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If you tried to calculate the average distance data is away from the population mean $\mu$ you will find that the positive differences cancel with the negative ones. To be specific $\sum (x_i-\mu)=0$ This shouldn't be that surprising because that's why one reason why we liked the mean in the first place. The squaring first essentially turns all the ...


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Well, you have largely solved it. The length of an interval $[a, b]$ is just $b-a$. So if the 95% CI for $\mu$ is $$ \left[ \bar x - 1.96 \frac{\sigma}{\sqrt{100}}, \bar x + 1.96 \frac{\sigma}{\sqrt{100}} \right] $$ the length is just $$ 2 \cdot 1.96 \frac{\sigma}{\sqrt{100}} $$ Now for b the length of the 99% CI is going to be $$ 2 \cdot 2.58 ...


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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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I just found this wikipedia page discussing data of equal significance vs weighted data. The correct way to calculate the biased weighted estimator of variance is:


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Bayes Theorem can help you out of this dilemma, but we do not even have to go that far to understand what is going on. You can calculate the standard deviation for any two datasets and compare them. But it depends on what you want to know and what your assumptions are about the data. For example if you want to know what you can measure more exactly - the ...


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Your problem is that you are subtracting the full standard deviation from the mean. When you do that, it means that you are multiplying the standard deviation by 2, since you are subtracting $1*std(x)$ and adding $1*std(x)$. Instead, add/subtract $\frac{1}{2}*std(x)$. Thus, your new normals are: $$lowerNormal=26.82−\frac{41.16}{2}=6.24$$ ...


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At the root, the issue here seems to be whether to use the z-statistic or the t-statistic in finding a confidence interval for the population mean $\mu$ or in testing a hypothesis about $\mu.$ Suppose $X_1, X_2, \dots, X_n$ is a random sample from a normal population of which both the mean $\mu$ and the standard deviation $\sigma$ are unknown. We wish to ...


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Neither of the two methods of estimating the population standard deviation from the sample produces an unbiased estimate, though the $\frac{1}{n-1}$ method does produce an unbiased estimate of the variance. If you compare the two estimates of the variance $$s_s^2 = \frac{\sum_i^n (x_i - \bar{x})^2}{n-1}$$ with $$s_p^2 = \frac{\sum_i^n (x_i - ...


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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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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}} = ...


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What I've got isn't pretty, but here goes (note: I use $\varphi(\cdot)$ and $\Phi(\cdot)$ to denote the standard normal density and CDF, resp.): First, let's put everything in terms of standard normals: $y_1=\frac{x_1-\mu_1}{\sigma_1},y_2=\frac{x_2-\mu_2}{\sigma_2},C_1=\frac{C-\mu_1}{\sigma_1},C_2=\frac{C-\mu_2}{\sigma_2}$ The first key is drawing proper ...


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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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For $1 \leq i \leq 200$, let $x_i$ be the grade of the $i$th 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 ...


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If all the $x_i$ satisfy $0 \le x_i \le 2m$ where $m$ is the mean, then $|x_i-m| \le m$ so that the variance is $\begin{array}\\ v &=\frac1{n}\sum (x_i-m)^2\\ &\le\frac1{n}\sum m^2\\ &=m^2 \end{array} $ so the standard deviation $\sigma \le \sqrt{m^2} = m $.


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For a non-negative random variable over (here) $[0,100]$, you can write $$\mathop{Var}[X] = \mathbb{E}[X^2]-\mathbb{E}[X]^2 \leq \max_{x\in[0,100]} x\cdot \mathbb{E}[X]-\mathbb{E}[X]^2 = 100\cdot 50 - 50^2 = 2500,$$ so the standard deviation is at most $\sqrt{2500} = 50$. Here, we used the fact that $X$ is non-negative to write $$ \mathbb{E}[X^2] \leq ...


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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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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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you can use ANOVA for two categories. ANOVA is generally used when we have more than two categories and we want to know if there is a difference between (or among) them. If there are only 2 categories, the usually a means test like the t-test is used. sometimes there are only two categories, but one needs to know if the variance between the two groups is ...


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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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Look carefully at the last sentence in the quote: in it 'standard deviation' refers to that of the sample mean. Thus one is essentially looking at all possible samples of 200 students, given that the population's standard deviation is 5. wythagoras' answer provides the formula for the sample mean's standard deviation.


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First, the standard deviation is not the average distance to the mean, that is always zero. It is however, a value to measure how far the points are from the mean or not. Assuming the values are normally distributed, we know that 68% of the values are between $\mu-\sigma$ and $\mu+\sigma$, for example. Suppose we weigh potatoes with average weight 100 g ...


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Sample vs. population standard deviation. I'm not exactly sure what you mean by sample vs. population SD. It is unusual to know the actual population standard deviation $\sigma,$ but if you do know it, then by all means use it. Why would you use the sample SD to estimate something you already know. (If you knew the population SD, you would use a normally ...



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