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<> math is so bad that I forget it.

The mean is denoted by x*, and x*=(xi) n).

xi obeys the normal distribution n( ,2), then.

xi- ) obeys the standard normal distribution n(0,1).

According to the definition of chi-square distribution, it can be seen that (xi- )2 2 obeys the 2(n) distribution.

x* obeys the normal distribution n( , 2 n), then.

x*- ( n1 2) obeys the standard normal distribution n(0,1).

xi-μ)2/σ2

1/σ2)∑[xi- x*)2+μ2- x*2-2xix*+2xiμ]

1/σ2)∑(xi-x*)2+(1/σ2)∑(2-x*2+2xix*-2xiμ)

1/σ2)∑(xi-x*)2+(1/σ2)[n(μ-x*)(x*)-2(μ-x*)∑xi]

1/σ2)∑(xi-x*)2+(n/σ2)(μx*)[x*)-2(∑xi)/n]

1/σ2)∑(xi-x*)2+(n/σ2)(μx*)2

1/σ2)∑(xi-x*)2+[(x*-μ/ (σ/n1/2)]2

Write it out in full, as follows:

xi-μ)2/σ2=(1/σ2)∑(xi-x*)2+[(x*-μ/ (σ/n1/2)]2

x*- ( n1 2) obeys the standard normal distribution n(0,1).

x*- ( n1 2)]2 obeys a 2(1) distribution.

and (xi- )2 2 obeys the 2(n) distribution.

1/σ2)∑(xi-x*)2=∑(xi-μ)2/σ2-[(x*-μ/ (σ/n1/2)]2

Obedience obeys 2 (n-1) distribution.

First, find the size relationship between the fitting radius of x1 and x2 and 1, and r 0 then x1 and x2 have a strong linear correlation and a negative correlation. The linear regression equation must pass through the center point of the sample; In the previous group of models, the smaller the sum of squares of the residuals, the better the fitting effect, and the correlation index indicates the quality of the fitting effect, and the smaller the index, the stronger the correlation. The correlation index R2 is used to measure the strength of the linear relationship between two variables, and the closer R2 is to 1, the stronger the correlation, and conversely, the smaller the correlation.

The method of correlation between two variables, to know the precise degree of confidence in the correlation or irrelevance between two variables, can only be judged by using the relevant calculations of the independence test.

Extension: Mathematical statistics is a branch of mathematics that is divided into descriptive statistics and inferential statistics. It is based on probability theory and studies the statistical regularity of a large number of random phenomena.

The task of descriptive statistics is to collect data, organize and group them, compile the number distribution table, draw the number distribution only manuscript search curve, and calculate various characteristic indicators to describe the concentrated trend, the deviation trend and the skew of the data distribution. Inferential statistics is based on the description of statistics, according to the regularity of the sample data, the population is inferred and **.

Both probability values are the same, 1 in 16

p(max<0)=p(x1<0,x2<0,x3<0,x4<0)=p(x1<0)p(x2<0)p(x3<0)p(x4<0)=(1/2)^4=1/16.

p(max≤0)=p(x1≤0,x2≤0,x3≤0,x4≤0)=p(x1≤0)p(x2≤0)p(x3≤0)p(x4≤0)=(1/2)^4=1/16.

1: Yes; 2: Estimators and statistics are both functions of samples, so they are both random variables;

Statistic: Samples (x1, x2...... that do not directly contain (population) unknown parametersxn), as long as it is a qualified sample function, it is a statistic, and the estimator: in fact, it is a statistic that can be used to estimate unknown parameters; No unknown parameters!

3: No: Estimation is divided into two types: point estimation and interval estimation;

The commonly used methods of point estimation are: moment estimation and likelihood estimation;

4: Yes, sample mean and sample variance are important statistics and random variables;

Parameters of the distribution: Parameters are the characteristic quantities used to characterize the overall distribution.

For the commonly used sampling distribution statistic, degrees of freedom are parameters of the distribution, such as the "three major distributions".

About degrees of freedom: you can encyclopedia about this, degrees of freedom = the number of variables in the system that can change freely;

6: Yes, yes;

If x1, x2 ,..., xn is a sample of f: it means that the population distribution is f, and xi is a sample of capacity n drawn from the population, please note my next sentence "sampling does not refer to a fixed value at one time, but refers to the whole process of extraction." "A specified sampling result is called a sampling realization or sample observation, it can only represent a specific sampling, not the entire sampling, the result of sampling from the population is not fixed, but each sampling realization is a change, sampling can be regarded as a set of random numbers from n random variables that obey the same distribution.

For example, if the population x n(0,1) is taken from it and a sample with a capacity of n=3 is taken, then the sampling refers to (x1 x2 x3), and the specific implementation means that for example, I now get a sample and observe it as (,, and I sample it again, it may become (,,1), and it will change again if I sample it again. That is to say, sampling (x1 x2 x3) is a 3-dimensional random vector that has a joint probability distribution, and because assuming that they are independent of each other, more independent formulas, then their joint probability distributions are equal to the product of their respective distributions.

So why isn't it equal to f(xi) n? Because f(xi) n represents the same product of f(x) to be expressed as a power form, and x1 x2 ......Just obey the same distribution and not mean that they are equal, only in the case of equality can be written as power.

In fact, the population distribution is usually not fully known, and often means that it is possible to assume its distribution form, but do not know the specific parameters, for example, knowing that the population is a normal distribution but not knowing the mean and variance, then its distribution is in the form of conditional density, and if the form of probability density is used, the likelihood function is obtained, f(xi; mu,sigma)。The parameter values are then reversed by the maximum likelihood estimation.

The first question, whether the question says "upper a-quantile", if not, is wrong.

Is the aquantile what you teach in class? This concept is not commonly used, and this table is commonly used.

The answer is false....But why not? (Push by Figure 2) should be.

x~n（u，σ^2)

x(i+1)-xi] n(0,2 2), so [x(i+1)-xi] ( 2 )n(0,1).

then 2 2(n-1).

e∑^2=n-1

c [ x(i+1)-xi] 2=2c 2 2 from the question: ec [ x(i+1)-xi] 2= 2 i.e. ec [ x(i+1)-xi] 2=2c 2* e {[x(i+1)-xi] 2=2c 2 (n-1)= 2

Therefore c = 1 [2(n-1)].

Please don't forget to adopt it, and I wish you a happy study.

Finding Sigma from Poisson Distribution Using Poisson's formula to find (stunned) = sigma (variance).

Then use the formula of the statistic u to find it out and see if it is between plus or minus.

hypothesis testing questions).