# Tag Info

2

Would someone please tell me what an F test is and what is shows? The term "F test" may be any test whose sampling distribution under the null hypothesis has an F-distribution. There are several quite distinct tests. Here are a couple of the more common ones: (i) an F-test for equality of means of multiple (more than two) groups (also called ANOVA, ...

1

This sounds to me like a standard problem in faint disguise. Observed frequencies are total frequency 182, achieved 31, so not achieved 151; expected frequencies are achieved 19.87179, so not achieved is 182 $-$ 19.87179. Your chi-square statistic must be calculated from both pairs of observed and expected, with 1 d.f. I get Pearson chi-square statistic ...

0

Since your sample size is small, I think you can perform a wilcoxon rank test. For example, to test the laptop performance: > y<-c(2.3,4.1, 5.6, 2.3, 10.5, 3.4, 15.2, 4.6, 11.3, 3.1) > laptop<-c(1, 0, 1, 0, 1, 0, 1, 0, 1, 0) > wilcox.test(y, laptop) Wilcoxon rank sum test with continuity correction data: y and laptop W = 100, ...

0

You could try: 1. Subtract mean 2. Ridge regression 3. principal component methods 4. some other regression method for handling highly co-linear data.

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 ...

-1

There is no need to include indicator variables for each of the categories. Let's say category A is coming out significant. Your results are suggesting that you consider collapsing the categories into "category A" and "all other categories". Of course, you should perform an F-test for nested model vs. full model to check if removing indicator variables for ...

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 ...

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

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

You can find the formula for estimating them (a cubic regression in $n$, the sample size) in this paper. Dixon's Q is a small sample test: as discussed here and here there are few reasons to use the Q-test and even less when the sample size is that large. [1]Rorabacher, D.B. (1991) "Statistical Treatment for Rejection of Deviant Values: Critical Values of ...

0

Assuming a large sample, I suggest using cross-validation. Randomly split your data set in two (or more) sub-sets Build your model on one sub-set of the data using the Bonferroni correction you suggested. Use another sub-set of the data to test your explorative hypotheses built on the first set of the data. Now use a new Bnonferroni correction, e.g. by ...

0

Welcome to the site. This means that, after controlling for the effects of the big exhibit, the significance of the ad is lessened. You should also look at the effect sizes (parameter estimates). It also implies that there is some relationship between the ad and the exhibit. In your case, this makes perfect sense: Quite likely, different sorts of ads are ...

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 ...

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 ...

0

I think your confusion lies in framing your problem incorrectly. Think less in terms of your raw data and more in terms of the entities that characterize your study. You have two populations. These populations are defined by some features. You want to know if these populations are different. Let's consider first a simpler case where each population is ...

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 ...

0

1) If you haven't done it already your cost data should be expressed in values of some base year common for all 218 companies, using some inflation/deflation index 9it doesn't matter which year). This is the least you can do to deal with the fact that data come from different years. 2) Considering the percentage change, it is a sensible thing to do to ...

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

For a simple test of the difference between populations I would consider using the mean for each individual, and a Mann-Whitney U or t-test to test for a difference between the 15 individuals in each population (since you are only comparing a single variable). You could also do glm where you include 'individual' as a random effect, and test for the fixed ...

0

According to the ACF graphs, it is obviously that the fit 1 is better since the correlation coefficient at lag k(k>1) drops sharply, and close to 0.

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Under certain regularity conditions, the maximum likelihood estimates follow asymptotically a normal distribution with mean the true parameter values and covariance matrix the inverse of the Fisher information matrix also evaluated at the true parameter values. The Delta method is typically used to derive standard errors for a nonlinear function of the MLEs ...

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 ...

0

It sounds like you should be thinking about likelihoods rather than significance. The likelihood function for each proportion shows the strength of evidence in the data for different true values of the binomial proportion. See my answer to this similar question for the easy calculation: How to combine likelihoods from multiple binomial trials. Can we also ...

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 ...

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 $1/\mu_x ... 8 These are two different phenomena:$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$will get bigger as well (albeit more slowly), because the square root is a monotonic ... 2 Using multiple testing correction as advocated by Corone is ok, but it will cost you mountains of power as your p-values will generally be well correlated, even using Hommel correction. There is a solution which is computation demanding but will do much better in term of power. If$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 ...

0

Some idea. Hopefully you'll get a few responses and all together there will be something useful. (1) I usually use the term "main effects" to refer to a multivariate regression model before I've addressed interactions, or that part of a multivariate model that includes the variables but not the interactions. Thus in y ~ x + z + x*z, x and z are the main ...

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

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 ...

0

(1) You can't use lowess as it is a data exploration tool. The degree of flexibility of the function is set by the user. So if you play around with the function you'll see that you get dramatically different results depending on the bandwidth you select. (in R this is "f" the smoother span) (2) There are a number of smoothing functions that derive the ...

0

Welcome to the site. You can say that smaller chromosomes get more Y-event than larger ones based on this graph; if you want to say that they get significantly more then you need some sort of statistical test. If you are using R then the rms package offers tests of splines and loess fits. If you are using SAS there is PROC LOESS. Several interesting ...

0

However, the drug dose is not constant, as it needed to be reduced at later time points, which caused a convergence in the treatment and control by the end First, this is an assumption you are making. I would recommend treating the control as 0 dose and fitting a dose response curve at each time point. Then we could attempt to determine whether the ...

0

"Clearly, there is some difference between the groups." - Never ever do such statement before you calculate the actual probability of such event. Things can be extremely deceiving sometimes (and this is one of those cases). Yes, you do need the actual number. Look at your first example - one out of 35 and 3 out of 35 is pretty much random (one sick person ...

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