# Tag Info

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Given the restriction that there can not be different prices for different customers of the same country. There are two primary options: Sequential testing - For each country, all customers get the same price but the price systematically varies over time. This is often called pre and post testing. Cluster testing - Partition countries into two relatively ...

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You cannot. You either remove the pair(s) where one value is missing (which might introduce bias - especially in this case where only the high values in df1 are missing their pairs in df2) or you impute them (e.g., by using multiple imputation).

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Mann Whitney U Test (Wilcoxon Rank Sum Test) shall enable us to compare the ratings on System A and system B. This test compares the shape of each population and tells whether the two samples differ. If the shape is same, the null hypothesis is accepted. If the shape of one of distribution is different from the another distribution, the two systems have a ...

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You need to perform some tests to identify the appropriate statistical measure for comparing the two distributions accurately. For each group/system, run a normality test to make sure that you are not dealing with an ultra exotic distribution to which central limit theorem does not apply (very unlikely). Calculate the variance for each group, in order to ...

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Since you want to build a binary classifier based on time-ordered tabular data, I see two possible approaches among others: as you suggest, split your dataset in ordered train-test folds, so you reproduce the "real" situation of having, at each time interval, a historic dataset to train on and a test (and later evaluation) set; you can use the ...

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Do you mean when to compare to a t-distribution as opposed to a standard normal distribution? That is because, in the former case, you have to estimate the standard deviation. Chugging through the theory gives you that you compare to $t$, not standard normal. The details of that should be covered in a statistics textbook. The intuition might not be. The way ...

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You simply cannot use these data points for this test, you have to use only the data points for which both variables have a value. This will make the sample smaller, so it's less likely to result in a significant difference.

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(0.0053726552728412666, 0.9415686658133314, 1), that 0.005 represents the test statistic Yes, 0.94 represents the P-value - Yes. and 1 the degrees of freedom? No. And 1,2,3 and 4 are ordinary serial numbers and not, degree of freedom. At 10% level of confidence, you need at least 2.71(value of Chi-square). You have computed the chi-square = 2.576 and ...

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Yes it does, you resample from the whole population without replacement $N_A$ samples, which are associated to group $A$, and the rest of the samples are associated to group $B$. As I understand, you compute the statistic for every possible permutation of the total population and you measure how the actual statistic compares in the distribution of all the ...

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I recently wrote about this in a blog post. Given this is a rate evaluation metric, you will want to use the z-test. The basic steps are (more details in the blog post) calculate the pooled standard error of your pairwise comparison calculate z-statistic by normalizing the delta or lift by the standard error look up the cdf value of the normalized delta p-...

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