1. S S Lewis 2/15/051 Confidence Intervals and Maximum Errors By Sidney S. Lewis For Baltimore Section, ASQ February 15, 2005

2. S S Lewis 2/15/052 CONFIDENCE INTERVAL A confidence interval expresses our belief, or confidence, that the interval we construct from the data will contain the mean, µ, (for example) of the population from which the data were drawn. Confidence Intervals can be computed on any population parameter: µ, , p’, c’, and even on complex parameters, such as Cp and Cpk. 812.1

3. S S Lewis 2/15/053 CONFIDENCE INTERVAL EXAMPLE Example of a confidence Interval (C.I.) on the population mean µ: Statement: The interval 38.0 - 42.0 contains µ with 90% confidence. Alternatively, there is a 5% chance that the C.I. falls entirely below µ (µ above 42.0), and likewise, a 5% chance that the C.I. is entirely above µ (µ below 38.0). 812.1

4. S S Lewis 2/15/054 Example Pollsters report that 55% of a sample of 1005 members of the voting population support Proposition A. The Margin of Error is 3.1% There is an implied risk of being wrong, usually 5% Calcs.: 812.8

5. S S Lewis 2/15/055 POPULATION PROCESS SAMPLE PROCESS POPULATION vs. SAMPLES 810

6. S S Lewis 2/15/056  /2  = 2 n = 4  (xbar) = 1  = 5%  810.05

7. S S Lewis 2/15/057 Logic:  is known to be 2.0 A sample of 4 yields X-bar = 42.0. Question? What values of  are probable, with 95% Confidence (  = 5%). Solution:  Xbar =  /on = 1.0 Z  /2 = 1.96, or about 2 ME = Z  /2  Xbar = ± 2.0 or from 40.0 to 44.0 MAXIMUM ERROR 810.15

8. S S Lewis 2/15/058 MAXIMUM ERROR of the MEAN 812.3

9. S S Lewis 2/15/059 C. I. on the MEAN Calculation of the C.I. on the mean µ typically uses either the population standard deviation, , if known, or if not, the sample standard deviation, s. If X-bar is the sample mean, then a 1-  confidence interval on µ is: C.I. =X-bar ± Maximum Error (ME) = X-bar ± z  /2  /%n if  is known, or = X-bar ± t a/2 s/%n if  is unknown. 812.2

10. S S Lewis 2/15/0510 Diameters of 3/4" HR Bars Diameters of 3/4" HR barsX-barR 0.74660.74570.75240.74950.74890.74860.0067 0.74960.75490.75420.75660.74930.75290.0073 0.75630.74360.74750.75250.74920.74980.0127 0.74910.75080.75120.74820.75200.75020.0038 0.74980.75520.75080.74770.74530.74970.0099 0.75080.74800.74980.75260.75320.75090.0052 0.74980.74910.75070.75140.75270.75070.0037 0.75260.75210.74960.75070.75330.75160.0037 0.75200.74700.75500.75170.74040.74920.0146 0.74630.75540.74830.75070.74740.74960.0091 0.75370.75200.75010.75220.75240.75210.0036 0.74760.75350.75420.75480.75150.75230.0072 0.75160.74420.74990.75090.74720.74880.0074 940

11. S S Lewis 2/15/0511 840.1

12. S S Lewis 2/15/0512 840.1

13. S S Lewis 2/15/0513 DATA STATISTICS X-barRs 0.74860.00670.0026 0.75290.00730.0033 0.74980.01270.0048 0.75020.00380.0016 0.74970.00990.0037 0.75090.00520.0021 0.75070.00370.0014 0.75160.00370.0015 0.74920.01460.0057 0.74960.00910.0036 0.75210.00360.0013 840.01

14. S S Lewis 2/15/0514 3/4" HR Bars – MEANS and CONFIDENCE INTERVALS  z(.90) = 1.645t(4,.90) = 2.132 SampleX-barsLCI(z)UCI(z)LCI(t)UCI(t) 10.74860.002630.74640.75080.74610.7511 20.75290.003280.75070.75510.74980.7561 30.74980.004840.74760.75200.74520.7544 40.75020.001570.74800.75250.74880.7517 50.74970.003710.74750.75190.74620.7533 60.75090.002090.74870.75310.74890.7529 70.75070.001420.74850.75290.74940.7521 80.75160.001480.74940.75390.75020.7531 90.74920.005680.74700.75140.74380.7546 100.74960.003600.74740.75180.74620.7530 110.75210.001290.74990.75430.75090.7533 840.41

15. S S Lewis 2/15/0515 840.5

16. S S Lewis 2/15/0516 840.6

17. S S Lewis 2/15/0517 FACTORS AFFECTING THE WIDTH OF A CONFIDENCE INTERVAL 812.4 Factors:  or s, n, 

18. S S Lewis 2/15/0518 FACTORS AFFECTING THE WIDTH OF A CONFIDENCE INTERVAL  or s:...................Width increases as  or s increases; Sample size, n.....Width decreases as n increases; C. I. is proportional to 1/on Confidence level 1 - , or risk  : Width increases as confidence increases, or as risk  decreases. 812.4

19. S S Lewis 2/15/0519 CONFIDENCE INTERVALS ON  Small samples (n<30) : Large samples (n>30): 812.7

20. S S Lewis 2/15/0520 CONFIDENCE INTERVALS ON p’ Large samples (np>5) : 812.7

21. S S Lewis 2/15/0521 CONFIDENCE INTERVAL ON p’, SMALL n Upper C.I. on p'Lower C.I. on p'  cP’n  cP’n 27.9%15.0%5082.7%11.5%50 3.38%110.0%5091.1%11.0%50 5.32%19.0%5097.4%10.5%50 4.25%19.5%5096.4%10.6%50 4.87%19.2%5095.2%10.7%50 5.00%19.14%5095.0%10.72%50 ExcelFunction:=BINOMDIST(c,n,p’,1) 861 90% C.I. on p’ for n=50, c=1

22. S S Lewis 2/15/0522 STATISTICAL CALCULATION EXAMPLE A TECHNIQUE:1-SAMPLE TEST OF THE MEAN, SIGMA UNKNOWN: t-TEST SUBJECT:Machinability: Increased by a New Practice? GOAL: Determine whether the average machinability of steel made using a new practice in the Melt Shop can increase the machinability, with 95% CONFIDENCE (5%  ). HISTORIC DATA: The recent past average machinability is 85.0 =  0 ;  = 12.2. DATA: Machinability data of steel made with the new practice are 91 99 83 87 98 94 86 92 85 81 890.53

23. S S Lewis 2/15/0523 STATISTICAL CALCULATION EXAMPLE A Calcs.: X-bar = 89.6; s = 6.196; n = 10; Maximum Error, ME = t   df) * S X-bar = 1.833 * 1.957 = 3.6 units 90% C.I. = X-bar ± ME = 89.6 ± 3.6 = 86.0 to 93.2. That means that the machinability should increase by at least 1.0 units, and may increase by 8 units. 890.53

24. S S Lewis 2/15/0524 STATISTICAL CALCULATION EXAMPLE B TECHNIQUE: 1-SAMPLE TEST OF PROPORTIONS – Z-TEST SUBJECT:Cap leakers – reduction trial GOAL: Determine whether the rates of leaking caps are lower if a new cap design is used, with 95% CONFIDENCE (  = 5%). HISTORIC DATA: Cap leaker rate = 1.2% = p’. DATA: A trial using 2000 caps of a new design found 18 leaking caps. p = 0.90%. 890.83

25. S S Lewis 2/15/0525 STATISTICAL CALCULATION EXAMPLE B FORMULAS: where p is in percent. CALCS: p = 18/2000 = 0.90%;  = p 0 – p = 1.2 - 0.90 = 0.30% Maximum Error, ME = Z.05 *  p = 1.645 * 0.243 = 0.400% 90% C.I. (2-tail) on the difference = (  -  0 ) ± ME = (0.90 – 1.20) ± 0.400 = +0.10% to -0.70%; 90% C.I. on p’: p ± ME = 0.90 ± 0.40 = 0.5% to 1.3% 890.83

26. S S Lewis 2/15/0526 CONCLUSION: The long term leaker rate of the new caps may be 0.7% lower than the old caps, but it may also be 0.1% higher, which if true, says to avoid the new caps. Therefore the data are insufficient to show, with 95% confidence, that the new caps are definitely better, which confirms the test of hypothesis. STATISTICAL CALCULATION EXAMPLE B 890.83

27. S S Lewis 2/15/0527 STATISTICAL CALCULATION EXAMPLE C TECHNIQUE: 1-SAMPLE TEST OF A SAMPLE STANDARD DEVIATION–CHI-SQUARED TEST SUBJECT: XYZ Digital Blood Pressure Monitor measures of systolic blood pressure GOAL: To determine if the monitor has become more variable than when new. HISTORIC DATA: Early evaluation of this monitor found the standard deviation to be 2.5 units. DATA : Using the monitor, the systolic blood pressure of a patient was measured 7 times over a ten minute period. The patient sat quietly throughout the testing. The results were: 144, 147, 147, 149, 140, 140, 144, from which s = 3.51. 890.72

28. S S Lewis 2/15/0528 STATISTICAL CALCULATION EXAMPLE C CONFIDENCE INTERVAL: a 2-tail, 90% confidence interval will be calculated The critical values of  2 are: With 90% confidence, the true standard deviation lies between 1.83 and 5.05 units, which includes the earliest determined standard deviation of 2.5. 890.72