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What If There Are More Than Two Factor Levels?

What If There Are More Than Two Factor Levels?. The t -test does not directly apply There are lots of practical situations where there are either more than two levels of interest, or there are several factors of simultaneous interest

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What If There Are More Than Two Factor Levels?

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  1. What If There Are More Than Two Factor Levels? • The t-test does not directly apply • There are lots of practical situations where there are either more than two levels of interest, or there are several factors of simultaneous interest • The analysis of variance (ANOVA) is the appropriate analysis “engine” for these types of experiments – Chapter 3, textbook • The ANOVA was developed by Fisher in the early 1920s, and initially applied to agricultural experiments • Used extensively today for industrial experiments

  2. An Example (See pg. 61) • Consider an investigation into the relationship between the RF (radio-frequency) power setting and the etch rate for a wafer etching tool. • The response variable is etch rate. • The experimenter wants to determine the power setting that will give a desired target etch rate. • Other variables are fixed (gas, gap, etc.). • RF power levels: 160, 180, 200, and 220 W. • The experiment is replicated 5 times – runs made in random order

  3. An Example (cont’ed) The run order should be randomized. First, we need to assign numbers to the experimental run, e.g., as follows: RF Power Experimental Run Number

  4. An Example (cont’ed) Choose a random number between 1 and 20, and assign it as the first to run/test, then the next. Until all 20 runs are assigned a test number. E.g.: Test Sequence Run Number Power Level What if the test was run in the original nonrandomized order?

  5. An Example (cont’ed)

  6. The Analysis of Variance (Sec. 3-2, pg. 63) • In general, there will be alevels of the factor, or atreatments, andnreplicates of the experiment, run in randomorder…a completely randomized design (CRD) • N = an total runs • We consider the fixed effects case…the random effects case will be discussed later • Objective is to test hypotheses about the equality of the a treatment means

  7. The Analysis of Variance • The name “analysis of variance” stems from a partitioning of the total variability in the response variable into components that are consistent with a model for the experiment • The basic single-factor ANOVA model is

  8. Models for the Data There are several ways to write a model for the data:

  9. The Analysis of Variance • Total variability is measured by the total sum of squares: • The basic ANOVA partitioning is:

  10. The Analysis of Variance • A large value of SSTreatments reflects large differences in treatment means • A small value of SSTreatments likely indicates no differences in treatment means • Formal statistical hypotheses are:

  11. The Analysis of Variance • While sums of squares cannot be directly compared to test the hypothesis of equal means, mean squares can be compared. • A mean square is a sum of squares divided by its degrees of freedom: • If the treatment means are equal, the treatment and error mean squares will be (theoretically) equal. • If treatment means differ, the treatment mean square will be larger than the error mean square.

  12. The Analysis of Variance is Summarized in a Table • Computing…see text, pp 70 – 74 • The reference distribution for F0 is the Fa-1, a(n-1) distribution • Reject the null hypothesis (equal treatment means) if F0 > Fa,a-1, a(n-1) => an upper tail, one tail critical region.

  13. Calculation of Sum of Squares The error sum of squares is SSE = SST – SSTreatments Usually these are done using a computer program.

  14. Confidence Intervals in ANOVA • Can be established based on t-distribution • A 100(1- a)% confidenceinterval on the ith treatment mean miis: • A 100(1- a)% confidenceinterval on the difference in any two treatments means is:

  15. Simultaneous Confidence Intervals • Can be established based on one-at-a-time confidence intervals. • If there are r 100(1- a)% confidence intervals of interest, the probability that the r intervals will simultaneously be correct is at least 1-ra. • E.g. r = 5, a = 0.05, 1 – ra = 0.75 • r = 10, a = 0.05, 1 – ra = 0.50 • Bonferroni method: • Replacing a in the equations with a /r, then the simultaneous confidence intervals on treatment means/differences have a confidence level of at least 100(1- a)%.

  16. ANOVA Computer Output (Design-Expert) Response:etch rate ANOVA for Selected Factorial ModelAnalysis of variance table [Partial sum of squares] Sum ofMeanFSourceSquaresDFSquareValueProb > F Model 66870.55 3 22290.18 66.80 < 0.0001A66870.55 3 22290.1866.80< 0.0001 Pure Error 5339.20 16 333.70 Cor Total 72209.75 19 Std. Dev. 18.27 R-Squared 0.9261 Mean 617.75 Adj R-Squared 0.9122 C.V. 2.96 Pred R-Squared 0.8845 PRESS 8342.50 Adeq Precision 19.071

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