Probability statistics: Sixth chapter Sample and sampling distribution _ probability statistics

Sixth chapter sample and sampling distribution

Content Summary

First, the overall

In mathematical statistics, the whole of the research object is called the whole, and each element that makes up the whole is called an individual. The overall use of a random variable x represents. If the distribution function of X is f (x), the F (x) is called the distribution function of the total x.

Second, the sample

If x is a random variable with a distribution function f (x), if it has an independent random variable of the same distribution function f (x), it is called a sample from the total X, and the sample size is N. For a set of observations, the joint distribution function is

F () =f () f () ... F () =

Third, statistical volume

Set is a sample from the total X, G () is a function, if the continuous function and G does not contain any unknown parameters, then the G () is a statistic. For a set of observations, it is said that G () is a G () observed value.

Iv. Statistics of common usage

1, the sample mean value =

2, sample Variance = =

Sample Standard deviation s=

3, the sample K-order (origin) moment = k=1,2 ....

4, sample K-Order Center moment k=2,3 ....

Distribution of common statistics

1, the distribution: set is from the overall n () a sample, then

N ();

2. Distribution: Set x1,x2,..., xn is a sample from the total n (0,1), then the statistic

=x12+x22+...+xn2=

Obey the distribution of the degrees of freedom N, remember ~.

1 if the x~,y~,x,y are independent of each other, then x+y~

2 if x~, E (χ2) =n. D (χ2) =2n.

3 x1,x2,..., xn is a sample from the total N (μ,σ2), and the S2 are sample mean and sample variance, then ~; Independent with S2.

4) Set x~, the point of the location to meet the conditions

= =, (0<<1)

3.t Distribution: X~n (0.1), Y~,x and y independent of each other, called random variables

t=

Obeys the T distribution of the freedom of N, and T~t (n).

1) Set x1,x2,..., xn is a sample from the normal population n (μ,σ2), S2 is the sample mean and the variance of the sample, then ~t (n-1)

2) Set T~t (n), t (n) to meet the conditions of the points

p{t>}==, (0<<1)

3) Setting up x1,x2,..., Xn1; Y1,y2, ..., Yn2 are the normal population n (μ1,σ2), N (μ2,σ2) samples, and they are independent of each other, then

~t,

of which: =,

, the sample mean value of two normal population, and the sample variance of two normal population were respectively.

4. F Distribution: Set x~,y~, and x,y mutually independent, then called random variable

f=

Obey the F distribution of degrees of freedom (), remember F~f ()

1) Setting up x1,x2,..., Xn1; Y1,y2, ..., Yn2 are independent samples from normal population n,n, respectively, the sample mean of two normal population, and the sample variance of two normal population respectively. The

F=~f,

In particular, f=~f.

2) Set F~f (), F () to meet the conditions of the points

= =, (0<<1).

Basic requirements

1. Understand the concept of totality, individual, sample and statistic, master the calculation of sample mean value, sample variance and sample moment.

2, understand the distribution, distribution, distribution and the definition and nature, understand the concept of the point and table calculation.

3. Master the distribution of some common statistical quantities of normal population.

4, understand, the distribution.

This chapter focuses on the concept and distribution of statistics.

Analysis of typical examples

Example 1. Set X1,X2, ... Xn is a sample from the overall x, in the following three cases, the E (), D (), E (S2) are respectively obtained.

(1) x~b (1,p), (2) x~exp (λ), (3) x~u (0,θ);

Analysis: It can be solved by using the expectation of common distributions, variance, and the property of S2 definition and expectation variance.

Solution: (1) due to x~b (1,p), E (x) =p, D (x) =p (1-p).

So E () =ex=p,

D () = (1/n) *d (X) =p (1-p)/N,

E (S2) =p (1-p)

(2) due to x~exp (λ), E (x) =λ, D (x) =λ2

So E () =λ

D () = (1/n) *d (X) =λ2/n,

E (S2) =λ2

(3) due to X~u (0,θ), E (x) =Θ/2, D (x) =Θ2/12

So E () =Θ/2,

D () =θ2/(12n),

E (S2) =Θ2/12

Example 2, in the total N (7.6,4) of the sample capacity of N, if the average number of samples required to fall in (5.6,9.6) probability is not less than 0.95, then n at least how much.

Analysis: Because the sample mean ~n (7.6,4/n). To solve the deformation of P (5.6<<9.6), on behalf of

P (a< ( -7.6)/<b) Form, and then using standard normal distribution look-up table can solve n

Solution: Because ~n (7.6,4/n). So

P (5.6<<9.6) =p{< ( -7.6)/<}≥0.95

namely p{-< ( -7.6)/<}≥0.95,

i.e. 2φ () -1≥0.95,φ () ≥0.975

By the table φ (1.96) = 0.975,

So ≥1.96 or n≥3.84, that is, sample size n is at least 4

Example 3, two independent samples were extracted from the normal population n (100,4), and the samples were of 15 and 20 respectively. Try to find P (-|>0.2)

Analysis: First find out the distribution, and then use P (|-|>0.2) on behalf of the standard normal distribution in the interval probability, you can solve

Solution: Due to ~n (100,4/15), ~n (100,4/20), with Independent

So ~, ie ~

So

Example 4, the sample of capacity is extracted from normal distribution, try to find

Analysis: Because ~, try to get the identity of the same deformation, and then find the definition of distribution and look-up table can be obtained

Solution: Because ~,~

so =

=

Example 5. Set is the sample from the known

Please

Analysis: Because ~,

, the identical deformation and distribution definition can be used to obtain

Solution: Because ~

So

Example 6. Set random variable ~

Analysis: Using the relationship between the two can be verified

Solution: Because

So

and.

Therefore there are

Example 7. Set a sample from N (0,). Try to find the distribution of y=.

Analysis: y== is the quotient of the square sum of two normal distributions. If you can convert to two distribution quotient, you can prove that Y obeys F distribution.

Solution: Because,

So.

Because Cov (,) =d ()-d () = 0,

and obey two yuan normal distribution, so, independent.

So, y==.

Example 8. Set up a ... A sample from N (,). =, the sample mean of the first n samples and the variance of the sample. Try to find constant c. Causes the distribution to be subject to T distributions and indicates the degree of freedom of distribution.

Analytic: First find out-obey the normal distribution, then have ~ (n-1)

Finally, the T-distribution is defined to be solved.

Solution: Because

So

So

Instant,;

The degree of freedom is.

Example 9. The following probabilities are obtained.

(1) (2)

Analysis: Use, and standardize the normal distribution, you can solve

Solution: =

.

Example 10. A sample with a capacity of 16 is set in the population, which is known here.

1). 2) Seek

Analysis: Because of the known here, so can solve 1)

Use and distribution variance can be obtained 2)

Solution: 1) because, so

=

2) because ~;

That

So

Self-Test questions

Fill in a blank question

1. Set the random variable and the independent all obeys the normal distribution, but and separately is from the overall and the sample, then the statistic

Obey distribution, freedom is.

2. is a sample from the normal population,

，

At that time, the statistics were subject to distribution and the degree of freedom was

3. Set the general to obey the normal distribution, but from the overall sample, then the statistics

Obey distribution, freedom is.

Choice questions

1. is a sample from a normal population, a sample mean,

The random variable that obeys the distribution of degrees of freedom is ()

A) B)

C) D)

2. Set the general to obey the normal distribution, which is known, unknown, is the sample, then the following expression is not a statistic is ()

A) (B) min ()

C) D)

Answer question:

1. Weighing a heavy item on the balance, assuming that each weighing result is independent of each other and obeys normal distribution, if the arithmetic average of the result of n weighing is expressed, if the requirement n is at least how much.

2. Set is a sample from the normal population, please

。

3. is a sample from the normal population,

It is shown that the statistic obeys the T distribution of 2 degrees of freedom.

4. (For normal overall