# Tag Info

7

Quick answer The reason is because, assuming the data are i.i.d. and $X_i\sim N(\mu,\sigma^2)$, and defining \begin{eqnarray*} \bar{X}&=&\sum^N \frac{X_i}{N}\\ S^2 &=& \sum^{N} \frac{(\bar{X}-X_i)^2}{N-1} \end{eqnarray*} when forming confidence intervals, the sampling distribution associated with the sample variance ($S^2$, remember, a ...

4

As already pointed out in my comment referring to the original question, your preferred null hypothesis "color distribution in Urn 1 is equal to color distribution in all Urns combined" is equivalent to the null hypothesis "color distribution in Urn 1 is equal to color distribution in Urn 2-7". The former recycles observations in Urn 1, destroying ...

4

Unless your observations have a cost-benefit tradeoff of some kind (e.g. paying subjects) then there's not really any such thing as too many. More observations give better parameter estimates. The test you used handled 10,000,000 observations just fine. Your "problem" isn't a problem at all. The estimate of a parameter becomes very good when N is very ...

4

Yes, if you collect more data, eventually any effect size will become significant. This isn't cheating, this is part of the problem of using p-values instead of focusing on effect sizes. Also, with chisquare, there really is no dependent and independent variable - you may want to consider a logistic regression.

2

The chisquare is a hypothesis test for differences from independence in the counts in your table. If you want to test that you're probably not doing anything wrong. You can produce a table of contribution to chi-square or a table of Pearson residuals which help to identify which parts of the table contribute most to the differences. However, it sounds ...

2

You correctly performed a $\chi^2$-test of independence, so the only problem is in the formulation of its hypotheses and the interpretation of the test result: The $\chi^2$-test of independence tests the null hypothesis "The two color distributions are equal" versus the working hypothesis of any difference. The p value is smaller than the prespecified level ...

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It is not actually true that the power of an approximate test, or, more specifically, an approximate $\chi^2$ test, cannot be much. One can easily construct cases with very high power and not very large sample sizes, for example the following, where $N=30$ observations are drawn from a $\text{Binomial}(5,0.75)$ distribution and compared to a null hypothesis ...

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 ...

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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

You would only ignore the 0's if there is some reason (not a statistical one) to do so; but including it would only change the degrees of freedom since (0-0) is, of course, 0. However, I am not sure you want chi-square here at all. It would depend on why you expected only AA genotype. If you do want chi-square, it would be \$\frac{(2-0)^2}{0} + ...

1

Both the tests for homogeneity & for independence test association in contingency tables: the test of homogeneity is for when you have one set of marginal totals fixed; the test of independence for when you have only the total sample size fixed. The exact distribution of Pearson's chi-squared statistic will be different in the two cases (unless you ...

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These sound like paired data, since you've accessed different language versions of the same sites. Paired sites mean that the presence or absence of some item in one language version of one site is more closely related to presence or absence of that same item in the same version of the site but for a different language. To account for data of that structure, ...

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I don't think you want chi-square at all. I think you want some form of count regression (possibly Poisson or negative binomial). You have a model Number of plants ~ environment + species (maybe + interaction) where both of the independent variables are categorical. The above model would allow you to see overall effects of environment and species as well ...

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I think you probably want a regression model. Surely mortality is a dependent variable here, and it seems like you want to examine the effect of race on it. What sort of regression model depends on how mortality is recorded. It could be a (regular) logistic regression, an ordered logistic or possibly a survival model. As for a simple display, it seems ...

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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 ...

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Let me emphasize that I'm a newcomer to clustering, and am not sure of the right answer in this case. That said, my first thought would be to fit a logistic model with random effects for family. Here is a tutorial from UCLA statistics on estimating these models in R.

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