# Tag Info

30

Hypothesis testing versus parameter estimation Typically, hypotheses are framed in a binary way. I'll put directional hypotheses to one side, as they don't change the issue much. It is common, at least in psychology, to talk about hypotheses such as: the difference between group means is or is not zero; the correlation is or is not zero; the regression ...

10

Just to add to the existing answers (which are great, by the way). It is important to be aware that statistical significance is a function of sample size. When you get more and more data, you can find statistically significant differences wherever you look. When the amount of data is huge, even the tiniest effects can lead to statistical significance. This ...

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 ...

6

If there was a reasonable basis for suspecting your hypothesis might be true before you ran your study; and you ran a good study (e.g., you didn't induce any confounds); and your results were consistent with your hypothesis and statistically significant; then I think you are fine, as far as that goes. However, you shouldn't think that significance is all ...

6

Before reporting this and this and this and this, start by formulating what do you want to learn from you experimental data. The main problem with usual hypothesis tests (these tests we learn at school...) is not the binarity: the main problem is that these are tests for hypotheses which are not hypotheses of interest. See slide 13 here (download the pdf to ...

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 ...

3

I ran a survey on pricing and asked respondents to choose between option A or Option B. 61% chose option A and 39% chose Option B. The sample size is 90. How do I determine if the % selection Option A is statistically significant? What test do I run and how? Note that with selection among options, the proportions selecting the options are not ...

3

The proper solution depends on the nature of your dependent variable: donation amount. If this takes on a small number of integer values (in number of dollars or Euros or whatever) and those integers are all small (or adjacent) then some sort of count model is appropriate (as @Tomas notes). This could be Poisson or negative binomial regression, or perhaps ...

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 ... 2 Amount (donation) is a count so first thing that comes to my mind is a Poisson regression. This actually uses the log link function. The fact that the relationship is no longer significant does not necessarily mean that the model is less accurate. This is also related to the difference between accuracy and precision. I cannot say more until I know more ... 2 It is certainly possible to have a very large sample and a non-significant result. Perhaps this is simplest to demonstrate with categorical data. Suppose you have data on one condition on two groups of data, e.g. Men and women and whether last name ends in a vowel or consonant. Suppose you find, among 1,000,000 people, that exactly equal proportions of men ... 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 If you have a y and a y2, and you want to form an interaction, you need to add both x:y and x:y2 (see here, but moreso the links therein). You may also be interested in reading this great answer about interactions and curvilinear terms: Either quadratic or interaction term is significant in isolation, but neither are together. To test the interaction, ... 1 I would suggest to not use the word "significant" alone. Speak of "statistical significance" because this is how the term was historically settled, and speak of substantive "importance", like for example "economic importance", (this will depend on context) to talk about whether an effect or an association that appears to exist (i.e. given that it is ... 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$-$19.87179. Your chi-square statistic must be calculated from both pairs of observed and expected, with 1 d.f. I get Pearson chi-square statistic ... 1 The basic chi-square statistic for a test of a proportion being from a population with an expected of 19.8 is (O-E)^2/E = 6.52. (We do need to ask you here whether that was a numeric expected of 19.87 of a proportion or a percentage. If it's a percentage then you need to compare 31 to N*E or 36.) The statistic given is the normal approximation to an exact ... 1 I'm not convinced this will work, but maybe it will stimulate someone who knows more R/statistics to provide a better answer. This code assumes you've created a new time variable called "time" which is centered on the time-point dividing periods: attach(data) time0 <- time*I(time>=0) library(nlme) model1 <- lme(outcome ~ time + group*time0, ... 1 You can find the formula for estimating them (a cubic regression in$n$, the sample size) in this paper. Dixon's Q is a small sample test: as discussed here and here there are few reasons to use the Q-test and even less when the sample size is that large. [1]Rorabacher, D.B. (1991) "Statistical Treatment for Rejection of Deviant Values: Critical Values of ... 1 This seems to be a somewhat strange design. It does not make much sense in an industrial setting: do you really want to generalize to the population of papers to compare the effect of two very specific pencils? You could not say anything about the pencil brand (unless the pencils of the brand are completely identical, but then the variance would have to ... 1 Caveat emptor: I am NOT a biostatistician...but I play one on TV ;-) In all seriousness, I have a strong statistics background, but it is not medically oriented. However, your question was simply stated, so my suggestions utilize a general approach, not one that is domain specific to biomedical research. First, you have a very small sample size. I hope your ... 1 Firstly, this is a very small sample size (4 observations), hopefully you have more, or at least have the same four observations for many different patients. If not, it will be difficult to find a model. Generally it is good to have a sample size greater than 100, or at a bare minimum, 20. Secondly, the survey (6 questions) is also small. Does the patient ... 1 I will denote$\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

I agree with what others have said -- namely that "variance" is probably the wrong word to use (seeing as the function you are considering isn't a probability distribution but a time-series). I think you may want to approach this problem from a different perspective -- just fit the two time series with LOWESS curves. You can calculate 95% confidence ...

Only top voted, non community-wiki answers of a minimum length are eligible