# Tag Info

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If you roll a die, in order to test whether all six faces are equally likely, then the number of rolls $n$ required for a useful test depends on the amount of unfairness we want to be reasonably sure to detect. A chi-squared test of goodness-of-fit is often used to test whether a die is fair. Suppose a die is biased in favor of 1's and against 6's so that ...

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You might try a chi-square test of homogeneity, with the null hypothesis that the distribution of sentences is the same for the A's and the B's. Just 2 statements in R. Thanks for the cute example; I'm swiping it (attributed of course) for my introductory stats students.

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I see this as a multi step process to show significance. Perform exploratory analysis. Is the data is normal, is the differences between plots look normal, are there trends in yield per year, etc. Perform a matched pair t-test between plot 1 and plot 2 for the first six years. This test will show that both plots are similar (or not) or at least have a ...

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I'm assuming that you are asking whether there is a significant difference between plot 1 and plot 2 (with and without fertilizer). In order to solve that you need at least the variances of your yields of both plots. However: Which test to apply depends on several assumptions. So in general (if you have the raw data) you should proceed as follows: Test ...

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You did not know a priori that the specific analysis would render insufficient numbers of cases, or else you would have excluded the tests before you ever collected the data. Alternately, consider if you did the experiment again, you can't be certain that that specific test would meet the criteria for exclusion again. In some large dimensional analyses I ...

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As stated in other answers, $p$-values don't assume that an unlikely event occurred. The $p$-value is a probability that is conditioned only on the null hypothesis being true, $$P(\text{Obtaining data as or more extreme than that observed} \, | \, H_0)$$ and not on anything about the observed data. Some people's logic concerning $p$-values, though, can be ...

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Answering without equations: p value is a measure of surprise. Given that the null hypothesis is true (Design of the experiment) what is the chances that you stumble upon a value at least this extreme in your data. You compute p on the test data. You never can include the actual data. If you can get the actual data there is no need of hypothesis testing ...

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If your data is $\mathcal{D}$, and your hypothesis $H_0$ then the p-value is $p = \mathbb{P}(\mathcal{D}\mid H_0)$. The $p$ value tells you the following: If $H_0$ is true, how likely is the data I'm currently observing ? So if $p$ is very low, it only means the data cannot easily happen in a world in which $H_0$ is true. This does not mean $H_0$ is wrong:...

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I strongly suspect that this particular part of the table, explicitly based on a multi-predictor logistic regression model stratified by sex, has an error in the p-value entry. The odds ratios are presented with respect to a reference category for each of the predictors, so that the p-values should represent the p-value corresponding to a null hypothesis of ...

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You didn't specify how you did the sampling, but if simple random sampling, the number of M in a sample of size $n$ from a population with equal number of M, F would be $\mathcal{Binomial}(n, p=0.5)$. So you can do a test based on the binomial distribution. That is probably what the paper you referenced did. For an example see Exact Binomial Test p-values?, ...

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I haven't yet sit down to work the math, but from the few simulations I run it seems that replacing the number of samples $n$ in the formulas with the effective number of samples $n_\text{eff}$ yields very good approximations. $n_\text{eff} = \exp(H)$, where $H=-\sum_{i=1}^n w_i \ln w_i$ is the entropy of weights (normalized to $\sum_{i=1}^n w_i = 1$). For ...

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In short: ask a machine. Say your t-statistic is 2.14 on 39 degrees of freedom. The table probably gives you t for conventional values of alpha (here, three of them). They're called 'p' in R's quantile function for the t-distribution, qt alphas <- c(0.001, 0.01, 0.05) qt(p = alphas, df = 39, lower.tail = FALSE) [1] 3.312788 2.425841 1.684875 whereas you ...

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It is a bit difficult to see what they mean by at least as much in favour of $H_1$ One could view this definition in terms of the likelihood ratio test which is (for simplicity we use simple hypotheses): $$P \left ( \frac{\mathcal{L}(H_1|X)}{\mathcal{L}(H_0|X)} \geq \frac{\mathcal{L}(H_1|x_{observed})}{\mathcal{L}(H_0|x_{observed})} \right)$$ $p$-value is ...

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The rejection region is fixed beforehand. If the null hypothesis is true then some $\alpha \%$ of the observations will be in the region. The p-value is not the same as this $\alpha \%$. The p-value is computed for each separate observation, and can be different for two observations that both fall inside the rejection region. The p-value indicates how ...

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I think this will be best understood with an example. Let us solve a hypothesis test for the mean height of people in a country. We have the information about the heights of a sample of people in that country. First, we define our null and alternative hypthesis: $H_0: \mu \geq a$ $H_1: \mu < a$ And (let me change your notation) we have our test ...

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The p-value is used to express the outcome in a test of a model and it's parameters (to test a hypothesis). Typically it relates to some statistic that measures a discrepancy (e.g. distance from the expected mean). $P(T> t_{observed}|H_0)$ The probability that an observation of the statistic $T$ given the null hypothesis $H_0$ is larger than the observed ...

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The key point here is that the $\chi^2$ test is homogeneous. Ok, the p-value is overemphasized in research, but here the core idea of the test is "wrong". Multiplying your table by $\lambda$ (ideally, taking $\lambda$ identical samples) you'll see your statistic grow by a $\lambda$ factor. Try to conduct a $\chi^2$ on a (pseudo)random matrix with &...

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