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I don't know if mathy enough (but it should have the right references to get you started): Gelman, A. (2005). Analysis of Variance: Why It Is More Important than Ever. The Annals of Statistics, 33(1), 1–31. doi:10.2307/3448650 Should be available from Andrew Gelman's website.

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Try: Linear Models by Shayle Searle

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Drop Day 8. Then you have a standard $n \times 2 \times 7$ repeated measures design that lets you test both the Stage and Day main effects and the Stage $\times$ Day interaction.

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You can do a chisq.test with 6 df. For example, if you have 100 surveys: IV<-sample(1:3,100,T) DV<-sample(1:4, 100,T) chisq.test(table(IV,DV))

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It rather depends on what you mean by "by hand". There is more than one way to do it. You can use the residuals: > etasq(xyaov) Partial eta^2 x 0.4854899 Residuals NA > 1 - var(xyaov$residuals)/var(y) [1] 0.4854899 (You didn't set a seed, so we don't have exactly the same result). Almost equivalently, you can use ... 4 eta-squared ($\eta^2$), is a measure of effect size for ANOVA models that is analogous to$R^2$. That is, it gives the proportion of the variability in$Y$that can be accounted for by knowledge of$X$. There is a 'regular'$\eta^2$, and a partial$\eta^2$. This distinction only comes into play when you have an ANOVA with multiple factors. Here are the ... 1 In a (one-way) MANOVA, you study the effect of a single categorical variable (e.g. treatment yes/no) on the averages of two or more continuous response variables (e.g. diastolic and systolic blood pressure). In a two-way ANOVA, you study the effects of two categorical variables (e.g. treatment yes/no and and sex) on the average of a single continuous ... 1 When you add variables to an ANOVA (or any regression model) you are controlling for it. You added two variables (amount of distractors and an interaction) so, of course all the values changed. The only way they could not is if there was no relation between the independent variables. The two models ask different questions so they get different answers. ... 0 In the one-way model you look at only the effect of the sound system, whereas in the two-way model you look als at the effect of distractors and the interaction between distractors and the soundsystem. When you change the model, it is not unusual that the values of the model change as wel. Your two-way anova model seems to explain your data better than the ... 1 I'm certainly no ANOVA expert but I guess the other way to do this analysis is to switch to a regression framework and use lme4 which doesn't mind unbalanced data and will itself work out what it 'between' and what is 'within'. I believe the relevant line for an additive model would be mod0 <- lmer(top_start ~ (1 | id) + task + org + sex, data=df) ... 3 If you use type 3 for ANOVAs it is critical in R that you set the contrast to effect coding (i.e., "contr.sum"). The default contrast in R is dummy coding (or in R parlance, treatment coding) in which 0 represents the first factor level. This doesn't make too much sense when having interactions as explaind on the page I linked to. To set effect coding, ... 2 car does support the F-statistic for analysis of deviance tables: Anova(m, test.statistic="F") Resulting in Analysis of Deviance Table (Type II Wald F tests with Kenward-Roger df) Response: some_response F Df Df.res Pr(>F) some_factor1 0.30 1 3.4 0.6196 some_factor2 26.94 1 ... 2 When most people think of a non-parametric equivalent of ANOVA, they think of the Kruskal-Wallis test. The Kruskal-Wallis test cannot be applied to a factorial structure, however. The first workaround to this is to run all of your conditions as a one-way analysis. This does not let you test your factors individually, but you may be able to get what you ... 1 You don't have to run an ANOVA first, but most people do out of habit. (Whether reviewers will give you a hard time about not having done so is a separate issue.) Note that the original Dunnett's test required that the conditions have equal$n$s. The test has since been generalized, so it is fine if you do not have equal$n$s, just be sure you are running ... 0 If you have three groups you should do an ANOVA (after checking assumptions of normality etc of course) which will test if the three groups differ overall. If that is the case you can then either do contrasts or post-hoc tests to test your hypotheses directly, e.g. does group 1 differ from group 2. How to do contrasts or post-hoc tests depends on the ... 0 This is an old post, but I'll try to answer it anyway for other people who see this. This message means that there is at least one participant who has some double value for repetition (within each participant, the values for repetition must be unique, for example 1,2,3 while your data might have 1,1,2 for some participant). So the problem is most likely in ... 2 Under the assumption that your data are continuous, you can use an ANOVA for this situation. 1 You should not use any form of ANOVA since the number of goals per game is going to be a count, probably with a fairly small mean (I presume you are talking about what Americans call soccer, not what Americans call football). Therefore, you need some sort of count regression model: Candidates could be Poisson, negative binomial and possibly zero-inflated ... 2 The student wrote because the effect of task is significant in the treatment group but non-significant in the control group, there's some evidence that there really is an interaction effect. By itself, this is not very good evidence of an interaction effect, for reasons others have identified: p in treatment group could be 0.049 and in control ... 1 Scheffe's test allows you to do more than compare group means. You can compare the average of groups A and B with the average of groups C, D and E. Etc. The only real reason to choose Scheffe's test for multiple comparisons is to do that kind of contrast. If you only compare group means, this test has less power than others (Tukey, Dunnett). The Scheffe's ... 0 Given your comments, yes, the other paper seems to be using the mean value. This isn't wrong, but doesn't let you look at change over time. There's a nice writeup of this sort of analysis near the beginning of Hedeker & Gibbons 0 How about this: Take the null model that at each timepoint there is 1/3 chance the score will be higher, lower, or the same as the previous timepoint. The probability you would get no scores worse than the previous week 3 weeks in a row is (2/3)X(2/3)X(2/3) = 8/27 = 0.296 Edit: Here is one way to look at it using an approach similar to that proposed by ... 1 Caveat emptor: I am NOT a biostatistician...but I play one on TV ;-) In all seriousness, I have a strong statistics background, but it is not medically oriented. However, your question was simply stated, so my suggestions utilize a general approach, not one that is domain specific to biomedical research. First, you have a very small sample size. I hope your ... 1 Firstly, this is a very small sample size (4 observations), hopefully you have more, or at least have the same four observations for many different patients. If not, it will be difficult to find a model. Generally it is good to have a sample size greater than 100, or at a bare minimum, 20. Secondly, the survey (6 questions) is also small. Does the patient ... 0 I think there are a few approaches. I haven't looked at them all and not sure which is the best: (1) -sandwich- package library(sandwich) coeftest(model, vcov=sandwich) But doesn't give me the same answers I get from Stata for some reason. I've never tried to work out why, I just don't use this package. (2) -rms- package. I find this a bit of a pain to ... 0 The R function 'fligner.test' takes into account only the first factor, that's right. A multiway anova with interaction could be handled in 'fligner.test' by creating a single variable distinguishing all groups, e.g. using fligner.test(a ~ interaction(b, c, d)). Tests for equal variances are generally not considered as being very useful, e.g. since null ... 0 Every statistical method (and every other method of making decisions as well) can give false negatives and false positives. The only way to avoid mistakes is to know in advance what is correct, but, if we already knew what was correct, we wouldn't need to do any analysis! One advantages of statistical methods is that, if you follow the rules and meet the ... 2 The analyses being presented aren't being fairly compared so there's no way to know if they're actually saying different things. They could be perfectly compatible. You could use lm (what underlies aov) and get the treatment effects just as you have for lme. You can also find omnibus F's using lme. Then you could compare things a little better but that's not ... 2 (Converted from a set of comments.) You don't have to choose because the two methods do not contradict each other in any way. One presents the omnibus test while the other tests the individual contrasts. It is a well-known fact that a non-significant omnibus test does NOT imply that none of the comparisons subsumed under that omnibus test are significant. ... 0 aov is designed for balanced data (link). Balanced design is: An experimental design where all cells (i.e. treatment combinations) have the same number of observations (link). 1 COIem-easy is absent because the effect is 0. What R is doing is setting that as the reference value that all other COI's are compared to. The p-value for the t-test is the test that the effect of that level is equal to 0 or not. In other words is that level of COI equal to COIem-easy. The ANOVA is testing if there is some difference between the ... 2 The answer to this question holds for all model types. The key distinction is order. Do the categories have an order? If so, and there are many levels, using integers is valid. If there is order and there is a small number of levels, it would be better to use a factor (in R) or a string because the small number of unique values will make it harder for ... 0 Venables and Ripley explain how to do residual diagnostics for a repeated-measures design later in their book (p. 284), in the section on random and mixed effects. The residuals function (or resid) is implemented for the aov results for each stratum: from their example: oats.aov <- aov(Y ~ N + V + Error(B/V), data=oats, qr=T) To get the fitted values ... 5 this looks like that actually the intercepts are compared and not the slopes? Your confusion there relates to the fact that you must be very careful to be clear about which intercepts and slopes you mean (intercept of what? slope of what?). The role of a coefficient of a 0-1 dummy in a regression can be thought of both as a slope and as a difference ... 1 As far as I know (but please correct me if I'm wrong cause I'm not sure), the Kruskal-Wallis test is constructed in order to detect a difference between two distributions having the same shape and the same dispersion, that is, one is obtained by translating the other by a difference$\Delta$, such as: Let's call$(*)\$ this assumption. The KW test tests the ...

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There are differences in the assumptions and the hypotheses that are tested. The ANOVA (and t-test) is explicitly a test of equality of means of values. The Kruskal-Wallis (and Mann-Whitney) can be seen technically as a comparison of the mean ranks. Hence, in terms of original values, the Kruskal-Wallis is more general than a comparison of means: it tests ...

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Kruskal-Wallis is rank based, rather than value-based. This can make a big difference if there are skewed distributions or if there are extreme cases

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Yes there is. The anova is a parametric approach while kruskal.test is a non parametric approach. So kruskal.test does not need any distributional assumption. From practical point of view, when your data is skewed, then anova would not a be good approach to use. Have a look at this question for example.

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