Week 20 Flashcards
1) In the structural model for a 2-way between-subjects ANOVA having factors A and B, what does b1 represent on?
a) the difference between the marginal mean at level 1 of factor B and the grand mean
b) the difference between the marginal mean at level 1 of B and the marginal means of A
c) differences between the cell means at level 1 of factor B
d) differences between the grand mean and the cell means at level 1 of factor B
e) difference between observations at level 1 of B and the grand mean
a) the difference between the marginal mean at level 1 of factor B and the grand mean
Explanation … The a and b terms in a 2-way structural model quantify the main effects of the factors or treatments in a 2-way design. In particular, b1 quantifies the effect, on a participant, of being exposed to level 1 of factor (or treatment) B, b2 quantifies the effect of being exposed to level 2 of factor B, and so on.
For real data, these effects are estimated by determining how far the appropriate marginal mean is pushed away for the grand mean. Thus, b1 is calculated from the difference between the marginal mean for level 1 of B and the grand mean. The equation is b1 = M.1 – M..
2) In the structural model for a 2-way between-subjects ANOVA having factors A and B, how would you interpret b1?
a) the effect of treatment B for participants at level 1 of A
b) the effect of the interaction between A and B on participants at level 1 of treatment B
c) the effect of being exposed to level 1 of treatment B regardless of what level of A participants were exposed to
d) the overall effect of treatment B on the dependent variable
c) the effect of being exposed to level 1 of treatment B regardless of what level of A participants were exposed to.
Explanation … The terms in a structural model have interpretations that involve explained or unexplained effects of the treatments on the dependent variable. In the structural model for a 2-way ANOVA, we try to explain observed variation between experimental groups by differences in treatment A or in treatment B or their interaction, whereas differences within the experimental groups is treated as error.
In this scheme, the bj terms have to do with the pure effect of treatment B on the dependent variable, regardless of what level of A participants are exposed to. Specifically, the term b1 is supposed to capture the effect, on participant’s dependent variable values, of being exposed to level 1 of B while disregarding (“collapsing across”) the levels of treatment A.
In the structural model for a 2-way between-subjects ANOVA having factors A and B, what does Ekji depend on?
a) the difference between a marginal mean and the grand mean
b) the difference between a cell mean and the grand mean
c) the difference between an individual observation and the grand mean
d) the difference between an individual observation in a cell and the cell’s mean
d) the difference between an individual observation in a cell and the cell’s mean
Explanation … The terms Ekji are error terms. A structural model always tries to offer explanations as to why individual observations deviate from the grand mean. As usual, however, after accounting for all of the effects of treatments and factors in the design, the question still remains as to why dependent-variable observations within a particular cell in a design differ from each other. We really have no explanations for this. After all, the participants within a cell have all been treated in exactly the same way (they all experience see the same level of A and the same level of B) so why should they differ? Whatever is causing these differences is not explained by the factors or their interactions. These within-cell differences are therefore exactly what we mean by error and the way we keep track of them is to measure the deviation of each observation in a cell from the cell mean.
4) Fill in the blanks in the following ANOVA results table for a 2-way factorial balanced design with a = 4 and n = 4 (b is unknown).
MSA = 8
MS A X B = 2
SSError = 144
SSTotal = 248 TOTAL df = 95
dfA = 3 MSA = 8 dfB = 5 dfAxB=15 MSAxB=2 dfError=72 dfTotal=95
Explanation … In problems like this, always try to fill in the df column first because it only depends on the experimental design. We are told that a = 4 and so immediately know that dfA = 3. We also see from the table that dfTotal = 95. All other degrees of freedom will have to be worked out. To do this we need to understand more about the experimental design.
If only we know the number of levels in factor B, everything would fall into place. But there are N = 96 participants total in the design (because dfTotal = 95) and 16 participants at every level of A (because n = 4 and a = 4). Thus, there must be 96/16 = 6 levels of B. This automatically means that dfB = 5 and that dfError (the only df term that is left) must be 72. These results have all been insterted in the output table below and you can verify for yourself that dfTotal is the sum of all the df terms, as it has to be.
Having filled in the df column we can now tackle the SS column. We know that SS = MS x df therefore we must have SSA = 3 x 8 = 24 and SSAxB = 2 x 15 = 30. This means we have filled in all the SS entries except SSB. But there is a relationship between all the SS terms so we can always find the final unknown one SSB = SSTotal –(SSA + SSAxB + SSError) = 248 – (24+30+144) = 50. After filling in all the SS values the table now looks like this…
SSA=24 SSB=50 SSAxB=30 dfB=5 dfAxB=15 dfError=72
It is now easy to fill in the rest of the MS values. Only 2 are missing. They are MSB = SSB/dfB = 50/5 = 10, and MSError = SSError/dfError = 144/72 = 2. Updating the table with these values gives…
SSA=24 SSB=50 SSAXB=30 SSERROR=144 SSTOTAL=248 DFA=3 DFB=5 DFAXB=15 DFERROR=72 DFTOTAL=95 MSA=8 MSB=10 MSAXB=2 MSERROR=2 FA=4 FB=5 FAXB=1
For a 1-way ANOVA design, 2 = SSTreatment/SSError.
a) true
b) false
b) false
Explanation … The correct formula is n^2 = SSTreatment/SSTotal. In a 1-way ANOVA, n^2 represents the proportion of observed variation in the dependent variable that is explained by differences in the independent variable. SSTreatment quantifies explainable variation in the dependent variable because it is composed of nothing but deviations of group means from the grand mean. Dividing SSTreatment by SSTotal simply expresses this explained variation as a proportion of total variation.
For a 1-way ANOVA, the possible values of n^2 range between 0 and 1.
a) true
b) false
a) true
Explanation … By definition, n^2 = SSTreatment/SSTotal. But SSTotal = SSTreatment + SSError, and so you can see that another way to write n^2 is… n^2 = SSTreatment/ (SSTreatment + SSError). From this you can see that the numerator can never be larger than the denominator and hence n^2 will never be larger than 1. On the other hand, since n^2 is composed exclusively of sums of squares it can never be be negative. Hence n^2 is always between 0 and 1. This is what allows us to interpret n^2 as a proportion … the proportion of variation in the dependent variable that is explained by differences in the independent variable.
For a 1-way ANOVA design, the possible values of Cohen’s f range between 0 and 1.
a) true
b) false
b) false
Explanation … Cohen’s f the ANOVA counterpart of Cohen’s d. Its interpretation is the standard deviation of the group means in a 1-way design standardized by the within-groups pooled standard deviation. The smallest value for Cohen’s f is 0 (in the unlikely circumstance that all of the observed group means are identical) but there is no upper limit on it. This is just the way that Cohen’s d worked for t-tests. Thus, whereas in most experimental situations you would expect to see Cohen’s f values below 1, in theory they can attain any size.
The value of Cohen’s f depends on the sample size.
a) true
b) false
b) false
Explanation … Cohen’s f is intended as an effect size. It is supposed to capture the magnitude of the effect of the independent variables on the dependent variable while being independent of any arbitrary choices a researcher might make in terms of measurement units, samples sizes, or whether to employ a within-subjects or between-subjects analysis. Thus, Cohen’s f is independent of sample size.
In a 1-way ANOVA design, if 50% of the observed variation in dependent variable values is explained by differences in the independent variable, what is the value of Cohen’s f?
a) f = 0
b) f = 0.25
c) f = 0.5
d) f = 1
d) f = 1
Explanation … For this design, f^2 = n^2/(1-n^2). If 50% of variation in the dependent variable has been explained then n^2= 0.5. Plugging this into the expression converting n^2 to f^2 gives f^2 = 0.5/(1 - 0.5)= 0.5/0.5 = 1. Therefore f = 1.
Suppose that in the previous problem you were told that there was 3 levels of treatment with 5 observations at each level. What would the value of the noncentrality parameter be?
a) L-symbol = 1
b) L-symbol = 3
c) L-symbol = 5
d) L-symbol = 15
d) L-symbol = 15
Explanation … It is easy to calculate the value of the noncentrality parameter for a 1-way ANOVA. The formula is L-symbol = f^2N. But in the previous you were told that the value of Cohen’s f is f = 1 and in this problem you have been additionally told that k = 3 and k = 5 making N = k × n = 3 × 5 = 15. Therefore L-symbol = 1^2 × 15 = 15.
In a 1-way between-subjects factorial design with k = 4 levels … if n^2= 0.2 and SSTotal = 100 then what is the value of SSTreatment?
a) 0
b) 0.2
c) 20
d) 20/4
c) 20
Explanation … In a 1-way between-subject design (the simplest ANOVA design you know) the formula for n^2 is n^2 = SSTreatment/SSTotal . Inserting the information from the problem into this relationship gives 0.2 = SSTreatment/100. You can see that only if SSTreatment = 20 is the equation true. The answer doesn’t depend on k.
For the same design as in the previous question, what is the value of Cohen’s f?
a) 0
b) 1/2
c) 1
d) 2
b) 1/2
Explanation … There are 2 ways to calculate Cohen’s f for the design. First, in the previous question we were told that SSTotal = 100 and n^2= 0.2, then calculated from that that SSTreatment = 20. For a 1-way between subjects design this must mean that SSError = 80. Substituting the SSTotal and SSTreatment values into the formula for Cohen’s f gives, f=√(〖SS〗_Treatment⁄〖SS〗_Error )=√(20⁄80)=√(1⁄4)=1/2.
The other way to calculate Cohen’s f is to use the formula that connects Cohen’s f values with n^2. As part of the original information in the previous problem, we were told that n^2= 0.2. Using this in the formula for Cohen’s f gives, f=√(n^2⁄((1-n^2 ) )) = √(0.2⁄((1-0.2) ))= √(0.2⁄0.8)= √(1⁄4)=1/2.
For a 1-way ANOVA, the larger the value of the noncentrality parameter the greater the power.
a) true
b) false
a) true
Explanation … For 2-sample t-tests there was a noncentrality parameter called L-symbol. It was easy to calculate because it depended only on effect size and sample size, and it had an overall “monotonic” relationship with power meaning that the larger the value of L-symbol, the greater the power. The relations was a bit complicated to calculate exactly, however, and that is why the L-symbol was generally used as the input to a lookup table. Alternatively, software like G*Power could calculate both L-symbol and power.
The situation for ANOVA is the same. Instead of L-symbol there is a different noncentrality parameter called . It is easy to calculate because it depends only on effect size (Cohen’s f) and sample size and it has a monotonic relationship with power meaning that larger the value of L-symbol the greater the power for the ANOVA.
For a 2-way ANOVA design , n^2A + n^2B +n^2AxB =n^2Total
a) true
b) false
a) true
n^2A + n^2B + n^2AxB = SSA/SStotal + SSa/SStotal+SSaxb/SStotal
= (SSa + SSb + SSaxb)/SStotal = n^2total
Source SS df MS F A 10 1 10 5 B 30 3 10 5 AxB 12 3 4 2 Error 64 32 2 Total 116 39
What is the value of f^2a?
What is the value of f2A?
a) 10 / 116
b) 10 / 64
c) 10 / 30
d) 64 / 116
b) 10 / 64
Explanation … The question asks for the value of the square of Cohen’s effect size (f^2 is used in power calculations) for factor A in this 2-way between-subjects design. In a 2 way design calculating either the fA or n^2A values is easy, particularly if you have the ANOVA output table. For Cohen’s f the formula is f^2 = SSTreatment / SSError, where SSTreatment is the sum-of-squares for the factor or interaction you are interested in. Here we are interested in factor A and looking at the output table we see SSTreatment = 10 for factor A and SSError = 64. Therefore f^2A = 10/64.
Source SS df MS F A 10 1 10 5 B 30 3 10 5 AxB 12 3 4 2 Error 64 32 2 Total 116 39
What is the value of n^2A?
a) 10 / 116
b) 10 / 64
c) 10 / 30
d) 64 / 116
a) 10/116
Explanation … For each factor or interaction, the formula for n^2 is always n^2= SSTreatment / SSTotal, where SSTreatment is the sum-of-squares for the factor or interaction you are interested in. Here we are interested in factor A and looking at the output table we see SSTreatment = 10 for factor A and SSTotal = 116. Therefore n^2A = 10/116.
Source SS df MS F A 10 1 10 5 B 30 3 10 5 AxB 12 3 4 2 Error 64 32 2 Total 116 39
What is the value of partial n^2A?
a) 10 / 64
b) 10 / 74
c) 10 / 86
d) 64 / 86
b) 10 / 74
Explanation … To calculate partial n^2 for any factor or interaction in a 2-way design, start with the corresponding n^2 value and replace the denominator with the sum of SSTreatment + SSError where SSTreatment is the sum-of-squares for the factor or interaction you are interested in. For this question we are interested in factor A. Looking at the output table we see SSTreatment = 10 for factor A and SSError = 64 so we should replace the denominator in n^2 by 74 = 10 + 64. This yields partial n^2A = 10/74
