Design of Engineering Experiments Part 2 Basic Statistical Concepts Simple comparative experiments

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1 Design of Engineering Experiments Part 2 Basic Statistical Concepts Simple comparative experiments The hypothesis testing framework The two-sample t-test Checking assumptions, validity Comparing more that two factor levels the analysis of variance ANOVA decomposition of total variability Statistical testing & analysis Checking assumptions, model validity Post-ANOVA testing of means Sample size determination DOX 6E Montgomery 1

2 Portland Cement Formulation (page 23) 2 different formulations referred to as 2 treatments or 2 levels of the factor formulations. New formulation is able reduce cure time. However, need to evaluate the bond strength of the formulations. DOX 6E Montgomery 2

3 Graphical View of the Data Each observations is called a run. Each strength value differ so there s some fluctuation or noise or experimental error which is generally unavoidable. This implies bond strength is a random variable. DOX 6E Montgomery 3

4 Box Plots, Fig. 2-3, pp. 26 DOX 6E Montgomery 4

5 The Hypothesis Testing Framework Need to evaluate the bond strength of the formulations. Statistical hypothesis testing is a useful framework for many experimental situations Origins of the methodology date from the early 1900s Use a procedure known as the two-sample t- test DOX 6E Montgomery 5

6 The Hypothesis Testing Framework Sampling from a normal distribution Statistical hypotheses: H H : μ = μ : μ μ DOX 6E Montgomery 6

7 Estimation of Parameters 1 y = n n i= 1 y 1 S y y i estimates the population mean μ n = ( i ) estimates the variance σ n 1 i= 1 DOX 6E Montgomery 7

8 Summary Statistics (pg. 36) Formulation 1 New recipe y S S n = = = = 10 Formulation 2 Original recipe y S S n = = = = 10 DOX 6E Montgomery 8

9 1 2 How the Two-Sample t-test Works: Use the sample means to draw inferences about the population means y y = = 0.28 Difference in sample means Standard deviation of the difference in sample means 2 2 σ σ y = n This suggests a statistic: Z 0 = y σ n y 1 2 σ + n DOX 6E Montgomery 9

10 How the Two-Sample t-test Works: Use S and S to estimate σ and σ The previous ratio becomes However, we have the case where p = n1+ n Pool the individual sample variances: S ( n 1) S + ( n 1) S y S n DOX 6E Montgomery 10 y S n σ = σ = σ

11 How the Two-Sample t-test Works: The test statistic is t 0 = S p y n n 1 2 Values of t 0 that are near zero are consistent with the null hypothesis Values of t 0 that are very different from zero are consistent with the alternative hypothesis t 0 is a distance measure-how far apart the averages are expressed in standard deviation units Notice the interpretation of t 0 as a signal-to-noise ratio y DOX 6E Montgomery 11

12 The Two-Sample (Pooled) t-test S S ( n 1) S + ( n 1) S 9(0.100) + 9(0.061) = = = p n1 n2 p = t = = = S p y y n n The two sample means are a little over two standard deviations apart Is this a "large" difference? DOX 6E Montgomery 12

13 The Two-Sample (Pooled) t-test So far, we haven t really done any statistics We need an objective basis for deciding how large the test statistic t 0 really is In 1908, W. S. Gosset derived the reference distribution for t 0 called the t distribution Tables of the t distribution - text, page 606 t 0 = DOX 6E Montgomery 13

14 The Two-Sample (Pooled) t-test A value of t 0 between and is consistent with equality of means It is possible for the means to be equal and t 0 to exceed either or 2.101, but it would be a rare event leads to the conclusion that the means are different Could also use the P-value approach t 0 = DOX 6E Montgomery 14

15 The Two-Sample (Pooled) t-test t 0 = The P-value is the risk of wrongly rejecting the null hypothesis of equal means (it measures rareness of the event) The P-value in our problem is P = DOX 6E Montgomery 15

16 Minitab Two-Sample t-test Results DOX 6E Montgomery 16

17 Checking Assumptions The Normal Probability Plot DOX 6E Montgomery 17

18 Importance of the t-test Provides an objective framework for simple comparative experiments Could be used to test all relevant hypotheses in a two-level factorial design, because all of these hypotheses involve the mean response at one side of the cube versus the mean response at the opposite side of the cube DOX 6E Montgomery 18

19 Confidence Intervals (See pg. 43) Hypothesis testing gives an objective statement concerning the difference in means, but it doesn t specify how different they are General form of a confidence interval L θ U where P( L θ U) = 1 α The 100(1- α)% confidence interval on the difference in two means: y y t S (1/ n ) + (1/ n ) μ μ 1 2 α /2, n + n 2 p y y + t S (1/ n ) + (1/ n ) 1 2 α /2, n + n 2 p DOX 6E Montgomery 19

20 Substituting the values into the equation, we get (2.101 ) μ1 μ (2.101 ) μ1 μ μ1 μ Thus, the 95 % confidence interval estimate on the difference in means extends from kgf/cm 2 to kgf/cm 2 or CI is μ 1 -μ 2 = ± 0.27 kgf/cm 2 or the difference in the mean strengths is kgf/cm 2 and the accuracy of the estimate is ±0.27 kgf/cm 2. Note that because μ 1 -μ 2 = 0 is not included in this interval, the data does not support the hypothesis that μ 1 = μ 2 at 5 % level of significance. DOX 6E Montgomery 20

21 Comparing a single mean to a specified value Some experiments involve comparing only one population mean μ to a specified value, say, μ 0, then the hypotheses are H 0 : μ = μ 0 H 1 : μ = μ 0 If the population is normal with known variance or if the population is nonnormal but the sample size is large enough so that central limit theorem applies, then hypothesis tested using a direct application of the normal distribution. The test statistic is Z 0 = y σ μ 0 n DOX 6E Montgomery 21

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25 Figure 2.13 (p. 50) The reference distribution (t with 9 degrees of freedom) for the hardness testing problem. DOX 6E Montgomery 25

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