Learning Statistics with R - A tutorial for psychology students and other beginners, 2018a

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etaSquared( mod, type=2 ) eta.sq eta.sq.part sugar 0.22537682 0.4925493 milk 0.07019886 0.2321436 sugar:milk 0.43640732 0.6527155 and out pops the η 2 and partial η 2 values, as requested. However, when you’ve got an unbalanced design, there’s a bit of extra complexity involved. To see why, let’s expand the output from the etaSquared() function so that it displays the full ANOVA table: > es es eta.sq eta.sq.part SS df MS F p sugar 0.22537682 0.4925493 3.0696323 2 1.5348161 5.823808 0.017075099 milk 0.07019886 0.2321436 0.9561085 1 0.9561085 3.627921 0.081060698 sugar:milk 0.43640732 0.6527155 5.9438677 2 2.9719339 11.276903 0.001754333 Residuals 0.23219530 NA 3.1625000 12 0.2635417 NA NA Okay, if you remember back to our very early discussions of ANOVA, one of the key ideas behind the sums of squares calculations is that if we add up all the SS terms associated with the effects in the model, and add that to the residual SS, they’re supposed to add up to the total sum of squares. And, on top of that, the whole idea behind η 2 is that – because you’re dividing one of the SS terms by the total SS value – is that an η 2 value can be interpreted as the proportion of variance accounted for by a particular term. Now take a look at the output above. Because I’ve included the η 2 value associated with the residuals (i.e., proportion of variance in the outcome attributed to the residuals, rather than to one of the effects), you’d expect all the η 2 values to sum to 1. Because, the whole idea here was that the variance in the outcome variable can be divided up into the variability attributable to the model, and the variability in the residuals. Right? Right? And yet when we add up the η 2 values for our model... > sum( es[,"eta.sq"] ) [1] 0.9641783 . . . we discover that for Type II and Type III tests they generally don’t sum to 1. Some of the variability has gone “missing”. It’s not being attributed to the model, and it’s not being attributed to the residuals either. What’s going on here? Before giving you the answer, I want to push this idea a little further. From a mathematical perspective, it’s easy enough to see that the missing variance is a consequence of the fact that in Types II and III, the individual SS values are not obliged to the total sum of squares, and will only do so if you have balanced data. I’ll explain why this happens and what it means in a second, but first let’s verify that this is the case using the ANOVA table. First, we can calculate the total sum of squares directly from the raw data: > ss.tot ss.tot [1] 13.62 Next, we can read off all the SS values from one of our Type I ANOVA tables, and add them up. As you can see, this gives us the same answer, just like it’s supposed to: > type.I.sum type.I.sum [1] 13.62 - 550 -
However, when we do the same thing for the Type II ANOVA table, it turns out that the SS values in the table add up to slightly less than the total SS value: > type.II.sum type.II.sum [1] 13.1321 So, once again, we can see that there’s a little bit of variance that has “disappeared” somewhere. Okay, time to explain what’s happened. The reason why this happens is that, when you have unbalanced designs, your factors become correlated with one another, and it becomes difficult to tell the difference between the effect of Factor A and the effect of Factor B. In the extreme case, suppose that we’d run a 2 ˆ 2 design in which the number of participants in each group had been as follows: sugar no sugar milk 100 0 no milk 0 100 Here we have a spectacularly unbalanced design: 100 people have milk and sugar, 100 people have no milk and no sugar, and that’s all. There are 0 people with milk and no sugar, and 0 people with sugar but no milk. Now suppose that, when we collected the data, it turned out there is a large (and statistically significant) difference between the “milk and sugar” group and the “no-milk and no-sugar” group. Is this a main effect of sugar? A main effect of milk? Or an interaction? It’s impossible to tell, because the presence of sugar has a perfect association with the presence of milk. Now suppose the design had been a little more balanced: sugar no sugar milk 100 5 no milk 5 100 This time around, it’s technically possible to distinguish between the effect of milk and the effect of sugar, because we have a few people that have one but not the other. However, it will still be pretty difficult to do so, because the association between sugar and milk is still extremely strong, and there are so few observations in two of the groups. Again, we’re very likely to be in the situation where we know that the predictor variables (milk and sugar) are related to the outcome (babbling), but we don’t know if the nature of that relationship is a main effect of one predictor, or the other predictor or the interaction. This uncertainty is the reason for the missing variance. The “missing” variance corresponds to variation in the outcome variable that is clearly attributable to the predictors, but we don’t know which of the effects in the model is responsible. When you calculate Type I sum of squares, no variance ever goes missing: the sequentiual nature of Type I sum of squares means that the ANOVA automatically attributes this variance to whichever effects are entered first. However, the Type II and Type III tests are more conservative. Variance that cannot be clearly attributed to a specific effect doesn’t get attributed to any of them, and it goes missing. 16.11 Summary • Factorial ANOVA with balanced designs, without interactions (Section 16.1) and with interactions included (Section 16.2) • Effect size, estimated means, and confidence intervals in a factorial ANOVA (Section 16.3) - 551 -
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Learning statistics with R: A tutor
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This book is published under a Crea
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Preface Table of Contents I Backgro
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8.6 Summary . . . . . . . . . . . .
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VI Endings, alternatives and prospe
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February 16, 2015 Preface to Versio
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and density is discussed. A detaile
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1. Why do we learn statistics? “T
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And finally, an invalid argument wi
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Admission rate (both genders) 0 20
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experiment, and they don’t get an
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2. A brief introduction to research
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different ways: the length of time
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that when I ask 100 people how they
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Table 2.1: The relationship between
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2.3 Assessing the reliability of a
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2.5.1 Experimental research The key
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things “inside” the study. Let
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they lack face validity, you’ll f
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that age is growing at an incredibl
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then it can be a big problem. 2.7.7
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2.7.10 Placebo effects The placebo
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it doesn’t, which one do you thin
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Part II. An introduction to R - 35
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go out into the real world. It’s
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download Rstudio. To understand why
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Most of this text is pretty uninter
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citation() To cite R in publication
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Table 3.1: Basic arithmetic operati
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Division / and Multiplication *, th
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sales * royalty [1] 2450 As far as
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x to the power of 0.5”. Personall
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As you can see, specifying the argu
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Figure 3.3: If you’ve typed the n
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sales.by.month sales.by.month [1]
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profit / days.per.month [1] 0.00000
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not true of R. R is not infinitely
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Table 3.3: Some more logical operat
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In other words, any.sales.this.mont
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and gotten exactly the same result.
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Figure 3.6: The options window in R
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- 72 -
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print( keeper ) [1] 8.539539 So, fr
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to import an SPSS data file into R,
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The output here is telling you that
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Figure 4.3: The Rstudio dialog box
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Figure 4.5: The Rstudio “Environm
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this should be obvious already. But
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Figure 4.6: The “file panel” is
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from this file into my workspace. T
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2 February 28 100 high 3 March 31 2
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Figure 4.9: The Rstudio window for
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4.6.1 Special values The first thin
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x y x * y Error in x * y : non-nu
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4.7.1 Introducing factors Suppose,
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factors in 7.11.2, but for now you
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An alternative method is to use the
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4.11 Generic functions There’s on
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load (base) R Documentation The R D
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Warning Saved R objects are binary
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- 110 -
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5. Descriptive statistics Any time
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mean of these observations is just:
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mean( afl.margins[1:5] ) [1] 36.6 A
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the median rises only to $62,500. I
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Next, let’s calculate means and m
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5.2 Measures of variability The sta
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English: which game value deviation
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and you get the same... no, wait...
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Mean − SD Mean Mean + SD Figure 5
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Negative Skew No Skew Positive Skew
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Platykurtic ("too flat") Mesokurtic
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e useful, let’s try this for a ne
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argument specifies the grouping var
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grumpiness score. 13 If the mean is
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Frequency 0 5 10 15 20 Frequency 0
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My grumpiness 40 50 60 70 80 90 4 6
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following extract illustrates the b
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Y1 4 5 6 7 8 9 10 Y2 3 4 5 6 7 8 9
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Okay, so let’s have a look at our
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pay 0.760 0.720 . 0.137 . 0.196 . d
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5 6.68 9.75 NA 5 6 5.99 5.04 72 6 B
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• Measures of variability. In con
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6. Drawing graphs Above all else sh
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delete the plot and then type a new
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Fibonacci 2 4 6 8 10 12 1 2 3 4 5 6
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high-level functions all rely on th
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type = ’p’ type = ’o’ type
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Fibonacci 2 4 6 8 10 12 1 2 3 4 5 6
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Figure 6.8: Altering the scale and
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y the height of the leftmost bar in
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6.3.1 Visual style of your histogra
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0 20 40 60 80 100 120 (a) (b) Figur
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0 20 40 60 80 100 120 Winning Margi
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0 50 100 150 200 250 300 Winning Ma
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6.5.3 Drawing multiple boxplots One
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Winning Margin 0 50 100 150 1987 19
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The second way do to it is to use a
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cor( x = parenthood ) # calculate c
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0 10 20 30 0 10 20 30 Adelaide Fitz
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value for mar is c(5.1, 4.1, 4.1, 2
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This takes the “active” figure
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7. Pragmatic matters The garden of
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in the command above I didn’t nam
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speaker ee onk oo pip makka-pakka 0
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2 7 3 3 1 3 3 -1 1 -1 BLAH BLAH BLA
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seen it before. In any case, those
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Table 7.2: Two more arithmetic oper
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fact R does provide a function for
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But suppose, on the other hand, tha
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is select several different variabl
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column number. So, if we want to pi
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case.2 upsy-daisy pip 2 case.4 makk
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Note that R will only allow you to
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7.6.2 Sorting a factor You can also
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Two other methods that I want to br
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At this point you should have two q
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(alcohol, caffeine or no drug). Bec
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3 3 female 4.6 643 7.4 226 7.3 412
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discuss melt() and cast() in a fair
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paste( hw, ng, sep = ".", collapse
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Table 7.3: The ordering of various
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Table 7.4: Standard escape characte
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sub( pattern = "a", replacement = "
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Figure 7.1: The booksales2.csv data
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note that if you don’t have the R
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etween the two. However, there are
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a third variable to the cross-tabs
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today print(today) # display the d
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encode numbers in binary, 29 0.1 is
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environment. And so when you type i
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8. Basic programming Machine dreams
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for ages, you figure out some reall
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Figure 8.2: A screenshot showing th
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8.1.5 Differences between scripts a
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to the second value in vector; and
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crunching, we tell R (on line 21) t
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What this does is create a function
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hundreds of other things besides. H
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Part IV. Statistical theory - 269 -
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me: I guess I don’t see that. Sur
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discussion of probability theory is
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In this case 11 of these 20 coin fl
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frequentists do this sometimes. And
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Probability of event 0.0 0.1 0.2 0.
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proceed to roll all 20 dice, what
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(a) Probability 0.00 0.05 0.10 0.15
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In other words, there is a 76.9% ch
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Probability Density 0.0 0.1 0.2 0.3
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Shaded Area = 68.3% Shaded Area = 9
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Probability Density 0.0 0.1 0.2 0.3
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• The χ 2 distribution is anothe
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work rather than rchisq() - the obs
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- 300 -
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10.1.1 Defining a population A samp
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Figure 10.2: Biased sampling withou
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a stratified sampling technique you
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10.2 The law of large numbers In th
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Table 10.1: Ten replications of the
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Sample Size = 1 Sample Size = 2 Sam
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Sample Size = 1 Sample Size = 2 0.0
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efer to them. For instance, if true
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Average Sample Mean 96 98 100 102 1
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10.5 Estimating a confidence interv
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sense idea of what it means to say
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Average Attendance 0 10000 30000 19
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Here’s how to plot the means and
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Let’s suppose that this glorious
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Dan’s research hypothesis: “ESP
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the type I error rate. As Blackston
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Critical Regions for a Two−Sided
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Critical Region for a One−Sided T
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one based on Neyman’s approach to
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alternative hypothesis is a near ce
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Sampling Distribution for X if θ=.
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Table 11.2: A crude guide to unders
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ureaucratic process. It’s not par
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(binomial test was non significant)
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12. Categorical data analysis Now t
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observations that fall within the i
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it is when it predicts too many (wh
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df = 3 df = 4 df = 5 0 2 4 6 8 10 V
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That’s why I told R to use the up
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clubs 35 40 0.2 diamonds 51 60 0.3
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only one of many (albeit one of the
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That’s more or less what we’re
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Well, since these probabilities hav
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Chapek 9 has an unfortunate tendenc
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12.5 Assumptions of the test(s) All
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To get the test of independence, al
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This is a bit more output than we g
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subj.1 : 1 no :70 no :90 subj.10 :
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13. Comparing two means In the prev
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40 50 60 70 80 90 Grades Figure 13.
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Two Sided Test One Sided Test −1.
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lower.area print( lower.area ) [1]
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df = 2 df = 10 −4 −2 0 2 4 valu
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than I’ve done here. For instance
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Anastasia’s students Bernadette
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null hypothesis μ alternative hypo
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ut once again I’m going to start
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for the difference “(mean 1) minu
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null hypothesis μ alternative hypo
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13.5.1 The data The data set that w
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ciMean( x = chico$improvement ) 2.5
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Paired samples t-test Variables: gr
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Grouping variable: ID variable: tim
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In a one-sided confidence interval,
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are different tests. Secondly, I wa
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Table 13.1: A (very) rough guide to
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ut you no longer believe that the c
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Frequency 0 5 10 15 20 Normally Dis
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Sampling distribution of W (for nor
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We then count up the number of chec
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• A paired samples t-test is used
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14. Comparing several means (one-wa
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Mood Gain 0.5 1.0 1.5 placebo anxif
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This formula looks pretty much iden
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is true, then you’d expect all th
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mean square, MS w , can be viewed a
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3 placebo 0.1 0.45 -0.350 0.1225 4
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is choose an α level (i.e., accept
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names( my.anova ) [1] "coefficients
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etaSquared( x = my.anova ) eta.sq e
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One thing that bugs me slightly abo
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If we compare these three p-values
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• Homogeneity of variance. Notice
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Secondly, I did mention that it’s
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Frequency 0 1 2 3 4 5 6 7 Sample Qu
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Looking at this table, notice that
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• Post hoc analysis and correctio
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15. Linear regression The goal in t
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The Best Fitting Regression Line No
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• data. The data frame containing
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Figure 15.4: A 3D visualisation of
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Wonderful. A big number that doesn
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If our regression model has K predi
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You can see why this is handy, sinc
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what it is. The test for the signif
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Fortunately, confidence intervals f
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• Uncorrelated predictors. The id
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Outlier Outcome Predictor Figure 15
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Outcome High influence Predictor Fi
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Cook’s distance 0.00 0.02 0.04 0.
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Standardized residuals −2 −1 0
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78 Residuals vs Fitted Residuals
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In a lot of cases, the solution to
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Signif. codes: 0 *** 0.001 ** 0.01
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15.10.1 Backward elimination Okay,
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+ day 1 1.55760 1837.2 297.08 + bab
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etween those two SS values as a sum
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16. Factorial ANOVA Over the course
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Page 519 and 520: attributable to the “ANOVA model
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Page 537 and 538: tfm.2 grade attend reading 1 90 yes
Page 539 and 540: line, we obtain the following: Ŷ 7
Page 541 and 542: Estimate Std. Error t value Pr(>|t|
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Page 547 and 548: R comes with a variety of functions
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Page 551 and 552: 16.8 Post hoc tests Time to switch
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Page 555 and 556: 1 yes none 3.700 2 no none 5.550 3
Page 557 and 558: Analysis of Variance Table Response
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Page 561 and 562: sugar 2.13 2 4.0446 0.045426 * milk
Page 563: even simpler. The main effect of su
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Page 580 and 581: Cumulative Probability of Type I Er
Page 582 and 583: are wrong. Worse yet, because we do
Page 584 and 585: library( BayesFactor ) # ...because
Page 586 and 587: make sense to people who already un
Page 588 and 589: -------------- [1] Non-indep. (a=1)
Page 590 and 591: Before moving on, it’s worth high
Page 592 and 593: 17.6.2 The Bayesian version Okay, s
Page 594 and 595: [2] day : 0.2724027 ˘0% [3] baby.s
Page 596 and 597: therapy 0.4672 1 8.5816 0.01262 * d
Page 598 and 599: command that I use when I want to g
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Page 602 and 603: • Other types of correlations. In
Page 604 and 605: measurements mean that independence
Page 606 and 607: the list that I’ve given above is
Page 608 and 609: So if you were forced to guess my w
Page 610 and 611: to be able to do more than ANOVA, m
Page 612 and 613: Fisher, R. A. (1922a). On the inter
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