# Tag Info

1

According to the gambler's fallacy shouldn't it be expected the same probability for the scoring and not necessarily more likely close to 50? No. You trapped in Gambler's fallacy. Assuming large enough number of question asked in the second time, the ratio of their success will still remain the same for high scorers from the first test. Your expectation ...

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A random walk will depart from the starting point increasingly as a measure of distance in time and decreasingly so in proportion to the total distance traveled. There is no controversy therein. For example, gambling on coin tosses will exaggerate the total amount won or lost with time, but the relative amount will tend to 50-50. I wonder how one eliminates ...

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What you describe is a 2 x 2 design, the simplest form of 2-way analysis of variance (ANOVA). You have 2 environmental conditions. In each environmental condition, you have one control experiment and one experiment in which you add a compound. An important consideration in this type of design is whether the effect of the compound differs between ...

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I’m going to try to simplify. After flipping 10 heads; Gambler’s fallacy says that the next flip is more likely to be tails. Regression to the mean says the next flip is 50/50, BUT the following series of tosses should go back to an even distribution. In other words: RTTM isn’t making a specific occurrence prediction. Did I get that right?

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Brute-force search can find these parameters easily. Here's my code (R): # Find the parameters of a Weibull distribution with a given mean and sd library(nloptr) objective_mean = 8.4 objective_sd = 3.8 objective = function(weibull_params) { scale = weibull_params[1] shape = weibull_params[2] weibull_mean = scale * gamma(1 + 1/shape) weibull_sd = ...

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It seems to be a self-study question so let me just give you some hints to figure it out by yourself. Mean is sensitive to outliers. MAD is an acronym generally used for median of deviations from median. MAD is a robust measure of variability, so it is insensitive to outliers (see also other questions tagged as mad). You may want to use it if you have clear ...

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Answering my own question a year later after the system is in production. Following @George Savva's suggestions we looked at a number of different smoothed fits. We eventually decided on LOESS (locally estimated scatterplot smoothing). However we only want to smooth when the error for the specific point is high. To accommodate this we took the average value ...

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I've always wondered this myself. My statistics background is limited, but here are the two different thoughts that made the difference clear to me. If you flip a fair coin 20 times and get 18 heads. Does your confidence interval have a 95% chance of containing 10? Obviously not. The probability only works the other way. Second example. You run one ...

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Since $X$ conditional on $Y$ is not truncated, the usual formula for $EX|Y$ holds. Hence, using the law of total expectation, \begin{align} EX&=EEX|Y\\&=E(\mu_x+\rho\frac{\sigma_x}{\sigma_y}(Y-\mu_y))\\&=\mu_x+\rho\frac{\sigma_x}{\sigma_y}(EY-\mu_y). \end{align} Since $Y$ marginally has is univariate truncated normal distribution, $$EY=\mu_y - \... 7 The answer is yes, it is possible for all the sample mean distributions to be multimodal. The idea is to exhibit a distribution of the X_i that has rapidly increasing gaps in its support, so that no matter how many of the X_i you might sum, there eventually will be a gap so large that the sum cannot fill it: that is, there will be places within the gap ... 2 Dave was alluding to this in the comments, but I think it's important to spell this out in black and white: The probability that these results are due to chance is not the p-value. The p-value is the probability of observing these data, or data more extreme, IF (and only if) the null hypothesis is true. The distinction may appear to be subtle but it is ... 1 This type of coefficient of variation is generally only useful if the mean is not 'arbitrary', that is, the data means something very different if the means is artificially raised. An example would be count data, but there are many practical situations where the coefficient of variation could be sensible with continuous data as well. For circular data, ... 1 Say that you want to measure a voltage V(t) that is varying in time and has a certain additive noise e(t) (a random variable), for example,$$ V(t) = \sin(ft) + e(t)  Often one assumes that $e(t)$ has a constant zero mean and constant positive variance. One is interested in $\sin(ft)$, and the variance of $e(t)$ is unknown, how should one proceed? One ...

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In the real world, these types of problems primarily happen in manufacturing. You usually see them when there are strong constraints on the behavior of a variable. For example, the normal distribution assumes that a value can take on any value over $(-\infty,\infty)$ but if you are building cars, it is never going to happen that a tire will be larger than ...

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A practical example: suppose I have a thermometer and want to build a picture of how accurate it is. I test it at a wide variety of different known temperatures, and empirically discover that if the true temperature is $T$ then the temperature that the thermometer displays is approximately normally distributed with mean $T$ and standard deviation 1 degree ...

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Suppose we have a random variable $X\sim N(\mu, \sigma^2)$ where $\mu$ is the mean and unknown but with $\sigma^2=1$ variance. Now to answer your question how this could be possible, we can consider one way to relate the mean to the variance and see if we can use this method to back out what the mean should be. I'll use the fact that \$Var(X)=\mathbb{E}[(X-\...

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I don't think you need any statistical test. You have five values. Three are positive; two are negative. There are fewer negative tests, but their absolute value (on average) is bigger. About half the time, the accuracy goes up; the other half, it goes down. What more is there to say? (As is often the case, the word "significant" obscures clear ...

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Comment continued: In case you want to check your answer against computer printout for a Welch t test in R, here are fictitious data and relevant computer printout. set.seed(2021) x1 = rnorm(30, 50, 7) summary(x1); length(x1); sd(x1) Min. 1st Qu. Median Mean 3rd Qu. Max. 36.54 47.66 51.10 51.06 57.38 62.11 [1] 30 [1] 7.651895 x2 = ...

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