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There are a number of 'measures of diversity' (see Wikipedia). Perhaps the simplest is called the Simpson Index (often denoted $\lambda$). It is the sum of squares of relative frequencies for the categories. In your case: $$\lambda = \sum_{i=1}^3 r_i^2 = .3^2 + .5^2 + .2^2 = .38.$$ Very roughly speaking, this is the probability that two people chosen at ...

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ANOVA should only be run on homoscedastic data. The first test you reference above is a test for heteroscedasticity. The test should be run prior to the ANOVA, which tests for a difference in means. Since the first test is a prerequisite to the second test, they do not fall in the same category. Prior to running the second ANOVA, you should also test for ...

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Let $X_i$ be the random variable which equals $1$ when box $i$ is filled, and $0$ otherwise. Then the variable you want to know about is $X = \sum_{i = 1}^{n}X_i$. We have: $$E(X) = E(\sum_{i = 1}^{n}X_i) = \sum_{i = 1}^{n}E(X_i)$$ And, since the $X_i$ are independent: $$\sigma^{2} = Var(X) = Var(\sum_{i = 1}^{n}X_i) = \sum_{i = 1}^{n}Var(X_i)$$ Since ...

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The distribution is not a normal distribution, it is a binomial distribution. But for large $N$ the normal distribution is an excellent approximation to the binomial distribution. At any rate, the exact answer for the binomial distribution is that the mean is $\frac{n}{r}$ and the standard deviation is $$\sqrt{npq} = \sqrt{n \frac{1}{r} \frac{r-1}{r}} = ... 1 For 1 \leq i \leq 200, let x_i be the grade of the ith student. Let y_i = x_i - 50. Then$$\sqrt{\frac{\sum{(x_i - 50)}^2}{N}} = \sqrt{\frac{\sum y_i^2}{N}}.$$But since 0 \leq x_i \leq 100, it follows that |y_i| \leq 50 and y_i^2 \leq 2500. Therefore$$ \frac{\sum y_i^2}{N} \leq \frac{N \cdot 2500}{N} = 2500,$$and so ... 1 The confusion seems to be over operations on fractions. In general, consider the two fractions$$\frac AB \quad \mbox{and} \quad \frac{A+D}{B+1}.$$Your question is essentially asking whether the fraction on the right is necessarily greater than the fraction on the left if D > 1. The answer is, no, it is not. For example:$$\frac{100}{10} > ...

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Let me write $$S(x):=\sum_{i=1}^L ( \bar x - x_i) = \sum_{i=1}^L (\sum_{j=1}^K K^{-1}x_j - x_i) = \sum_{i=1}^L (L/K-1)x_i + \sum_{i=L+1}^K K^{-1}x_i.$$ Thus, $S(x)$ is maximal if $x_i$, $i=1\dots L$ are at the lower bound, while $x_i$ are at the uppper bound $i=L+1\dots K$. Hence the maximum is $$S_{max}=L(L/K-1)a + (K-L)/K b = \frac{K-L}{K}(-La + b).$$

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You are computing the standard deviation of the mean, $\sigma_{\bar X}$, not that of the individual samples, $\sigma_X$. When the variables are independent, the variances do add up. So $$\text{var}_{\sum_i Xi}=n\text{var}_X,$$ and dividing by $n^2$ (the variance is quadratic), $$\text{var}_{\bar X}=\frac1n\text{var}_X.$$ Hence taking the square root ...

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The issue that you rised with this question is in the area of robust statistics. In the case of estimating a parameter, it is called robust parameter estimation. There is a good book by Huber. I think this one will help alot. The idea is as follows. When you are estimating a parameter, the regular process first finds the log likelihood ratio of the density ...

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