1. EGR 252 Ch. 9 Lecture1 JMB 2014 9th edition Slide 1 Chapter 9: One- and Two- Sample Estimation  Statistical Inference  Estimation  Tests of hypotheses  Interval estimation: (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.

2. EGR 252 Ch. 9 Lecture1 JMB 2014 9th edition Slide 2 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 1 -  

3. EGR 252 Ch. 9 Lecture1 JMB 2014 9th edition Slide 3 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 seconds 2. Based on this data, what is the 95% confidence interval (C.I.) around the mean stop time during a weekend day?

4. EGR 252 Ch. 9 Lecture1 JMB 2014 9th edition Slide 4 Example (cont.) X = ______________ σ = _______________ α = ________________ α/2 = _____________ Z 0.025 = _____________ Z 0.975 = ____________ Solution: 12.416 < μ STOP TIME < 13.984 Z 0.025 = -1.96 Z 0.975 = 1.96 13.2-(1.96)(2/sqrt(25)) = 12.416 13.2+(1.96)(2/sqrt(25)) = 13.984

5. EGR 252 Ch. 9 Lecture1 JMB 2014 9th edition Slide 5 Your turn …  What is the 90% C.I.? What does it mean? 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. 90% 5%

6. EGR 252 Ch. 9 Lecture1 JMB 2014 9th edition Slide 6 What if σ 2 is unknown?  For example, what if the traffic engineer doesn’t know the variance of this population? 1.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: 2.If σ 2 is unknown and n is not “large”, we must use the t-statistic.

7. EGR 252 Ch. 9 Lecture1 JMB 2014 9th edition Slide 7 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:

8. EGR 252 Ch. 9 Lecture1 JMB 2014 9th edition Slide 8 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, s 2, is found to be 4 seconds 2. Based on this data, what is the 95% confidence interval (C.I.) around the mean stop time during a weekend day?

9. EGR 252 Ch. 9 Lecture1 JMB 2014 9th edition Slide 9 Small Sample Example (cont.) n = _______ df = _______ X = ______ s = _______ α = _______ α/2 = ____ t 0.025,24 = _______ _______________ < μ < ________________ 13.2 - (2.064)(2/sqrt(25)) = 13.374 13.2 + (2.064)(2/sqrt(25)) = 14.026

10. EGR 252 Ch. 9 Lecture1 JMB 2014 9th edition Slide 10 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.

11. EGR 252 Ch. 9 Lecture1 JMB 2014 9th edition Slide 11 Solution X = ______________ s = _______________ α = ________________ α/2 = _____________ (draw the picture) t ___, ____ = _____________ __________________ < μ < ___________________ X = 59.81s = 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

12. EGR 252 Ch. 9 Lecture1 JMB 2014 9th edition Slide 12 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

13. EGR 252 Ch. 9 Lecture1 JMB 2014 9th edition Slide 13 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, x 0 is if σ is known, and if σ is unknown

14. EGR 252 Ch. 9 Lecture1 JMB 2014 9th edition Slide 14 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.

15. EGR 252 Ch. 9 Lecture1 JMB 2014 9th edition Slide 15 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.

16. EGR 252 Ch. 9 Lecture1 JMB 2014 9th edition Slide 16 Summary Confidence interval  population mean μ Prediction interval  a new observation x 0 Tolerance interval  a (1-α) proportion of the measurements can be estimated with 100( 1- Ƴ )% confidence