Chapter 9 One and Two Sample Estimation Point Estimates Unknown Parameter Symbol Point Estimator Population mean m Sample mean Population standard deviation s Sample standard deviation Population proportion p Sample proportion     Point estimates are computed from sample information and used to estimate population parameters. EGR 252 Ch. 9 Lecture1 MDH 2015 9th edition

Interval Estimation Interval estimation: The interval in which you would expect to find the value of a parameter. By taking a random sample, we can compute an interval with upper and lower limits called a (1-a) 100% confidence interval for the unknown parameter In most a cases the parameter for intervals to be created will be m Updated for 8th edition by JMB EGR 252 Ch. 9 Lecture1 MDH 2015 9th edition 252Chapter9 JMB F06 8th edition

Confidence Intervals Single value Mean Degree of uncertainty Point Estimates Single value Mean Intervals Degree of uncertainty Range of certainty around the point estimate Based On Point Estimates (mean) Confidence level (1-a) Standard deviation Expressed As The mean score of the students was 80.1 with a 95% CI of 77.4 – 82.5 EGR 252 Ch. 9 Lecture1 MDH 2015 9th edition

Confidence Intervals (1 – α) 100% confidence interval for the unknown parameter. Example: if α = 0.01, we develop a 99% confidence interval. Example: if α = 0.05, we develop a 95% confidence interval. Updated for 8th edition by JMB EGR 252 Ch. 9 Lecture1 MDH 2015 9th edition 252Chapter9 JMB F06 8th edition

Single Sample: Estimating the Mean Given: σ is known and X is the mean of a random sample of size n, Then, the (1 – α)100% confidence interval for μ is show α/2 above and below and 1-α in the center. Values obtained from Z-table 1 - a a/2 a/2 EGR 252 Ch. 9 Lecture1 MDH 2015 9th edition 252Chapter9 JMB F06 8th edition

Example A traffic engineer is concerned about the delays at an intersection near a local school. The intersection is equipped with a fully actuated (“demand”) traffic light and there have been complaints that traffic on the main street is subject to unacceptable delays. To develop a benchmark, the traffic engineer randomly samples 25 stop times (in seconds) on a weekend day. The average of these times is found to be 13.2 seconds, and the variance is known to be 4 seconds2. Based on this data, what is the 95% confidence interval (C.I.) around the mean stop time during a weekend day? EGR 252 Ch. 9 Lecture1 MDH 2015 9th edition 252Chapter9 JMB F06 8th edition

Example (cont.) X = ______________ σ = _______________ α = ________________ α/2 = _____________ Z0.025 = _____________ Z0.975 = ____________ Enter values into equation EGR 252 Ch. 9 Lecture1 MDH 2015 9th edition 252Chapter9 JMB F06 8th edition

Example (cont.) X = ______13.2________ σ = _____2________ α = ______0.05________ α/2 = __0.025_______ Z0.025 = ___-1.96______ Z0.975 = ____1.96____ Solution: 12.416 < μSTOP TIME < 13.984 X = 13.2 σ = 2 α = .05 α/2 = .025 Draw picture z.025 = -1.96 z.975 = 1.96 13.2-(1.96)(2/sqrt(25)) = 12.416 13.2+(1.96)(2/sqrt(25)) = 13.984 Interpretation: there is a 95% chance that the “true mean” of traffic delay time is somewhere between 12.416 and 13.984 Z0.025 = -1.96 Z0.975 = 1.96 13.2-(1.96)(2/sqrt(25)) = 12.416 13.2+(1.96)(2/sqrt(25)) = 13.984 EGR 252 Ch. 9 Lecture1 MDH 2015 9th edition 252Chapter9 JMB F06 8th edition

Your turn … What is the 90% C.I.? What does it mean? Z = + 1.645 12.542 < μ < 13.858 5% 5% Z(.05) = + 1.645 All other values remain the same. The 90 % CI for μ = (12.542,13.858) Note that the 95% CI is wider than the 90% CI. EGR 252 Ch. 9 Lecture1 MDH 2015 9th edition 252Chapter9 JMB F06 8th edition

What if σ 2 is unknown? For example, what if the traffic engineer doesn’t know the variance of this population? If n is sufficiently large (n > 30), then the large sample confidence interval is calculated by using the sample standard deviation in place of sigma: If σ 2 is unknown and n is not “large”, we must use the t-statistic. EGR 252 Ch. 9 Lecture1 MDH 2015 9th edition 252Chapter9 JMB F06 8th edition

Single Sample: Estimating the Mean (σ unknown, n not large) Given: σ is unknown and X is the mean of a random sample of size n (where n is not large), Then, the (1 – α)100% confidence interval for μ is: show α/2 above and below and 1-α in the center. Values obtained from t-table 1 - a a/2 a/2 EGR 252 Ch. 9 Lecture1 MDH 2015 9th edition 252Chapter9 JMB F06 8th edition

Recall Our Example A traffic engineer is concerned about the delays at an intersection near a local school. The intersection is equipped with a fully actuated (“demand”) traffic light and there have been complaints that traffic on the main street is subject to unacceptable delays. To develop a benchmark, the traffic engineer randomly samples 25 stop times (in seconds) on a weekend day. The average of these times is found to be 13.2 seconds, and the sample variance, s2, is found to be 4 seconds2. Based on this data, what is the 95% confidence interval (C.I.) around the mean stop time during a weekend day? EGR 252 Ch. 9 Lecture1 MDH 2015 9th edition 252Chapter9 JMB F06 8th edition

Small Sample Example (cont.) n = _______ v = _______ X = ______ s = _______ α = _______ α/2 = ____ t0.025,24 = _______ Find interval using equation below EGR 252 Ch. 9 Lecture1 MDH 2015 9th edition 252Chapter9 JMB F06 8th edition

Small Sample Example (cont.) n = __25___ v = _24___ X = __13.2_ s = __2__ α = __0.05___ α/2 = 0.025_ t0.025,24 = _2.064_ __12.374_____ < μ < ___14.026____ X = 13.2 s = 2 α = .05 α/2 = .025 Draw picture t.025,24 = 2.064 13.2-(2.064)(2/sqrt(25)) = 12.374 13.2+(2.064)(2/sqrt(25)) = 14.026 Interpretation: there is a 95% chance that the “true mean” of traffic delay time is somewhere between 13.374 and 14.026 13.2 - (2.064)(2/sqrt(25)) = 13.374 13.2 + (2.064)(2/sqrt(25)) = 14.026 EGR 252 Ch. 9 Lecture1 MDH 2015 9th edition 252Chapter9 JMB F06 8th edition

Your turn A thermodynamics professor gave a physics pretest to a random sample of 15 students who enrolled in his course at a large state university. The sample mean was found to be 59.81 and the sample standard deviation was 4.94. Find a 99% confidence interval for the mean on this pretest. EGR 252 Ch. 9 Lecture1 MDH 2015 9th edition 252Chapter9 JMB F06 8th edition

Solution X = ______________ s = _______________ α = ________________ α/2 = _____________ (draw the picture) t___ , ____ = _____________ __________________ < μ < ___________________ X = 59.81 s = 4.94 α = .01 α/2 = .005 Draw picture t.005,14 = 2.977 59.81-(2.977)(4.94/sqrt(15)) = 56.01 59.81-(2.977)(4.94/sqrt(15)) = 63.61 Interpretation: X = 59.81 s = 4.94 α = .01 α/2 = .005 t (.005,14) = 2.977 Lower Bound 59.81 - (2.977)(4.94/sqrt(15)) = 56.01 Upper Bound 59.81 + (2.977)(4.94/sqrt(15)) = 63.61 EGR 252 Ch. 9 Lecture1 MDH 2015 9th edition 252Chapter9 JMB F06 8th edition

Standard Error of a Point Estimate Case 1: σ known The standard deviation, or standard error of X is Case 2: σ unknown, sampling from a normal distribution The standard deviation, or (usually) estimated standard error of X is ???? Z = + 1.645 12.542 < μ < 13.858 EGR 252 Ch. 9 Lecture1 MDH 2015 9th edition 252Chapter9 JMB F06 8th edition

Prediction Intervals Used to predict the possible value of a future observation Example: In quality control, an experimenter may need to use the observed data to predict a new observation. EGR 252 Ch. 9 Lecture1 MDH 2015 9th edition

9.6: Prediction Interval For a normal distribution of unknown mean μ, and standard deviation σ, a 100(1-α)% prediction interval of a future observation, x0 is if σ is known, and if σ is unknown EGR 252 Ch. 9 Lecture1 MDH 2015 9th edition 252Chapter9 JMB F06 8th edition

Tolerance Limits (Intervals) What if you want to be 95% sure that the interval contains 95% of the values? Or 90% sure that the interval contains 99% of the values?  These questions are answered by a tolerance interval. To compute, or understand, a tolerance interval you have to specify two different percentages. One expresses how sure you want to be, and the other expresses what fraction of the values the interval will contain. EGR 252 Ch. 9 Lecture1 MDH 2015 9th edition

9.7: Tolerance Limits For a normal distribution of unknown mean μ, and unknown standard deviation σ, tolerance limits are given by x + ks where k is determined so that one can assert with 100(1-γ)% confidence that the given limits contain at least the proportion 1-α of the measurements. Table A.7 (page 745) gives values of k for (1-α) = 0.9, 0.95, or 0.99 and γ = 0.05 or 0.01 for selected values of n. EGR 252 Ch. 9 Lecture1 MDH 2015 9th edition 252Chapter9 JMB F06 8th edition

Tolerance Limits How to determine 100(1-γ)% and 1-α. For a sample size of 8, find the tolerance interval that gives two-sided 95% bounds on 90% of the distribution or population. X is 15.6 and s is 1.4 From table on pg. 745, find the corresponding value: n = 8, g = .05, a = 0.1 corresponding k…k = 3.136 x + ks = 15.6 + (3.136)(1.4) Tolerance interval 19.99 – 11.21 We are 95% confident that 90% of the population falls within the limits of 11.21 and 19.99 1-g (boundary or the limits) 1-a (proportion of the distribution) EGR 252 Ch. 9 Lecture1 MDH 2015 9th edition

Case Study 9.1c (Page 281) Find the 99% tolerance limits that will contain 95% of the metal pieces produced by the machine, given a sample mean diameter of 1.0056 cm and a sample standard deviation of 0.0246. Table A.7 (page 745) (1 - α ) = 0.95 (1 – Ƴ ) = 0.99 n = 9 k = 4.550 x ± ks = 1.0056 ± (4.550) (0.0246) We can assert with 99% confidence that the tolerance interval from 0.894 to 1.117 cm will contain 95% of the metal pieces produced by the machine. EGR 252 Ch. 9 Lecture1 MDH 2015 9th edition 252Chapter9 JMB F06 8th edition

Summary Confidence interval  population mean μ Prediction interval  a new observation x0 Tolerance interval  a (1-α) proportion of the measurements can be estimated with 100( 1-Ƴ )% confidence EGR 252 Ch. 9 Lecture1 MDH 2015 9th edition 252Chapter9 JMB F06 8th edition