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| visits | member for | 1 year, 11 months |
| seen | May 18 at 2:53 | |
| stats | profile views | 3,717 |
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May 18 |
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Interpreting the regression output from a mixed model when interactions between categorical variables are included @crash, I guess it depends on what you mean by the "overall effect of all predictors". If you have several predictors $(x_1, ..., x_p)$, one way to interpret your question is to suggest comparing the "saturated" model, which allows the most general possible relation between the outcome, $y$, and the predictors: $$E(y|x_1, ..., x_p) = f(x_1,...,x_p)$$ with the model where this expectation does not depend on the predictors at all, i.e. $f(x_1,...,x_p) \equiv c$ for some constant $c$. How to operationalize this depends on various things, e.g. whether your predictors are continuous or categorical. |
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May 18 |
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$r^2 = 32\%$, $r = 0.59$ How does a (pro) statistician formally interpret this correlation? Strong? Weak? A pro statistician may point out that $.59^2 \neq .32$ :-) |
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Apr 30 |
awarded | Nice Answer |
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Apr 29 |
awarded | Enlightened |
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Apr 29 |
awarded | Nice Answer |
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Apr 19 |
awarded | Enlightened |
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Apr 19 |
awarded | Nice Answer |
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Apr 6 |
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How to calculate the percentage of a sample that is delayed However, you could probably learn something by looking at a histogram of the second sample. For example, if the sample is roughly normally distributed, the delay is large relative to the variance, and a non-negligible proportion were subjected to the delay, then you should see a bimodal distribution. The relative heights of the two modes should give you a good idea of what proportion of the points were delayed. To actually estimate that proportion in a principled way, you'd probably need a mixture model. |
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Apr 6 |
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How to calculate the percentage of a sample that is delayed I'm saying that if only some of the people in the second sample's times were delayed, then the two sample difference in means does not reflect the average time delay - to estimate the average delay you will need to know how many times were delayed. Do you have that information? If so, just calculate the percentage like Peter Flom says. If not, I don't think it's possible - the two sample difference in means only estimates $\delta(1-p)$ - the average delay multiplied by the proportion of times that were delayed - you have to fix either $p$ or $\delta$ to estimate the other. |
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Apr 6 |
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How to calculate the percentage of a sample that is delayed The way you've described it, the second sample is contaminated by some delay, the first sample is not and to estimate the average delay (call it $\delta$), you take the difference of the means. If that's the setup, the answer to "Now I'd also like to know how much percentage of the second sample has a delay." is trivial - 100%. If you change it so that some proportion (say, $p$) of the times in the second sample are not delayed, then the two sample comparison of means does not estimate the average delay - it estimates $\delta(1-p)$. |
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Apr 6 |
reviewed | Close R: Assigning variable to quintile on monthly basis |
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Apr 6 |
reviewed | Approve suggested edit on Which statistical technique to use |
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Apr 6 |
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What are my choices for transforming data that are not normally distributed? What is the end goal here? i.e. How exactly do you plan to compare them? |
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Apr 6 |
reviewed | Approve suggested edit on How would you write the model? |
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Apr 5 |
revised |
Questions about the sampling distribution of the sample mean changed generic title |
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Apr 5 |
reviewed | Approve suggested edit on Overdispersion parameter |
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Apr 5 |
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Determining variance of an U.M.V.U.E I'm glad to see you figured out the answer, @SammyJ. If you get a chance, please consider posting your answer below for future reference and so the question comes off of the "unanswered" list. |
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Apr 4 |
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Swapping X and Y in a regression that contains a grouping predictor? Hi @whuber, I thought it was sufficiently vague that it did not qualify as an answer, particularly since the two hypothesis tests often do agree. This is something that interests me and I actually spent some time on this trying to characterize situations where the two are almost guaranteed to disagree. I found some examples but not a characterization and then I got sidetracked. When/if I get a chance to come back to that, I'll post an answer. Maybe in the meantime someone else will post a good answer :) |
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Apr 4 |
reviewed | Approve suggested edit on Clustering with “MCLUST” in R |
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Apr 4 |
reviewed | Approve suggested edit on EM algorithm manually implemented? |
