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1 Chapter 3 Experiments with a Single Factor: The Analysis of Variance

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2 3.1 An Example Chapter 2: A signal-factor experiment with two levels of the factor Consider signal-factor experiments with a levels of the factor, a 2 Example: –The tensile strength of a new synthetic fiber. –The weight percent of cotton –Five levels: 15%, 20%, 25%, 30%, 35% –a = 5 and n = 5

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3 Does changing the cotton weight percent change the mean tensile strength? Is there an optimum level for cotton content?

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An Example (See pg. 61) An engineer is interested in investigating the relationship between the RF power setting and the etch rate for this tool. The objective of an experiment like this is to model the relationship between etch rate and RF power, and to specify the power setting that will give a desired target etch rate. The response variable is etch rate. 4

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She is interested in a particular gas (C2F6) and gap (0.80 cm), and wants to test four levels of RF power: 160W, 180W, 200W, and 220W. She decided to test five wafers at each level of RF power. The experimenter chooses 4 levels of RF power 160W, 180W, 200W, and 220W The experiment is replicated 5 times – runs made in random order 5

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Does changing the power change the mean etch rate? Is there an optimum level for power? We would like to have an objective way to answer these questions The t-test really doesn’t apply here – more than two factor levels 8

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9 3.2 The Analysis of Variance a levels (treatments) of a factor and n replicates for each level. y ij : the jth observation taken under factor level or treatment i.

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10 Models for the Data Means model: –y ij is the ijth observation, – i is the mean of the ith factor level – ij is a random error with mean zero Let μ i = μ + τ i, is the overall mean and τ i is the ith treatment effect Effects model:

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11 Linear statistical model One-way or Signal-factor analysis of variance model Completely randomized design: the experiments are performed in random order so that the environment in which the treatment are applied is as uniform as possible. For hypothesis testing, the model errors are assumed to be normally and independently distributed random variables with mean zero and variance, σ 2, i.e. y ij ~ N(μ + τ i, σ 2 ) Fixed effect model: a levels have been specifically chosen by the experimenter.

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Analysis of the Fixed Effects Model Interested in testing the equality of the a treatment means, and E(y ij ) = μ i = μ + τ i, i = 1,2, …, a H 0 : μ 1 = … = μ a v.s. H 1 : μ i ≠ μ j, for at least one pair (i, j) Constraint: H 0 : τ 1 = … = τ a =0 v.s. H 1 : τ i ≠ 0, for at least one i

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13 Notations: Decomposition of the Total Sum of Squares Total variability into its component parts. The total sum of squares (a measure of overall variability in the data) Degree of freedom: an – 1 = N – 1

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14 SS Treatment : sum of squares of the differences between the treatment averages (sum of squares due to treatments) and the grand average, and a – 1 degree of freedom SS E : sum of squares of the differences of observations within treatments from the treatment average (sum of squares due to error), and a(n - 1) = N – a degrees of freedom.

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15 A large value of SS Treatments reflects large differences in treatment means A small value of SS Treatments likely indicates no differences in treatment means df Total = df Treatment + df Error No differences between a treatment means: variance cane be estimated by

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16 Mean squares: Statistical Analysis Assumption: ε ij are normally and independently distributed with mean zero and variance σ 2 Cochran’s Thm (p. 69)

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17 SS T /σ 2 ~ Chi-square (N – 1), SS E /σ 2 ~ Chi-square (N – a), SS Treatments /σ 2 ~ Chi-square (a – 1), and SS E /σ 2 and SS Treatments /σ 2 are independent (Theorem 3.1) H 0 : τ 1 = … = τ a =0 v.s. H 1 : τ i ≠ 0, for at least one i

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18 Reject H 0 if F 0 > F α, a-1, N-a Rewrite the sum of squares: See page 71 Randomization test

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ANOVA Table of Example

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Estimation of the Model Parameters Model: y ij = μ + τ i + ε ij Estimators: Confidence intervals:

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22 Example 3.3 (page 74) Simultaneous Confidence Intervals (Bonferroni method): Construct a set of r simultaneous confidence intervals on treatment means which is at least 100(1- ): 100(1- /r) C.I.’s Unbalanced Data Let n i observations be taken under treatment i, i=1,2,…,a, N = i n i,

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23 1. The test statistic is relatively insensitive to small departures from the assumption of equal variance for the a treatments if the sample sizes are equal. 2. The power of the test is maximized if the samples are of equal size.

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Model Adequacy Checking Assumptions: y ij ~ N(μ + τ i, σ 2 ) The examination of residuals Definition of residual: The residuals should be structureless.

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The Normality Assumption Plot a histogram of the residuals Plot a normal probability plot of the residuals See Table 3-6

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26 May be –the left tail of error is thinner than the tail part of standard normal Outliers The possible causes of outliers: calculations, data coding, copy error,…. Sometimes outliers are more informative than the rest of the data.

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27 Detect outliers: Examine the standardized residuals, Plot of Residuals in Time Sequence Plotting the residuals in time order of data collection is helpful in detecting correlation between the residuals. Independence assumption

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29 Nonconstant variance: the variance of the observations increases as the magnitude of the observation increase, i.e. y ij 2 If the factor levels having the larger variance also have small sample sizes, the actual type I error rate is larger than anticipated. Variance-stabilizing transformation PoissonSquare root transformation, LognormalLogarithmic transformation, BinomialArcsin transformation,

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30 Statistical Tests for Equality Variance: –Bartlett’s test: –Reject null hypothesis if

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31 Example 3.4: the test statistic is Bartlett’s test is sensitive to the normality assumption The modified Levene test: –Use the absolute deviation of the observation in each treatment from the treatment median. –Mean deviations are equal => the variance of the observations in all treatments will be the same. –The test statistic for Levene’s test is the ANOVA F statistic for testing equality of means.

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32 Example 3.5: –Four methods of estimating flood flow frequency procedure (see Table 3.7) –ANOVA table (Table 3.8) –The plot of residuals v.s. fitted values (Figure 3.7) –Modified Levene’s test: F 0 = 4.55 with P-value = Reject the null hypothesis of equal variances.

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33 Let E(y) = and y Find y * = y that yields a constant variance. y* + -1 Variance-Stabilizing Transformations y and = 1 - Transformation y constant 01No transformation y 1/2 ½½Square root y y 10Log y 3/2 3/2-1/2Reciprocal square root y 2y 2 2Reciprocal

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34 How to find : Use See Figure 3.8, Table 3.10 and Figure 3.9

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Practical Interpretation of Results Conduct the experiment => perform the statistical analysis => investigate the underlying assumptions => draw practical conclusion A Regression Model Qualitative factor: compare the difference between the levels of the factors. Quantitative factor: develop an interpolation equation for the response variable.

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The Regression Model 36

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Comparisons Among Treatment Means If that hypothesis is rejected, we don’t know which specific means are different Determining which specific means differ following an ANOVA is called the multiple comparisons problem Graphical Comparisons of Means

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Contrast A contrast: a linear combination of the parameters of the form H 0 : = 0 v.s. H 1 : 0 Two methods for this testing.

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39 The first method:

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40 The second method:

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41 The C.I. for a contrast, Unequal Sample Size

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Orthogonal Contrast Two contrasts with coefficients, {c i } and {d i }, are orthogonal if c i d i = 0 For a treatments, the set of a – 1 orthogonal contrasts partition the sum of squares due to treatments into a – 1 independent single-degree- of-freedom components. Thus, tests performed on orthogonal contrasts are independent. See Example 3.6 (Page 90)

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Scheffe’s Method for Comparing All Contrasts Scheffe (1953) proposed a method for comparing any and all possible contrasts between treatment means. See Page 91 and 92

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Comparing Pairs of Treatment Means Compare all pairs of a treatment means Tukey’s Test: –The studentized range statistic: –See Example 3.7

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46 Sometimes overall F test from ANOVA is significant, but the pairwise comparison of mean fails to reveal any significant differences. The F test is simultaneously considering all possible contrasts involving the treatment means, not just pairwise comparisons. The Fisher Least Significant Difference (LSD) Method For H 0 : i = j

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47 The least significant difference (LSD): See Example 3.8 Duncan’s Multiple Range Test The a treatment averages are arranged in ascending order, and the standard error of each average is determined as

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48 Assume equal sample size, the significant ranges are Total a(a-1)/2 pairs Example 3.9 The Newman-Keuls Test Similar as Duncan’s multiple range test The critical values:

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Comparing Treatment Means with a Control Assume one of the treatments is a control, and the analyst is interested in comparing each of the other a – 1 treatment means with the control. Test H 0 : i = a v.s. H 1 : i a, i = 1,2,…, a – 1 Dunnett (1964) Compute Reject H 0 if Example 3.9

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Determining Sample Size Determine the number of replicates to run Operating Characteristic Curves (OC Curves) OC curves: a plot of type II error probability of a statistical test,

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51 If H 0 is false, then F 0 = MS Treatment / MS E ~ noncentral F with degree of freedom a – 1 and N – a and noncentrality parameter Chart V of the Appendix Determine Let i be the specified treatments. Then estimates of i : For 2, from prior experience, a previous experiment or a preliminary test or a judgment estimate.

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52 Example 3.11 Difficulty: How to select a set of treatment means on which the sample size decision should be based. Another approach: Select a sample size such that if the difference between any two treatment means exceeds a specified value the null hypothesis should be rejected.

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Specifying a Standard Deviation Increase Let P be a percentage for increase in standard deviation of an observation. Then For example (Page 110): If P = 20, then

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Confidence Interval Estimation Method Use Confidence interval. For example: we want 95% C.I. on the difference in mean tensile strength for any two cotton weight percentages to be 5 psi and = 3. See Page 110.

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3.8 A Real Application 55

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The Regression Approach to the Analysis of Variance Model: y ij = μ + τ i + ij

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60 The normal equations Apply the constraint Then estimations are Regression sum of squares (the reduction due to fitting the full model)

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61 The error sum of squares: Find the sum of squares resulting from the treatment effects:

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62 The testing statistic for H 0 : 1 = … = a

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