1. 1 Design and Analysis of Experiments (2) Basic Statistics Kyung-Ho Park

2. 2 Descriptive Statistics : deals with procedures used to summarize the information contained in a set of measurements. Inferential Statistics : deals with procedures used to make inferences (predictions) about a population parameter from information contained in a sample.

3. 3 Population –Totality of the observations with which we are concerned Sample –A subset of observations selected from a population

4. 4 Population Sample Mean μ Variance σ 2 Standard deviation σ Mean x Variance S 2 Standard deviation S

5. 5 Descriptive statistics Numerical Methods Graphical Methods

6. 6 Measures of Central Tendency (Location) 1)sample mean: 2)sample median: the middle number when the measurements are arranged in ascending order 3)sample mode: most frequently occurring value Numerical methods i)sample is sensitive to extreme values ii)the median is insensitive to extreme values

7. 7 Measures of Dispersion (Variability) 1)range: max – min 2)sample variance: 3)sample standard deviation: Numerical methods

8. 8 Measures of Central Measures of Dispersion Tendency (Location) (Variability) 1. Sample mean 1. Range 2. Sample median 2. Mean Absolute Deviation (MAD) 3. Sample mode 3. Sample Variance 4. Sample Standard Deviation

9. 9 Graphical Methods 105221183186121181180143 97154153174120168167141 245228174199181158176110 163131154115160208158133 207180190193194133156123 13417876167184135229146 218157101171165172158169 199151142163145171148158 16017514987160237150135 196201200176150170118149 Table1: Compressive Strength (in psi) of 80 Aluminum-Lithium Alloy Speciments

10. 10 Graphical Methods Histogram Dot Plot Stem-and-Leaf Display: c1 Stem-and-leaf of c1 N = 80 Leaf Unit = 1.0 LO 76, 87 3 9 7 5 10 15 8 11 058 11 12 013 17 13 133455 25 14 12356899 37 15 001344678888 (10) 16 0003357789 33 17 0112445668 23 18 0011346 16 19 034699 10 20 0178 6 21 8 5 22 189 HI 237, 245

11. 11 second quartile first quartile third quartile whisker extends to smallest data point with 1.5 interquartile ranges from first quartile Extreme outliers whisker extends to largest data point with 1.5 interquartile ranges from third quartile outliers IQR1.5 IQR Box Plot

12. 12 Probability Plots Graphical method for determining whether sample data conform to a hypothesized distribution based on a subjective visual examination of the data 10 observations on the effective service life in minutes of batteries in a portable personal computer 176, 191, 214, 220, 205, 192, 201, 190, 193, 185 jX(j)(j-0.5)/10Zj 11760.05-1.64 21830.15-1.04 31850.25-0.67 41900.35-0.39 51910.45-0.13 61920.550.13 72010.650.39 82050.750.67 92140.851.04 102200.951.64 (j-0.5)/n=P(Z ≤ z i )

13. 13 Probability Plots (table 1)

14. 14 Population Sample Mean μ Variance σ 2 Standard deviation σ Mean x Variance S 2 Standard deviation S Estimation

15. 15 Normal Distribution Distribution of a random variable (sampling): Normal distribution y: a normal random variable the probability distribution of y

16. 16

17. 17 Standard Normal Distribution random variable

18. 18 Ex.1 Suppose the current measurement in a strip of wire are assumed to follow a normal distribution with a mean of 10 milliamperes and a variance of 4 (milliamperes) 2. What is the probability that a measurement will exceed 13 milliamperes? Cumulative Distribution Function Normal with mean = 10 and standard deviation = 2 x P(?X?<=?x?) 13 0.933193 MiniTab Cal – Probability distribution – Normal Mean=10.0, S.D=2, Input Constant=13.0

19. 19 Confidence Interval (CI)

20. 20 Confidence Interval (CI) sampling variability : Interval estimate for a population parameter : confidence interval CI is constructed so that we have high confidence that it does contain the unknown population parameter If is the sample mean of a random sample of size n from a normal population with known variance σ 2, a 100(1-α)% CI on μ is given by Where z α/2 is the upper 100 α/2 percentage point of the standard normal distribution

21. 21 Ex 2. Ten measurements of impact energy(J) on specimens of A238 steel cut at 60 ℃ as a follows: 64.1, 64.7, 64.5, 64.6, 64.5, 64.3, 64.6, 64.8, 64.2 and 64.3. Assume that impact energy energy is normally distributed with σ=1J. We want to find a 95% CI for μ, the mean impact energy z α/2 = z 0.025 =1.96 n=10, σ=1

22. 22 64.1 64.7 64.5 64.6 64.5 64.3 64.6 64.8 64.2 64.3 One-Sample Z: C1 The assumed standard deviation = 1 Variable N Mean StDev SE Mean 95% CI C1 10 64.4600 0.2271 0.3162 (63.8402, 65.0798) Stat -> Basic Stat -> 1 sample Z (Example 2)

23. 23 Confidence Interval (CI) If and s are the mean and standard deviation of of a random sample from a normal population with unknown variance σ 2, a 100(1-α)% CI on μ is given by Where t α/2,n-1 is the upper 100 α/2 percentage point of the t distribution with n-1 degrees fo freedom

24. 24 Ex. 3 An article describes the results of tensile adhesion tests on 22U-700 alloy specimens. The load at specimen failure is as follows (in megapascals): 19.8 10.1 14.9 7.5 15.4 15.4 15.4 18.5 7.9 12.7 11.9 11.4 11.4 14.1 17.6 16.7 15.8 19.5 8.8 13.6 11.9 11.4 We want to find a 95% CI for μ n=22, n-1=21, t 0.025,21 = 2.080

25. 25 Variable N Mean StDev SE Mean 95% CI C1 22 13.7136 3.5536 0.7576 (12.1381, 15.2892)

26. 26 Hypothesis Test

27. 27 Hypothesis Test We illustrated how to construct a confidence interval estimate of a parameter from sample data Many problems in engineering require that we decide whether to accept or reject a statement about some parameter : Hypothesis Decision-making procedure about the hypothesis : hypothesis testing Hypothesis testing and CI estimation of parameters : Data analysis stage of a comparative experiment

28. 28 Tensile adhesion tests on 22U-700 alloy specimens (Example.3) We are interested in deciding whether or not the tensile adhesion is 14 megapascals H 0 : μ= 14 megapascals Null hypothesis H 1 μ≠14. megapascals Alternative hypothesis H 1 μ≠14 Two-sided alternative hypothesis H 1 μ<>14 One-sided alternative hypothesis

29. 29 Probability of making a type I error: significance level, (α-error) α=0.05, 0.01 (confidence level : 95.0, 99.0) α = P(type I error) = P(reject H 0 when H 0 is true) β = P(type II error) = P(fail to reject H 0 when H 0 is false)

30. 30 One-Sample T: C1 Test of mu = 15 vs not = 15 Variable N Mean StDev SE Mean 95% CI T P C1 22 13.7136 3.5536 0.7576 (12.1381, 15.2892) -1.70 0.104 MiniTab Stat-Basic statistics -1t Test mean=15 Option Confidence level:95.0, Alternative: not equal Hypotheses Tests for a Single Sample

31. 31 Hypotheses Tests for Two Samples NumberCatalyst 1Catalyst 2 191.5089.19 294.1890.95 392.1890.46 495.3993.21 591.7997.19 689.0797.04 794.7291.07 889.2192.75 Average92.25592.733 s2.392.98 Table. Catalyst Yield Data

32. 32 MiniTab Hypotheses Tests for Two Samples Two-Sample T-Test and CI: Catalyst 1, Catalyst 2 Two-sample T for Catalyst 1 vs Catalyst 2 N Mean StDev SE Mean Catalyst 1 8 92.26 2.39 0.84 Catalyst 2 8 92.73 2.98 1.1 Difference = mu (Catalyst 1) - mu (Catalyst 2) Estimate for difference: -0.477500 95% CI for difference: (-3.394928, 2.439928) T-Test of difference = 0 (vs not =): T-Value = -0.35 P-Value = 0.729 DF = 13 Stat-Basic statistics -2t(2-sample t) Sample in different columns Option -Confidence level:95.0, - Alternative: not equal

33. 33 Hypotheses Tests for Two Paired Samples SpecimenTip1Tip2 176 233 335 443 588 632 724 899 954 1045 Ex.5 Data for Hardness testing Experiment

34. 34 MiniTab Hypotheses Tests for Two Samples Stat-Basic statistics t-t (paired t) Sample in different columns Option -Confidence level:95.0, - Alternative: not equal

35. 35 Hypotheses Tests for Two Paired Samples Paired T for Tip1 - Tip2 N Mean StDev SE Mean Tip1 10 4.80000 2.39444 0.75719 Tip2 10 4.90000 2.23358 0.70632 Difference 10 -0.100000 1.197219 0.378594 95% CI for mean difference: (-0.956439, 0.756439) T-Test of mean difference = 0 (vs not = 0): T-Value = -0.26 P-Value = 0.798 Two-sample T for Tip1 vs Tip2 N Mean StDev SE Mean Tip1 10 4.80 2.39 0.76 Tip2 10 4.90 2.23 0.71 Difference = mu (Tip1) - mu (Tip2) Estimate for difference: -0.100000 95% CI for difference: (-2.284675, 2.084675) T-Test of difference = 0 (vs not =): T-Value = -0.10 P-Value = 0.924 DF = 17