# Tag Info

## New answers tagged statistics

0

No. By your definition, $U_1,U_2$ will be both uniform on the unit interval. Then, by Holder's inequality $E[U_1U_2]\leq (E[U_1^p])^{1/p}(E[U_2^{(1-1/p)^{-1}}])^{1-1/p} = \left( \frac{1}{p+1} \right)^{1/p} \left( \frac{1}{\left(1-1/p \right)^{-1} +1} \right)^{1-1/p}$ for all $p>0$. Now, plot the upper bound, and see that there exists a $p$ such that it ...

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The random variables $U_i = \Phi(X_i)$ are $U(0,1)$ uniformly distributed since $\Phi$ is one-to-one and $$P(U_i \leqslant x ) = P(\Phi(X_i) \leqslant x ) = P(X_i \leqslant \Phi^{-1}(x)) = x.$$ We have $E(U_i) = 1/2$ and $$cov(U_1,U_2) = E(U_1U_2) - 1/4 = \int_{-\infty}^\infty\int_{-\infty}^\infty\Phi(x_1)\Phi(x_2)n(x_1,x_2,0.5) \, dx_1\, dx_2 - 1/4,$$ ...

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In order to correctly construct confidence interval, first of all, you need a "pivot" T such that T is a function of all data $X_1,X_2,...,X_n$ T is a function of $\mu$ The distribution of T is know, and it is not a function of $\mu$ For example, $T=\frac{\bar{x}-\mu}{\sigma/\sqrt{n}}$ is a pivot, and $T$~$N(0,1)$ regardless of the value of $\mu$ (if the ...

1

Hints (by no means complete solutions): by definition, $$\boldsymbol{\Sigma}_{Z} = \text{Var}(Z) = \mathbb{E}\left[(Z-\mu_{Z})(Z-\mu_{Z})^{T} \right]\text{.}$$ (This is just the covariance matrix of $Z$. Some people use $\text{Cov}(X)$ instead.) Independence can't be inferred without knowing the joint distribution and the marginal distributions of $Z_1$ ...

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For a simple example: Let $S = \{\mathrm{heads},\mathrm{tails}\}.$ Then: the information "heads has probability $1/3$, tails has probability $2/3$" yields a probability distribution. the information "heads has probability $p \in [0,1/2]$, tails has probability $1-p$" yields a statistical model. You could say that statistical models generalize ...

1

Strictly speaking, a probability distribution is a function (more precisely, a measure) that assigns to each event some real number in $[0,1]$. Whenever $X$ is a random variable, giving its probability distribution is giving the probabilities attached to the values that $X$ can take. For example if $X$ is the number given when you roll a die, and if $P_X$ is ...

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The highlighted points do not tell the whole story. And your point is correct about always using z tests and confidence intervals when the population standard deviation $\sigma$ is known. The distinction between t and z methods (for confidence intervals and tests of hypotheses) is very simple. If the population standard deviation $\sigma$ is known, ...

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is there a formal measure of bitwise entropy that takes into account these factors? You're using the term entropy in reference to some given finite string, whereas entropy is a function defined on probability distributions. One way of reconciling this is to suppose that the relevant probability distributions are the actual ("empirical") distributions of ...

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\begin{align} & \sum_i \Big( (x_i-\bar x)^2 - 2(x_i-\bar x)(\bar x-\mu_0)+(\bar x-\mu_0)^2 \Big) \\ = {} & \left( \sum_i (x_i-\bar x)^2 \right) - 2 \sum_i \Big( (x_i-\bar x)\overbrace{(\bar x - \mu_0)} \Big) + \sum_i \overbrace{(\bar x - \mu_0)^2}. \end{align} Observe that the two expressions under the $\overbrace{\text{overbraces}}$ do not change ...

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You have that $\mathbb{E}(X_1X_2 + X_3 - X_1^2) = \mathbb{E}(X_1X_2) + \mathbb{E}(X_3) + \mathbb{E}(X_1^2) = \mathbb{E}(X_1)\mathbb{E}(x_2) + \mathbb{E}(X_3) + [var(X_1) + (\mathbb{E}(X_1))^2] = 1 + 1 - (1 + 1) = 0$

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The aim is to calculate the 95% confidence interval of $\mu$ “using $\hat{\mu}$”. What is the sampling distribution of $\hat{\mu}$? $\hat{\mu}$ is a sum of normal random variables, as a result, the distribution of $\hat{\mu}$ will also be normal with unknown mean $\mu$ and variance $V(\hat{\mu}) = 1.04$ (as you calculated). Accordingly, and since the ...

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The p-value is somewhere in $[0,0.10)$. It is not necessarily less than $0.05$

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When you go for t statistic, it is assume that you do not know the population standard deviation. To answer your question, the sample standard deviation is calculated based on the below formula $$s = \sqrt{\frac{\sum_{1}^{n}(x_i -\bar x)^2}{n-1}}$$

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Proposition: Let $a,b,c,d$ be positive real numbers such that $ad-bc<0$. Then $$\frac{a}{b}<\frac{a+c}{b+d}<\frac{c}{d}$$ First, notice that the requirement is equivalent to $a/b<c/d$, so we can use either as the supposition in the proposition. $$ad-bc=ad+ac-(ab+bc)<0\Rightarrow\frac{a}{b}<\frac{a+c}{b+d}$$ by dividing both sides by ...

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$\dfrac{\alpha+y}{\alpha+n+\beta}=\dfrac{\alpha+\beta}{\alpha+n+\beta}\cdot\dfrac{\alpha}{\alpha+\beta}+\dfrac{n}{\alpha+n+\beta}\cdot\dfrac{y}{n}$ Take $\gamma=\dfrac{n}{\alpha+n+\beta}\in(0,1)$ then $\dfrac{\alpha+y}{\alpha+n+\beta}=(1-\gamma)\dfrac{\alpha}{\alpha+\beta}+\gamma\dfrac{y}{n}$. It is a convex combination of $\dfrac{\alpha}{\alpha+\beta}$ ...

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Many, many thanks to Michael Hardy for answering my question. The idea is this: let $$\mathbf{y} = \begin{bmatrix} d_1 \\ \vdots \\ d_n \end{bmatrix}$$ and $\boldsymbol{\beta} = [\mu_1 - \mu_2]$. Then our linear model is then $$\mathbf{y} = \mathbf{1}_{n \times 1}\boldsymbol{\beta} + \boldsymbol{\epsilon}$$ where $\mathbf{1}_{n \times 1}$ is the $n$-vector ...

0

It is $\sum (x_i-\overline x)^2$ First you have multiply out the brackets: $\sum (x_i^2- 2\left( x_i\right)\overline x+\overline x ^2)$ Each summand in the brackets gets a sigma sign. The constants $2, \overline x$ and $\overline x^2$ can factored out. $= \sum x_i^2-2\overline x\sum x_i+\overline x^2\sum 1$ $\sum x_i=n\cdot \overline x$ and $\sum 1=n$ ...

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I saw the question and answer from one of the website about the b) from https://www.algebra.com/algebra/homework/Probability-and-statistics/Probability-and-statistics.faq.question.408149.html Note: The probability of x successes in n trials is: In this case p = .05 & q = .95 a) P(exactly 3 of 20 being defective computers) = 1140(.05)^3(.95)17 = ...

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There are two reasons not to do a two-sample t test on paired data. First, while scrambling the observations may obscure some of the dependence inherent in pairing, scrambling does not address the difficulty that you have only one random sample of subjects from the population of interest; not two independent random samples 'treated' in different ways. ...

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Hints, in something approaching plain English. I hope you will learn something by expanding them carefully to give details and use formal notation as in your text: In (a) there is really only one LR test at level 10%, and you have found it. Thus "uniformly" is easily satisfied in (b). However in (c), there are two LR tests at level 5%, having rejection ...

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Setting. In a cross classification with $r$ rows and $c$ columns, the usual null hypothesis for a chi-squared test is that the row categorical variable with $r$ levels is $independent$ of the column categorical variable with $c$ levels. Finding expected cell counts to match 'independence'. The task is to find expected counts $E_{ij}$ for each of the $rc$ ...

1

This is calculated in two parts: Probability of a particular group attending. Average age of that group. For a list of $N$ guests, there are $N \choose k$ ways for $k$ people to attend. For each such group, you need to calculate the average age and the probability of that group attending. Assuming each person attends independently of the others, the ...

1

This implies that sales equal demand if demand is less than the quantity of loaves available, i.e. $Y=X$ when $X\le k-1$ (noting that $k$ is integer valued). But for demand greater than $k$, demand cannot be met, hence sales then equal the total supply of loaves, $k$, i.e. $Y = k$ for $X\ge k$. I don't know how to use this to find a pmf for $Y$. ...

1

$S = \{1,2,3,\ldots,20\}$ $A_n = \{x \in S: x \: \text{is a multiple of} \:n\}$ $\:\:\:n=1,2,3,\ldots$ $A_1 = \{1,2,3,\ldots,20\}$ $A_2 = \{2,4,6,\ldots,20\}$ because the products of $n=2$ with the elements in $S$ are $2,4,6,\ldots,20$ up to $20$ because we have the restriction that $x \in S$. So $1 \leq x \leq 20$ $A_3 = \{3,6,9,\ldots,18\}$ and ...

1

$A_2=\left\{2,4,6,8,\ldots,20 \right\}$, the even numbers. $$P(A_2)=\frac{10}{20}=\frac{1}{2}$$

0

Arithmetic mean of sample size follows Normal Distribution with $\mu{}, \sigma{}^2/n$. Calculate the probability of $\bar{x} > 2$ given that $X \sim Normal(\mu = \mu, \sigma^2 = \sigma^2/n)$ (by standardizing it to standard normal).

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Under $H_0$, $\bar{x}$ follows $N(0,\frac{\sigma^2}{n})=N(0,1)$. So $$P(\bar{x}>2)=1-\Phi(2)$$ where $\Phi$ is the cdf of a standard normal. There is no explicit way to compute $\Phi(2)$, so you have to look it up on a table.

0

Suppose your surveys come back, 5,5,6,7,7,7,8,8,9,9 You could say: (5+5+6+7+7+7+8+8+9+9)/10, or you could say:(2/10)*5+(1/10)6+(3/10)7+(2/10)*8+(1/10)*9 And either way you will get the same result.

1

Well, the ratio is simply $13:69$, as they have no common factors. A decimal approximation of $13\over69$ would be $0.1884$, to 4 significant figures, or $18.84\%$

1

To find the UMP test you should construct the likelihood ration test for the corresponding MP test, i.e., $$\frac{L_(\theta_1;X)}{L_(\theta_0;X)} = \frac{\theta_0^c}{\theta_1 ^c}\exp\{\sum x_i^c(1/\theta_0 - 1/\theta_1\} > k$$ where $\theta_1 > \theta_0$. So the LR is monotonic strictly increasing function of the MMS $\sum x_i^c$, hence the UMP ...

0

You can use the normal distribution to approximate the poisson distribution. Let $X \sim Poi(6)$. The approximation then is $P(X\leq x)=\Phi \left(\frac{x+0.5-\mu}{\sigma} \right)$ $\Phi(z)$ is the cdf of the standard normal distribution. And $+0.5$ is the continuity correction factor. The expected value of $X$ is $\lambda=6$ and the standard deviation of ...

1

You have $\mu = \dfrac{100}{6}$ and $\sigma = \sqrt{\dfrac{500}{36}}$, and that is okay. $\checkmark$ You are thus looking for $\mathsf P\left(Z\gt \dfrac{15.5-\mu}{\sigma}\right)$, where $Z\sim\mathcal N(0,1^2)$, as your Normal Approximation value. Plug in your values and use your Z-tail tableau, online calculator, mathematical package, or whatever.

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Let random variable $X$ be the number of $6$'s. Then $X$ has mean $\mu=100\cdot \frac{1}{6}$ and variance $\sigma^2=100\cdot \frac{1}{6}\cdot\frac{5}{6}$. The random variable $X$ is a sum of a fairly large number of reasonably nice independent identically distributed random variables, so the cdf of $X$ is reasonably well approximated by the cdf of a normal ...

2

Using Bayes' rule, for each value $x$ you have $$P(x | data) \propto P(data | x) prior(x)$$ Therefore, $P(x|data)$ is zero if $x \neq 0$ and so $P(0|data) = 1$, so your posterior is the same as the prior. No, the posterior is the same as the prior. Using this prior says that you have decided that the answer is zero and no evidence can convince you ...

0

For your edited question, the chance of 0 is .6^3 or .216. The flip side is a 79.4 chance of 1 or more guests. So the 1.2 projection is more accurate than defaulting to zero.

2

If the probability is more than 0.5, expect the guest to be present in the most likely set. Otherwise, expect the guest to be absent. This will maximize the likelihood.

0

My way to approach this problem is to assume the distribution of the score is like a normal distribution (may not be the case in reality). This is a reasonable assumption especially when the total number of samples is large, like 1000 as you said. Then suppose your score is $x$ and the mean and standard deviation of the sample are $\mu$ and $\sigma$. Then ...

0

The covariance is $s_{xy} = \frac{\sum (x_i - \bar x)(y_i - \bar y)}{n-1},$ where the sum is taken over $i = 1, \dots, n$ and $n$ is the sample size. Then the correlation is $r_{xy} = \frac{s_{x,y}}{s_x s_y},$ where $s_x$ and $x_y$ are the two standard deviations. If you have the regression line y = 13.555 -0.1688842 x. then you might say (over the ...

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This is Markov Switched Poisson Process, a special case of Markovian Arrival Process. Specifically, Markov Switched Poisson Process is a point process consisting of geometric runs of intervals that are exponentially distributed with rates depending on the state of the underlying discrete time Markov chain with m transient states and with transition ...

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You have not said what significance level you are using, but it seems that what you did for one tail used $5\%$ and standard error of the mean $\dfrac{250}{\sqrt{16}} = 62.5$ If the true population mean is $6500$ then there is a $5\%$ probability that the the sample mean will be below $6500 + 62.5 \Phi^{-1}(0.05) \approx 6397$ If the true population mean ...

0

According to this Stat Exchange post, a "z-test" and Chi-square can be compared to one another given they both focus on the same relationships, but you can't repeatedly run a Chi-square over and over without potentially having false-positives (or results the don't "preserve the alpha" to quote the link). Chi-square tests analyze what you observe with what ...

4

Informally, if we know that $X_1$ is small in absolute value, then $X_2$ cannot be small in absolute value. More formally, let $A$ be the event $|X_1|\le \frac{1}{2}$. A look at the sine curve shows that $\Pr(A)=\frac{1}{3}$. Similarly, let $B$ be the event $|X_2|\le \frac{1}{2}$. We have $\Pr(B)=\frac{1}{3}$. But since $X_1^2+X_2^2=1$, the event $A\cap ... 0 This fact is sometimes called the Fundamental Theorem of Simulation. Basically, you can consider sampling from some univariate distribution$F_X$as the same as sampling from the bi-variate distribution$(U,X)$where$U\sim\text{Uniform}(0,1)$and discarding$U$, since the marginal distribution of this will have distribution$F_X$. The same applies to any ... 2 The chance of second to last is zero. If you were second to last, the last was$46$. There were a total of$660$points scored. The average of other seven is then$660-81-46\approx 76$and at least one more is below your score. You cannot use the same reasoning for second place as the average of the six other than you, top, and bottom is$72.5$You can ... 0 Hint: As a simple example, suppose$X_1$and$X_2$independently each take the values$-1,+1$each with probability$\frac12$. Then the equally probable possibilities are X1 X2 Xbar X1-Xbar X2-Xbar -1 -1 -1 0 0 -1 +1 0 -1 +1 +1 -1 0 +1 -1 +1 +1 +1 0 0 How would you answer your questions for ... 1 I will show two calcultions. Example: w = (4.52 ± 0.02) cm, x = ( 2.0 ± 0.2) cm, y = (3.0 ± 0.6) cm. Find z = x + y - w and its uncertainty. z = x + y - w = 2.0 + 3.0 - 4.5 = 0.5 cm METHOD 1 Delta z = Delta x + Delta y + Delta w = 0.2 + 0.6 + 0.02 = 0.82 rounding to 0.8 cm So z = (0.5 ± 0.8) cm METHOD 2 Solution with standard deviations Delta z ... 0 Never plot a box-whisker plot beyond min and max of data. Upper and lower bound are just used to find out outliers and extreme outliers. In this case you have to take lower bound as 172. You can use MATLAB command : boxplot(x) to check the same rule. 1 To Find a critical value for a 90% confidence level. Step 1: Subtract the confidence level from 100% to find the α level: 100% – 90% = 10%. Step 2: Convert Step 1 to a decimal: 10% = 0.10. Step 3: Divide Step 2 by 2 (this is called “α/2”). 0.10 = 0.05. This is the area in each tail. Step 4: Subtract Step 3 from 1 (because we want the area in the middle, ... -1 The table you have been provided gives the probability$Z\leq z$, where$Z$follows a standard normal distribution. Loosely speaking it gives you the "area to the left" of$z$. Some tables (not this one) give the "area in the middle". Hence for d) Notice that is is asking for exactly that, the area to the left of$3$, represented by$P(Z\leq 3)$. It is not ... 1 Here's why what you did didn't work. Let$X,Y$be two independent normally distributed random variables with mean 2 and variance 1 (this is what you sampled from with your R code). This means that$X+Y\$ is normally distributed with mean 4 and variance 2, which implies that $$\frac{X+Y - 4}{\sqrt{2}}$$ is a standard normal random variable, so it follows that ...

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