# Tag Info

1

If you're allowed to take that sample repeatedly, it's basically bootstrapping. Procedure: Draw 100 points Calculate standard deviation Repeat Steps 1 & 2 a lot of times (empirically, I've found 5-10,000 to be enough), keeping track of the results of step 2. Examine the distribution of estimates from Step 2 with whatever tools you'd like -- ...

3

If you want to find out the uncertainty or standard error (SE) in the standard deviation of a chosen sample, then you can simply use $SE(\sigma) = \frac{\sigma}{\sqrt{2N - 2}}$, where $N$ is the number of data points in your sample. Hope that helps!

1

You're probably familiar with the interpretation of the $z$ score as "the number of standard deviations the test statistic is from the mean". This interpretation can be obtained from the formula for the $z$ score. Consider, $$z = \frac{x-\mu}{\sigma},$$ if we multiply $\sigma$ to the left hand side we get, $$z \sigma = x-\mu,$$ here we see that ...

0

I assume each coin toss is independent. Let each of the 3600 outcomes be numbered, so that you have 3600 independent Bernoulli random variables $X_1,\dots, X_{3600}$ (where $X_i$ equals 1 if the $i$-th coin toss is Heads, and $0$ otherwise). In particular, for all $1\leq i\leq 3600$, we have $\mathbb{P}\{X_i=1\} = \frac{1}{2}$ (the coins are fair), and $$X ... 0 In OLS models the covariance matrix of the coefficients is given by \sigma^2 (X'X)^{-1}. So, in your case,$$ X = \begin{pmatrix} 1 & x_1 \\ : & :\\1 & x_n \end{pmatrix} $$hence,$$ (X'X)^{-1} = \frac{1}{n\sum_{i=1}^n(x_i - \bar{x}_n)^2}\begin{pmatrix} \sum_{i=1}^nx^2_i & -\sum_{i=1}^nx_i \\-\sum_{i=1}^nx_i & n \end{pmatrix},  ...

1

The sample coefficient of variation is the sample standard deviation divided by the sample mean. Here is an illustration: Which has greater variability, weights of elephants or weights of ants. In terms of the standard deviation the answer has to be elephants because they weigh more. Dividing by the mean tends to put the two measures of variability on the ...

0

Technically, the standard error is the standard deviation of an estimator. Most commonly, this refers to sample mean $\bar X$ as an estimator of the population mean $\mu.$ So the 'standard error of the mean' is $SD(\bar X) = \sigma/\sqrt{n}.$ If $\sigma$ is unknown, it is estimated as the sample standard deviation $S.$ This means that the '(estimated) ...

-1

If a negative number is not a possible/feasible data point than your lower limit is zero in this particular example. It will stay zero until you get more data that shifts your mean higher.

0

By definition, $M(t)=\int_\Omega e^{tX}dP$. Then, $M^{(n)}(t)=\int_\Omega X^ne^{tX}dP$ (it follows by differentiated $n$ times) Then, $M^{(n)}(0)=\int_\Omega X^ndP$, and the last is precisly $E[X^n]$.

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