New answers tagged

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Quantile regression allows you to test any quantile. Here is an example in R > library(quantreg) > summary(rq(mpg~cyl+disp+hp,c(0.25,0.5,0.75),data=mtcars)) Call: rq(formula = mpg ~ cyl + disp + hp, tau = c(0.25, 0.5, 0.75), data = mtcars) tau: [1] 0.25 Coefficients: coefficients lower bd upper bd (Intercept) 26.53473 26.32579 ...


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You're right, this is definitely a good situation for a mixed model. The random intercepts/slopes you use depends on a few factors, such as the desired interpretation and power. I would start at the simplest (random intercept) model, and add your random effects in terms of importance; because without a huge sample you can run into issues with convergence. So ...


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As I understood, you have two binary independent variables named Group and Condition and one binary dependent variable named Correct_Humorous. You are going to need a glm with the binomial family and logit link: fit <- glm( Correct_Humorous ~ Group + Condition, family = binomial(link = logit), # logistic regression data = data4 ) summary(fit) ...


2

Your problem isn't really whether the t-tests work, the problem is in interpreting the p-values. (That's always a problem, but people don't realize that.) If I do 1000 completely independent t-tests, and 50 are statistically significant, this is likely to have occurred by chance. But what if 100 are significant? Are these paired t-tests? If so you have 6 ...


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First, you should not remove non-significant terms from a model simply based on p-values, that is not their purpose. Second, to get means for each combination of groups (and comparisons) use a linear model.


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It really depends on your scenario, which there's very minimal information on. If you do end up running the ANOVA, make sure you use type III SS; as these tests are not reliant on cell size or order of parameters. I am not quite sure how to get type III for the individual parameters without manual multiplication by contrast (will add example within a day or ...


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You could explain that ANOVA is a decomposition of the data as components that correspond to different groups or variables or sources of variation. An example is $$ {\tiny \begin{pmatrix} 89 & 88 & 97 & 94 \\ 84 & 77 & 92 & 79 \\ 81 & 87 & 87 & 85 \\ ...


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No, it does not mean that BL is significantly different from treatment. Treatment and BL are two different variables, the first is a factor the second is numeric. An ANOVA tells you, that a model with both variables treatment and BL is a significantly better model than a simpler model with just treatment variable (in terms of fit). These two variables ...


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I found David Lane's online book very useful. In a more fundamental way, there's an invited paper in Annals of Statistics by T.P. Speed called "What is Analysis of Variance?". It took me a few attempts, but at the end it was very informative. The essence of the paper is to show that ANOVA is simply a decomposition of variance into a summation of variances ...


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I don't think the arcsine tranformation is appropriate when you're comparing your exploration indexes (proportions) against a hypothetical value. Suppose you have data as follows, which may be somewhat similar to those you describe in your Question. x [1] 0.4592 0.2577 0.1213 0.4931 0.1618 0.4592 [7] 0.2216 0.4176 0.1370 0.2855 0.2973 0.6395 [13] 0.1795 ...


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Yes, that would be appropriate. You can run a two way ANOVA as follows (using the built in mtcars data set as an example). summary(aov(mpg ~ cyl + gear, data = mtcars)) which yields ... Df Sum Sq Mean Sq F value Pr(>F) cyl 1 817.7 817.7 78.29 9.82e-10 *** gear 1 5.4 5.4 0.52 0.477 Residuals 29 ...


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It's good to be concerned about multiple testing, but it's possible to go overboard with it. I'd say this is a case of that. I'd also not recommend doing a test for every single model assumption. Given sufficient data, you will find that many assumptions do not hold according to standard tests. This does not necessarily mean that the statistical procedure ...


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For the first part of your question, reporting p values by themselves provides little useful information. This thread is a good introduction to the limits, even the dangers, of relying on p values. A reader is going to want to know details about the size of the sample on which the results are based, the magnitudes of individual effects, of differences among ...


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Figured it out! I used stack on my data, and then separate (i.e. s = separate(stack(data), "ind", c("IV1", "IV2"). Then I could do the ANOVA by aov(values ~ IV1 * IV2, data = s) Hope this helps someone!


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To simplify the discussion here, suppose you have $k = 3$ groups instead of six groups. If subjects are assigned at random into the three groups, then you would expect the 'Before' scores to be about the same in the three groups. Even so, you might want to test for that as you begin the study. Suppose the numbers of subjects in the three groups are 50, 60, ...


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I think your confusion arises around the point that you consider pre- vs. post-test as the dependent variable which is not wrong but this is how a MANOVA would conceptualize it not how a rmANOVA does it. A rmANOVA would consider knowledge as the one and only dependent variable with 3 predictors: time point (pre vs. post), textbook type, and campus type - ...


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You can easily compute Cohen's $d$ using the three means and ANOVA output. The square-root of the MS-error (or MS-residual depending on which program you used) is the pooled within-groups standard deviation. Thus, simply divide each mean difference by this value.


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One of the difficulties of departing from the original measurement scale to find a suitable test is figuring out how to understand results on an unfamiliar, possibly unintuitive, measurement scale. If you get a suitably small P-value, then you can assert that there are significant differences among groups, but it can be difficult to assess whether these ...


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Your question has many different aspects: Kruskall Wallis test is not for multiple independent variables Note that the Kruskall Wallis H-test is a one-way test (it relates to a single independent variable $X_1$, albeit with multiple levels). This is already a crucial differentiation with the situation that you are looking at (multiple regressors $X_1, X_2, ...


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Kruskal-Wallis takes N > 2 independent samples. With two groups, it reduces to the Mann-Whitney U. So the blog is accurate enough. To my knowledge, there is no analytical non-parametric solution when you begin adding covariates. In this case, I would consider: Ordinal logistic regression. It's an extension of non-parametric models to more predictor ...


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If you do not have any other independent variables (e.g. age controls or something like that) and you only have a limited number of unique treatments (e.g. Condition A, B, C or AB BC etc.) it seems most logical to 'just' compare mean performance per group. If there aren't any other shared regressors, there isn't much reason to complicate matters: a model ...


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First, you should be using a binary regression model (e.g., logistic regression) to get correct inferences. ANOVA is designed for continuous outcomes (not binary), and the standard errors and p-values will only be correct under normally distributed outcomes. You can continue to use ANOVA with binary data as long as you use a robust standard error. There is ...


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None of the methods you mention are used for feature selection. You have high-dimensional data (where $p>n$), so regular methods go out the window. Given your data, and the fact that you want to keep your variables intact and reduce the dimensionality, then something like LASSO is more appropriate, or a Random Forest.


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You can look at coefficient estimates and choose the two categories which have the largest negative coefficients


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I think since the models are nested, your comparison makes sense. Resid. Dev. is the residual deviance, which is given in comparison to the Null model. The lower number of your Model2 indicates that it was a better fit. If you are looking for the likelihoods, you do have this output under LR stat, it is the results from the likelihood ratio test.


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Whether you view ANOVA as testing hypotheses as parameters or view ANOVA as comparing models, you perform the same exact test. If you understand how to use ANOVA to test differences between means, then you already understand how to use ANOVA to compare models. When you test $H_{0}$ : there is no difference in means, you're comparing a simple model (whereby ...


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I think this is a good question but needs some fleshing out in terms of whether order matters. If you don't care about the amounts of time or order then, yes, an RM ANOVA will work, but only if your data meets the assumptions of the ANOVA and garners relatively normally distributed residuals. The benefit is that the RM ANOVA only assumes that the repeated ...


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The F test is the test that you use to understand whether a model as whole is statistically significant in predicting y, and/or (in case the model as a whole is significant) to test whether a subset of the model variables may be not significant. For a simple description check this . Remember also that there are criteria like the R2 adjusted for the number of ...


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ANOVA and correlation in this context do completely different things. ANOVA tests whether there is a difference in the mean scores on the three tests. Correlation tests whether there is a linear relationship between pairs of tests.


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In some situations, researchers know in advance (i.e., when they design their study) what groups they would like to compare with respect to the mean value of an outcome variable. In that case, they would not perform an omnibus ANOVA F-test but rather focus directly on performing the desired a priori group comparisons. In other situations, researchers will ...


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Why did you conduct this experiment on 15 different days, rather than just on day 1 and again on day 15? Are you interested in how the intervention plays out over time? (Is most of the change early on and then steady? Is there not much change at the start and then the importance of intervention becomes clear? Do some rats change immediately, while others ...


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I think both approaches are not recommendable. Let's first look at your second approach: Here, you violate an extremely important assumption of ANOVA, namely independence of observations in the rows. The replicates from the same rats are necessarily correlated because they stem from the same rats. The consequences of this violation are dramatic. The ...


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ANOVA doesn't have a minimum sample (or category) size requirement, as long as it can be estimated. The only difference between an ANOVA and Kruskal-Wallis is that KW is the non-parametric version of ANOVA, used when the assumptions of ANOVA do not hold. The reason why someone could suggest using a KW instead of an ANOVA in such cases is if you cannot ...


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Yes, it's fine to do an ANOVA. A Kruskal-Wallis test is basically an ANOVA on the ranks instead of the values themselves. If you do the ANOVA and it turns out that the residual distributions are a problem (and cannot be dealt with via a transformation or generalized linear model), you can do a Kruskal-Wallis instead. I would not do a Bonferroni, but I also ...


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If you want to control the type 1 error rate across the three null hypotheses, then an adjustment is necessary in the scenario you describe.


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With repeated tests the probability of committing a type 1 error increases. This is because, by definition, we reject a null hypothesis if there is less than a 5% chance of obtaining that result assuming the null hypothesis were true. By making 3 comparisons to a control group, we have 3 chances to obtain an unusual result. For this reason, we perform ANOVA ...


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If I am understanding this correctly, the goal is to compare the degree of knowledge of different diseases. Rather than ANOVA, I would use a count regression model, as each disease can only get integer values from 0 to 10 - with, hopefully, no 0s. You could try Poisson regression to start, but negative binomial might be needed. The "check all that apply" ...


1

Analysis of variance (ANOVA) is really a misnomer. ANOVA is really fitting a linear model with the group as a categorical variable. Hence the benefits of ANOVA are the things you get from a linear model, which is more than just the variances and means of each group (the latter is descriptive, not a “model”. Specifically, as mentioned in the answer above, ...


2

ANOVA extends the t-test to more than two groups. Doing so, it asks a slightly different question to the t-test (phrasing it liberally): "Does putting the data into the different factor levels (groups) make sense?" In a predictive setting, this would mean "Does forming groups make my prediction better (particulary reduce prediction error)?" In an explanatory ...


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If I have two dummy independent variables (both binary, with two levels each) in a regression, then how do I construct the the interaction variable? By coding both variables as (1,0) and multiplying them? Yes. When both are 1, then the interaction is 1, else 0. As for the second part of your post, there is a lot of unpack there. ...the main effects ...


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A significant effect in an ANOVA analysis is interpreted as: At least one level of the categorical variable has a mean, which is significantly different from at least one other level. It tells you nothing regarding which group is different from any other group. When a categorical by categorical interaction is added to the model the groups are broken down ...


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