Estimation of Standard Deviation & Percentiles

1. Systems Engineering Program Department of Engineering Management, Information and Systems EMIS 7370/5370 STAT 5340 : PROBABILITY AND STATISTICS FOR SCIENTISTS AND ENGINEERS Estimation of Standard Deviation & Percentiles Dr. Jerrell T. Stracener, SAE Fellow Leadership in Engineering Stracener_EMIS 7370/STAT 5340_Sum 08_07.10.08

2. Estimation of Standard Deviation & Percentiles • Estimation of Standard Deviation - Normal Distribution • Estimation of Ratio of Standard Deviations - Normal Distributions • Estimation of Percentiles - Tolerance Intervals

3. Estimation of Standard Deviations

4. Estimation of Standard Deviation - Normal Distribution • Point Estimate of • (1 -)  100% Confidence Interval foris, • where • and

5. Example: Estimation of  The following are the weights, in decagrams, of 10 packages of grass seed: 46.4, 46.1, 45.8, 47.0, 46.1, 45.9, 48.8, 46.9, 45.2, 46.0 Estimate  in terms of a point estimate and a 95% confidence interval for the standard deviation of all such packages of grass seed, assuming a normal population.

6. Example - Solution First we find

7. Example - Solution Then a point estimate of  A (1 -)  100% Confidence Interval for is ,

8. Example - Solution Where

9. Example - Solution and , and where and and is the value of X ~ x2n-1 such that

10. Estimation of Ratio of Two Standard Deviations Normal Distribution • Let X11, X12, …, , and X21, X22, …, be • random samples from N(1, 1) and N(2, 2), • respectively • Point estimation of • where • for i = 1, 2

11. Estimation of Ratio of Two Standard Deviations Normal Distribution • (1 -)  100% Confidence Interval for • where • and

12. Estimation of Ratio of Two Standard Deviations Normal Distribution where is the value of the F-Distribution with and degrees of freedom for which

13. A standardized placement test in mathematics was given to 25 boys and 16 girls. The boys made an average grade of 82 with a standard deviation of 8, while the girls made an average grade of 78 with a standard deviation of 7. Find a 98% confidence interval for , where and are the variances of the populations of grades for all boys and girls, respectively, who at some time have taken or will take this test. Example - Estimation of1/2

14. We have n1 = 25, n2 = 16, s1 = 8, s2 = 7. For a 98% confidence interval,  = 0.02. So that F0.01 (24, 15) = 3.29, and F0.01(15, 24) = 2.89. Then a 98% confidence interval for is where Example - solution and

15. Estimation of Percentiles - Tolerance Intervals

16. If X1, X2, …, Xn is a random sample of size n from a normal distribution with unknown mean  and unknown standard deviation , tolerance limits are given by , where k is determined so that one can assert with ()100% confidence that the given limits contain at least the proportion p of the measurements, i.e., the tolerance interval is (LTL, UTL), where and where k may be obtained from a table of tolerance factors which is located on the Information and Resources page of the website. Tolerance Limits

17. Example - Tolerance Interval A machine is producing metal pieces that are cylindrical in shape. A sample of these pieces is taken and the diameters are found to be: 1.01, 0.97, 1.03, 1.04, 0.99, 0.98, 0.99, 1.01, and 1.03 centimeters. Find tolerance limits that will contain at least 95%, with a 99% confidence, of the metal pieces produced by this machine, assuming a normal distribution.

18. Solution The sample mean and standard deviation for the given data are x = 1.0056 and s = 0.0245 From the Tolerance Factors Table for n = 9, 1 -  = 0.99, and 1 -  = 0.95 we find k = 4.550 for two-sided limits. Hence the tolerance limits are and

19. Solution That is, we are 99% confident that the tolerance interval from 0.894 to 1.117 will contain at least 95% of the metal pieces produced by this machine.