Presentation on theme: "Analysis of Variance Outlines: Designing Engineering Experiments Completely Randomized Single-Factor Experiment Random Effects Model Randomized."— Presentation transcript:
Analysis of Variance Outlines: Designing Engineering Experiments Completely Randomized Single-Factor Experiment Random Effects Model Randomized Complete Block Design
Designing Engineering Experiments Factor: Parameter of interest Levels: Possible value of Factors. Analysis of Variance (ANOVA): Analysis the effects of different factor levels to the response. Randomization: Random running order. Controllable variables: Other parameters that involve in the experiment Experiment activities: 1. Conjecture 2. Experiment 3. Analysis 4. Conclusion
Completely Randomized Single-Factor Experiment Ex: A manufacturer of paper is interested in improving the tensile strength of the product. The product engineer thinks that tensile strength is a function of the hardwood concentration in the paper. The range of hardwood concentrations of practical interest is between 5 and 20%. The engineers design to investigate four levels of hardwood concentration: 5% 10% 15% and 20%. They design to make up six test specimens at each concentration level.
Completely Randomized Single-Factor Experiment Randomizing order: randomly select the order for each run to reduce the effect of nuisance variable such as the warm-up effect. Box plot: Represent the variability within a treatment and the variability between treatments.
Analysis of Variance Typical data for single-factor experiment Linear Statistical Model For each treatment For each treatment, y ij is normal distribution with µ i and E ij has a normal distribution with mean 0 and sd =
Analysis of Variance We are interested in the equality of the a treatment means µ 1, µ 2,…, µ a If H 0 is true, each observation consists of the overall mean µ plus the random error E ij => Changing the level of the factor has no effect on the mean response. ANOVA partitions the total variability in the sample data into two components 1. The variation between treatments 2. The variation within treatment If there are no differences between treatments => 1=2
Analysis of Variance Total variation is described by the total sum of squares (SS T )
Analysis of Variance Degree of freedom Mean square for treatment Mean square for error To verify hypothesis, we compare By using F test statistic
Analysis of Variance If H0 is reject => MS treatment > MS E => H1 should be upper-tail test Reject H0 if Computing formulas for ANOVA ANOVA Table
Analysis of Variance Ex. Consider the tensile strength. We can use ANOVA to test the hypothesis that different hardwood concentrations do not affect the mean tensile strength of the paper. Use α=0.01 For at least one i
Analysis of Variance f 0 = >f 0.01,3,20 =4.94, reject H 0 Conclusion: hardwood concentration significantly affects the mean strength of the paper
Analysis of Variance Confidence interval on the mean of the ith treatment Ex. Find the 95% CI of the mean strength of 20% hardwood concentration?
Analysis of Variance Confidence interval on a difference in treatment means Ex. Find a 95% CI on the difference in means µ 3 -µ 2
Multiple comparisons following ANOVA When is rejected in ANOVA, we know that some of the treatment are different. ANOVA doesn’t identify which means are different. Methods for investigating this issue is called multiple comparisons methods. Fisher LSD : compares all pairs of the means (µ i and µ j ) H0: µ i = µ j with test-statistic The pair of means µ i and µ j would be declared significantly different if
Multiple comparisons following ANOVA Ex. Apply the Fisher LSD to the hardwood concentration experiment. a= 4 levels, n=6 replicates, and t 0.025,20 = The treatment means are: The value of LSD, Compare the difference for every pairs of treatments and LSD, % 10% 15% 20%
Model Adequacy Checking Residual Analysis and Model Checking Residual VS time: Test independence assumption of error Residual VS fitted values: Test constant variance assumption of error Normality Plot: test the normal distribution assumption of error.
Randomized Complete Block Design An extension of the paired t-test but with more than 2 treatments. Reduce the nuisance factor. Ex. 3 methods could be used to evaluate the strength reading on steel plate girders. If there are 4 plates and each plate is large enough to hold all the treatment, the experimental design would be appear as Figure.
Randomized Complete Block Design ANOVA Sums of square
Randomized Complete Block Design ANOVA for a Randomized Complete Block Design
Randomized Complete Block Design Ex. Fabric Strength Data from a randomized complete block design can be shown in table. We want to test the effect of chemical type to the strength of fabric by using α=0.01 H0:all types provide identical strength H1: not equal strength
Randomized Complete Block Design f 0 =75.13 > f 0.01,3,12 =5.95, reject Ho and conclude that there is a significant difference in the chemical types
Randomized Complete Block Design Multiple comparison Similar to simple ANOVA but Ex. Refer to previous example, use Fisher’s LSD method to analyze the difference between each pair of treatment. type 4 results in significantly different strengths than the other three types. types 2 and 3 do not differ, and types 1 and 3 do not differ. There may be a small difference in strength between types 1 and 2.
Exercise 1. A civil engineer is interested in determining whether four different methods of estimating flood flow frequency produce equivalent estimates of peak discharge when applied to the same watershed. Each procedure is used six times on the watershed and the resulting discharge data are shown in table Estimation method Observation
Exercise 2. An experiment was conducted to investigate leaking current in a SOS MOSFETS device. The purpose of the experiment was to investigate how leakage current varies as the channel length changes. Four channel lengths were selected. For each channel length, five different widths were also used, and width is to be considered a nuisance factor. The data are as follows:
Exercise 3. An article in the American Industrial Hygiene Association Journal (Vol. 37, 1976, pp. 418– 422) describes a field test for detecting the presence of arsenic in urine samples. The test has been proposed for use among forestry workers because of the increasing use of organic arsenics in that industry. The experiment compared the test as performed by both a trainee and an experienced trainer to an analysis at a remote laboratory. Four subjects were selected for testing and are considered as blocks. The response variable is arsenic content (in ppm) in the subject’s urine. The data are as follows:
Exercise 4. An article in the Food Technology Journal (Vol. 10, 1956, pp. 39–42) describes a study on the protopectin content of tomatoes during storage. Four storage times were selected, and samples from nine lots of tomatoes were analyzed. The protopectin content (expressed as hydrochloric acid soluble fraction mg/kg) is in the following table.
Exercise 5. An article in the IEEE Transactions on Components, Hybrids, and Manufacturing Technology (Vol.15, No. 2, 1992, pp. 146–153) describes an experiment in which the contact resistance of a brake-only relay was studied for three different materials (all were silver-based alloys). The data are as follows.
Exercise 6. An article in Lubrication Engineering (December 1990) describes the results of an experiment designed to investigate the effects of carbon material properties on the progression of blisters on carbon face seals. The carbon face seals are used extensively in equipment such as air turbine starters. Five different carbon materials were tested, and the surface roughness was measured. The data are as follows:
Exercise 7. article in Communications of the ACM (Vol. 30, No. 5, 1987) studied different algorithms for estimating software development costs. Six algorithms were applied to eight software development projects and the percent error in estimating the development cost was observed. The data are in the table at the bottom of the page.