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4. 4 4-1 Statistical Inference The field of statistical inference consists of those methods used to make decisions or draw conclusions about a population. These methods utilize the information contained in a sample from the population in drawing conclusions. Example: Estimate the average height of the class.

5. 5 4-1 Statistical Inference

6. 6 4-2 Point Estimation

7. 7 An estimator should be close in some sense to the true value of the unknown parameter. Θ is an unbiased estimator of  if E(Θ) =  If the estimator Θ is not unbiased, then the difference E(Θ)   is called the bias of the estimator Θ 4-2 Point Estimation

8. 8 Suppose that X is a random variable with mean  and variance  2. Let X 1, X 2,.., X n be a random sample of size n from the population represented by X. Show that the sample mean X and sample variance S 2 are unbiased estimators of  and variance  2, respectively. 4-2 Point Estimation Example 4-1

9. 9 4-2 Point Estimation Example 4-1

10. 10 Sample variance S 2 are unbiased estimators of  variance  2. Sample standard deviation S are not unbiased estimators of  population standard deviation . 4-2 Point Estimation Example 4-1

11. 11 Sometimes there are several unbiased estimators of the sample population parameter. Example: a random sample of size n =10. 1.Sample mean 2.Sample median 3.The first observation X 1 All are unbiased estimator of population X. Cannot rely on the property of unbiasedness alone to select the estimator. 4-2 Point Estimation Different Unbiased Estimators

12. 12 Suppose that Θ 1 and Θ 2 are unbiased estimators of  The variances of these two distribution of each estimator may be different. Θ 1 has a smaller variance than Θ 2 does. Θ 1 is more likely to produce an estimate close to the true value . 4-2 Point Estimation Different Unbiased Estimators

13. 13 A logical principle of estimation is to choose the estimator that has minimum variance. 4-2 Point Estimation How to Choose an Unbiased Estimator

14. 14 In practice, one must occasionally use a biased estimator. For example, S for . What is the criterion? Mean square error 4-2 Point Estimation How to Choose a Good Estimator

15. 15 MSE(Θ) = E[Θ  E(Θ) 2 ] + [  E(Θ)] 2 = V(Θ)+(bias) 2 Unbiased estimator  bias = 0 A good estimator is the one that minimizes V(Θ)+(bias) 2 4-2 Point Estimation Mean Square Error

16. 16 Given two estimator Θ 1 and Θ 2, The relative efficiency of Θ 1 and Θ 2 is defined as r = MSE(Θ 1 )/MSE(Θ 2 ) r < 1  Θ 1 is better than Θ 2. Example: Θ 1 = X: the sample mean of sample size n. Θ 2 = X i : the i-th observation. 4-2 Point Estimation Mean Square Error

17. 17 r = MSE(Θ 1 )/MSE(Θ 2 ) = (  2 /n)/  2 = 1/n Sample mean is a better estimator than a single observation. The square root of the variance of an estimator,  V(Θ), is called the standard error of the estimator. 4-2 Point Estimation Mean Square Error

18. 18 4-2 Point Estimation Methods of obtaining an estimator: 1. Method of Moments Estimator (MME) Example: If the random sample is really from the population with pdf, the sample mean should resemble the population mean and the MME of can be obtained by solving. Therefore, the MME of is.

19. 19 MME – cont’d Example 2: The MME of and can be obtained by solving and. Therefore, the MMEs of and are and

20. 20 The MMEs of the population parameters can be obtained by: 1.Express the k population moments as functions of parameters 2.Replace the notation for population parameters by those of the MMEs 3.Equate the sample moments to the population moments 4.Solve the k equations to obtain the MMEs of the parameters

21. 21 4-2 Point Estimation Methods of obtaining an estimator: 2. Maximum Likelihood Estimator (MLE) The likelihood function of, given the data is the joint distribution of The idea of ML estimation is to find an estimator of which maximizes the likelihood of observing

22. 22 4-2 Point Estimation 2. Maximum Likelihood Estimator (MLE) Example: Maximizing is equivalent to maximizing We obtain the MLE by solving the first derivative equation:, hence the MLE of is

23. 23 It can be shown that the MLEs of the mean and variance of the normal distribution are:

24. 24 4-3 Hypothesis Testing The purpose of hypothesis testing is to determine whether there is enough statistical evidence in favor of a certain belief about a parameter. Examples Is there statistical evidence in a random sample of potential customers, that support the hypothesis that more than p% of the potential customers will purchase a new products? Is a new drug effective in curing a certain disease? A sample of patient is randomly selected. Half of them are given the drug where half are given a placebo. The improvement in the patients conditions is then measured and compared.

25. 25 4-3 Hypothesis Testing We like to think of statistical hypothesis testing as the data analysis stage of a comparative experiment, in which the engineer is interested, for example, in comparing the mean of a population to a specified value (e.g. mean pull strength). 4-3.1 Statistical Hypotheses

26. 26 4-3 Hypothesis Testing 4-3.1 Statistical Hypotheses The critical concepts of hypothesis testing. There are two hypotheses (about a population parameter) The null hypothesis [ for example m = 5] The alternative hypothesis [m > 5] Assume the null hypothesis is true. Build a statistic related to the parameter hypothesized. Pose the question: How probable is it to obtain a statistic value at least as extreme as the one observed from the sample

27. 27 4-3 Hypothesis Testing 4-3.1 Statistical Hypotheses The critical concepts of hypothesis testing-Continued Make one of the following two decisions (based on the test): Reject the null hypothesis in favor of the alternative hypothesis. Do not reject the null hypothesis in favor of the alternative hypothesis. Two types of errors are possible when making the decision whether to reject H 0 Type I error - reject H 0 when it is true. Type II error - do not reject H 0 when it is false.

28. 28 4-3 Hypothesis Testing For example, suppose that we are interested in the burning rate of a solid propellant used to power aircrew escape systems. Now burning rate is a random variable that can be described by a probability distribution. Suppose that our interest focuses on the mean burning rate (a parameter of this distribution). Specifically, we are interested in deciding whether or not the mean burning rate is 50 centimeters per second. 4-3.1 Statistical Hypotheses

29. 29 4-3 Hypothesis Testing 4-3.1 Statistical Hypotheses Two-sided Alternative Hypothesis One-sided Alternative Hypotheses

30. 30 4-3 Hypothesis Testing 4-3.1 Statistical Hypotheses Test of a Hypothesis A procedure leading to a decision about a particular hypothesis Hypothesis-testing procedures rely on using the information in a random sample from the population of interest. If this information is consistent with the hypothesis, then we will conclude that the hypothesis is true; if this information is inconsistent with the hypothesis, we will conclude that the hypothesis is false.

31. 31 4-3 Hypothesis Testing 4-3.2 Testing Statistical Hypotheses Based on the hypotheses and the information in the sample, the sample space is divided into two parts: Reject Region and Accept Region.

32. 32 4-3 Hypothesis Testing 4-3.2 Testing Statistical Hypotheses Rejection region (RR) is the subset of sample space leads to the rejection of the null hypothesis. RR in Fig4-3. The complement of the rejection region is the acceptance region. AR.

33. 33 4-3 Hypothesis Testing 4-3.2 Testing Statistical Hypotheses Sometimes the type I error probability is called the significance level, or the  -error, or the size of the test. The Probabilities of committing errors are calculated to determine the performance of a test.

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35. 35 4-3 Hypothesis Testing 4-3.2 Testing Statistical Hypotheses If the mean is 50, the probability of obtaining a sample mean less than 48.5 or greater than 51.5 is 0.0576.

36. 36 Since the alternative hypothesis is composite, there are many distributions in the corresponding subset of parameter space. Therefore, the probability of committing type II error depends on the distribution chosen to calculate.

37. 37 4-3 Hypothesis Testing 4-3.2 Testing Statistical Hypotheses If the mean is actually 52, the probability of falsely accept is 0.2643

38. 38 4-3 Hypothesis Testing 4-3.2 Testing Statistical Hypotheses Similarly, the probability of falsely accept when is 0.8923, which is much higher than the previous case. It is harder to detect the difference if two distribution are close.

39. 39 4-3 Hypothesis Testing 4-3.2 Testing Statistical Hypotheses The probabilities of committing errors depends also on the sample size n, the amount of information. decreases as n increases.

40. 40 4-3 Hypothesis Testing 4-3.2 Testing Statistical Hypotheses 1. can be decreased by making the AR larger, at the price that will increase. 2. The probability of type II error is a function of population mean 3. The probabilities of committing both types of error can be reduced at the same time only by increasing the sample size.

41. 41 4-3 Hypothesis Testing 4-3.2 Testing Statistical Hypotheses The power is computed as 1 - , and power can be interpreted as the probability of correctly rejecting a false null hypothesis. We often compare statistical tests by comparing their power properties. For example, consider the propellant burning rate problem when we are testing H 0 :  = 50 centimeters per second against H 1 :  not equal 50 centimeters per second. Suppose that the true value of the mean is  = 52. When n = 10, we found that  = 0.2643, so the power of this test is 1 -  = 1 - 0.2643 = 0.7357 when  = 52.

42. 42 Example #4-19.(p157) = P(Z > 1.58) = 1  P(Z  1.58) = 1  0.94295 = 0.057 = P(Z   2.37) = 0.00889. c) 1 -  = 1 – 0.00889 = 0.99111 (The power of the test when mean=200)

43. 43 Example #4-20 (p157). a)Reject the null hypothesis and conclude that the mean foam height is greater than 175 mm. The probability that a value of at least 190 mm would be observed (if the true mean height is 175 mm) is only 0.0082. Thus, the sample value of = 190 mm would be an unusual result. b)

44. 44 4-3 Hypothesis Testing 4-3.3 P-Values in Hypothesis Testing

45. 45 Example #4-21 (p157). Using n = 16:

46. 46 Example #4-22(a) (p157). n = 16: a) 0.0571 =

47. 47 4-3 Hypothesis Testing 4-3.3 One-Sided and Two-Sided Hypotheses Two-Sided Test: One-Sided Tests:

48. 48 4-3 Hypothesis Testing 4-3.5 General Procedure for Hypothesis Testing

49. 49 4-4 Inference on the Mean of a Population, Variance Known Assumptions

50. 50 4-4 Inference on the Mean of a Population, Variance Known 4-4.1 Hypothesis Testing on the Mean We wish to test : The test statistic is :

51. 51 The analyst selects critical values according to a preassigned, The rejection of H 0 is a “strong conclusion”. depends on the sample size and the true value of the parameter, “fail to reject H 0 ” is a “weak conclusion”. To accept H 0, we need more evidence. The critical values are determined by 2-sided hypothesis: 1-sided hypothesis:, or 4-4 Inference on the Mean of a Population, Variance Known

52. 52 4-4 Inference on the Mean of a Population, Variance Known 4-4.1 Hypothesis Testing on the Mean Reject H 0 if the observed value of the test statistic z 0 is either: z 0 > z  /2 or z 0 < -z  /2 Fail to reject H 0 if -z  /2 < z 0 < z  /2 where is the observation of

53. 53 4-4 Inference on the Mean of a Population, Variance Known 4-4.1 Hypothesis Testing on the Mean Reject H 0 if the observed value of the test statistic z 0 is either: or Fail to reject H 0 if

54. 54 4-4 Inference on the Mean of a Population, Variance Known 4-4.1 Hypothesis Testing on the Mean

55. 55 4-4 Inference on the Mean of a Population, Variance Known 4-4.1 Hypothesis Testing on the Mean

56. 56 Example #4-33 (p173) a) 1) The parameter of interest is the true mean yield, . 2) H 0 :  = 90 3) H 1 :   90 4)  = 0.05 5) 6) Reject H 0 if z 0 <  z  /2 where  z 0.025 =  1.96 or z 0 > z  /2 where z 0.025 = 1.96 7),  = 3 8) Since  1.96 < 0.36 < 1.96 do not reject H0 and conclude the yield is not significantly different from 90% at  = 0.05.

57. 57 4-4 Inference on the Mean of a Population, Variance Known 4-4.1 P-Values in Hypothesis Testing

58. 58 2-sided 1-sided P-value P-value if the critical value is

59. 59 The p-value of a test is the probability of observing a test statistic at least as extreme as the one computed, given that the null hypothesis is true. The p - value provides information about the amount of statistical evidence that supports the alternative hypothesis. We can conclude that the smaller the p-value the more statistical evidence exists to support the alternative hypothesis.

60. 60 4-4 Inference on the Mean of a Population, Variance Known 4-4.2 P-Values in Hypothesis Testing

61. 61 Example #4-33 (p173)-continued a) P-value = Interpreting the p-value Because the probability that the absolute value of the standardized test statistics assume a value of more than 0.36 when is so large (0.7188), there are reasons to believe that the null hypothesis is true.

62. 62 The p-value and rejection region methods The p-value can be used when making decisions based on rejection region methods as follows: Define the hypotheses to test, and the required significance level Perform the sampling procedure, calculate the test statistic and the p-value associated with it. Compare the p-value to Reject the null hypothesis only if ; otherwise, do not reject the null hypothesis.

63. 63 Conclusions of a test of Hypothesis If we reject the null hypothesis, we conclude that there is enough evidence to infer that the alternative hypothesis is true. If we do not reject the null hypothesis, we conclude that there is not enough statistical evidence to infer that the alternative hypothesis is true.

64. 64 4-4 Inference on the Mean of a Population, Variance Known 4-4.3 Type II Error and Choice of Sample Size Finding The Probability of Type II Error 

65. 65 4-4 Inference on the Mean of a Population, Variance Known 4-4.2 Type II Error and Choice of Sample Size Finding The Probability of Type II Error 

66. 66 For a 2-sided test, recall

67. 67 If, the second term can be neglected, Solving for n, we have

68. 68 4-4 Inference on the Mean of a Population, Variance Known 4-4.2 Type II Error and Choice of Sample Size Sample Size Formulas

69. 69 4-4 Inference on the Mean of a Population, Variance Known 4-4.2 Type II Error and Choice of Sample Size Sample Size Formulas

70. 70 We have

71. 71 4-4 Inference on the Mean of a Population, Variance Known 4-4.2 Type II Error and Choice of Sample Size

72. 72 Example #4-33(p173)-continued,, b) n = n  5. c) If n=5, =  (1.96 – 1.49)   (  1.96 – 1.49) =  (0.47)   (–3.45) = 0.680822  0.000280 = 0.680542

73. 73 4-4 Inference on the Mean of a Population, Variance Known 4-4.2 Type II Error and Choice of Sample Size

74. 74 4-4 Inference on the Mean of a Population, Variance Known 4-4.3 Large Sample Test We assume that σ 2 is known In most practical situation, σ 2 is unknown

75. 75 4-4 Inference on the Mean of a Population, Variance Known 4-4.3 Large Sample Test In general, if n  30, according to the Central Limit Theorem, the sample variance s 2 will be close to σ 2 for most samples, and so s can be substituted for σ in the test procedures with little harmful effect.

76. 76 4-4 Inference on the Mean of a Population, Variance Known 4-4.4 Some Practical Comments on Hypothesis Testing The Seven-Step Procedure Only three steps are really required:

77. 77 4-4 Inference on the Mean of a Population, Variance Known 4-4.4 Some Practical Comments on Hypothesis Testing Statistical versus Practical Significance

78. 78 4-4 Inference on the Mean of a Population, Variance Known 4-4.4 Some Practical Comments on Hypothesis Testing Statistical versus Practical Significance If the sample size becomes very large, it is almost sure that the null hypothesis will be rejected and conclude that the true mean is not equal to  0.

79. 79 4-4 Inference on the Mean of a Population, Variance Known 4-4.5 Confidence Interval on the Mean An interval estimator draws inferences about a population by estimating the value of an unknown parameter using an interval.

80. 80 4-4 Inference on the Mean of a Population, Variance Known 4-4.5 Confidence Interval on the Mean Two-sided confidence interval: One-sided confidence intervals: Confidence coefficient:

81. 81 4-4 Inference on the Mean of a Population, Variance Known 4-4.5 Confidence Interval on the Mean

82. 82 4-4 Inference on the Mean of a Population, Variance Known How is an interval estimator produced from a sampling distribution ? To estimate, a sample of size n is drawn from the population, and its mean is calculated. Under certain conditions, is normally distributed (or approximately normally distributed.), thus 4-4.5 Confidence Interval on the Mean

83. 83 4-4 Inference on the Mean of a Population, Variance Known 4-4.6 Confidence Interval on the Mean

84. 84 –We know that –This leads to the relationship --1 -  of all the values of obtained in repeated sampling from this distribution, construct an interval that includes (covers) the expected value of the population. --1 -  of all the values of obtained in repeated sampling from this distribution, construct an interval that includes (covers) the expected value of the population.

85. 85 4-4 Inference on the Mean of a Population, Variance Known 4-4.5 Confidence Interval on the Mean

86. 86 Example Discuss Example 4-5

87. 87 Example #4-33(p173)–continued (d) Find a 95% two-sided CI on the true mean yield. For  = 0.05,, Therefore, is the 95% CI on the true mean yield. With 95% confidence, we believe the true mean yield of the chemical process is between 87.85% and 93.11%.

88. 88 4-4 Inference on the Mean of a Population, Variance Known 4-4.5 Confidence Interval on the Mean Relationship between Tests of Hypotheses and Confidence Intervals If [l,u] is a 100(1 -  ) percent confidence interval for the parameter, then the test of significance level  of the hypothesis will lead to rejection of H 0 if and only if the hypothesized value is not in the 100(1 -  ) percent confidence interval [l, u].

89. 89 Interval estimators can be used to test hypotheses. Calculate the confidence level interval estimator, then if the hypothesized parameter value falls within the interval, do not reject the null hypothesis, if the hypothesized parameter value falls outside the interval, conclude that the null hypothesis can be rejected (m is not equal to the hypothesized value).

90. 90 Example #4.33(p.173)-continued (e) Use the CI found in (d) to test the hypothesis. The 95% CI on the true mean yield is [87.85,93.11]. The null hypothesis cannot be rejected since the hypothesized value, 90%, lies within this interval.

91. 91 4-4 Inference on the Mean of a Population, Variance Known 4-4.5 Confidence Interval on the Mean Confidence Level and Precision of Estimation The length of the two-sided 95% confidence interval is whereas the length of the two-sided 99% confidence interval is Because the 99% confidence interval is wider, it is more likely to include the value of .

92. 92 Interpreting the interval estimate wrong It is wrong to state that the interval estimator is an interval for which there is chance that the population mean lies between the and the. This is so because the is a parameter, not a random variable. Note that and are random variables. Thus, it is correct to state that there is chance that will be less than and will be greater than.

93. 93 4-4 Inference on the Mean of a Population, Variance Known 4-4.5 Confidence Interval on the Mean Choice of Sample Size The width of the interval estimate is a function of: the population standard deviation, the confidence level, and the sample size. We can control the width of the interval estimate by changing the sample size. Thus, we determine the interval width first, and derive the required sample size.

94. 94 The phrase “estimate the mean to within E units”, translates to an interval estimate of the form

95. 95 4-4 Inference on the Mean of a Population, Variance Known 4-4.5 Confidence Interval on the Mean Choice of Sample Size

96. 96 4-4 Inference on the Mean of a Population, Variance Known 4-4.5 Confidence Interval on the Mean Choice of Sample Size

97. 97 4-4 Inference on the Mean of a Population, Variance Known 4-4.5 Confidence Interval on the Mean Choice of Sample Size

98. 98 4-4 Inference on the Mean of a Population, Variance Known 4-4.5 Confidence Interval on the Mean One-Sided Confidence Bounds

99. 99 Example #4.36(p174) a) 1) The parameter of interest is the true mean life, . 2) H 0 :  = 540 3) H1:   540 4)  = 0.05 5) 6) Reject if where 7), 8) Since 2.19 > 1.65, reject the null hypothesis and conclude there is sufficient evidence to support the claim the life exceeds 540 hrs at  = 0.05.

100. 100 b) P-value = P(Z > 2.19) = 1  P(Z  2.19) c)  = =  (1.65  3.873) =  (  2.223) = 0.0131 d) n = n  9.

101. 101 e) 542.83   With 95% confidence, the true mean life is at least 542.83 hrs. f) Since 540 does not fall within this interval, we can reject the null hypothesis in favor of the alternative.

102. 102 4-4 Inference on the Mean of a Population, Variance Known 4-4.6 General Method for Deriving a Confidence Interval

103. 103 4-5 Inference on the Mean of a Population, Variance Unknown 4-5.1 Hypothesis Testing on the Mean When is unknown, we will replace it by an estimator of it. Since S is a random variable, that increases the variation in.

104. 104 Theorem Let be a random sample of size n from. Then (1) (2) (3) and are independent.

105. 105 Definition Let and be independent. Then the random variable defined by has the Student’s t-distribution with degrees of freedom. The t probability density function is

106. 106 can be rewritten as and we have the following result.

107. 107 4-5 Inference on the Mean of a Population, Variance Unknown 4-5.1 Hypothesis Testing on the Mean

108. 108 4-5 Inference on the Mean of a Population, Variance Unknown 4-5.1 Hypothesis Testing on the Mean

109. 109 Example

110. 110 4-5 Inference on the Mean of a Population, Variance Unknown 4-5.1 Hypothesis Testing on the Mean Calculating the P-value

111. 111 4-5 Inference on the Mean of a Population, Variance Unknown 4-5.1 Hypothesis Testing on the Mean

112. 112 4-5 Inference on the Mean of a Population, Variance Unknown 4-5.1 Hypothesis Testing on the Mean

113. 113 Discuss Example 4-7 Parameter of interest: the mean coefficient of restitution  Null Hypothesis, H 0 :  =0.82 Alternative Hypothesis, H 1 :  >0.82 Test statistic: 4-5 Inference on the Mean of a Population, Variance Unknown 4-5.1 Hypothesis Testing on the Mean Example 4-7

114. 114 4-5 Inference on the Mean of a Population, Variance Unknown 4-5.1 Hypothesis Testing on the Mean

115. 115 Reject H 0 : if the P-value < 0.05 Computations: Conclusion: REJECT 4-5 Inference on the Mean of a Population, Variance Unknown 4-5.1 Hypothesis Testing on the Mean Example 4-7

116. 116 Example #4.50 (p.185) A particular brand of diet margarine was analyzed to determine the level of polyunsaturated fatty acid. A sample of 6 packages resulted in the following data: 16.8, 17.2, 16.9, 17.4, 16.5, 17.1. The normality assumption appears to be satisfied.

117. 117 a) 1) The parameter of interest is the true mean level of polyunsaturated fatty acid, . 2) H0:  = 17 3) H1:   17 4)  = 0.01 5) 6) Reject if where or 7) = 16.98 s = 0.318 n = 6 8) Since  4.032 <  0.154 < 4.032, do not reject the null hypothesis and conclude the true mean level is not significantly different from 17% at  = 0.01.

118. 118 P-value = 2P(t > 0.154): for degrees of freedom of 5 we obtain 2(0.40) < P-value 0.80 < P-value

119. 119 4-5 Inference on the Mean of a Population, Variance Unknown 4-5.1 Hypothesis Testing on the Mean

120. 120 4-5 Inference on the Mean of a Population, Variance Unknown 4-5.2 Type II Error and Choice of Sample Size Fortunately, this unpleasant task has already been done, and the results are summarized in a series of graphs in Appendix A Charts Va, Vb, Vc, and Vd that plot for the t-test against a parameter  for various sample sizes n. Where has a noncentral t-distribution. =0+=0+

121. 121 4-5 Inference on the Mean of a Population, Variance Unknown 4-5.2 Type II Error and Choice of Sample Size These graphics are called operating characteristic (or OC) curves. Curves are provided for two-sided alternatives on Charts Va and Vb. The abscissa scale factor d on these charts is d efi ned as

122. 122 b) Using the OC curves on Chart Vb, with d = 1.567, n = 10, when   0.1. Therefore, the current sample size of 6 is inadequate.

123. 123 4-5 Inference on the Mean of a Population, Variance Unknown 4-5.3 Confidence Interval on the Mean Rearrange the above equation

124. 124 4-5 Inference on the Mean of a Population, Variance Unknown 4-5.3 Confidence Interval on the Mean

125. 125 c) For  = 0.01, n=6, we have 16.455    17.505. With 99% confidence, we believe the true mean level of polyunsaturated fatty acid is between 16.455% and 17.505%.

126. 126 4-5 Inference on the Mean of a Population, Variance Unknown 4-5.3 Confidence Interval on the Mean

127. 127 4-5 Inference on the Mean of a Population, Variance Unknown 4-5.4 Confidence Interval on the Mean

128. 128 4-6 Inference on the Variance of a Normal Population 4-6.1 Hypothesis Testing on the Variance of a Normal Population Test Statistic:

129. 129 4-6 Inference on the Variance of a Normal Population 4-6.1 Hypothesis Testing on the Variance of a Normal Population

130. 130 4-6 Inference on the Variance of a Normal Population 4-6.1 Hypothesis Testing on the Variance of a Normal Population

131. 131 4-6 Inference on the Variance of a Normal Population 4-6.1 Hypothesis Testing on the Variance of a Normal Population Chi-squared distribution is a skewed distribution.

132. 132 4-6 Inference on the Variance of a Normal Population 4-6.1 Hypothesis Testing on the Variance of a Normal Population

133. 133 4-6 Inference on the Variance of a Normal Population 4-6.1 Hypothesis Testing on the Variance of a Normal Population

134. 134 4-6 Inference on the Variance of a Normal Population 4-6.1 Hypothesis Testing on the Variance of a Normal Population Discuss Example 4-10

135. 135 4-5 Inference on the Mean of a Population, Variance Unknown Rearrange the above equation 4-6.2 Confidence Interval on the Variance of a Normal Population

136. 136 4-6 Inference on the Variance of a Normal Population 4-6.2 Confidence Interval on the Variance of a Normal Population

137. 137 4-6 Inference on the Variance of a Normal Population 4-6.2 Confidence Interval on the Variance of a Normal Population Discuss Example 4-11

138. 138 Example 4-59(p.191) n=15, s=0.016mm (a) In order to use  2 statistic in hypothesis testing and confidence interval construction, we need to assume that the underlying distribution is normal. 1)The parameter of interest is the true standard deviation of the diameter,. 2) vs. 3)  = 0.05 4) 5) Reject if 6) Since 8.96 < 23.685 do not reject and conclude there is insufficient evidence to indicate the true standard deviation of the diameter exceeds 0.02 at  = 0.05.

139. 139 P-value = for 14 degrees of freedom: 0.5 < P-value < 0.9 b) 95% lower confidence interval on For  = 0.05 and n = 15, The 95% lower confidence interval on is c) Based on the lower confidence bound, we cannot reject the null hypothesis.

140. 140 4-7 Inference on Population Proportion 4-7.1 Hypothesis Testing on a Binomial Proportion We will consider testing:

141. 141 4-7 Inference on Population Proportion 4-7.1 Hypothesis Testing on a Binomial Proportion

142. 142 4-7 Inference on Population Proportion 4-7.1 Hypothesis Testing on a Binomial Proportion

143. 143 4-7 Inference on Population Proportion 4-7.1 Hypothesis Testing on a Binomial Proportion

144. 144 4-7 Inference on Population Proportion 4-7.2 Type II Error and Choice of Sample Size

145. 145 4-7 Inference on Population Proportion 4-7.2 Type II Error and Choice of Sample Size

146. 146 4-7 Inference on Population Proportion 4-7.2 Type II Error and Choice of Sample Size Discuss Example 4-13

147. 147 Example #4-75(p201) In the process of manufacturing lens, a machine will be qualified if the percentage of the polished lens contains surface defects less than 4%. A random sample of 300 lens contains 11 defective lenses. (a) Formulate and test the hypotheses at 1) The parameter of interest is the true percentage of polished lenses that contain surface defects, p. 2) H0 : p = 0.04 vs. H1 : p < 0.04 3)  = 0.05 4) 6) Reject if 7) Since do not reject the null hypothesis and conclude the machine cannot be qualified at the 0.05 level of significance.

148. 148 b) P-value =  (-0.295) = 0.386 = 0.614 c) d) (4-69) Take n=764.

149. 149 4-7 Inference on Population Proportion 4-7.3 Confidence Interval on a Binomial Proportion Rearrange the above equation

150. 150 4-7 Inference on Population Proportion 4-7.3 Confidence Interval on a Binomial Proportion Idea: Use instead of p! But we don’t know p! (Cannot construct CI!)

151. 151 4-7 Inference on Population Proportion 4-7.3 Confidence Interval on a Binomial Proportion

152. 152 4-7 Inference on Population Proportion 4-7.3 Confidence Interval on a Binomial Proportion

153. 153 4-7 Inference on Population Proportion 4-7.3 Confidence Interval on a Binomial Proportion Choice of Sample Size is the point estimator of p E = | - p |

154. 154 4-7 Inference on Population Proportion 4-7.3 Confidence Interval on a Binomial Proportion Choice of Sample Size p+1/p  4

155. 155 Example Round up to n=1068

156. 156 In some situation, we are interested in predicting a future observation of a random variable Find a range of likely values for the variable associated with making the prediction Given X 1,..,X n, to predict X n+1   [?,?] 4-8 Other Interval Estimates for a Single Sample 4-8.1 Prediction Interval

157. 157 4-8 Other Interval Estimates for a Single Sample 4-8.1 Prediction Interval To predict the value of a single future value,. The point estimator is. The expected value of the prediction error, The variance of the prediction error, When is unknown, has distribution.

158. 158 4-8 Other Interval Estimates for a Single Sample 4-8.1 Prediction Interval Discuss Example 4-16

159. 159 4-8 Other Interval Estimates for a Single Sample 4-8.2 Tolerance Intervals for a Normal Distribution

160. 160 4-10 Testing for Goodness of Fit So far, we have assumed the population or probability distribution for a particular problem is known. (parametric approach). There are many instances where the underlying distribution is not known, and we wish to test a particular distribution. (nonparametric approach 無母數統計 ). Use a goodness-of-fit test procedure based on the chi- square distribution.

161. 161 4-10 Testing for Goodness of Fit O i : the observed frequency in the i-th class interval E i : the expected frequency in the i-th class interval from the hypothesized probability distribution Test Statistics:

162. 162 Where is the expected number of observations in ith group, and is the observed number observations in the ith group. Then Theorem: Where k is the number of groups p is the number of parameters estimated by the sample.

163. 163 4-10 Testing for Goodness of Fit Discuss Example 4-18

164. 164 Value012345 Observed Frequency 8252321167 Expected Frequency 9.121.826.220.912.56.02 Example #4-85(p207) P(X=0)=0.091, P(X=1)=0.218, P(X=2)=0.262, P(X=3)=0.209, P(X=4)=0.125, P(X=5)=0.06, P(X>5)=0.035 k=6, p=1 df=4 Do not reject P-value = lies between 0.5 and 0.9 (p.437).