 at = Mom s claim is not true. / /

92 P-VALUE p-value =. P(z < -.89) =(0.011) = 0.0 The smallest value of to reject H 0 is 0.0. Since p-value = 0.0 < = 0.08, reject H

93 CONFIDENCE INTERVAL APPROACH Find the 9% CI for p. pq (0.7)(0.3) p z / n 110 9% CI for p: 0.63 p Since ˆp 0.6 is not in the above interval, reject H 0. Mom has underestimated the cost of her meal. 93

94 EXAMPLE H 0 : p =.10 H A : p >.10 Data: x=5 (number of visitors in sample that would rent the device) in a sample of 400 visitors surveyed Test Statistic: pˆ - p z pq (.10)(.90) n 400 P-value = P( z > ) = 0.08 > = 0.05 Not Reject H 0 at =

95 EXCEL OUTPUT Test of Hypothesis About p Test of p = 0.1 Vs p greater than 0.1 Sample Proportion = 0.13 Test Statistic = P-Value =

96 Frequency Testing the Normality Assumption Miles

97 SAMPLING DISTRIBUTION OF s The statistic (n 1)s is chi-squared distributed with n-1 d.f. when the population random variable is normally distributed with variance. 97

98 CHI-SQUARE DISTRIBUTION f( ) A A A A 98

99 Inference about the Population Variance ( ) Test statistic (n 1)s which is chi-squared distributed with n - 1 degrees of freedom LCL = (n - 1) s Confidence interval estimator: / UCL = (n - 1) s 1 - / 99

100 Testing the Population Variance ( ) EXAMPLE Proctor and Gamble told its customers that the variance in the weights of its bottles of Pepto-Bismol is less than 1. ounces squared. As a marketing representative for P&G, you select 5 bottles and find a variance of 1.7. At the 10% level of significance, is P&G maintaining its pledge of product consistency? H 0 : = 1. H A : <

101 ANSWER n=5, s =1.7, =0.10, Test Statistics: 0.90, (n 1)s (4) Decision Rule: Reject H 0 if,n Conclusion: Because =34 > , do not reject H 0. The evidence suggests that the variability in product weights exceed the maximum allowance. 101

102 EXAMPLE A random sample of observations from a normal population possessed a variance equal to Find 90% CI for. 90% CI for : 0.05,1 0.95,1 (n 1)s (n 1)s (1)37.3 (1)

103 INTERPRETATION OF THE CONFIDENCE INTERVAL We are 90% confident that the population variance is between and

104 INFERENCE ABOUT THE DIFFERENCE BETWEEN TWO SAMPLES INDEPENDENT SAMPLES POPULATION 1 POPULATION PARAMETERS: 1, 1 PARAMETERS:, Sample size: n 1 Sample size: n Statistics: 1 1 Statistics: x, s x, s 104

105 SAMPLING DISTRIBUTION OFX X 1 Consider random samples of n 1 and n from two normal populations. Then, X X ~ N(, 1 ) 1 1 n n 1 For non-normal distributions, we can use Central Limit Theorem for n 1 30 and n

106 INFERENCE ABOUT 1-106

107 CONFIDENCE INTERVAL FOR 1-1 AND ARE KNOWN FOR NORMAL DISTRIBUTION OR LARGE SAMPLE A 100(1- C.I. for 1 is given by: 1 x x z 1 / n 1 If 1 and are unknown and unequal, we can replace them with s 1 and s. s 1 s x x z 1 / n n 1 n 107

108 EXAMPLE n 00, x 15530,s n 50, x 16910,s 5840 Set up a 95% CI for - 1. z / z x x s1 s x x x x n1 n s s % CI for - 1 : x x 1.96(s ) 1 xx

109 INTERPRETATION With 95% confidence, mean family income in the second group exceeds that in the first group by between $363 and $

110 Test Statistic for 1 - when 1 and are known Test statistic: (x1 x ) ( 1 ) z= 1 n1 n If 1 and are unknown and unequal, we can replace them with s 1 and s. 110

111 EXAMPLE Two different procedures are used to produce battery packs for laptop computers. A major electronics firm tested the packs produced by each method to determine the number of hours they would last before final failure. n 150, x 81hrs,s n 00, x 789hrs,s 7440 The electronics firm wants to know if there is a difference in the mean time before failure of the two battery packs.=

112 SOLUTION STEP 1: H 0 : 1 = H 0 : 1 - = 0 H A : 1 H A : 1-0 STEP : Test statistic: (x x ) 0 (81 789) 0 1 z s s n n STEP 3: Decision Rule = Reject H 0 if z<-z / = or z>z / = STEP 4: Not reject H 0. There is not sufficient evidence to conclude that there is a difference in the mean life of the types of battery packs. 11

113 1 AND ARE UNKNOWN 1 = A 100(1- C.I. for 1 is given by: where 1 1 x x t s 1 /,n1n p n n s (n 1)s (n 1)s 1 1 p n1n 1 113

114 Test Statistic for 1 - when 1 = and unknown Test Statistic: where t = (x x ) ( ) 1 1 s p 1 1 n n 1 s (n 1)s (n 1)s 1 1 p n1n 114

115 EXAMPLE The statistics obtained from random sampling are given as n 8, x 93,s n 9, x 19,s 4 It is thought that 1 <. Test the appropriate hypothesis assuming normality with =

116 SOLUTION n 1 < 30 and n < 30 t-test Because s 1 and s are not much different from each other, use equal-variance t-test. H 0 : 1 = H A : 1 < ( 1 - <0) 116

117 (n 1)s (n 1)s (7)0 (8)4 s p n1 n 8 9 (x x ) 0 (93 19) 0 1 t s p ( 15) n n Decision Rule: Reject H 0 if t < -t 0.01,8+9- =-.60 Conclusion: Since t = < -t 0.01,8+9- =-.60, reject H 0 at =

118 Test Statistic for 1 - when 1 and unknown Test Statistic: t = (x x ) ( ) 1 1 s n with the degree of freedom s n 1 1 (s / n s / n ) s / n s / n n 1 n

119 EXAMPLE Does consuming high fiber cereals entail weight loss? 30 people were randomly selected and asked what they eat for breakfast and lunch. They were divided into those consuming and those not consuming high fiber cereals. The statistics are obtained as x s 595.8; x ; s

120 SOLUTION Because s 1 and s are too different from each other and the population variances are not assumed equal, we can use a t statistic with degrees of freedom df 35.7 / / / /

121 DECISION RULE: H 0 : 1 - = 0 H A : 1 - < 0 (x x ) ( ) ( ) 0 t = s1 s n n Reject H 0 if t < -t,df = -t 0.05, 5 = CONCLUSION: Since t =-.31< -t 0.05, 5 =-1.708, reject H 0 are =

122 MINITAB OUTPUT Two Sample T-Test and Confidence Interval Twosample T for Consmers vs Non-cmrs N Mean StDev SE Mean Consmers Non-cmrs % C.I. for mu Consmers - mu Non-cmrs: ( -13, -7) T-Test mu Consmers = mu Non-cmrs (vs <): T= -.31 P=0.015 DF= 5 1

123 EXAMPLE 1 =mean assembly time (in minutes) using Design A =mean assembly time (in minutes) using Design B H 0 : ( 1 - ) = 0 H A : ( 1 - ) 0 To decide the correct test, calculate the sample standard deviations. s 1 =0.91 and s = 1.14 Because s 1 approximately equal to s, use equalvariance t-test.

124 Pooled variance: S p (n 1)S 1 (n 1)S n n (5 1)(.91) (5 1)(1.14) 5 5 Test Statistic: (assuming equal variances) 1 Rejection region: t >t.05,48 = t = (x 1 S x p n ) ( ) ( ) n Conclusion: Do not reject H 0

125 Frequency Frequency Normality Assumption Design A Design B Although the histograms are not bell-shaped, given the robustness of the t-test the conclusion (not rejecting H 0 ) may be accurate

126 Inference about the Difference of Two Means: Matched Pairs Experiment Data are generated from matched pairs not independent samples. Let X i and Y i denote the measurements for the i-th subject. Thus, (X i, Y i ) is a matched pair observations. Denote D i = Y i -X i or X i -Y i. If there are n subjects studied, we have D 1, D,, D n. Then, n n D i Di nd i1 i1 sd D D D and s s n n 1 n

127 CONFIDENCE INTERVAL FOR D = 1 - A 100(1- C.I. for D = 1 is given by: x t D /, n-1 s D n For n 30, we can use z instead of t.

128 HYPOTHESIS TESTS FOR D = 1 - The test statistic for testing hypothesis about D is given by x t = D s / n D with degree of freedom n-1. D

129 EXAMPLE Sample data on attitudes before and after viewing an informational film. Subject Before After Difference i X i Y i D i =Y i -X i

130 x 7.64,s 1,57 D 90% CI for D = 1 - : D xd t/,n D s 1.57 n 10 9 D 1 t 0.05, With 90% confidence, the mean attitude measurement after viewing the film exceeds the mean attitude measurement before viewing by between 0.36 and 14.9 units.

131 EXAMPLE How can we design an experiment to show which of two types of tires is better? Install one type of tire on one wheel and the other on the other (front) wheels. The average tire (lifetime) distance (in 1000 s of miles is: X with a sample D 4.55 difference s.d. of s 7. There are a total of n=0 observations D

132 Test Statistics: SOLUTION H 0 : D =0 H A : D >0 xd D t =.8 s / n 7./ 0 D Rejection H 0 if t>t.05,19 =1.79, Conclusion: Reject H 0 at =0.05

133 Inference About the Difference of Two Population Proportions Population 1 Population PARAMETERS: p 1 PARAMETERS: p Sample size: n 1 Sample size: n Statistics: ˆp 1 Statistics: ˆp

134 SAMPLING DISTRIBUTION OF pˆ pˆ 1 A point estimator of p 1 -p is x1 x pˆ1 pˆ n1 n The sampling distribution of pˆ1 pˆ is p1q1 pq pˆ1 p ˆ ~ N(p1 p, ) n n if n i p i 5 and n i q i 5, i=1,. 1

135 STATISTICAL TESTS Two-tailed test (pˆ1 p ˆ) 0 x1 x H o :p 1 =p z where p= ˆ 1 1 n 1 n H A :p 1 p pq ˆˆ n 1 n Reject H 0 if z < -z / or z > z /. One-tailed tests H o :p 1 =p H o :p 1 =p H A :p 1 > p H A :p 1 < p Reject H 0 if z > z Reject H 0 if z < -z

136 EXAMPLE A manufacturer claims that compared with his closest competitor, fewer of his employees are union members. Of 318 of his employees, 117 are unionists. From a sample of 55 of the competitor s labor force, 109 are union members. Perform a test at = 0.05.

137 ˆp 1 SOLUTION H 0 : p 1 = p H A : p 1 < p x1 117 x 109 and ˆp, so pooled n1 318 n 55 sample proportion is x1 x ˆp 0.39 n n Test Statistic: 1 (117 / / 55) 0 z (0.39)(1 0.39)

138 Decision Rule: Reject H 0 if z < -z 0.05 = Conclusion: Because z = > -z 0.05 =-1.96, not reject H 0 at =0.05. Manufacturer is wrong.

139 Example In a study, doctors discovered that aspirin seems to help prevent heart attacks. Half of,000 male physicians took aspirin and the other half took a placebo. After 3 years, 104 of the aspirin and 189 of the placebo group had heart attacks. Test whether this result is significant. p 1 : proportion of all men who regularly take aspirin and suffer from heart attack. p : proportion of all men who do not take aspirin and suffer from heart attack 104 pˆ pˆ ; Pooled sample proportion: p= ˆ

140 Test of Hypothesis for p 1 -p Test Statistic: z= pˆ pˆ pq ˆˆ n n 1 H 0 : p 1 -p = 0 H A : p 1 -p < (.0133)(.98668) 11,000 11, Conclusion: Reject H 0 since p-value=p(z<-5.0) 0

141 Confidence Interval for p 1 -p ˆ A 100(1- C.I. for p 1 p is given by: ˆ pˆ qˆ (pˆ1 p ˆ ) z / n pˆ qˆ pˆ qˆ pˆ qˆ 1 1 n (p1 p ) z / * n1 n.08 p p.074 1

142 TEST OF HYPOTHESIS HOW TO DERIVE AN APPROPRIATE TEST 14

143 MOST POWERFUL TEST (MPT) H 0 := 0 Simple Hypothesis H 1 := 1 Simple Hypothesis Reject H 0 if (x 1,x,,x n )C The Neyman-Pearson Lemma: Reject H 0 if 143 k L L x x x C n :,,, k L L L k L P k k L P

144 EXAMPLES X~N(, ) where is known. H 0 : = 0 H 1 : = 1 Find the most powerful test of size. 144

145 EXAMPLES On the basis of a r.s. of size 1 from the pdf H 0 := 0 H 1 := 1 f 1 x; x,0 x 1, 1 Use the Neyman-Pearson lemma to derive MPT of size. 145

146 UNIFORMLY MOST POWERFUL (UMP) TEST If a test is most powerful against every possible value in a composite alternative, then it will be a UMP test. To be able to find UMPT, we use Monotone Likelihood Ratio (MLR). If L=L( 0 )/L( 1 ) depends on (x 1,x,,x n ) only through the statistic y=u(x 1,x,,x n ) and L is an increasing function of y for every given 0 > 1, then we have a monotone likelihood ratio (MLR) in statistic y. If L is an decreasing function of y for every given 0 > 1, then we have a monotone likelihood ratio (MLR) in statistic y. 146

147 UNIFORMLY MOST POWERFUL (UMP) TEST Theorem: If a joint pdf f(x 1,x,,x n ;) has MLR in the statistic Y, then a UMP test of size for H 0 : 0 vs H 1 :> 0 is to reject H 0 if Yc where P(Y c 0 )=. for H 0 : 0 vs H 1 :< 0 is to reject H 0 if Yc where P(Y c 0 )=. 147

148 EXAMPLE X~Uniform(0,) a) Single observation b) Random sample of size n H 0 :=3 H 1 :>3 Find UMPT of size. 148

149 EXAMPLE X~Exp() H 0 : 0 H 1 :> 0 Find UMPT of size. 149

150 GENERALIZED LIKELIHOOD RATIO TEST (GLRT) Derives a test when we have two-sided composite alternative or when we have unknown nuisance parameters GLRT is the generalization of MPT and provides a desirable test in many applications but it is not necessarily a UMP test. 150

151 GENERALIZED LIKELIHOOD RATIO TEST (GLRT) L H 0 : 0 H 1 : 1 r. s. 1, n 1 n f x x,, x ; f x ;, f x ;,, f x ; Let Lˆ max L L ˆ L ˆ max L L ˆ 0 0 MLE of 0 and MLE of under H 0 151

152 GENERALIZED LIKELIHOOD RATIO TEST (GLRT) L L ˆ 0 ˆ,0 1 The Generalized Likelihood Ratio GLRT: Reject H 0 if 0 15

153 EXAMPLE X~N(, ) H 0 : = 0 H 1 : 0 Derive GLRT of size. X~N(, ) H 0 : = 0 H 1 : > 0 Derive GLRT of size. 153

154 EXAMPLE X~Exp() H 0 := 0 H 1 : 0 a) Find GLR for this test. b) Determine the rejection region c) Find the power function d) From the acceptance region of this test, find a 100(1-)% CI for e) Write a general expression for the p-value of this test for a given observed value x of X 154

155 ASYMPTOTIC DISTRIBUTION OF ln GLRT: Reject H 0 if 0 GLRT: Reject H 0 if -ln>-ln 0 =c under H0 ln ~ k asympt. where k is the number of parameters to be tested. Reject H 0 if -ln>,k 155

156 EXAMPLE X~N(, ) H 0 : = 0 H 1 : 0 Derive approximate GLRT of size. 156

157 TWO SAMPLE TESTS ,,..,, ~,,..,, ~ y x s y n s r N Y s x n s r N X : : H H Derive GLRT of size.