2. Learning Objectives for Chapter 8 After careful study of this chapter, you should be able to do the following: 1.Construct confidence intervals on the mean of a normal distribution, using normal distribution or t distribution method. 2.Construct confidence intervals on variance and standard deviation of normal distribution. 3.Construct confidence intervals on a population proportion. 4.Constructing an approximate confidence interval on a parameter. 5.Prediction intervals for a future observation. 6.Tolerance interval for a normal population. 2Chapter 8 Learning Objectives

3. Statistical Intervals For A Single Sample Confidence Interval Confidence Interval. We have learned in the previous chapter how a parameter can be estimated from sample data It is important to understand how good is the estimate obtained. Confidence Interval An interval estimate for population parameter is called a Confidence Interval. Tolerance interval is another important type of interval estimate. Introduction The confidence interval is a random interval The appropriate interpretation of a confidence interval (for example on  ) is: The observed interval [l, u] brackets the true value of , with confidence 100(1-  ).

4. Confidence interval A range of values, derived from sample statistics, that is likely to contain the value of an unknown population parameter. Because of their random nature, it is unlikely that two samples from a given population will yield identical confidence intervals. But if you repeated your sample many times, a certain percentage of the resulting confidence intervals would contain the unknown population parameter. The percentage of these confidence intervals that contain the parameter is the confidence level of the interval. For example, suppose you want to know the average amount of time it takes for an automobile assembly line to complete a vehicle. You take a sample of completed cars, record the time they spent on the assembly line, and use the 1-sample t procedure to obtain a 95% confidence interval for the mean amount of time all cars spend on the assembly line. Because 95% of the confidence intervals constructed from all possible samples will contain the population parameter, you conclude that the mean amount of time all cars spend on the assembly line falls between your interval's endpoints, which are called confidence limits.

5. 5 A confidence interval estimate for  is an interval of the form l ≤  ≤ u, where the end-points l and u are computed from the sample data. There is a probability of 1  α of selecting a sample for which the CI will contain the true value of . The endpoints or bounds l and u are called lower- and upper- confidence limits,and 1  α is called the confidence coefficient. Sec 8-1 Confidence Interval on the Mean of a Normal, σ 2 Known 8-1.1 Confidence Interval and its Properties

6. Confidence interval Creating confidence intervals is analogous to throwing nets over a target with an unknown, yet fixed, location. Consider the graphic below, which depicts confidence intervals generated from 20 samples from the same population. The black line represents the fixed value of the unknown population parameter; the blue confidence intervals contain the value of the population parameter; the red confidence interval does not. A 95% confidence interval indicates that 19 out of 20 samples (95%) from the same population will produce confidence intervals that contain the population parameter.

7. Confidence Interval on the Mean of a Normal Distribution, Variance Known Figure 8-1 Repeated construction of a confidence interval for .

8. Guidelines for Constructing Confidence Intervals

9. Prediction Interval: Provides prediction on future observations.

12. Hw 2 groups Probability Tables Description, Objective and uses …5 ++ slides

14. 8-2.1 Development of the Confidence Interval and its Basic Properties The endpoints or bounds l and u are called lower- and upper-confidence limits, respectively. Since Z follows a standard normal distribution, we can write: Confidence Interval on the Mean of a Normal Distribution, Variance Known

15. One sided confidence Interval Confidence Interval on the Mean of a Normal Distribution, Variance Known

18. Example 8-1 Confidence Interval on the Mean of a Normal Distribution, Variance Known

19. Confidence interval A range of values, derived from sample statistics, that is likely to contain the value of an unknown population parameter. Because of their random nature, it is unlikely that two samples from a given population will yield identical confidence intervals. But if you repeated your sample many times, a certain percentage of the resulting confidence intervals would contain the unknown population parameter. The percentage of these confidence intervals that contain the parameter is the confidence level of the interval. For example, suppose you want to know the average amount of time it takes for an automobile assembly line to complete a vehicle. You take a sample of completed cars, record the time they spent on the assembly line, and use the 1-sample t procedure to obtain a 95% confidence interval for the mean amount of time all cars spend on the assembly line. Because 95% of the confidence intervals constructed from all possible samples will contain the population parameter, you conclude that the mean amount of time all cars spend on the assembly line falls between your interval's endpoints, which are called confidence limits.

22. Confidence Interval on the Mean of a Normal Distribution, Variance Known Figure 8-1 Repeated construction of a confidence interval for .

23. Confidence Level and Precision of Error The length of a confidence interval is a measure of the precision of estimation. Confidence Interval on the Mean of a Normal Distribution, Variance Know n Figure 8-2 Error in estimating  with.

24. 8-2.2 Choice of Sample Size Sample Confidence Interval on the Mean of a Normal Distribution, Variance Known

25. Example 8-2 Confidence Interval on the Mean of a Normal Distribution, Variance Known EXAMPLE 8-2 Metallic Material Transition

26. 8-1.3 One-Sided Confidence Bounds 26 A 100(1 − α)% upper-confidence bound for  is (8-3) and a 100(1 − α)% lower-confidence bound for  is (8-4) Sec 8-1 Confidence Interval on the Mean of a Normal, σ 2 Known

27. Example 8-3 One-Sided Confidence Bound 27 The same data for impact testing from Example 8-1 are used to construct a lower, one-sided 95% confidence interval for the mean impact energy. Recall that z α = 1.64, n = 10,  = l, and. A 100(1 − α)% lower-confidence bound for  is Sec 8-1 Confidence Interval on the Mean of a Normal, σ 2 Known

28. 8-2.5 A Large-Sample Confidence Interval for  Confidence Interval on the Mean of a Normal Distribution, Variance Known See example 8-3

29. 8-2.5 A Large-Sample Confidence Interval for  Definition Confidence Interval on the Mean of a Normal Distribution, Variance Known

30. Example 8-3 Confidence Interval on the Mean of a Normal Distribution, Variance Known, Example 8-5 Mercury Contamination

31. 31 The summary statistics for the data are as follows: VariableNMeanMedianStDevMinimumMaximumQ1Q1Q3Q3 Concentration530.52500.49000.34860.04001.33000.23000.7900 Sec 8-1 Confidence Interval on the Mean of a Normal, σ 2 Known Example 8-5 Mercury Contamination (continued) Because n > 40, the assumption of normality is not necessary to use in Equation 8-5. The required values are, and z 0.025 = 1.96. The approximate 95  CI on  is Interpretation: This interval is fairly wide because there is variability in the mercury concentration measurements. A larger sample size would have produced a shorter interval.

33. Text and Material Exam 1 Probability Tables Home Work Policy Attendance policy

34. t-distribution In probability and statistics, Student's t-distribution (or simply the t-distribution) is any member of a family of continuousprobability distributions that arises when estimating the mean of a normally distributed population in situations where thesample size is small and population standard deviation is unknown.probabilitystatisticsprobability distributionsmeannormally distributedpopulationsample sizestandard deviation he t-distribution plays a role in a number of widely used statistical analyses, including Student's t-test for assessing thestatistical significance of the difference between two sample means, the construction of confidence intervals for the difference between two population means, and in linear regression analysis. The Student's t-distribution also arises in theBayesian analysis of data from a normal family.Student's t-teststatistical significancemeansconfidence intervalsregression analysisBayesian analysis The t-distribution is symmetric and bell-shaped, like the normal distribution, but has heavier tails, meaning that it is more prone to producing values that fall far from its mean. This makes it useful for understanding the statistical behavior of certain types of ratios of random quantities, in which variation in the denominator is amplified and may produce outlying values when the denominator of the ratio falls close to zero. The Student's t-distribution is a special case of the generalised hyperbolic distributionnormal distributiongeneralised hyperbolic distribution Probability density function Cumulative distribution function

35. Confidence Interval on the mean of a normal distribution, Variance Unknown

36. 8-2.2 The Confidence Interval on Mean, Variance Unknown Table III 36 If and s are the mean and standard deviation of a random sample from a normal distribution with unknown variance  2, a 100(1   )  confidence interval on  is given by (8-7) where t  2,n  1 the upper 100  2 percentage point of the t distribution with n  1 degrees of freedom. One-sided confidence bounds on the mean are found by replacing t  /2,n-1 in Equation 8-7 with t ,n-1. Sec 8-2 Confidence Interval on the Mean of a Normal, σ 2 Unknown

37. Example

39. Example 8-6 Alloy Adhesion 39 Construct a 95% CI on  to the following data. The sample mean is and sample standard deviation is s = 3.55. Since n = 22, we have n  1 =21 degrees of freedom for t, so t 0.025,21 = 2.080. The resulting CI is Interpretation: The CI is fairly wide because there is a lot of variability in the measurements. A larger sample size would have led to a shorter interval. 19.810.114.97.515.4 18.57.912.711.911.4 14.117.616.715.8 19.58.813.611.911.4 Sec 8-2 Confidence Interval on the Mean of a Normal, σ 2 Unknown

40. Sec 8-7 to write40

41. Sec 8-7 to write41

44. Exam : First week on March No cell phones Signature Cheat Sheet First three Chapters + Sat 1 (Introduction)

49. (Chi-square or χ²-distribution) The Chi Square test is the most important and most used method in statistical tests. The purpose of Chi Square test A-Test how well a sample fits a theoretical distribution B- The difference between an observed frequency and expected frequency. (differences between the two or more observed data). C- Its value can be calculated by using the given observed frequency and expected frequency. D- Test the independence between categorical variables. For example, a manufacturer wants to know if the occurrence of four types of defects (missing pin, broken clamp, loose fastener, and leaky seal) is related to shift (day, evening, overnight). The chi-squared distribution (also It is a special case of the gamma distribution and is one of the most widely used probability distributions in inferential statistics,gamma distributionprobability distributionsinferential statistics For example, you can use a goodness-of-fit test of an observed distribution to a theoretical one and classification of qualitative data, and in confidence interval estimation for a population standard deviation of a normal distribution from a sample standard deviation. also to determine whether your sample data fit a Poisson distribution.qualitative dataconfidence intervalstandard deviation The shape of the chi-square distribution depends on the number of degrees of freedom. The distribution is positively skewed, but skewness decreases with more degrees of freedom. When the degrees of freedom are 30 or more, the distribution can be approximated by a normal distribution.

50. Goodness of Fit Test (observed and expected test) Failure times,Bulbs

51. Poisson Distribution (no of defects in finite space A/C Mech sys..lamda represent mean and variance binomial and standard no upper bound of calls

52. Chi Square Formula The Chi Square is denoted by X2and the formula is given as: Here, O = Observed frequency E = Expected frequency ∑ = Summation X2 = Chi Square value

53. Confidence Interval on the (Variance and Standard Deviation of a Normal Distribution) Figure 8-8 Probability density functions of several  2 distributions.

54. 54 Let X 1, X 2, , X n be a random sample from a normal distribution with mean  and variance  2, and let S 2 be the sample variance. Then the random variable (8-8) has a chi-square (  2 ) distribution with n  1 degrees of freedom. Sec 8-3 Confidence Interval on σ 2 & σ of a Normal Distribution

55. Confidence Interval on the Variance and Standard Deviation 55 If s 2 is the sample variance from a random sample of n observations from a normal distribution with unknown variance  2, then a 100(1 –  )% confidence interval on  2 is (8-9) where and are the upper and lower 100  /2 percentage points of the chi-square distribution with n – 1 degrees of freedom, respectively. A confidence interval for  has lower and upper limits that are the square roots of the corresponding limits in Equation 8–9. Sec 8-3 Confidence Interval on σ 2 & σ of a Normal Distribution

56. Sec 8-7 to write56

57. Confidence Interval on the Variance and Standard Deviation of a Normal Distribution One-Sided Confidence Bounds

58. Example 8-7 Detergent Filling 58 An automatic filling machine is used to fill bottles with liquid detergent. A random sample of 20 bottles results in a sample variance of fill volume of s 2 = 0.0153 2. Assume that the fill volume is approximately normal. Compute a 95% upper confidence bound. A 95% upper confidence bound is found from Equation 8-10 as follows: A confidence interval on the standard deviation  can be obtained by taking the square root on both sides, resulting in Sec 8-3 Confidence Interval on σ 2 & σ of a Normal Distribution

59. Sec 8-7 to write59

60. 8-6.1 Prediction Interval for Future Observation 8-6 Tolerance and Prediction Intervals The prediction interval for X n+1 will always be longer than the confidence interval for . 60 A 100 (1   )% prediction interval (PI) on a single future observation from a normal distribution is given by (8-15) Sec 8-6 Tolerance & Prediction Intervals

61. Repeated for comparison Example 8-11 Alloy Adhesion 61 The load at failure for n = 22 specimens was observed, and found that and s  3.55. The 95% confidence interval on  was 12.14    15.28. Plan to test a 23rd specimen. A 95% prediction interval on the load at failure for this specimen is Interpretation: The prediction interval is considerably longer than the CI. This is because the CI is an estimate of a parameter, but the PI is an interval estimate of a single future observation. See next slide Sec 8-6 Tolerance & Prediction Intervals

62. Example 8-6 Alloy Adhesion 62 Construct a 95% CI on  to the following data. The sample mean is and sample standard deviation is s = 3.55. Since n = 22, we have n  1 =21 degrees of freedom for t, so t 0.025,21 = 2.080. The resulting CI is Interpretation: The CI is fairly wide because there is a lot of variability in the measurements. A larger sample size would have led to a shorter interval. 19.810.114.97.515.4 18.57.912.711.911.4 14.117.616.715.8 19.58.813.611.911.4 Sec 8-2 Confidence Interval on the Mean of a Normal, σ 2 Unknown

63. 63 A tolerance interval for capturing at least  % of the values in a normal distribution with confidence level 100(1 –  )% is where k is a tolerance interval factor found in Appendix Table XII. Values are given for  = 90%, 95%, and 99% and for 90%, 95%, and 99% confidence. Sec 8-6 Tolerance & Prediction Intervals 8-6.2 Tolerance Interval for a Normal Distribution

65. Example 8-12 Alloy Adhesion 65 The load at failure for n = 22 specimens was observed, and found that and s = 3.55. Find a tolerance interval for the load at failure that includes 90% of the values in the population with 95% confidence. From Appendix Table XII, the tolerance factor k for n = 22,  = 0.90, and 95% confidence is k = 2.264. The desired tolerance interval is Interpretation: We can be 95% confident that at least 90% of the values of load at failure for this particular alloy lie between 5.67 and 21.74. Sec 8-6 Tolerance & Prediction Intervals

67. Important Terms & Concepts of Chapter 8 Chi-squared distribution Confidence coefficient Confidence interval Confidence interval for a: – Population proportion – Mean of a normal distribution – Variance of a normal distribution Confidence level Error in estimation Large sample confidence interval 1-sided confidence bounds Precision of parameter estimation Prediction interval Tolerance interval 2-sided confidence interval t distribution Chapter 8 Summary67

68. Guidelines for Constructing Confidence Intervals

69. 8-5 Guidelines for Constructing Confidence Intervals 69 Table 8-1 provides a simple road map for appropriate calculation of a confidence interval. Sec 8-5 Guidelines for Constructing Confidence Intervals

70. Normal Approximation for Binomial Proportion 8-4 A Large-Sample Confidence Interval For a Population Proportion The quantity is called the standard error of the point estimator. 70 If n is large, the distribution of is approximately standard normal. Sec 8-4 Large-Sample Confidence Interval for a Population Proportion

71. Approximate Confidence Interval on a Binomial Proportion 71 If is the proportion of observations in a random sample of size n, an approximate 100(1   )% confidence interval on the proportion p of the population is (8-11) where z  /2 is the upper  /2 percentage point of the standard normal distribution. Sec 8-4 Large-Sample Confidence Interval for a Population Proportion

72. Example 8-8 Crankshaft Bearings 72 In a random sample of 85 automobile engine crankshaft bearings, 10 have a surface finish that is rougher than the specifications allow. Construct a 95% two-sided confidence interval for p. A point estimate of the proportion of bearings in the population that exceeds the roughness specification is. A 95% two-sided confidence interval for p is computed from Equation 8-11 as Interpretation: This is a wide CI. Although the sample size does not appear to be small (n = 85), the value of is fairly small, which leads to a large standard error for contributing to the wide CI. Sec 8-4 Large-Sample Confidence Interval for a Population Proportion

73. Sample size for a specified error on a binomial proportion : If we set and solve for n, the appropriate sample size is The sample size from Equation 8-12 will always be a maximum for p = 0.5 [that is, p(1 − p) ≤ 0.25 with equality for p = 0.5], and can be used to obtain an upper bound on n. 73 Sec 8-4 Large-Sample Confidence Interval for a Population Proportion Choice of Sample Size (8-12) (8-13)

74. Example 8-9 Crankshaft Bearings 74 Consider the situation in Example 8-8. How large a sample is required if we want to be 95% confident that the error in using to estimate p is less than 0.05? Using as an initial estimate of p, we find from Equation 8-12 that the required sample size is If we wanted to be at least 95% confident that our estimate of the true proportion p was within 0.05 regardless of the value of p, we would use Equation 8-13 to find the sample size Interpretation: If we have information concerning the value of p, either from a preliminary sample or from past experience, we could use a smaller sample while maintaining both the desired precision of estimation and the level of confidence. Sec 8-4 Large-Sample Confidence Interval for a Population Proportion

75. Approximate One-Sided Confidence Bounds on a Binomial Proportion 75 The approximate 100(1   )% lower and upper confidence bounds are (8-14) respectively. Sec 8-4 Large-Sample Confidence Interval for a Population Proportion

76. Example 8-10 The Agresti-Coull CI on a Proportion 76 Reconsider the crankshaft bearing data introduced in Example 8-8. In that example we reported that. The 95% CI was. Construct the new Agresti-Coull CI. Interpretation: The two CIs would agree more closely if the sample size were larger. Sec 8-4 Large-Sample Confidence Interval for a Population Proportion

77. 7: Normal Probability Distributions77 Probabilities Normal Distributions Normal Distributions Determining Normal Probabilities Finding Values That Correspond to Normal Probabilities Assessing Departures from Normality

78. 7: Normal Probability Distributions78 §7.1: Normal Distributions This pdf is the most popular distribution for continuous random variables First described de Moivre in 1733 Elaborated in 1812 by Laplace Describes some natural phenomena More importantly, describes sampling characteristics of totals and means

79. 7: Normal Probability Distributions79 Normal Probability Density Function Recall: continuous random variables are described with probability density function (pdfs) curves Normal pdfs are recognized by their typical bell-shape Figure: Age distribution of a pediatric population with overlying Normal pdf

80. 7: Normal Probability Distributions80 Area Under the Curve pdfs should be viewed almost like a histogram Top Figure: The darker bars of the histogram correspond to ages ≤ 9 (~40% of distribution) Bottom Figure: shaded area under the curve (AUC) corresponds to ages ≤ 9 (~40% of area)

81. 7: Normal Probability Distributions81 Parameters μ and σ Normal pdfs have two parameters μ - expected value (mean “mu”) σ - standard deviation (sigma) σ controls spreadμ controls location

82. 7: Normal Probability Distributions82 Mean and Standard Deviation of Normal Density μ σ

83. 7: Normal Probability Distributions83 Standard Deviation σ Points of inflections one σ below and above μ Practice sketching Normal curves Feel inflection points (where slopes change) Label horizontal axis with σ landmarks

84. 7: Normal Probability Distributions84 Two types of means and standard deviations The mean and standard deviation from the pdf (denoted μ and σ) are parameters The mean and standard deviation from a sample (“xbar” and s) are statistics Statistics and parameters are related, but are not the same thing!

85. 7: Normal Probability Distributions85 68-95-99.7 Rule for Normal Distributions 68% of the AUC within ±1σ of μ 95% of the AUC within ±2σ of μ 99.7% of the AUC within ±3σ of μ

86. 7: Normal Probability Distributions86 Example: 68-95-99.7 Rule Wechsler adult intelligence scores: Normally distributed with μ = 100 and σ = 15; X ~ N(100, 15) 68% of scores within μ ± σ = 100 ± 15 = 85 to 115 95% of scores within μ ± 2σ = 100 ± (2)(15) = 70 to 130 99.7% of scores in μ ± 3σ = 100 ± (3)(15) = 55 to 145

87. 7: Normal Probability Distributions87 Symmetry in the Tails … we can easily determine the AUC in tails 95% Because the Normal curve is symmetrical and the total AUC is exactly 1…

88. 7: Normal Probability Distributions88 Example: Male Height Male height: Normal with μ = 70.0˝ and σ = 2.8˝ 68% within μ ± σ = 70.0  2.8 = 67.2 to 72.8 32% in tails (below 67.2˝ and above 72.8˝) 16% below 67.2˝ and 16% above 72.8˝ (symmetry)

89. 7: Normal Probability Distributions89 Reexpression of Non-Normal Random Variables Many variables are not Normal but can be reexpressed with a mathematical transformation to be Normal Example of mathematical transforms used for this purpose: – logarithmic – exponential – square roots Review logarithmic transformations…

90. 7: Normal Probability Distributions90 §7.2: Determining Normal Probabilities When value do not fall directly on σ landmarks: 1. State the problem 2. Standardize the value(s) (z score) 3. Sketch, label, and shade the curve 4. Use Table B

91. 7: Normal Probability Distributions91 §7.2: Determining Normal Probabilities When value do not fall directly on σ landmarks: 1. State the problem 2. Standardize the value(s) (z score) 3. Sketch, label, and shade the curve 4. Use Table B

92. 7: Normal Probability Distributions92 Step 1: State the Problem What percentage of gestations are less than 40 weeks? Let X ≡ gestational length We know from prior research: X ~ N(39, 2) weeks Pr(X ≤ 40) = ?

93. 7: Normal Probability Distributions93 Step 2: Standardize Standard Normal variable ≡ “Z” ≡ a Normal random variable with μ = 0 and σ = 1, Z ~ N(0,1) Use Table B to look up cumulative probabilities for Z

94. 7: Normal Probability Distributions94 Example: A Z variable of 1.96 has cumulative probability 0.9750.

95. 7: Normal Probability Distributions95 Step 2 (cont.) z-score = no. of σ-units above (positive z) or below (negative z) distribution mean μ Turn value into z score:

96. 7: Normal Probability Distributions96 3. Sketch 4. Use Table B to lookup Pr(Z ≤ 0.5) = 0.6915 Steps 3 & 4: Sketch & Table B

97. 7: Normal Probability Distributions97 a represents a lower boundary b represents an upper boundary Pr(a ≤ Z ≤ b) = Pr(Z ≤ b) − Pr(Z ≤ a) Probabilities Between Points

98. 7: Normal Probability Distributions98 Pr(-2 ≤ Z ≤ 0.5) = Pr(Z ≤ 0.5) − Pr(Z ≤ -2).6687=.6915 −.0228 Between Two Points See p. 144 in text.6687.6915.0228 -2 0.5 -2

99. 7: Normal Probability Distributions99 §7.3 Values Corresponding to Normal Probabilities 1.State the problem 2.Find Z-score corresponding to percentile (Table B) 3.Sketch 4. Unstandardize:

100. 7: Normal Probability Distributions100 z percentiles  z p ≡ the Normal z variable with cumulative probability p  Use Table B to look up the value of z p  Look inside the table for the closest cumulative probability entry  Trace the z score to row and column

101. 7: Normal Probability Distributions101 Notation: Let z p represents the z score with cumulative probability p, e.g., z.975 = 1.96 e.g., What is the 97.5 th percentile on the Standard Normal curve? z.975 = 1.96

102. 7: Normal Probability Distributions102 Step 1: State Problem Question: What gestational length is smaller than 97.5% of gestations? Let X represent gestations length We know from prior research that X ~ N(39, 2) A value that is smaller than.975 of gestations has a cumulative probability of.025

103. 7: Normal Probability Distributions103 Step 2 (z percentile) Less than 97.5% (right tail) = greater than 2.5% (left tail) z lookup: z.025 = −1.96 z.00.01.02.03.04.05.06.07.08.09 –1.9.0287.0281.0274.0268.0262.0256.0250.0244.0239.0233

104. 7: Normal Probability Distributions104 The 2.5 th percentile is 35 weeks Unstandardize and sketch

105. Example: A Z variable of 1.96 has cumulative probability of …… z lookup: z.025 = ----- What is the 97.5 th percentile on the Standard Normal curve? z.975 =

106. 7: Normal Probability Distributions106 Between Two Points.6687.6915.0228 -2 0.5 -2 Q 2:A Z variable of 1.96 has cumulative probability of …… Q3: Find P(Z less than 1.96) Q1: z lookup: z.025 = ----- q2 What is the 97.5 th percentile on the Standard Normal curve? z.975 = Find P(Z greater than 1.96)