# Tag Info

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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 ...

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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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Transformation that will change the shape leaves you no longer comparing means. If you really want to compare means you may want to avoid transform (there can be some particular exceptions where, at least with some accompanying assumptions, you can compute or approximate the means on the original scale as well). If you don't need an estimate of the ...

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You are correct in assuming that you can't (shouldn't, really) analyse the data with the controls having zero variance. It sounds like you should consider using a two-way ANOVA on the raw data with the within day variance accounted for in the manner of a paired test. I wrote about the approach in this paper that is intended for pharmacologists with little ...

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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 ...

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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 ...

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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 ...

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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 ...

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(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. ...

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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 ...

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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 ...

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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 ...

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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 ...

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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 ...

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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 ...

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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 ...

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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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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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This is a more a partial practical answer, but it works for me to do some exercises before getting deeply into theory. This ats.ucla.edu link is a reference that might help beggining to understand about multinomial logistic regression (as pointed out by Bill), in a more practical way. It presents reproducible code to understand function multinom from ...

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