One-Way Analysis of Variance Stat 700 Lecture 12 11/29/2001

Illustrative Example Four chemical plants, producing the same product and owned by the same company, discharge effluents into streams in the vicinity of their locations. To check the extent of the pollution created by the effluents and to determine whether this varies from plant to plant, the company collected random samples of liquid waste, five specimens for each of the four plants. The data is presented in the table of the next slide. 11/13/2018

The Data 11/13/2018

Another Example College students were assigned to various study methods in an experiment to determine the effect of study technique on learning. The data presented in the next table was generated to be consistent with summary quantities found in the paper ``The Effect of Study Techniques, Study Preferences and Familiarity on Later Recall.’’ The study methods compared were reading only, reading and underlining, and reading and taking notes. One week after studying the paper ``Love in Infant Monkeys’’ students were given an exam on the article. 11/13/2018

The Data on Learning 11/13/2018

Testing Equality of Several Population Means The setting is that there are p normal populations each with variance . This assumption of equal variances is called the homoscedasticity assumption. The ith normal population has mean mi, i=1,2,…p. The goal is to test the null hypothesis that the p population means are all identical, versus the alternative hypothesis that there are at least two means which are different. 11/13/2018

Sample Data For the ith population, we observe a random sample of size ni, i=1,2,…,p. The data for this sample therefore consists of Yi1, Yi2, …, Yin(i). The sample data for the p samples could then be summarized in tabular form as follows: 11/13/2018

Data in Tabular Form 11/13/2018

Test Procedure The test procedure for testing the equality of the p population means is based on the F-distribution, and is usually called the one-way analysis of variance. The test statistic is given by: If the value of this test statistic is larger than Obtained from the F-distribution table, then the null hypothesis of equal population means is rejected. 11/13/2018

ANOVA Representation 11/13/2018

Formulas 11/13/2018

Example Using Effluents Data We present and illustrate the analysis using Minitab. In the next slide is the output from the Minitab analysis. We will illustrate how this is done in class. 11/13/2018

Output from Minitab One-way Analysis of Variance Analysis of Variance Source DF SS MS F P Factor 3 0.4649 0.1550 5.20 0.011 Error 16 0.4768 0.0298 Total 19 0.9417 Individual 95% CIs For Mean Based on Pooled StDev Level N Mean StDev -+---------+---------+---------+----- A 5 1.5680 0.1366 (-------*--------) B 5 1.7720 0.2160 (--------*-------) C 5 1.5460 0.1592 (-------*-------) D 5 1.9160 0.1689 (-------*-------) -+---------+---------+---------+----- Pooled StDev = 0.1726 1.40 1.60 1.80 2.00 Conclusion: The p-value of .011 is quite small, so there is indication that at least two population means are different. 11/13/2018

Calculations from Excel 11/13/2018

Example for RCB (Two-Analysis of Variance) A study was conducted to compare the effects of three levels of digitalis on the levels of calcium in the heart muscles of dogs. A description of the actual experimental procedure is omitted, but it is sufficient to note that the general level of calcium uptake varies from one animal to another so that comparison of digitalis levels (treatments) had to be blocked on heart muscles. That is, the tissue for a heart muscle was regarded as a block, and comparisons of the three treatments were made within a given muscle. The calcium uptakes for the three levels of digitalis, A, B, and C, were compared based on the heart muscles of the four dogs. 11/13/2018

The Raw Data from Study 11/13/2018

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