If we cross-tabulate drug by therapy, using the xtabs() function (see Section 7.1), we get the following table: 230 load("./rbook-master/data/clinicaltrial.rdata") We refer to this as a 3×2 factorial design. For this analysis each person is cross-classified by the drug they were given (a factor with 3 levels) and what therapy they received (a factor with 2 levels). Maybe there actually is an effect of therapy on mood gain, but we couldn’t find it because it was being “hidden” by the effect of drug? In other words, we’re going to want to run a single analysis that includes both drug and therapy as predictors. We didn’t find one, but there’s something a bit worrying about trying to run two separate analyses trying to predict the same outcome. In that chapter we did find a significant effect of drug, but at the end of the chapter we also ran an analysis to see if there was an effect of therapy. Another example appears in Chapter 14, in which we were looking at the effect of different drugs on the mood.gain experienced by each person. I gave one example of how this kind of design might arise above. In this section, I’ll discuss a broader class of experimental designs, known as factorial designs, in we have more than one grouping variable. When we discussed analysis of variance in Chapter 14, we assumed a fairly simple experimental design: each person falls into one of several groups, and we want to know whether these groups have different means on some outcome variable. Images not copyright InfluentialPoints credit their source on web-pages attached via hypertext links from those images.\) Remember this only provides an estimate of the variance you would obtain from the original data - and is dependent upon the choice of midpoints, and the number of class intervals used.Įxcept where otherwise specified, all text and images on this page are copyright InfluentialPoints, all rights reserved.sum(f*(y-ybar)^2) / (sum(f)-1) calculates the sample variance from the frequencies, f, midpoints, y, and the mean estimated from them, ybar.Īlternately, you could combine two of these instructions as: sum(f*(y-sum(y*f)/sum(f))^2)/(sum(f)-1).If you do not do this your estimated variance will be too high - because this formula gives the mean based upon the same assumptions as your variance will be calculated. However, even if you have a more accurate arithmetic mean, calculated directly from the observations themselves, you need to use this formula. ybar=sum(y*f)/sum(f) creates a variable called ybar, containing the arithmetic mean - as calculated from these frequencies and midpoints.f=c(23, 15, 6, 2) copies the frequency of each class into a variable called f.y=c(110, 125, 135, 155) copies the class interval midpoints into a variable called y.R can calculate the variance from the frequencies ( f) of a frequency distribution with class midpoints (y) using these instructions: Similarly, to obtain the 'population' standard deviation, use:.Remember if n=1 the second variance formula will always yield zero, because the mean of y will equal y, whereas the first formula will always yield NA, because 0/(1-1) = 0/0 and cannot be evaluated.But, there are 2 simple ways to achieve that: This var function cannot give the 'population variance', which has n not n-1 d.f.In other words, this is the uncorrected sample standard deviation. sd(y) instructs R to return the sample standard deviation of y, using n-1 degrees of freedom. In other words it uses n-1 'degrees of freedom', where n is the number of observations in Y. var(y) instructs R to calculate the sample variance of Y.R can calculate the sample variance and sample standard deviation of our cattle weight data using these instructions:
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