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Hint: if $X\sim \Gamma (a,\lambda)$ so $E(X^{k})=\frac{\Gamma(a+k)}{\Gamma(a)}\lambda^k$ so $X\sim \chi^{(2)}_{(n)}=\Gamma (\frac{n}{2},2)$ so $E(X^{k})=\frac{\Gamma(\frac{n}{2}+k)}{\Gamma(\frac{n}{2})}2^k$ $$F_{(n-1,m-1)}=\frac{\frac{\chi^{(2)}_{(n-1)}}{n-1}}{\frac{\chi^{(2)}_{(m-1)}}{m-1}}$$ $$E(F_{(n-1,m-1)})=E\left( \frac{\frac{\chi^{(2)}_{(n-1)}}{n-...


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The effect you are observing --- that the p value decreases with a larger sample size for the same effect size --- is a feature of p values. That's just the way p values work. In contrast, a proper effect size statistic will not increase as the sample size increases for the same effect size. It's a good practice to report effect size statistics, or some ...


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The relationship between the Wald test and the Pearson $\chi^2$ is a particular example of the relationship between Wald tests and score tests. The Wald test statistic for the difference between a value of a parameter $\hat \theta$ estimated from a data sample and a null-hypothesis value $\theta_0$ is: $$W = \frac{ ( \widehat{ \theta}-\theta_0 )^2 }{\...


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The most general answer: It depends. The crucial question is, what exactly do you mean by "voted similar". In statistics, often people tend to just use some given measure and then assume that it fits what they are testing. But it is important to first clarify what the terms mean. As a somewhat sophisticated example. Assume Party 1 and Party 2 are two types ...


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There is no unique answer to this. It depends on what you mean by "similarity", which one could say is implicitly defined by the measure you choose. Note that the data alone or statistical theory cannot tell you what "similarity" you should be interested in! Personally, for the given situation, I would not use Spearman or Kendall correlation, because this ...


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High $p$-values are no guarantees that there is no association between two variables. The high $p$-value just means that the evidence is not strong enough to indicate an association. In other words, the lack of evidence for a claim is not the same as evidence for the opposite of the claim. Also, it is not unusual for two tests to say differing things about a ...


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Since your research question involves a dependent variable (number of colonies) and an independent variable (temperature) you want regression. But you a) Should include all your plates in one regression and b) Will almost certainly need more data to get useful statistical tests, unless the relationship is very strong and clear. I am not sure what "sets of ...


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I'm no expert on this subject, but there seems to be quite some blog posts and even publications on this subject. I would suggest reading those, but in general it seems that the test might be biased and have low power when using small samples, and when the original distribution is short-tailed.


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The best method to achieve this would be a PCA or factor analysis on the variables before you do the regression. Think about it this way: If $A$ and $X$ are highly correlated, and $B$ and $X$ are also highly correlated, then very likely (but not necessarily) $A$ and $B$ are also correlated. So your variables $A,B,C,\ldots$ don't really capture independent ...


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Probably take a look at this answer, and consider van der Waerden scores before running EFA in order to address skewness. In some respects, skewness can also result in false results due to extreme values in tails of your distributions, which can bias the covariance to the point of claiming there is non-zero covariance, when in fact after skew-zeroing the ...


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It is not true that $\Pr(X|A, B) = \Pr(X|A)$ implies $\Pr(X|A, B, C) = \Pr(X|A, C)$: Given, $A$, a variable might predict $X$ together with another variable $C$ but not on it's own. And the same could hold for $C$. Here's a stark example illustrating the point. Let $B$ be a coinflip. $C$ is another, independent, one. $X$ is a light. It turns on if $B$ and $...


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What you apparently want to do is to start by evaluating the relationship between the 3 classes and each of your features individually. For each continuous feature you are proposing a one-way ANOVA of that feature against the known classification (3 classes). ANOVA is not appropriate for a similar evaluation of the binary features, as interpretation of ANOVA ...


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It sounds like you are checking each predictor separately against the binary outcome. That's not a good idea with logistic regression, as is has an inherent omitted-variable bias. Omitting from a logistic regression any predictor associated with outcome will bias the coefficients for the included predictors. Unlike in linear regression, the omitted predictor ...


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As you expected, the chi-squared test seems fit for your problem. The null hypothesis would be $$ H_0: p_1=p_2=p_3 $$ where $p_i$ is the proportion of people in group $i$ who decided to buy an item. The alternative would be not $H_0$.


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From comments: These are not counts, but rather measurements of physical quantities, The $\sum_i (O_i-E_i)^2/E_i$ formula is specifically for counts (in particular it relies on the way the variance works in the Poisson or the way variance and covariance work in the multinomial, which simplifies back to the Poisson formula). Don't use that statistic if ...


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The phi you are calculating is an effect size statistic. If you were to take a perfect subsample of your whole sample, the phi statistic would come out the same. So there is no benefit to doing such a sub-sampling when considering this statistic. That being said, it make be helpful to look at the confidence interval for phi. I don't think using the sub-...


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Remember that a p-value does not tell you how different the truth is from what the null hypothesis asserts. What you’re looking for is called effect size. You’re allowed to get a statistically significant result and say, “Okay, but it isn’t interesting enough for me to care,” and this does not make you a lazy scientist or negligent statistician. So you’re ...


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