# Tag Info

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These are two different phenomena: $t$-statistic Holding all else constant, if $N$ increases the $t$-value must increase as a simple matter of arithmetic. Consider the fraction in the denominator, $\hat\sigma/\sqrt{n}$, if $n$ gets bigger, then $\sqrt n$ will get bigger as well (albeit more slowly), because the square root is a monotonic ...

5

Categorical variables can be represented several different ways in a regression model. The most common, by far, is reference cell coding. From your description (and my prior), I suspect that is what was used in your case. The standard statistical output will give you two tests. Let's say that A is the reference level, you will have a test of B vs. A, and ...

4

You can't really even compare the two since the Kolmogorov-Smirnov is for a completely specified distribution (so if you're testing normality, you must specify the mean and variance; they can't be estimated from the data*), while the Shapiro-Wilk is for normality, with unspecified mean and variance. * you also can't scale by parameter estimates and test for ...

3

First of all, I think that you should look at the seasonal distributions separately, since the bimodal distribution is likely to be the outcome of two fairly separate processes. The two distributions might be controlled by different mechanisms, so that e.g. winter distributions could be more sensitive to yearly climate. If you want to look at population ...

3

Are these distributions of something over time? Counts, perhaps? (If so then you might need something quite different from the discussions here so far) What you describe doesn't sound like it would be very well picked up as a difference in variance of the distributions. It sounds like you're describing something vaguely like this (ignore the numbers on the ...

3

You can test equality of the mean parameters against the alternative that the mean parameters are unequal with a likelihood ratio test. (However, if the mean parameters do differ and the distribution is exponential, this is a scale shift, not a location shift.) Let's say we parameterize the $i$th observation in the first exponential as having pdf $1/\mu_x ... 3 Well, the short answer is that's what falls out of the math. The long answer would be to do the math$^3$. Instead I'll try to rephrase gung's explanation that these are two different (though related) things. You've collected a sample$X_1...X_n$that is normally distributed with unknown variance$^4$and want to know if its average is different from some ... 2 Using multiple testing correction as advocated by Corone is ok, but it will cost you mountains of power as your p-values will generally be well correlated, even using Hommel correction. There is a solution which is computation demanding but will do much better in term of power. If$p_1, p_2, \dots, p_n$are your p-values, let$p^* = \min (p_1, \dots, p_n)$. ... 2 This sort of thing would usually covered by multiple hypothesis testing, although it isn't quite a typical situation. You are correct in noting that this is different from meta-analysis, in that you are using the same data for multiple tests, but that situation is still covered by multiple-hypothesis testing. What is slightly odd here is that it is almost ... 2 1) I found this pre-print paper by @Michael Lew to clarify many things for me. In terms of calculating a p-value under the null hypothesis, it can be seen as more of a matter of convenience than anything else: the null hypothesis serves as little more than an anchor for the calculation|a landmark in parameter space To P or not to P: on the ... 2 The test statistic is chosen to be a measure of the discordance of the data with the null hypothesis, in some direction of interest (e.g. the difference of sample means between two groups, the correlation between successive observations in time, &c.). The bigger it gets, the more evidence against the null hypothesis. Well, we don't always. Sometimes the ... 1 The basic chi-square statistic for a test of a proportion being from a population with an expected of 19.8 is (O-E)^2/E = 6.52. (We do need to ask you here whether that was a numeric expected of 19.87 of a proportion or a percentage. If it's a percentage then you need to compare 31 to N*E or 36.) The statistic given is the normal approximation to an exact ... 1 I'm not convinced this will work, but maybe it will stimulate someone who knows more R/statistics to provide a better answer. This code assumes you've created a new time variable called "time" which is centered on the time-point dividing periods: attach(data) time0 <- time*I(time>=0) library(nlme) model1 <- lme(outcome ~ time + group*time0, ... 1 This seems to be a somewhat strange design. It does not make much sense in an industrial setting: do you really want to generalize to the population of papers to compare the effect of two very specific pencils? You could not say anything about the pencil brand (unless the pencils of the brand are completely identical, but then the variance would have to ... 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 ... 1 I will denote$\hat \theta$the maximum likelihood estimator, while$\theta^{\left(m+1\right)}$and$\theta^{\left(m\right)}$are any two vectors.$\theta_0$will denote the true value of the parameter vector. I am suppressing the appearance of the data. The (untruncated) 2nd-order Taylor expansion of the log-likelihood viewed as a function of ... 1 Update: I misunderstood what was the nature of the prior information. You have a problem with your experiment, because we can't disentangle the effect of the nationality from the effect of the banner. As far as I understood, banner_past is different from banner A and B. So, you know that, on banner-past, french are slightly more likely to click on the ... 1 As near as I can make out, you have 1000+ patients, one disease which the patients either have or do not have, and about 600 genetic markers, for which you have number of copies of that marker. What you have done with that is to run a polynomial regression of the number of repeats against the disease prevalence. You then have some kind of model (you don't ... 1 Given what you are trying to do, I am not sure a t-test is what you want. I am guessing that advertising for 7 days costs more than advertising for 5. So, let's look at cost per day: Week 1: 184,418/5 = 36,883 per day 202,316/7 = 28,902 per day Week 2: 179,650/5 = 35,930 196,395/7 = 28,056 Then what impresses me is that the difference is ... 1 I would NOT recommend the$R^2$as this measure increases as the number of variables increases. In other words, the$R^2$does not account for overfitting. Among the options you mentioned the adjusted$R^2$would be the best. If you take a look at the formula:$R^2_{adj} = 1 - \frac{(1-R^2)\cdot(n-1)}{n-p-1}$Since the number of parameters$p$is in the ... 1 Suppose that the populations are distributed as$N(\mu_1, \sigma^2)$and$N(\mu_2, \sigma^2)$, and the value is$x$, then $$P(\text{X is from population 1}|X=x) = {P(X=x|\text{X is from population 1})P(\text{X is from population 1}) \over P(X=x)}$$ But $$P(X=x) = \sum_{i=1}^2 P(X=x|\text{X is from population i})$$ Therefore$\$P(\text{X is from population ...

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