One-way Analysis of VarianceWe consider a simple example that is a study of the use of a semiautomated method for measuringthe amount of chlorpheniramine mealeate in tablets (Rice). We start by comparing measurementsfrom two labs:Lab 1 4.13, 4.07, 4.04, 4.07, 4.05, 4.04, 4.02, 4.06, 4.1, 4.04Lab 2 3.86, 3.85, 4.08, 4.11, 4.08, 4.01, 4.02, 4.04, 3.97, 3.95To study the consistency between labs of the measurement process, we could examine severalstatistics.BoxplotsA visual comparison could be made with box plots. For the first lab, the quartiles are: 4.04, 4.055,and 4.07. For the second lab, the quartiles are: 3.955, 4.015, 4.07. Make side-by-side boxplots forthese two labs.Two-sample testA two-sample test of the hypothesis that these labs have the same mean would be conducted asfollows:Lab1: mean = 4.062, sd = 0.03Lab2: mean = 3.997, sd = 0.09If the SDs are assumed to be different, then we would use the following test statistic:4.062 − 3.997p0.032/10 + 0.092/10= 2.17The p-value for a t-distribution with 9 degrees of freedom (two-sided) is 0.06. (Based on theassumption that the observations follow a normal distribution).If the SDs are assumed to be the same then we would compute a pooled estimate of the SD:0.06714.062 − 3.9970.067p1/10 + 1/10= 2.17The p-value for a t-distribution with 18 degrees of freedom (two-sided) is 0.04.Multiple comparisonsWe consider the case where we have 7 labs.NotationYi,j= jth measurement taken at the ith lab. Here j = 1, . . . , ni, and i = 1, . . . , I.¯Yi= the mean of the nimeasurements taken at the ith lab.BoxplotsWe can still make boxplots and compare them side-by-side.Pairs-wise comparisonsWe can still compare two means at a time. There are 21 pairs to compare.The statistic for the i1, i2comparison would b e :¯Yi1−¯Yi2spp1/10 + 1/10where s2p=Pi(ni− 1)s2i/(Pini− I). We would then use a t-distribution withPini− I degreesof freedom as our test statistic.To compensate for making 21 tests at once, our α-level would be